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[Articles Home]  [Add Article]  

Bandwidth versus Keying Speed Revisited

Mickey Cox (K5MC) on August 16, 2007
View comments about this article!


Bandwidth versus Keying Speed Revisited

(Theory and Experimental Data in Harmony)

In late May my eHam article "Bandwidth versus Keying Speed" (http://www.eham.net/articles/16649) pointed out that the FCC's definition of occupied bandwidth is equivalent to the 99% power bandwidth. Using the concept of Fourier series to model the output signal from a typical CW transmitter, my calculations demonstrated that the occupied/power bandwidth is a function of both the keying speed and the rise/fall characteristics of the keying envelope. Table 1 from my original article, which is reproduced below, lists the 99.1% power bandwidths for sinusoidal keying (5-ms rise and 5-ms fall times) and square-wave keying at 2.4 and 30 words per minute.

2.4 words per minute

30 words per minute

Sinusoidal keying (5 ms)

34 Hz

150 Hz

Square-wave keying

42 Hz

525 Hz

Table 1. 99.1% power bandwidth for CW/ASK transmitter for periodic signaling

Although the theoretical results presented in Table 1 are consistent with the professional electrical engineering literature, as well as the latest editions of The ARRL Handbook, a number of hams criticized my article's lack of experimental data on typical CW transmitters. Fortunately, my school recently purchased two Agilent Technologies N9020A signal analyzers (at a cost of approximately $26,000 each) and I now have occupied bandwidth data as a function of keying speed for four different transmitters.

The occupied bandwidths for four CW transmitters I've tested can be seen at http://www.arrl.org/sections/?sect=LA (Figures 6 through 15). These measurements are all consistent with the fact that the occupied bandwidth does vary with the keying speed, even in the case of constant rise/fall characteristics of the RF envelope. In other words, my measurements are consistent with the calculated results I presented in my original eHam article. The transmitters I tested were operating into a dummy load at 18.1 MHz with an output power of approximately 50 W. In all cases, the transmitters were sending a string of "dots" (dits) with a 50% duty cycle.

For example, Figures 6, 7, and 8 show that the occupied bandwidth (the 99% power bandwidth) for my Kenwood 940 transceiver is about 60 Hz, 190 Hz, and 260 Hz at 6, 30, and 53 words per minutes (wpm), respectively. (I observed the RF envelope on an oscilloscope to verify that the rise/fall characteristics are essentially constant at all the keying speeds used in my tests.) Measurements of the occupied bandwidths at similar speeds as the 940 are also presented for my Kenwood 2000, Ten Tec Corsair, and Ten Tec Orion II. The occupied bandwidths are noticeably smaller for my Orion II (Figures 14 and 15) compared to my other three rigs because the rise/fall times on the Orion were set to 8 ms (the normal setting I use on the air), whereas the rise/fall times on the other rigs are closer to 5 ms and are not readily adjustable. When I set the rise/fall times on the Orion II to 5 ms, the occupied bandwidths track the other three rigs rather closely.

As I tried to point out in my original article and follow-up comments, the occupied/power bandwidth is not the same concept as the "key-click" bandwidth. However, if the occupied bandwidth of a CW signal is "reasonably" close to the necessary bandwidth as calculated by the simple equation in Part 2 of the FCC's Rules, there should be little or no key-click interference on adjacent frequencies. For example, since the necessary bandwidth is 120 Hz at 30 wpm (for a fading circuit where K = 5), sinusoidal keying at 5-ms rise/fall times (with an occupied bandwidth of about 150 Hz at 30 wpm) should generate little key-click interference, whereas the key clicks generated by square-wave keying (with an occupied bandwidth of about 525 Hz at 30 wpm!) will be very objectionable for many kHz up and down the band.

In conclusion, I hope the fact that my measurements of occupied bandwidth using a state-of-the-art spectrum analyzer will convince practically everyone that the Fourier series models I assume in my original eHam article are correct. The occupied bandwidths of typical CW transmitters do, in fact, vary with the keying speed.

73, K5MC

Member Comments:
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Bandwidth versus Keying Speed Revisited  
by KA8I on August 16, 2007 Mail this to a friend!
You've discovered the Nyquist rate.

http://en.wikipedia.org/wiki/Nyquist_rate

73,

-Mark KA8I
 
RE: Bandwidth versus Keying Speed Revisited  
by W9PMZ on August 16, 2007 Mail this to a friend!
uhhhhhhhh

not this discussion again......


73,

Carl - W9PMZ
 
RE: Bandwidth versus Keying Speed Revisited  
by KB9CRY on August 16, 2007 Mail this to a friend!
The occupied bandwidths of typical CW transmitters do, in fact, vary with the keying speed.


Wow, what can I say. Just, Wow!
 
RE: Bandwidth versus Keying Speed Revisited  
by W9AC on August 16, 2007 Mail this to a friend!
> "The occupied bandwidths of typical CW transmitters do, in fact, vary with the keying speed."

Strictly speaking, yes. As a practical matter at traditional CW keying rates, the answer is no.

Typical CW keying rates are relatively low. As the keying rate increases, occupied bandwidth increases accordingly, but in relation to the bandwidth created by *slope* of the waveform, the resulting increase in bandwidth as a function of keying rate is masked by the waveform slope bandwidth.

At some faster key-rate, bandwidth becomes significant in relationship to the waveform slope bandwidth -- but that point is certainly above typical CW keying rates.
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 16, 2007 Mail this to a friend!

Mickey,

At the end of the last drawn out thread you concluded the interference level on the adjacent channel didn't really change with keying speed. Anyone with a transmitter and a receiver can PROVE this for themselves, as several people actually did.

Hams really have to realize the thing that causes keyclick bandwidth is exclusively the waveshape of the envelope, and NOT the speed. This of course assumes there is no frequency shift involved.

Bill Fisher, when eHam was starting, was instrumental in trying to get people with FT1000MP's and other radios to clean up the envelope shape so people up and down the band could operate without clicks. No matter how fast or slow we send the signal extends the same bandwidth with the same peak power so long as the rise and fall shapes aren't changed, and that's a fact.

It took Yaesu years to realize the rising and falling edges are everything, not the speed. Ten Tec and Elecraft both fixed their radios, and now even the ARRL has corrected the Handbook by showing the correct waveform necessary for minimum bandwidth.

Anyone with a radio or a device that measures occupied BW can see unless the envelope shape is changed, the interference off channel extends the same distance and the same peak level regardless of speed.

73 Tom
 
Bandwidth versus Keying Speed Revisited  
by K4LJA on August 16, 2007 Mail this to a friend!
Mickey --

Great article. Keep up the good work. We appreciate you.

Randy K4LJA
 
RE: Bandwidth versus Keying Speed Revisited  
by W4LGH on August 16, 2007 Mail this to a friend!
Geez...you can't send Morse Code fast enough to worry about bandwidth, even using a computer! So what's the concern?

73 de W4LGH - Alan
http://www.w4lgh.com
 
Bandwidth versus Keying Speed Revisited  
by WA1RNE on August 16, 2007 Mail this to a friend!

All your efforts in researching this issue are appreciated.

But when we begin to approach the next solar maximum in the coming years, I predict the scope of most articles on eHam will shift from Code vs No Code and occupied bandwidth on CW to articles describing amazing HF and VHF DX, New Digital Voice Modes and best tips for working DX without a G5RV or a J Pole.


Maybe we should start brainstorming now, as these events are 11 years in the making.


....WA1RNE
 
RE: Bandwidth versus Keying Speed Revisited  
by K1BXI on August 16, 2007 Mail this to a friend!
For the what ever it's worth department......Any cw signal I copy is a single tone. As I tune through it with a 250 Hertz wide filter, either I hear it or I don't. Makes no difference between 5 or 55 wpm. It occupies the same space on my receiver. So is this a function of my receiver, my old ears, or the transmitter?

Either way there still seems to be plenty of room for any bandwidth in the cw portion of the bands, except on a major cw contest weekend. Where does everybody go after the contest is over?

John.....K1BXI
 
RE: Bandwidth versus Keying Speed Revisited  
by AA4PB on August 16, 2007 Mail this to a friend!
As a practical matter at traditional CW keying rates, the answer is no
------------------------------------------------------
Either the keying speed does have an impact on BW or it doesn't.

From the chart you can see that going from 5mS to 0mS has the greater impact but you can also see that increasing the speed from 2.4 to 30 WPM (without changing the rise/fall times) also has a very significant impact on the occupied bandwidth.

The BW of a CW signal could become a concern if the FCC ever goes to designating band segments by bandwidth in lieu of mode.
 
RE: Bandwidth versus Keying Speed Revisited  
by N6KYS on August 16, 2007 Mail this to a friend!
W8JI wrote...
"Hams really have to realize the thing that causes keyclick bandwidth is exclusively the waveshape of the envelope..."

Excellent, excellent, excellent point! People would be astonished at what occurs in those sharp corners. Mathematically, the cosine content and many intermediate wierd combinations, distort the sine wave of the CW signal and splatter junk all over the place.

This should be a required topic of understanding for every ham.

Best,

Brad
N6KYS

 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
This is a debate between Engineers and never to be Engineers.

Math says yes to the post, but the Oscilloscope says, hey, its too fast for me to decide.

In Geometry of two vertical lines, what does this information tell you?

Now that this is settled, lets go for the bandwidth of single side band, bandwidth, which should be of more concern.

I can't visualize a ham operator using Morse code and into his 70 years, still having the pleasure of enjoying it, to this very day, mind you.

W6TH
.:
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 16, 2007 Mail this to a friend!
While this article’s analysis is technically correct when we look at power averaged over a very long time, it does NOT apply to the systems we use.

Here are (once again) the reasons why it doesn't apply:

1.) The receivers we use cannot have a bandwidth small enough to sort out the sidebands. The receivers MUST have a bandwidth and response time fast enough that the user can sort out the rising and falling edges without distorting the waveform. This is an absolute.

We can't make the system into something that does not and cannot work just to be right.

2.) The rise and fall (assuming no other defects like hum, noise, or FM) MUST generate at least one sideband above and below the carrier. The spacing of that sideband and the number of sidebands is set by the rise and fall waveshape and duration, just like the signal on an AM transmitter. The peak power is determined by the delta in amplitude at any area of slope.

The receiver responds to those sidebands, and that is what we hear. There is no energy or power storage over time except the hold of the AGC system as it charges. It is the PEAK envelope power of the sidebands that determines what bothers us, and even a fraction of a second later it is GONE.

Because there is no memory or storage over very long time periods, the very slow modulation by the keying rate isn't important. The only thing a faster or slower keying rate does is repeat the same pulse or click level at the same bandwidth spacing faster or slower.

Now granted we can extend the analysis to include days, weeks, months, or even years. If we do that, it appears that the really slow low frequency components get stronger. They actually do when we look at energy over very long periods.

The problem is neither our ears nor our minds nor do our receivers allow us to sort out these really slow variations, so they are meaningless.

An example of how meaningless they are is found by looking at speech. We can "prove", mathematically, that when we talk slower the bandwidth is less. That's a fact.

In the REAL world, it doesn't matter how slowly or fast we talk. The reason is the SYSTEM we use can't store or remember energy over seconds or minutes of time. It may be that the "splatter" happens more often when we talk faster, but so far as we are concerned the bandwidth isn't affected by the speech rate in words per minute.

We can prove many meaningless (or useless) things if we aren't careful to make sure we aren't going outside the bounds of the system.

People who insist on going outside the scope of what is meaningful will always disagree with the observations of those who talk about the real-world effects we can observe.

This will be true no matter how many times this thread is resurrected.

73 Tom






 
RE: Bandwidth versus Keying Speed Revisited  
by N6KYS on August 16, 2007 Mail this to a friend!
W6TH, I have NO idea what your bizarre rambling is supposed to imply. It's so easy to clean up kep clicks...ever hear of an invention called a capacitor?

Now we're supposed to veer off topic and talk about SSB bandwidths? Geez..... is your Ritalin wearing off or something?

Brad
N6KYS
 
RE: Bandwidth versus Keying Speed Revisited  
by W4VR on August 16, 2007 Mail this to a friend!
Who cares besides the enforcement agencies?
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
N6KYS on August 16, 2007

W6TH, I have NO idea what your bizarre rambling is supposed to imply. It's so easy to clean up kep clicks...ever hear of an invention called a capacitor?
.....................................................

""ever hear of an invention called a capacitor""?

No Fly Boy, tell me about it.

Fellas, we now have added to our eham another blowphart.

.:
 
RE: Bandwidth versus Keying Speed Revisited  
by N6NKN on August 16, 2007 Mail this to a friend!
Vito,

Why does everything you get involved in become a "urination" contest?
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
A capacitor and a resistor were never mentioned as an invention to my knowledge.

To eliminate key clicks is a simple matter of calculating the proper charge and discharge of electric motive force.

This can be done by a simple formula such as Frequency equals the reciprocal of Time or put as F=1/T.

This was my very first electronic math I used when at the age of eleven years (1933) when building my first transmitter, which was a TNT self excited oscillator.

This key click stuff is very exciting as some ears will detect and other ears will not detect, so the Manufacturers of many solid state radios have a problem to satisfy everyones needs. I am sure they do try their very best to please all and some will never be pleased.

.:
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
N6NKN
Vito,

Why does everything you get involved in become a "urination" contest?
................................................

Because of people like you.

W6TH

.:
 
RE: Bandwidth versus Keying Speed Revisited  
by N3JBH on August 16, 2007 Mail this to a friend!
Get the buckets boy's. It is going be a good one.
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 16, 2007 Mail this to a friend!
W8JI (May 31, 2007) - Since I can't seem to get the theorists to actually measure the bandwidth of a transmitter, I've measured it for them.

Please review this new page on my website:

http://www.w8ji.com/occupied_bw_of_cw.htm

These measurements were made with currently certified equipment that directly measures occupied bandwidth.

CW speed does NOT affect the occupied BW of a CW transmitter unless it changes the rising and falling edges of the waveform.

W8JI (May 31, 2007) - The problem is I have measured at least a dozen different models of radios and NONE of them follow their theory either, and I know other people who have done the same.

W8JI (June 12, 2007) - My analyzer is one of those very expensive analyzers that measures many things directly, and can do so in 10Hz bandwidth resolution. It can average multiple sweeps to weed out momentary glitches or things it might miss on a single sweep.

It calcultes occupied bandwidth according to FCC requirements, and so what you see on the display is really what it is.

Now it just happens that what is reads agrees very closely with Kevin W9AC and many other people, and disagrees with you and Tim.

W8JI (June 13, 2007) - Your math is flawless, but it includes things that are meaningless to the real application. As a result it is garbage in and garbage out. We can’t stretch the time of the analysis out so long that we are looking at things that change very fast over many dozens of slowly changing cycles when the fast things are exclusively cause the problems we are concerned with.

W8JI (June 14, 2007) - Let's not confuse power bandwidth, which really has no use in this type of communications system, with the occupied bandwidth.

W8JI (June 14, 2007) - By the way, The ARRL has re-written and corrected some earlier incorrect statements in the section on CW and bandwidth of CW.

People keep referring to earlier erroneous text to support their position, but the newer Handbooks have the correct information.

K5MC - I simply had to copy and paste above several comments posted by W8JI following my original article and before I had access to a high-quality RF spectrum analyzer. Perhaps W8JI now regrets his comments on June 13 about "garbage in, garbage out." More importantly, W8JI (as of June 14, at least) still doesn't understand that the 99% power bandwidth is equivalent to the FCC's definition of occupied bandwidth! Concerning the newer handbooks, on page 9.7 of the 2007 ARRL Handbook you will read the following:

"The bandwidth occupied by a CW signal depends on the keying rate, with higher speeds requiring a wider filter to pass the sidebands. In addition, occupied bandwidth depends on the rise and fall time and the shape of the keyed RF envelope." (These exact sentences also appear in the 2005 and 2006 editions as I pointed out previously.)"

By all means I hope everyone will review the "occupied" bandwidth data posted by W8JI at http://www.w8ji.com/occupied_bw_of_cw.htm for an IC-751A and a FT-1000 MP MK V. In referring to the IC-751A, W8JI even says, "We see the actual measured occupied BW of a relatively good CW transmitter is essentially the same and does not track CW speed." A relatively good transmitter? At 10 dits per second (24 wpm), W8JI's "measurement" shows an occupied bandwidth (99% power bandwidth) of 490 Hz! My four transmitters have a measured occupied bandwidth (99% power bandwidth) of well under 300 Hz at 50 wpm, plus my earlier article indicated an occupied bandwidth of about 525 Hz at 30 wpm for square-wave keying!

As I commented several times before, I believe W8JI is being misled by his spectrum analyzer. If the "dozens" of CW transmitters tested by W8JI show essentially no change of occupied bandwidth as a function of keying speed, then there is obviously something wrong with his spectrum analyzer measurements.

Once you truly understand the FCC's definition of the occupied bandwidth (assuming you also have a basic understanding of Fourier analysis!), you should have no trouble understanding that both the keying speed and the rise/fall characteristics of the keying waveform are factors in determining the occupied/power bandwidth of a CW transmitter.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by N7YA on August 16, 2007 Mail this to a friend!
Uh oh! I think the 6's are headed for a civil war.

 
RE: Bandwidth versus Keying Speed Revisited  
by W9OY on August 16, 2007 Mail this to a friend!
Since the Orion is adjustable in rise time from quite hard to soft, as well as keying speed, and you have proven consistency across transmitters with your spectrum analyzer. Why not do a spread sheet that shows how 99% power bandwidth varies holding keying speed constant and varying rise time, and then an entry that holds rise time constant and varies speed? You have already presented half the data. You might do rise time at a few keying speeds and keying speed at at a few rise times. You can then do a statistical correlation to determine the interdependence of the 2 variables. They are either independent variables or they are not. The experiment uses real data at real ham keying speeds from the same real transmitter into the same real receiver. That's as real world as it gets. You basically already have presented half the data, but you need a few more data points to do the regression.

73 W9OY
 
Bandwidth versus Keying Speed Revisited  
by K1CJS on August 16, 2007 Mail this to a friend!
Geez, this reminds me of another recent article...... What was it?? Oh, yes! Why a copper wire is a much better material than aluminum, steel or stainless steel for making antennas. Oh-oh, we're heading for another 'scientific' discussion here, aren't we?!?
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 16, 2007 Mail this to a friend!
W8JI - Hams really have to realize the thing that causes keyclick bandwidth is exclusively the waveshape of the envelope, and NOT the speed.

K5MC - I have never said otherwise. Once again I will quote myself from my original article on May 26:

"A crucial point to be made here is that the concept of occupied bandwidth does not say that the strength of the individual key clicks generated by a poorly designed transmitter is reduced when the sending speed is decreased! The information provided by the occupied bandwidth is exactly that described in its definition."

With hindsight, I concede that my comments above were probably too subtle for some folks. I certainly made the distinctions between occupied/power bandwidth and "key-click" bandwidth clear many times during my later comments.

A fundamental problem that W8JI apparently fails to understand is that the FCC doesn't have a quantitative definition of "key-click" bandwidth. As a matter of fact, I suggested to W8JI at one point during the earlier thread that he might submit a mathematically precise definition of "key-click" bandwidth, although I doubt that the FCC these days would show much interest in such a petition.

The FCC does define both the "necessary" bandwidth (which is, in essence, a legal definition given by a very simple equation for CW signals) and the "occupied" bandwidth. The occupied bandwidth of a CW signal is a mathematically precise concept and can be readily calculated and measured as I have demonstrated. The FCC's definition of occupied bandwidth is equivalent to the 99% power bandwidth for CW signals. The 99% occupied bandwidth measured by such spectrum analyzers as the Agilent N9020A that I now have access to is consistent with the FCC's definition of occupied bandwidth.

W8JI is trying very hard to convince other hams that only the "key-click" bandwidth is important and that the occupied/power bandwidth concept is of no importance in amateur radio. To my knowledge, nobody has argued that proper waveshaping to avoid excessive key clicks is not important. However, from a regulatory point of view, at least, it certainly would be nice for all hams to have a basic understanding of occupied bandwidth. Moreover, the relationship between the rate of information (keying speed) and the occupied/power bandwidth is a fundamental concept in all communication systems, including amateur radio. Just ask any ham active in EME or QRSS!

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
As I repeat myself;
This is a debate between Engineers and never to be Engineers.

Math says yes, to the post, but the Oscilloscope says, hey, its too fast for me to decide.

In Geometry of two vertical lines, what does this information tell you?

Perhaps this is just as well as the logic of the proof is subtle, (So slight as to be difficult to detect or describe), in any case some do not have a burning desire to see the proof..It is helpful to first go through arguments presented with specific numerical examples.

Theory, math and then design a piece of equipment to meet the needs. Then again, train to meet the goals of this new equipment.

73, W6TH......A Non Vanity Call.

.:

 
Bandwidth versus Keying Speed Revisited  
by KX0R on August 16, 2007 Mail this to a friend!
K5MC has done a nice presentation here to show that "Occupied/Power Bandwidth" does rise as the keying rate increases. The images from the 9020A analyzer are a tribute to the Agilent engineers who built the "Power Bandwidth" function into their machine.

This thread and the incredibly long and tedious earlier thread all revolve around the exact meaning of "occupied/power bandwidth". I don't think this number is very relevant to us as radio operators, never mind the fact that the FCC refers to the term, or something similar, because what we really care about is the instantaneous bandwidth of a CW signal during the "key down" and "key up" transitions.

I'm wondering if the term "occupied/power bandwidth" in K5MC's posts, also measured by the Agilent 9020A's function, really has the same meaning as what was intended when the FCC regulations were written. I have a hunch, but no proof, that the FCC used the term "occupied bandwidth" to mean the practical bandwidth of a signal, in other words, the stuff you can actually hear as you scan through the signal with a narrow filter. This is how signal bandwidth was determined before modern spectrum analyzers, and it's how the early spectrun analyzers worked. My HP141T/8552B/8553B has a single-scan mode that you can run very slowly with narrow bandwidth to see what's occupying a slice of spectrum. This old analog machine then integrates the information using a storage CRT.

One thing I can't see in the images, and would like to know, is how long the various scans took to run. Since the analyzer bandwidth is 9.1 Hz, and some of the scans are about 2 minutes apart, we can only guess within limits. I would like to know if the instrument does one sweep or an average of many to get these displays. The scans at the higher speeds show the periodicity of the keying .

The choice of "99% power bandwidth" seems somewhat arbitrary. Severe interference can occur well outside of this number if the interfering signal is much stronger than the one we wish to copy.

I hope that we can all take K5MC's nice measurements and analysis along with W8JI's precise, practical observations as opposite sides of the very same coin. What W8JI has to say about real radios and key clicks is very relevant today. There is still a lot of misinformation about keying being spread around the Internet. What comes out of the discussions on these topics is that keying and bandwidth are very technical areas not well understood by many.

I got into a thing with some QRP operators relating to key clicks being produced by a certain kit radio under certain conditions, and one of the views that came out of our various posts was that many operators believe short rise and fall times, like 1 ms or even less, are desirable for QRP work. There is a widespread view that a "sharp" signal is easier to copy through noise. I don't know if this idea is really true, but people believe it and use it to justify rigs with sharp keying. Of course, there is also the false idea that the keying sidebands of a QRP signal are so far down in the noise that no one can ever hear them anyhow - but if that were true, how would sharp keying help a person copy?

If we run a sharply-keyed signal through a narrow CW filter in a receiver, we strip off much of the sharpness, which appears as sidebands. So how then could the sharpness help anyone copy, unless they use a wide bandwidth receiver, which is impractical if the signal is really weak? (Most QRP receivers use IF filters that are only a few hundred Hz wide.)

This question of keying sharpness and whether it aids copy is really very interesting, much more so than the tedious posts about power bandwidth. What happens when we hear and copy a weak CW signal through a filter and then with our trained brain is really a very complex process, and it is not the same for all operators, especially experienced ops.

I wish someone here would write a post here on the question of whether CW copy is enhanced by shortening the rise and fall times (increasing the instantaneous bandwidth). Many rigs can vary the keying times. There are people deliberately building sharp keying into their gear, or not fixing it, because they think it gives them an advantage.

Does anyone know of any tests conducted, perhaps decades ago when CW was king, that show whether experienced operators can copy sharp signals better, either when they're strong, or when they're down in the noise?

My ears and brain think that a sharp signal is easier to copy than a softly-shaped one, at least when the signal-to-noise ratio is high, and the receiver bandwidth is set wide enough to hear the sharpness. Even though such a signal occupies more bandwdth, there is information in the sidebands, and our ears process it if it gets into our heads.

Of course we need shorter rise and fall times for really high-speed CW, like 50 wpm and up, but I'm not getting into that - I'm wondering about copying 20 wpm through noise.

Sorry for going into a related topic, but I think we’ve beat the “occupied bandwidth” topic to death, and the question of what works best for real CW is much more relevant to most of us. I doubt we know all the answers either.

73,

George Carey Fuller
KX0R
 
RE: Bandwidth versus Keying Speed Revisited  
by ONAIR on August 16, 2007 Mail this to a friend!
OK, all you bandwith mavens win! From now on, no one is permitted to send CW at more than 3 words per minute. Now let's get on to the really important stuff, like why so much VHF and UHF bandwith is wasted on FM? Can't we just build repeaters to work on AM or SSB? The drain on natural resources is just absurd.
 
RE: Bandwidth versus Keying Speed Revisited  
by K4GK on August 16, 2007 Mail this to a friend!
Oh!... No!... Here we go again, with the debate on CW bandwidth.

I'll be glad when this issue is settled once and for all!
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 16, 2007 Mail this to a friend!
"Necessary" bandwidth is used to give an indication of the minimum bandwidth necessary for the transmission in question. It is generally used to compare the relative merits of a given emission. It can also be used to gauge how much extra bandwidth a mode requires from the minimum bandwidth.

"Occupied" bandwidth is a measured quantity that is used by the FCC when determining channel widths and whether or not a given transmitter should be certified as meeting the requirements of the given service. It is also the reason most commercial rigs do not have operator controls such as power level and microphone gains that can affect the occupied bandwidths.

This is why the ARRL's recent proposal that started with "occupied" bandwidths was so dangerous. As you can see from these threads, it is obvious that not much thought was given when choosing this recommendation. Luckily, the comments from many of us convinced the ARRL that they should at least use "necessary" bandwidth since it is a calculated value, not measured. Otherwise, many of us would have been sending our rigs to Mickey to have them measured to see if they could meet the required occupied bandwidth.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
Hard keying or soft keying is the choice of the able to copy cw operator.

I myself am in favor of the hard copy as at the higher rate of keying there is a noticeable effect of space in between characters. I am talking of speeds of 50 wpm plus. I copy by words and not by characters as is done today.

To experience your choice, one must be a dye in the wool cw operator and my choice is hard keying. Soft keying is like hearing an unneutralized amplifier and keying the amplifier and hearing the oscillator running continuously, something you new to ham radio will never come across and understand.

When I mention copy by words, back in the old commercial and ham traffic system the operators would copy behind a word or two, if and when the sending operator sent an error, the operator would not have to stop his/her copy to erase the word. Copying behind was a normal procedure back then and trained themselves to be that kind of an operator with the greatest interest to be the best.

Today its a hobby and take it or leave it.

W6TH

.:
 
RE: Bandwidth versus Keying Speed Revisited  
by N6KYS on August 16, 2007 Mail this to a friend!
by W6TH on August 16, 2007:

>>>>As I repeat myself;
This is a debate between Engineers and never to be Engineers.

Math says yes, to the post, but the Oscilloscope says, hey, its too fast for me to decide.

In Geometry of two vertical lines, what does this information tell you?

Perhaps this is just as well as the logic of the proof is subtle, (So slight as to be difficult to detect or describe), in any case some do not have a burning desire to see the proof..It is helpful to first go through arguments presented with specific numerical examples.

Theory, math and then design a piece of equipment to meet the needs. Then again, train to meet the goals of this new equipment<<<<


OK, so NOW you want to stay on track here (don't see your insistance in beating up SSB this time....oh, don't forget AM Vito, I mean....let's really get this thing rolling, you troll). Well, let's stay on topic, shall we? If the entire gist of your remarks regard "vertical lines", then you need to adjust your horizontal gain and really have a look. Dismiss the math and science all you want, but it's what has produced the greatest levels of technology the world has ever known.....NOT from old trolls who think that the years have somehow provided them a greater insight of practical understanding...but in fact, who've never bothered to apply themselves with formal education in math or the engineering sciences. What a joke.

 
RE: Bandwidth versus Keying Speed Revisited  
by N6NKN on August 16, 2007 Mail this to a friend!
Yes,

Vito does have a knack for bringing out the best in people.
 
To key...or not to key...*that* is the question!  
by N4QA on August 16, 2007 Mail this to a friend!
Next time we tap out "FB", we should all stand for a salute to our timeless leader, Samuel Finley Breese Morse or, ol' 'FB', for short.

How could he know, yet I imagine he suspected, that his code would still be in use in AD 2007 and beyond ?
He *may* also have felt, at various times, a key-click or two...

No relay in QSK

72,
Bill, N4QA
ps
"Easy to be hard"
 
RE: Bandwidth versus Keying Speed Revisited  
by W6TH on August 16, 2007 Mail this to a friend!
.
n6nkn, n6kys

Be good boys and I will see what I can do to dumb the tests down for you two boys, to advance to extra.

Now give us your opinion on how to describe the effects of "Bandwidth versus Keying Speed".

I then will have a few questions to ask you.

Actually, I am not so fond of your education as I am in concern of your process of civilizing or reaching a civilized state.

.:
 
Bandwidth versus Keying Speed Revisited  
by AB0WR on August 16, 2007 Mail this to a friend!
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
This thread and the incredibly long and tedious earlier thread all revolve around the exact meaning of "occupied/power bandwidth". I don't think this number is very relevant to us as radio operators, never mind the fact that the FCC refers to the term, or something similar, because what we really care about is the instantaneous bandwidth of a CW signal during the "key down" and "key up" transitions.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

I had hoped that Mickey's prior thread had put to rest the concept of an "instantaneous bandwidth" for a CW signal.

The instantaneous bandwidth and the "long term" bandwidths are the same for a keyed signal with a regular period. Actually they are the same for a keyed signal with a non-regular period but the calculations get a lot messier.

The bandwidth of a keyed CW signal can be determined by a Fourier analysis. The frequencies (e.g. the odd harmonics of the keying speed) exist during the entire keying pulse. From the start to the finish.

I know this concept can be hard to believe but it is reality. Those harmonics don't magically appear at the start of a pulse, go away during the middle of a pulse, and then magically reappear at the end of a pulse.

What *does* magically appear and disappear are intermodulation products produced from keying pulses in a non-linear amplifier stage. It is *those* products that you hear khz away from a CW signal, not the odd harmonics of a shaped square wave. It is exactly the same thing you hear with a splattering SSB signal. If we started calling the typical key clicks "CW splatter" like we call the intermod that SSB transmitters generate "SSB splatter" we would be closer to understanding what is going on.

The number the FCC defines is quite relevant to us as amateurs. You *have* to pick some defintion for bandwidth in order for everyone to be working from the same baseline. The 99% figure puts the occupied bandwidth out to about the 25th harmonic of the base square wave keying rate if I remember correctly. The level at which that harmonic actually gets transmitted is very dependent on the shape of the keyed pulse (i.e. how narrow of a filter with what response slopes do you run it through?). In effect, the shaping moves the 99% point to lower harmonics making the bandwidth narrower.

Yes, those higher number harmonics can interfere with close in competing signals, even with shaped keying pulses. A 1.8khz SSB signal can interfere with adjacent signals if it overlaps with the spectrum occupied by the adjacent signal. We don't consider that 1.8khz bandwidth signal to have some "inherent problem" do we? Why would we consider a 30wpm CW signal with an occupied bandwidth of 150hz to have an "inherent problem"?

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by N9CYS on August 16, 2007 Mail this to a friend!
Amen to that!
 
RE: Bandwidth versus Keying Speed Revisited  
by N6NKN on August 16, 2007 Mail this to a friend!
Sorry Vito,

I'm keeping my Advanced ticket. It proves I passed a CW test to get licensed. Only Advanced class operators can make that claim.
 
RE: Bandwidth versus Keying Speed Revisited  
by W9PMZ on August 16, 2007 Mail this to a friend!
An intersting use of test equipment for sure. Seems to me that this test equipment has evolved over the past 10 years to measure occupied bandwidth for CDMA signals.

But then again I am in the busisness of testing CDMA telecommunication systems (your cell phone).

An isnsturement designed to measure multiple MHz's of bandwidth to measure an obselete (but none the less still viable) modulation scheme.

73,

Carl - W9PMZ
 
Bandwidth versus Keying Speed Revisited  
by N0AH on August 16, 2007 Mail this to a friend!
This is really a lot about nothing..............unless you have a whine complaint during a CQWW contest, or a pile up,this point is mute. Too many radios out there to make this a perfect world where this subject would even make the map on most reflectors- But I'm sure MFJ will come up with something to fix it for you for under $99.95. After all, they do make the MFJ-1026..........
 
Bandwidth versus Keying Speed Revisited  
by N0AH on August 16, 2007 Mail this to a friend!
Hey,

Maybe K5MC and W8JI can discuss this at Tom's nest on 1.824MHz-
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 16, 2007 Mail this to a friend!
KX0R - I'm wondering if the term "occupied/power bandwidth" in K5MC's posts, also measured by the Agilent 9020A's function, really has the same meaning as what was intended when the FCC regulations were written. I have a hunch, but no proof, that the FCC used the term "occupied bandwidth" to mean the practical bandwidth of a signal, in other words, the stuff you can actually hear as you scan through the signal with a narrow filter.

K5MC - Yes, the 99% occupied bandwidth measured by the 9020A really has the same meaning as what was intended when the FCC regulations were written. The FCC likely "borrowed" the definition of occupied BW from the ITU. In the ITU's "Spectra and Bandwidth of Emissions" document (Rec. ITU-R SM.328-10), the following definition appears:

"Occupied Bandwidth - The width of a frequency band such that, below the lower and above the upper frequency limits, the mean powers emitted are each equal to a specified percentage B/2 of the total mean power of a given emission. Unless otherwise specified by the Radiocommunication Assembly for the appropriate class of emission, the value of B/2 should be taken as 0.5%."

From the ITU definition above, we see that the occupied BW uses the concept of mean (average) power, not the peak or maximum instantaneous power. That doesn't mean, as W8JI continues to suggest, that we have to wait for minutes, hours, or even years to make the measurement! That is total nonsense for the keying speeds I have been using in my calculations and measurements. W8JI either is trying to mislead other unsuspecting hams or he is truly clueless himself. At 2.4 wpm (1 dit per second), the time period is exactly 1 second. At 30 wpm (12.5 dits per second), the time period is exactly 80 milliseconds. So the theoretical minimum times to make the measurements are 1 second and 80 milliseconds, respectively, for 2.4 wpm and 30 wpm.

The actual time to make each of my measurements with the 9020A was around 10 seconds or so; certainly much less than the 90 seconds used by W8JI. Averaging more than 10 "runs" did practically nothing to change the occupied BW values.

WA0LYK - This is why the ARRL's recent proposal that started with "occupied" bandwidths was so dangerous. As you can see from these threads, it is obvious that not much thought was given when choosing this recommendation. Luckily, the comments from many of us convinced the ARRL that they should at least use "necessary" bandwidth since it is a calculated value, not measured. Otherwise, many of us would have been sending our rigs to Mickey to have them measured to see if they could meet the required occupied bandwidth.

K5MC - Jim, I actually started out thinking that the ARRL proposal on bandwidth was probably OK, although I did note a number of serious concerns in my comments to the FCC. After reading the various comments posted following my earlier eHam article, however, I've pretty well changed my mind. With such "experts" as W8JI lecturing the rest of us on the meaning and significance of the FCC's occupied bandwidth, I think we had better leave well enough alone.

BTW, the Agilent technical support engineer I've corresponded with concerning W8JI's posted occupied BW data seems to be as puzzled as I am. I've also asked Ed Hare at ARRL Headquarters about any occupied BW data on CW transmitters tested by the ARRL Lab, but I've received no data from Ed so far.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 16, 2007 Mail this to a friend!
Keep in mind this data is only for strings of dots. It is confusing to describe this as a "CW signal" and talk in terms of "wpm". Most of us use CW to send text messages, not strings of dots.

-Send a string of dots at 30 wpm and you'll measure one bandwidth. Leave keyer at same speed setting and send a string of dashes and you'll measure a different bandwidth. Sqeeze an iambic paddle to send dot-dash-dot-.. and you'll measure something different again. Most importantly: Send a text message at 30 wpm with random spacing between characters & words and the bandwidth will be much less than a string of dots sent at equivalent keying rate.

This is why from practical standpoint many of us are more interested in the bandwidth increase caused by the wave shape. I find it more interestng to compare the rows instead of the columns in the table. The bandwidth increase with square wave keying is significant.

73, Chris VA3NR (now QSYing!)
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 17, 2007 Mail this to a friend!
VA3NR--Send a string of dots at 30 wpm and you'll measure one bandwidth. Leave keyer at same speed setting and send a string of dashes and you'll measure a different bandwidth. Sqeeze an iambic paddle to send dot-dash-dot-.. and you'll measure something different again. Most importantly: Send a text message at 30 wpm with random spacing between characters & words and the bandwidth will be much less than a string of dots sent at equivalent keying rate.

AB0WR-I had not considered this concept in my other reply. Is *this* the "instantaneous" bandwidth other posters are thinking of? That at any point in time during a transmission the bandwidth is dependent on whether a dit or dah is being sent?

VA3NR-This is why from practical standpoint many of us are more interested in the bandwidth increase caused by the wave shape. I find it more interestng to compare the rows instead of the columns in the table. The bandwidth increase with square wave keying is significant.

AB0WR - If a consistent element length is being used, the bandwidth will vary between two bandwidth points, the widest point determined by the dit length and the narrowest side determined by a dah followed by an interword length (about two dahs) coupled with a following dah. The shape of the keying pulse will determine what the actual 99% bandwidth points are for these two limits. (the actual lower limit will be determined by the time between your transmissions but we can't measure frequencies that low so from a "practical" standpoint this limit is irrelevant.)

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by N2EY on August 17, 2007 Mail this to a friend!
"I'm keeping my Advanced ticket. It proves I passed a CW test to get licensed. Only Advanced class operators can make that claim."

Well, actually, no.

All Novices can make that claim too.

Plus all US hams who can show they were tested before the various dates of code-test reduction and elimination.

And the test in question is the 5 wpm code test.

So what?

Which is the more definitive demonstration of a skill:

1) Having passed a one-time test of the skill under controlled, predictable conditions

or

2) Demonstrating the skill in actual use under a wide variety of real-world conditions?

I say 2)

btw, I passed all my FCC license exams except the Novice at the FCC office in front of an FCC examiner. 20 wpm one-minute-solid receiving with a pencil and paper, 20 wpm sending with a straight key. No big deal - actual on-the-air operating skill is what matters most.

The test is just the beginning.

73 de Jim, N2EY
 
RE: Bandwidth versus Keying Speed Revisited  
by WB2WIK on August 17, 2007 Mail this to a friend!
>RE: Bandwidth versus Keying Speed Revisited Reply
by N2EY on August 17, 2007 Mail this to a friend!
"I'm keeping my Advanced ticket. It proves I passed a CW test to get licensed. Only Advanced class operators can make that claim."

Well, actually, no.


And the test in question is the 5 wpm code test.<

::I don't think that's true. As far as I know, the Advanced test was 13 wpm and never changed to anything less. The license class was eliminated (for new applicants) when the overall code proficiency test was downgraded to 5 wpm. I believe every single Advanced Class licensee had to pass a 13 wpm test.

WB2WIK/6
 
RE: Bandwidth versus Keying Speed Revisited  
by W3HF on August 17, 2007 Mail this to a friend!
"I believe every single Advanced Class licensee had to pass a 13 wpm test."

Nope. There used to be a medical exemption for at least the "higher level" of CW tests. I don't think there was an exemption at the 5 wpm level, but some licensees never had to take the 13 wpm test to upgrade. And I don't believe it's possible to determine "who did and who didn't" from any easily-available data. (It might be in the FCC archives, and accessible via search by the FCC records contractor--I'm not sure.)

That may be what Jim was referring to. And I will leave it up to the reader's judgment whether EVERYONE who received this exemption was truly entitled to it.
 
RE: Bandwidth versus Keying Speed Revisited  
by W5HLH on August 17, 2007 Mail this to a friend!
"I'm keeping my Advanced ticket. It proves I passed a CW test to get licensed. Only Advanced class operators can make that claim."

Sorry, Mr. Rocket Scientist, but I can make that claim and you can verify it for yourself by tracing my "ancestry" through the files of Buckmaster and/or the Radio Amateur Callbook. . . . from W5HLH to W7HLH, AK6C, AA6FW, KR2H, KA5M, and finally to 1978 when I upgraded to Extra and received AK4P. That not only was back when you had a 20 wpm code test for Extra, it was when you actually had to learn the theory instead of memorizing the test questions and answers. Heck, you even had to go to a FCC office (such as Norfolk, VA, like I did) to take the General or higher class exam.

I don't believe we need a CW test for a ham license, but, based on remarks like yours, I sometimes think we need a logical thinking (or common sense) test!
 
RE: Bandwidth versus Keying Speed Revisited  
by W3HF on August 17, 2007 Mail this to a friend!
"And I will leave it up to the reader's judgment whether EVERYONE who received this exemption was truly entitled to it."

I would also point out that portions of the amateur community weren't thrilled about the ease with which someone could get a medical exemption. In 1997, the ARRL requested a rules change in the process.

http://www.arrl.org/w1aw/1997-arlb059.html

Note especially the usage of the word "abuse" [of the waiver process] in paragraphs 3, 4 and 6.

I believe this request was dismissed by the FCC without action.
 
RE: Bandwidth versus Keying Speed Revisited  
by W3HF on August 17, 2007 Mail this to a friend!
(I should learn to do all my research BEFORE I hit the send button.)

"I believe this request was dismissed by the FCC without action."

Their response was actually embedded in the massive restructuring that took effect in April 2000. The FCC specifically said that they would take no action on the ARRL's request, as the elimination of the higher-speed CW tests meant that there would no longer be a provision for medical waivers.
 
RE: Bandwidth versus Keying Speed Revisited  
by KC8VWM on August 17, 2007 Mail this to a friend!

Pfffft!

I never passed any CW test and yet I use CW...

Strange huh?

I don't use CW because of license class and I don't need a higher license class as authorization to use it either for that matter.

I was forced to learn and use CW out of sheer necessity for operating VHF DX. My incentive wasn't for a piece of paper to hang on the wall, it was simply to make more contacts.

I found out that license class means nothing about any individuals particular ability to copy and use Morse Code.

Heck, I can recall many instances were many pre no code hams with higher license classes could not carry on a conversation with me using CW.

??

So yes, I agree... The "badge of honor" and proficiency in CW is about actually using the skill on the air. It's not about a dated piece of paper hanging on the wall covered in cobwebs that indicates at one time or another someone learned Morse code to pass a test.

That paper only proves that at one time or another the individual arrived at the starting point in CW. It doesn't draw into any final conclusions concerning the individuals current ability to operate using CW.

--... ...--

Charles - KC8VWM
 
RE: Bandwidth versus Keying Speed Revisited  
by N2EY on August 17, 2007 Mail this to a friend!
">And the test in question is the 5 wpm code test."<

"I don't think that's true."

It's true.

"As far as I know, the Advanced test was 13 wpm and never changed to anything less. The license class was eliminated (for new applicants) when the overall code proficiency test was downgraded to 5 wpm. I believe every single Advanced Class licensee had to pass a 13 wpm test."

No, they did not. They all had to pass a 5 wpm code test, though - same as Novices.

In 1990, FCC created medical waivers (in the form of a note from a doctor) for the 13 and 20 wpm code tests. They did it because a now-dead King who was also a ham asked President Bush I for a favor, and the 'request' was passed to the FCC.

When the 13 and 20 wpm code tests went away in 2000, there was a window of opportunity where a person could get a 5 wpm Advanced *without* a waiver, using CSCEs.

All ancient history now.

73 de Jim, N2EY





 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on August 17, 2007 Mail this to a friend!
K5MC and W8JI and others of their ilk are doing the right thing. That is, they are keeping good technical dialog going with lots of interesting arguments back and forth. This is where we learn. No, I am not saying that we listen to the loudest voice but arguments and issues brought forth in the debate are good starting points for following up on the theory, the practice, the measurements, and so on.

I know some of the readers here don't care about this stuff but I do and so I enjoy reading these threads -- keep it up.
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on August 17, 2007 Mail this to a friend!
"I'm keeping my Advanced ticket. It proves I passed a CW test to get licensed."


Why is it necessary to "prove" you passed a test. Most of the CW operators on the bands that I hear are faster then me and I am comfortable now with 20 wpm. I have no idea whether any of them passed a test or not -- sure, I assume they did but really, does it matter?

I think that if you operate CW and never took a required CW test but rather taught yourself and operate because you want to says a lot.

I came back to CW this last January and I am loving it every day. It is now my main mode. I use SSB so infrequently at times that I stumble over my words and ask others to speak more slowly so I can keep up.
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 17, 2007 Mail this to a friend!
Let me make one last attempt at explaining where Mickey and I disagree. I'm not going to stay in this because it already ran its course once, and I'm SURE the very same people will all have the same view.

But for the lurkers, here is a fresh approach to understand the difference between my view and Mickey’s view.

Imagine a clean pure carrier on for an infinite time. If we look at the energy in that carrier is has some finite value over a given time period.

Now when we change the level of that carrier sidebands are generated. No matter how slow or how fast we do that, we generate at least two sidebands. One sideband is upper, and one is lower.

The frequency spacing of that sideband depends ONLY on how fast we slope the level of the carrier. Nothing else. If we sloped it up and down at a 1mS rise and fall rate, the complete cycle would take 2ms. The sidebands, if it was a raised sine wave shape, would appear 500Hz above and below the carrier.

This is true on the leading edge, and it is true on the falling edge. If the shape of level change is not a sin shape, there would be additional sidebands generated that would be wider.

I'm sure we all agree to this point, or we had better, because that's how an amplitude modulated carrier works.

Now here is where we part company.

If we look at that carrier over a long period of time and if the duty cycle does not change, and if we look at the accumulated energy over that period, the energy remains the same as long as the off and on times are the same ration. But the FASTER we key the carrier off and on the more ACCUMULATED energy appears in the sidebands over that that long window of time where we accumulated energy from the carrier.

If we do that, it APPEARS that the sideband energy increases in proportion to the carrier. Now if we take the bandwidth as the ratio of accumulated power in the carrier over a long time compared to accumulated energy in the sidebands, we see the power bandwidth INCREASES. Mickey and I both agree on this also, as near as I can tell.

But this is where we part in agreement. The FCC does NOT want us to use the accumulated energy over long periods of time; they are interested in PEAK envelope power on adjacent channels. If we look at peak envelope power it does not change with speed for a fixed shape and length of rise and fall no matter how fast or slow we send.

This is also why my measurements disagree with Mickey, because I'm using an instrument that measures what the FCC wants and what is important to us.

Our receivers, by definition, cannot store that energy. Some spectrum analyzers can, and I have a selective level meter that has an average power function with a long accumulation time that would show what Mickey claims.

But the equipment we use to receive doesn't work like the math Mickey uses.

This is why people who send dots fast and tune a receiver off hear the clicks just as objectionable and strong the same distance away whether their rig transmits at 10 WPM or 40 WPM. The receiver by definition cannot accumulate power over the length of a dot or dash, or anything remotely close to that time, or we would hear "mush" from the speaker.

This is why some arguments can NEVER be resolved. It isn't that anyone is wrong, it's just that some people deal with the real world and real world effects and others work in theory alone.

I invite everyone to resolve this for yourself.

Run a transmitter into a dummy load and loosely couple a receiver with a good CW S meter into the load. Now vary the speed as you tune off frequency where you ONLY hear the clicks.

What you will notice is the clicks remain the same level and you hear them the same distance off frequency as long as the speed is a reasonable CW speed, say from a few words per minute to 40 or 50 words per minute.

The only thing that can make the bandwidth you observe change is if the ALC or some other circuit in the transmitter modifies the rising and falling edge.

We can all argue and debate until the cows come home, but Mickey and I will likely always disagree because Mickey wants to use a long-term power comparison and I want to use something that represents the real world system and how it affects users of that system.

It's simply a difference between theory and practice.

73 Tom
 
RE: Bandwidth versus Keying Speed Revisited  
by KC8VWM on August 17, 2007 Mail this to a friend!
If we look at the accumulated energy over that period, the energy remains the same as long as the off and on times are the same ration. But the FASTER we key the carrier off and on the more ACCUMULATED energy appears in the sidebands over that that long window of time where we accumulated energy from the carrier.

-----------

This only seems reasonable and I agree it's related to the timing of the signal. However, could it also be how it's actually sampled by the measuring equipment that we are observing? Does it have any limitations or does it skew the observed results in this equasion?

...Just Thinking.

73 de Charles - KC8VWM
 
Bandwidth versus Keying Speed Revisited  
by AB0WR on August 17, 2007 Mail this to a friend!
I don't have much time till Sunday so I'll make this quick.

W8JI is mushing together an explanation from a time domain representation that really needs a frequency domain explanation.

You change the level of that carrier by mixing it with a keying waveform. It is that mixing process that generates the sidebands. Each of the harmonics making up the keying waveform is mixed with the carrier. It is a multiplication process in the time domain and a convolution in the frequency domain. It is the relationship of the power in the harmonics present in the sidebands that determine the slope of the envelope seen in a time domain respresentation (i.e. an oscilliscope)

Those sidebands don't have power only during the leading and trailing edge of the time domain envelope but during the *entire* duration of the *entire* time domain envelope -- i.e. the keying pulse.

You simply *have* to look at the frequency domain and the power existing in each of the harmonics of the signal in order to determine the 99% power bandwidth.

But if you have a truly linear system with constant inputs the harmonics of the keying waveform that make up the sidebands will be there all the time to be measured. It should make little difference whether you sum powers over 1 keying pulse, 10 keying pulses, or 100 keying pulses. The power relationship between the harmonics in those sidebands won't change and therefore your 99% power bandwidth won't change.

If your power bandwidth changes as you vary the sampling time you need to look elsewhere for an explanation of why the change is occuring. It isn't because the number of harmonics change over time and it isn't because the power relationship of those harmonics change over time.

tim ab0wr



 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 17, 2007 Mail this to a friend!
Let me add as a further explanation: Consider the modulation envelope of an AM signal. Put one tone in and you get an upper sideband, a lower sideband, and a carrier. Look at this on an oscilloscope (a time domain instrument) and you will see a modulation envelope just like you see a modulation envelope with carrier keyed by a keying pulse.

That modulating tone for the AM signal doesn't appear at the start of the envelope, disappear in the middle of the envelope, and then reappear at the end of the envelope. It is there the whole time.

So are the modulating harmonics associated with the keying waveform that are mixed with the CW carrier to, in essence, send an AM signal that we call CW.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by VK3GJZ on August 17, 2007 Mail this to a friend!
This is very sad. For those who are technically literate the subject is neither complex nor controversial.

If we were to design a CW system for machine copy (using the narrowest possible bandwidth) we would use an identical "matched filter" in
both TX and RX. The shape would probably be gausian and its bandwidth would most definitely change with keying speed.

The envelope of this "perfect signal" (when sending dits) would have the outline of a sinewave.
Unfortunately it would sound lousy to the ear. It would sound far too soft and mushy.

A more suitable waveshape for the ear is something like the exponential attack and decay seen in most text books.
This is much wider than is strictly necessary. It even has sharp corners.

The bottom line is this: If we designed a truly narrow-band system (AKA PSK31), then bandwidth would be exactly proportional to keying
speed.

In practice however we use a much sharper shape (our ears need the clicks!) and so the bandwidth is much wider than necessary, and of
course the B/W stays largely constant.

It is only at improbably high speeds that the keying rate can effect bandwidth in a conventional rig.


Regards ............... Zim ............ VK3GJZ
 
RE: Bandwidth versus Keying Speed Revisited  
by MACKAY3031 on August 17, 2007 Mail this to a friend!
K5MC is right.The knowledge is not new.The attacks on it are bizzare and somewhat scary.How does misunderstanding theory advance the state of the art?It just seems so basic to me, and far too boring to debate.
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 17, 2007 Mail this to a friend!
W8JI - Let me make one last attempt at explaining where Mickey and I disagree. I'm not going to stay in this because it already ran its course once, and I'm SURE the very same people will all have the same view.

K5MC - As far as I'm concerned, it would be great if this post was your "last attempt" to explain the concept of occupied bandwidth.

W8JI - But for the lurkers, here is a fresh approach to understand the difference between my view and Mickey’s view.

K5MC - It's not the difference between W8JI's view and my own, it's the difference between W8JI's view and essentially the rest of the world! I can assure you that the vast majority of electrical engineers who work in the communications field, including the engineers who work at the FCC, are in that latter group. The professional electrical engineering literature, along with recent editions of the ARRL Handbook, are also in that latter group.

A rather easy way out for W8JI would be for him to simply acknowledge the fact that the "key-click" bandwidth is not the same thing as the occupied bandwidth as defined by such agencies as the FCC and the ITU.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 17, 2007 Mail this to a friend!
MACKAY3031 - K5MC is right.The knowledge is not new.The attacks on it are bizzare and somewhat scary.How does misunderstanding theory advance the state of the art?It just seems so basic to me, and far too boring to debate.

K5MC - Thanks, Michael, for your comments. The "attacks" on my original eHam article by W8JI and his troops are rather scary. I find the apparent inability of these folks to grasp the important distinctions between the FCC's occupied bandwidth and the rather vague "key-click" bandwidth to be quite sad. I'm guessing that my posted spectrum plots of occupied bandwidth for my four transceivers have brought a few hams around, but there appears to be no hope for W8JI or his most dedicated admirers.

If the ARRL bandwidth proposal was still open for comments, I would feel compelled to tell the FCC that ham radio in this country isn't ready for such a move. All I would have to do is quote a few statements from W8JI's theory of signal analysis and the FCC engineers would understand exactly where I'm coming from.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
In case anyone is still interested, I've posted two additional spectrum plots at www.arrl.org/sections/?sect=LA Figure 16 shows the spectrum of my Kenwood 940 at approximately 2.5 dits per second (6 wpm) with the Agilent N9020A analyzer span set equal to 100 Hz. Figure 17 shows the spectrum of the 940 at approximately 22 dits per second (53 wpm) with the span set to 400 Hz.

At 2.5 dits per second, the spacing between adjacent sidebands should be 2.5 Hz. If one examines the plot in Figure 16 carefully, you will see that the first upper sideband is indeed 2.5 Hz higher than the carrier and the first lower sideband is 2.5 Hz lower than the carrier. At 22 dits per second, the spacing between adjacent sidebands should be 22 Hz, which agrees quite well with Figure 17.

Just as the occupied bandwidth measurements shown in Figures 6 through 15, the spectrum plots in Figures 16 and 17 are consistent with the Fourier series models that I used in my earlier eHam article.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by KC8VWM on August 18, 2007 Mail this to a friend!
Seems like figure 14 shows a nice clean slope and looking at figure 12 seems to indicate sending at 6wpm takes up much more bandwidth than sending at a higher 10wpm data rate.

What's up with figure 14 at 10wpm and why does it appear cleaner than 6wpm?

???


 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
KC8VWM - What's up with figure 14 at 10wpm and why does it appear cleaner than 6wpm?

K5MC - Figure 12 is the 99% occupied bandwidth (about 67 Hz) of my Ten Tec Corsair at 6 wpm. Figure 14 is the 99% occupied bandwidth (about 72 Hz) of my Ten Tec Orion II at 10 wpm. The rise/fall times on my Orion II, which can be adjusted by the operator, were equal to 8 milliseconds (the normal setting I use on the air) for Figure 14 and Figure 15. The rise/fall times on my old Corsair are fixed at around 4 or 5 milliseconds.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
Imagine the following scenario:

Joe is a new ham anxious to operate CW, but he's just getting his feet wet and can't manage much more than 5 wpm code speed. Joe happens to own an Icom 751A and is wondering how its "occupied" bandwidth compares to its "necessary" bandwidth. Joe understands the following from the FCC rules:

97.307(a) No amateur station transmission shall occupy more bandwidth than necessary for the information rate and emission type being transmitted, in accordance with good amateur practice.

Joe then does some further reading of the FCC Rules (Part 2) at http://www.access.gpo.gov/nara/cfr/waisidx_06/47cfr2_06.html There he reads the following definition:

Occupied bandwidth. The frequency bandwidth such that, below its lower and above its upper frequency limits, the mean powers radiated are each equal to 0.5 percent of the total mean power radiated by a given emission.

Now Joe is a pretty sharp guy, but he realizes that he doesn't really understand the definition of occupied bandwidth. Joe then reads the following definition of "necessary" bandwidth:

Necessary bandwidth. For a given class of emission, the minimum value of the occupied bandwidth sufficient to ensure the transmission of information at the rate and with the quality required for the system employed, under specified conditions.

Joe's not too sure about this definition, either, but he does see the following simple equation provided by the FCC to calculate the necessary bandwidth for CW telegraphy:

Bn = BK

where Bn is the necessary bandwidth

B is the modulation rate (symbol rate) in bauds

K is either equal to 5 (fading circuits) or 3 (non-fading circuits)

Since Joe is a new ham, he's interested in sending at the rate of 6 words per minute. He reads in a recent ARRL Handbook that the commonly accepted ratio for bauds to words per minute is 1.2 B = wpm. Therefore, B = 5 bauds when the code speed is 6 wpm. So Joe then calculates that the necessary bandwidth is (5)(5) = 25 Hz if K = 5 and the necessary bandwidth is (5)(3) = 15 Hz if K = 3. Let's summarize these numbers below:

Necessary bandwidth Bn = 25 Hz for fading circuits at 6 wpm

Necessary bandwidth Bn = 15 Hz for non-fading circuits at 6 wpm

Now Joe is informed by W8JI that the "99% occupied" bandwidth of an Icom 751A is about 500 Hz regardless of sending speed. What is Joe's conclusion? If Joe takes everything that we've just discussed at face value, then he had better decide to NOT operate his 751A at 6 wpm because the necessary bandwidth is, at most, 25 Hz, but the "99% occupied" bandwidth specified by W8JI is about 500 Hz! As a matter of fact, Joe calculates that the minimum speed that he must send to satisfy 97.307(a) is 120 wpm for K = 5 or 200 wpm for K = 3!

Did everybody understand that last sentence? Joe must send at least 120 wpm for fading circuits and at least 200 wpm for non-fading circuits if W8JI is correct.

Joe sadly decides that he will have to increase his code speed some more before he will be ready to send CW on the air and still satisfy 97.307(a).

However, after reflecting for a moment, Joe (who, fortunately, is an independent thinker) begins to question the "99% occupied" bandwidth values posted by W8JI. Joe then decides to read up on the issue of "occupied" bandwidth and sending speed in his 2007 ARRL Handbook. On page 9.7 he reads the following:

2007 ARRL Handbook: The bandwidth occupied by a CW signal depends on the keying rate, with higher speeds requiring a wider filter to pass the sidebands. In addition, occupied bandwidth depends on the rise and fall time and the shape of the keyed RF envelope.

Joe also reads the earlier eHam article (Bandwidth versus Keying Speed) by K5MC that appears to be consistent with the ARRL Handbook. Joe notices that K5MC is very careful in his article concerning the various definitions of "bandwidth" and Joe further notices that K5MC quotes the definition of "occupied bandwidth" directly from the FCC rules along with the definition of the "99% power" bandwidth from what appears to be a reputable book on communication systems, at least according to its title as referenced by K5MC.

Joe also reads the follow-up article on eHam (Bandwidth versus Keying Speed Revisited) where K5MC has posted measured occupied bandwidth data of four typical CW transmitters used by hams. Just as the 2007 ARRL Handbook and K5MC's earlier article reported, the measured occupied bandwidth for each of the four tested transmitters does vary with keying speed.

After all of this reading, Joe decides that W8JI's values of "99% occupied" bandwidth for an Icom 751A must be misleading, if not downright ridiculous. Ignoring W8JI, Joe fires up his 751A by calling CQ at 5 wpm on 3.573 MHz.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
BTW, if Joe is willing to bring his Icom 751A to Louisiana Tech in Ruston, I will be very happy to measure its occupied bandwidth as a function of keying speed.

73, K5MC, Ph.D., P.E.
Professor of EE
LA Tech University
 
Bandwidth versus Keying Speed Revisited  
by W2WO on August 18, 2007 Mail this to a friend!
Too many people use "power" and "energy" loosely. That seems to be the heart of this disagreement. There is power in the key-click sidebands, but only for a few milliseconds. The total energy in the key clicks is small when averaged over time. Nevertheless, the power in the clicks affects receiver circuits (including AGC) and the listener.

It seems to me that both sides are correct, but are talking about different things: power vs energy.

Or did I get it backwards?

Bill - W2WO
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 18, 2007 Mail this to a friend!
W2WO:Too many people use "power" and "energy" loosely. That seems to be the heart of this disagreement. There is power in the key-click sidebands, but only for a few milliseconds. The total energy in the key clicks is small when averaged over time. Nevertheless, the power in the clicks affects receiver circuits (including AGC) and the listener.

It seems to me that both sides are correct, but are talking about different things: power vs energy.

Or did I get it backwards?



AB0WR:Energy is a "currency" if you will. For instance, you can have potential energy, i.e. a certain height above ground level will give you a certain potential energy. You can have kinetic energy, i.e. traveling at a certain speed in a car will give you a certain kinetic energy.

Work is a force applied over a distance. It is how you gain that potential energy or kinetic energy, i.e. climbing up a ladder has gravity (the force) applied to you over the distance you climb. Force x Distance = Work which creates an energy currency measured in newton-meters or joules.

Power is the rate of using energy or the rate of doing work. It is joules per second. If you will, it is equivalent to how fast do you climb that ladder to gain that potential energy.

So when you speak of the "energy" in a sideband it's probably really meaningless for our purposes. It's the power in those sidebands that mean something. For a one second sample interval, a "larger" signal ( in joules) will be able provide a higher power level (in joules/sec). But if you don't sample that signal you'll never *have* a power level.

So you are right, there is a small amount of energy in those higher level harmonics surrounding the carrier (the sidebands) or in the intermod products that are found far away from the carrier. That small amount of energy can, therefore, only provide a small power level over a measurement interval. That's why the concept of measuring bandwidth based on power is so important.

For a constant signal, however, the measurement period *should* be irrelevant. The "energy" relationship among the carrier and the harmonics making up that CW signal *should* remain constant and, therefore, so should the "power" relationship that will determine the power bandwidth.

In a straight squarewave keying signal, the odd harmonics extend out forever with an "energy" relationship determined by 1/n (n is the number of the harmonic). So the 25th harmonic is 1/25 smaller than the 1st harmonic (the basic keying rate). Using a filter (i.e. a gaussian shaped filter) we can knock down the "energy" level of these higher numbered harmonics even more. As we knock down the "energy" level of these higher numbered harmonics we decrease the 99% power bandwidth since less "energy" exists in the higher numbered harmonics to create power over a measurement interval.

As we raise the keying speed, those higher numbered harmonics get further away from the carrier (e.g. the 25th harmonic of 10hz is 250hz as opposed to 125hz for a 5hz keying rate). So it gets more important to filter the keying shape as your speed goes up.

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by W8JI on August 18, 2007 Mail this to a friend!
Mickey,

I think the problem here is you are using a very long term accumulation or average of signal levels, and I am using the peak envelope power in a 10Hz bandwidth as the analyzer slowly sweeps across the frequency.

I really don't think we will ever agree, and you might continue to think I am doing something wrong. But the fact is I am measuring the system as the system behaves in real life.

Now if I set an instrument up to read average power over a long time, I can make it agree with the position you take.

The problem is the system. We, or at least I, are trying to define the bandwidth at which interference occurs.

The receiver, by definition, has no memory. It cannot have a memory or we would not be able to copy CW. As such it doesn't really matter how often the wide sidebands appear except they may be more of a nuisance at higher speeds, but they won't have a higher peak power level at a given spacing away from the carrier as speed is increased.

It's very easy to measure that, or to prove it on paper.

This is why we disagree.

I am talking about the system and measuring the system as it applies to a REAL receiver, which by definition cannot average or store energy over a long time. It (and our ears) responds to the peak envelope power at any frequency.

For example I can use my HP selective level meter with a long term average power function to measure the sidebands of a CW transmitter, and the average power INCREASE with speed. This is because for a given time window the clicks (each of the SAME peak envelope power) appear more frequently.

That is NOT a valid or meaningful measurement as applied to the CW system, because the CW system all the way from the transmitter to the brain of the operator at the receiver cannot accumulate power in the sidebands. The response of the system to a changing signal has to be faster than the rise and fall, or it modifies the sound of the CW.

The only thing I hope in all of this is that we don't undo all the good work that has been done. Prior to all the attention on this, many rigs had a 1mS rise and fall with poor shaping. These rigs, no matter how fast or slow the operator sent, had very strong keyclicks a few kHz away.

I'm not sure, but I suspect the rigs were made that way because the designers thought speed governed the keyclick bandwidth.

It's really only been within the past several years they have been correcting that.

I hope you spend some time looking at the SYSTEM and how the system works. I think if you do that, you will see why I do not use any power averaging or accumulation of power but rater look at the peak envelope power at any spacing.

That is what the user hears and it is what bothers the receiver.

We'll just have to always disagree if you don't think that is valid. I think how the system behaves is what defines interference bandwidth. Not how we want it to behave.


73 Tom





 
RE: Bandwidth versus Keying Speed Revisited  
by K4ZMV on August 18, 2007 Mail this to a friend!
Like others, I really appreciate the technical discourse. I also think that we have two very well educated and experienced EE guys looking at a specific issue from different perspectives. What I really DON'T appreciate is the off-topic comments from trolls and the ad hominum responses K5MC posts. I waded through the previous post on this topic and all of this one. W8JI's responses have been generally civil. Why K5MC resorts to attacks is a mystery to me. Is it because he holds a Phd. and no one is permitted to have a technical disagreement with him?. As far as harming someone as K5MC suggests, this is a hobby, the power levels we use are, relatively speaking, low, and keying speeds are low (don't hear much CW over 20-30 WPM and all I use is CW). W8JI's point about people not being careful with the shaping factors in a waveform is absolutely to the point. It really is a practice, not a theory issue. How about you trolls getting lost and we keep to the technical issue at hand in a civil manner.
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
W8JI - I'm not sure, but I suspect the rigs were made that way because the designers thought speed governed the keyclick bandwidth.

K5MC - Perhaps W8JI is now willing to concede the point that the "key-click" or "interference" bandwidth is not the same concept as the occupied bandwidth defined by such governmental/professional agencies as the FCC and the ITU; if he is ready to concede that point, then he should also be ready to acknowledge the fact that the occupied bandwidth of a typical CW transmitter does vary with the keying speed as pointed out in recent editions of the ARRL Handbook and my two eHam articles. If W8JI agrees with these statements, then there's little left to "debate" as far as I'm concerned.

Even though the above "concessions" are the most I expect to ever read from W8JI (and I will be very surprised if he simply states that he agrees with them), perhaps W8JI is even now willing to acknowledge that the relationship between the occupied/power bandwidth and the information rate is a fundamental concept in all communication systems, including ham radio. If W8JI's operating interests included such activities as EME or QRSS, I'm sure he would never have made such brash statements that it is a "misconception" that the "bandwidth" of a CW signal varies with the keying speed. He would have been very careful to distinguish between occupied/power bandwidth and the so-called "key-click" bandwidth.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
K5MC - Below is an email I received last night from AI8H. I really appreciate Jeff's comments and his permission to post them on eHam.

73, K5MC

AI8H - A hearty 'Well Done' for taking your time to present a well deserved rebuttal to unfounded comments on your original article ... along with your careful research!

However, it seems that most recent comments are quite off subject. Specifically that key click stuff. And now code, vs no code (!!).

The thing that really grinds me is that both of your articles have a very simple title/subject: 'Bandwidth versus Keying Speed'. They both pertain to object lesson theory in basic information transfer. What do key clicks have to do with it? One might as well blend-in power supply hum on the CW note, lightning static on the circuit, or ear wax for it's relevance to the basic principles stated.

I think that your original article was meant to stimulate awareness and (hopefully) an understanding of basic theory; and to negate a previous article's statement that information rate versus bandwidth is a "misconception" because only key clicks matter.

All the Best,

Jeff AI8H
 
Bandwidth versus Keying Speed Revisited  
by W9AC on August 18, 2007 Mail this to a friend!
Another noteworthy consideration in this discussion is that the term "occupied bandwidth" has no relevancy with the FCC under Part 97. In fact, for the amateur service, 47 CFR 97.3(a)8) defines "bandwidth" as...

"The width of a frequency band outside of which the mean power of the transmitted signal is attenuated at least 26 dB below the mean power of the transmitted signal within the band."

Is this “bandwidth,” or even “occupied bandwidth?” When the ARRL was considering “regulation by bandwidth,” someone at the League pointed out the difficulty in defining it, and later on plenty of folks became concerned about how we as amateurs would measure it. Interestingly, the 99% or 99.5% "occupied bandwidth" definition as used by the FCC in other service classes, is actually inconsistent across the FCC rules. Under other areas of 47 CFR, I have seen the definition used as 99%, 99.5%, and “99% through a 100 kHz filter.” There may be others.

I fully agree that waveform, rise-time, and rate of information all play a part in the actual transmitted bandwidth of a CW signal. I just happen to believe that the bandwidth created by the rate of information (e.g., "dit" speed) at relatively slow transmission rates is masked by a much more powerful bandwidth component at typical CW speeds -- the envelope shape, which comprises the rise/fall time.

Clearly, the data shown by K5MC in his Table 1 does reflect what we measure with our receivers and spectrum analyzers, even at low RBW -- not even remotely close. I am not stating that his math is wrong; I never have. But there is a chasm of information between the various methods of measurement and the underlying theory. Sadly, the truth is being lost in this discussion due to inconsistent definitions, inconsistent measurements, and perhaps inconsistent application of theorems.

This entire discussion is a looser: Until some common ground can be established in terms of *defining* and *measuring* bandwidth, nobody will win this argument. Even the FCC cannot standardize its own set of definitions for the term “occupied bandwidth.”

Paul, W9AC
(not Kevin, not W9AF, not any one else)


 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 18, 2007 Mail this to a friend!
K5MC (May 27, 2007) - It is also interesting to note that the definition of bandwidth found in 97.3 is, in essence, the 99.75% power bandwidth. The definition actually says the following: "Bandwidth. The width of a frequency band outside of which the mean power of the transmitted signal is attenuated at least 26 dB below the mean power of the transmitted signal within the band."

W9AC - Another noteworthy consideration in this discussion is that the term "occupied bandwidth" has no relevancy with the FCC under Part 97. In fact, for the amateur service, 47 CFR 97.3(a)8) defines "bandwidth" as...

K5MC - W9AC reminds us that the FCC's definition of "bandwidth" in 97.3(a)(8) is not the "occupied bandwidth" defined in Part 2. The FCC is careful not to include the adjective "occupied" in 97.3, but it is a shame that the definitions aren't identical. In 97.307(a), we read that "No amateur station transmission shall "occupy" more bandwidth than necessary …" and so, perhaps, that bandwidth could be interpreted as the "occupied bandwidth" defined in Part 2.

The 99.75% power bandwidth, the definition used in 97.3(a)(8), will likely be significantly higher than the 99.0% occupied/power bandwidth. For example, the 99.0% power bandwidth (the occupied bandwidth) of my Kenwood 940 at 30 wpm is about 190 Hz; the 99.75% power bandwidth for the 940 at 30 wpm is about 355 Hz.

The idea of going from our traditional mode approach to a bandwidth regulatory scheme needs to be considered very carefully from many different angles! The ARRL clearly did not do its homework the first time around.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 18, 2007 Mail this to a friend!
Tim said:

"Let me add as a further explanation: Consider the modulation envelope of an AM signal. Put one tone in and you get an upper sideband, a lower sideband, and a carrier. Look at this on an oscilloscope (a time domain instrument) and you will see a modulation envelope just like you see a modulation envelope with carrier keyed by a keying pulse.

That modulating tone for the AM signal doesn't appear at the start of the envelope, disappear in the middle of the envelope, and then reappear at the end of the envelope. It is there the whole time.

So are the modulating harmonics associated with the keying waveform that are mixed with the CW carrier to, in essence, send an AM signal that we call CW. "

Tim, I am NOT going to get sucked back into this pointless discussion again. I will simply say this and be done. If you put a tone into an AM transmitter, it produces sidebands. Those sidebands are there for the entire duration of the *INPUT TONE* because for the entire duration of the input tone the carrier is being modulated. With a CW keying waveform, you have an initial time-varying rising edge (where the carrier is modulated), a constant period where the carrier is constant (in other words, unmodulated), and a final time-varying falling edge where the carrier is once again modulated. This is nothing like putting a constant tone into an AM transmitter and saying that the sidebands are there for the entire duration of the signal. A more accurate analogy with an AM transmitter would be keying the transmitter and putting in a tone for 5 ms, then turning off the tone (while the transmitter remains keyed) for some period of time, then turning the tone back on for 5 ms. And yes, the sidebands will exist ONLY during the first 5 ms and the last 5 ms. During the middle period where the tone is off, the sidebands will not be there. THIS is the parallel between an AM transmitter and a CW transmitter.

Do you really believe that if you transmit with an AM transmitter and put a tone in for a brief period, then turn the tone off for some period, then turn the tone on again that the sidebands will be there during the entire transmission?

73 - Jim
 
Bandwidth versus Keying Speed Revisited  
by KX0R on August 18, 2007 Mail this to a friend!
Thanks to all for a lot of careful writing and less emotion as we move through this interesting topic. I especially appreciate the clear responses from AB0WR and K5MC to what I posted.

VK3GJZ's post is worth re-reading for his concise summary. I agree that we like the sharp sound that results from keying that is faster than necessary; and we're generally using a much greater occupied bandwidth with our rigs than the necessary bandwidth. I don't agree that we like "clicks" - I think that "clicks" are sidebands that are wider and stronger than what we have with rigs keyed with rise and fall times in the range from roughly 3 to 10 ms, and not run through nonlinear amplifiers, with intermod products, etc.

I actually use exponential keying circuits (modulators) to key several of my rigs, and I've adjusted various parameters while watching my scope and listening to the signal with a non-overloaded receiver. I do like a sharper wave and sound than what's necessary to send the information. I doubt many amateurs have had this interesting experience; I'm also aware that this kind of keying may not be quite state-of-the-art.

A couple of comments to K5MC's post seem relevant:
If we examine the FCC's formula Bn = BK for necessary bandwidth, then the rigs measured by Mickey all seem to be using a lot more bandwidth than necessary. There has to be some room to breathe and be reasonable here...besides, we do like those sharper sounds, up to a certain point, don't we?

Another little jewel that K5MC revealed, though he doesn't dwell on it, is that the FCC allows a larger necessary bandwidth for fading circuits than non-fading circuits. Bn = BK, and the FCC says K = 5 for fading circuits and K = 3 for non-fading circuits. Why is this? Why does the extra bandwidth help when the signal is dipping into the noise? This idea seems to support the notion that we need sharper keying when band conditions are rough, although the bandwidths that come from the formula Bn = BK are much narrower than what most of us use.

Maybe when a signal is weakened by fading, and our center carrier is down near the noise floor, then the sidebands associated with keying are already in the noise, and not really audible. If we then sharpen the envelope of the keyed wave and increase the strength of our keying sidebands, they rise above the noise floor, and we can copy easier.

I think that most of the serious posters here have all contributed to a better understanding of the issues underlying these related topics, and most of what has been put here has helped us better understand keying and bandwidth.

We still have a lot of room to talk about why we like sharper keying, and whether it really helps us copy through noise or QRM.

George Carey Fuller
KX0R
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 19, 2007 Mail this to a friend!
KE3HO - With a CW keying waveform, you have an initial time-varying rising edge (where the carrier is modulated), a constant period where the carrier is constant (in other words, unmodulated), and a final time-varying falling edge where the carrier is once again modulated. . . . And yes, the sidebands will exist ONLY during the first 5 ms and the last 5 ms. During the middle period where the tone is off, the sidebands will not be there. THIS is the parallel between an AM transmitter and a CW transmitter.

K5MC - If you will look at Figures 16 and 17 at http://www.arrl.org/sections/?sect=LA you will clearly see that the keying sidebands are located as a function of the keying speed, not the rise/fall times. In effect, Figures 16 and 17 are an experimental "proof" of the Fourier series concept I used in my first eHam article. The rise/fall times for my Kenwood 940 are around 4 or 5 ms, which implies a fundamental frequency of around 200 to 250 Hz. If you were correct, at both 6 wpm and 53 wpm the first keying sidebands (the two immediately adjacent to the carrier) would be 200 to 250 Hz away from the carrier and the spacing between adjacent sidebands would be 200 to 250 Hz. Figures 16 and 17 clearly show that that is not happening. (And it didn't require minutes, hours, or years to make the measurement!)

Even though the carrier amplitude is constant for much of the time during the key down period, that doesn't mean that the "bandwidth" of the CW signal magically disappears during that time and magically reappears during the rise/fall times. This isn't a matter of being Tim's opinion or my opinion - it is something one clearly learns after studying Fourier analysis! It might not agree with one's intuition, but intuition can often lead folks down blind alleys. (Recall the story of the FM pioneers about their intuition of the bandwidth of an FM signal.)

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 19, 2007 Mail this to a friend!
KX0R - We still have a lot of room to talk about why we like sharper keying, and whether it really helps us copy through noise or QRM.

K5MC - I well remember copying W5GHP 35 years ago on our section traffic net on 3615 kHz. Bob's transmitter probably was about as close to square-wave keying as you could get. During the winter his signal often sounded like a hammer hitting against my head, but on those nights when the static was really high, his signal was definitely easier to copy than the boys with softer keying.

In addition to the "hardness" of the keying, however, we all need to remember that when band conditions are tough, having the sending operator slow down and using a more narrow filter on our receivers can often make the difference between copying or missing the message. As an old CW traffic handler, I don't hesitate to tell the sending operator to QRS when QRN is giving me fits!

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 19, 2007 Mail this to a friend!
KE3HO
""snip""
A more accurate analogy with an AM transmitter would be keying the transmitter and putting in a tone for 5 ms, then turning off the tone (while the transmitter remains keyed) for some period of time, then turning the tone back on for 5 ms. And yes, the sidebands will exist ONLY during the first 5 ms and the last 5 ms. During the middle period where the tone is off, the sidebands will not be there. THIS is the parallel between an AM transmitter and a CW transmitter.

Do you really believe that if you transmit with an AM transmitter and put a tone in for a brief period, then turn the tone off for some period, then turn the tone on again that the sidebands will be there during the entire transmission?

73 - Jim

======================
Your analogy is incomplete and incorrect. What we are discussing is when you turn the AM carrier on and off, not when you modulate it with a tone or even a partial tone, i.e. your 5 ms tone burst.

I've done some experimenting using audio freqs where I could use a "spectrum analyzer" software program and a sound card. This consisted of a 10 kHz sine wave through a class A amp (2n2222) that is switched off/on with another 2n2222 in the emitter lead. The switching transistor was driven with a function generator rather than a "key". The basic assumption here is that the "carrier" frequency really doesn't matter. What we are looking for are the harmonics of the keying waveform, not the harmonics of the carrier.

A couple of observations. First, the "sidebands" or more precisely, harmonics, generated are exactly what a Fourier analysis of a periodic waveform predict. Second, I could never find any "key clicks" generated regardless of the keying waveform shape.

Leaving aside the debate about power bandwidths these experiments generate a couple of conclusions.

First, since K5MC's pictures (as did my experiment) depict how well Fourier analysis of a periodic waveform predicts the harmonics, one must have empirical proof (as he was asked to provide) to show that some other process is occurring, i.e. the rise times predict the harmonics that are actually generated.

Second, the absence of key clicks with an audio freq carrier run through a class A amp with an adequate slew rate indicate that other processes within transmitters are occurring that result in the generation of key clicks.

The Icom 751a (751 and 761 too and probably others) actually keys the cw oscillator off and on with an exponential keying circuit. God only knows what IMD is present on an oscillator spinning up/down with a shaping control circuit in its emitter lead. In addition, these rigs also use ALC to "shape" the upper part of the cw waveform on the oscillator buffer amp. These are all non-linear processes and will generate IMD products. I suspect that many of the "key clicks" and wider than predicted bandwidths being observed come from these processes.

It is useless to try and define the output of these rigs mathematically when there are non-linear processes occurring all throughout the transmitter chain. One would need to examine each stage and determine its contribution when trying to calculate the overall transfer function. One can not ignore the fact that fast rise times introduce key clicks, however, the process that generates them may be IMD results rather than some non-periodic waveform analysis as some claim.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 19, 2007 Mail this to a friend!
KX0R:Maybe when a signal is weakened by fading, and our center carrier is down near the noise floor, then the sidebands associated with keying are already in the noise, and not really audible. If we then sharpen the envelope of the keyed wave and increase the strength of our keying sidebands, they rise above the noise floor, and we can copy easier.

AB0WR: Noise is typically a statistical phenomenon. It is distributed both in time and in frequency. That means that at any specific point in time the noise floor is not "solid" but is different at different freqencies.

I suspect that what you say is true, as you add harmonics some of them may occur at points in the spectrum where the noise floor is not quite as high at any specific point in time, e.g. at 3590100hz vs 3590300hz. This gives the ear and brain more "clues" as to what is happening with the signal thus allowing a better analysis.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 19, 2007 Mail this to a friend!
KE3HO:With a CW keying waveform, you have an initial time-varying rising edge (where the carrier is modulated), a constant period where the carrier is constant (in other words, unmodulated), and a final time-varying falling edge where the carrier is once again modulated. This is nothing like putting a constant tone into an AM transmitter and saying that the sidebands are there for the entire duration of the signal.

AB0WR:Jim, I posted multiple web sites that show how a square wave is generated by combining multiple odd harmonics. Those odd harmonics *do* exist during the ENTIRE duration of the square wave. Therefore they modulate the carrier just exactly like the tone I mentioned - they exist during the rise time of the envelope, during the constant portion of the envelope, and during the fall time of the envelope. Therefore the sidebands exist for the rise time, the constant time, and the fall time of the envelope and those sidebands are made up of all of the odd harmonics making up the keying wave.

If any reader is interested in taking a look at those web sites you can either post a message in this thread or email me and I'll happily send them to you.

You will find on all of these web sites that the odd harmonics making up a square wave do NOT only exist during rise and fall time of the square wave but for the total duration of the square wave. In fact, the frequencies of the square wave are determined by the duration of the pulses in the square wave, i.e. the period of each cycle in the wave.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 19, 2007 Mail this to a friend!
AB0WR:Jim, I posted multiple web sites that show how a square wave is generated by combining multiple odd harmonics. Those odd harmonics *do* exist during the ENTIRE duration of the square wave. Therefore they modulate the carrier just exactly like the tone I mentioned - they exist during the rise time of the envelope, during the constant portion of the envelope, and during the fall time of the envelope. Therefore the sidebands exist for the rise time, the constant time, and the fall time of the envelope and those sidebands are made up of all of the odd harmonics making up the keying wave.

Tim (and Mickey),
The harmonics that make up the square wave do not modulate the carrier during the middle portion between the rise and fall. Take any keying waveform that you like and calculate its spectrum using the Fourier Series or the Fourier Transform. All of the frequency components calculated do indeed exist for all time (from t = -infinity to t = +infinity). However, those frequency components *ADD* in the time domain to create the keying waveshape. During the middle portion of the keying waveform, they add to one (or some constant value).

I can't post graphs and diagrams in my reply here, and I am not going to piece together another paper and then impose upon Tom to put it on his website. So I will have to try to describe some examples to illustrate my point. I will try to keep this as short as possible.

Consider the following finite duration non-periodic waveform:

For -infinity < t < 0, the signal is zero.

For 0 < t < 1 second, the signal is a pure sine wave at 10 Hz and amplitude 1.

For 1 second < t < 2 seconds, the signal is a pure sine wave at 20 Hz and amplitude 1.

For 2 seconds < t < infinity, the signal is zero.

If you sample this signal at some suitable rate and then calculate its spectrum with the Fourier Transform, you will find a peak at 10 Hz, a peak at 20 Hz, and many much smaller peaks from near DC all the way out to the upper limit of your FFT calculation. Fourier theory says that all of these frequency components exist for all time (-infinity to +infinity). If you looked at this signal with a sweeping spectrum analyzer (as opposed to an FFT spectrum analyzer) you would see absolutely nothing prior to t = 0. Why is that when Fourier theory says that the frequency components exist for all time? Remember what the FFT calculates? the necessary frequency components to produce the original signal, and the original signal is zero prior to t = 0. All of the frequency components ADD to ZERO in the time domain prior to t = 0. From 0 < t < 1 second, the spectrum analyzer would show a single peak at 10 Hz. Where is the 20 Hz peak and all of the smaller peaks? They ADD to ZERO in the time domain, leaving ONLY the 10 Hz component during this period. How about from 1 second < t < 2 seconds? The 10 Hz peak on the spectrum analyzer would disappear at t = 1 second and would be replaced by a single peak at 20 Hz. Again, what happened to the 10 Hz peak and all of the other smaller spectral features? They ADD to ZERO in the time domain. Now, what happens after t = 2 seconds? According to Fourier Theory, all of the frequency components exist for all time. Will you see them on the spectrum analyzer at, say, t = 10 seconds? No. They ADD to ZERO in the time domain.

Now, you might say that even though the frequency components add to zero, they are still there. I say no. At least, not in the real sense. If you passed this signal through a low pass filter with a sharp cut off at 15 Hz, would you suddenly see a non-zero signal at t = 10 seconds? Of course not. Yet Fourier Theory says that the components are there, so shouldn't filtering out most of the components necessary to make a zero signal mean that you would no longer have a zero signal? No, because the spectral components calculated by Fourier Theory exist as discrete frequency components ONLY in the frequency domain. In the time domain, they exist ONLY as a composite sum. If they add to zero at some point in the time domain, they have no spectral components. If they did, you could filter out some of them and make the signal non-zero at t = 10 seconds. How about it t = 1.5 seconds? Would the 10 Hz component still be there because you had filtered out most of the components necessary to cancel it out? No.

This notion that because Fourier Theory says that the frequency components are constant and exist for all time means that the sidebands are constant and exist for all time is just wrong.

Another example: Say I take a microphone and audio amplifier and record my voice using a digitizing A/D at a suitable sampling rate as I read a paragraph from a book. The recorded signal would be a finite duration non-periodic waveform. As such, its spectrum could be calculated by FFT. The calculated frequency components are constant and exist for all time. Would you argue, based on that, that the sidebands produced by playing that recording into an AM transmitter would also be constant and exist for all time? If they did, then you could capture, say, 20 mS of the transmitted signal and have captured the entire paragraph that I read.

If you take an RF carrier and modulate it, sidebands are produced. If your carrier is un-modulated, no sidebands are produced. During the rise and fall time of a CW transmitter, the carrier's amplitude is time varying. It is modulated and is producing sidebands. During the middle portion of a CW pulse, the carrier is at constant amplitude, frequency, and phase. It is un-modulated and NO sidebands are produced. This is not "magic", it is not "intuition". It is physics, and it is in complete agreement with Fourier Theory.

73 - Jim
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 19, 2007 Mail this to a friend!
KE3HO - If you looked at this signal with a sweeping spectrum analyzer (as opposed to an FFT spectrum analyzer) . . .

K5MC - KE3HO appears to be saying in his latest comments that the Agilent N9020A analyzer I'm using is misleading me because it is an FFT-based instrument rather than the older style analyzers! Figure 1 posted at http://www.arrl.org/sections/?sect=LA shows the spectrum plots from an old all-analog analyzer. The position of the keying sidebands along the frequency axis depends upon the keying speed as I explained back on June 9. I've pasted some of my comments below.

K5MC (June 9, 2007) - According to SM0AOM, the symbol rate of the transmitter shown in the Telefunken literature is 40 baud. That means the sending speed is actually 48 wpm, rather than the 50 wpm approximation I used in the caption of Figure 1. Since we know the exact sending speed is 48 wpm, we can calculate the frequency locations of the various keying sidebands.

The fundamental frequency of the keying waveform is equal to the number of dits per second being transmitted. At 48 wpm, the number of dits per second is 20. Therefore, the first pair of keying sidebands will be located plus and minus 20 Hz away from the carrier frequency. Continuing in this fashion, we know that the second pair of sidebands are located plus and minus 40 Hz from the carrier, the third pair of sidebands are located plus and minus 60 Hz from the carrier, and so forth.

A close inspection of the spectrum plot on the left side in Figure 1 shows that every other pair of keying sidebands are much lower in amplitude than the others. These "missing" sidebands are the "even" harmonic pairs, starting with the second pair of sidebands at plus and minus 40 Hz from the carrier. The reason the even frequency harmonic sideband pairs are very low in amplitude relative to the others is that the square-wave keying waveform has so-called half-wave symmetry. One of the fundamental results from Fourier analysis of periodic signals is that those signals having half-wave symmetry will contain only odd-frequency harmonic components in their Fourier series.

With all of this information, we can now determine the frequency span in the Telefunken plot for the "unfiltered" transmitter. If you do it carefully, you will see that the total span of the spectrum plot is 2 kHz. We can also readily determine the 99.1% power bandwidth of the CW signal from the "unfiltered" transmitter. From my square-wave keying waveform study, I found that 99.1% of the total average power residing in the square-wave output signal is the sum of the carrier power and the powers of the first 21 harmonics symmetrically located about the carrier. If you examine the amplitudes of the two very distinct 21st harmonic sidebands on either side of the carrier (these specific keying sidebands are located plus and minus 420 Hz from the carrier), you will see that they are each about 27 to 28 dB lower in amplitude than the carrier. The frequency locations of these two specific keying sidebands are plus and minus 420 Hz away from the carrier because 21 (the harmonic "order") multiplied by 20 Hz (the frequency displacement of the first pair of sidebands located on each side of the carrier) is equal to 420.

The relative amplitude of these two specific keying sidebands located 420 Hz from the carrier is found from the Fourier series representing the "raised" square wave that I used in my article. The (time averaged) powers of the DC and 21st harmonic components in the Fourier series on a 1-ohm basis are the squared values of 0.5 volts and (1.414)/(21 pi) volts, respectively. These squared values turn out to be 0.25 watts and approximately 0.0004595 watts. Now let's compare the ratio of these two average powers on a log scale to find how many dB "down" these two specific keying sidebands are from the DC (carrier) term:

10 log (0.0004595/0.25) = -27.4 dB

That is, the two keying sidebands plus and minus 420 Hz from the carrier are 27.4 dB lower in amplitude than the carrier. This is essentially a perfect fit to the same two sidebands shown in the Telefunken literature for the "unfiltered" transmitter. We would also say that the 99.1% power bandwidth of this particular CW signal is 840 Hz because 420 Hz multiplied by 2 equals 840 Hz.

K5MC (August 20) - Note that the positions of the keying sidebands along the frequency axis are the same for both plots shown in Figure 1. If the keying sidebands are only generated during the rise/fall times, then we should see a noticeable difference in the spacing between the sidebands in the two plots in Figure 1 because the rise/fall times are obviously much greater for the "filtered" transmitter compared to the "unfiltered" transmitter. However, there is no difference because the keying speed determines the location of the sidebands, not the rise/fall times, and for both plots the speed is 48 wpm (20 dits per second).

KE3HO - During the middle portion of a CW pulse, the carrier is at constant amplitude, frequency, and phase. It is un-modulated and NO sidebands are produced. This is not "magic", it is not "intuition". It is physics, and it is in complete agreement with Fourier Theory.

K5MC - KE3HO needs to look very carefully at Figures 1, 16, and 17 posted at http://www.arrl.org/sections/?sect=LA so that he can see that the keying sidebands are located based on the keying speed, not the rise/fall times. (Figure 1 is from an old analog analyzer and Figures 16 and 17 are from the N9020A.)

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by WB1AIW on August 19, 2007 Mail this to a friend!
Say, let's talk about key clicks and sideband power and instantaneous amplitude and stuff.

Tim, AB0WR said:

"I know this concept can be hard to believe but it is reality. Those harmonics don't magically appear at the start of a pulse, go away during the middle of a pulse, and then magically reappear at the end of a pulse."

Actually, the sidebands are created by the rise/fall of the signal so they are only there *during* the rise and fall of the signal. They DO go away during the middle of the pulse, that's why you hear only the dreaded key click each time the cw element is started and/or stopped, and not for the full duration of the pulse.

Tim, AB0WR said:
"What *does* magically appear and disappear are intermodulation products produced from keying pulses in a non-linear amplifier stage. It is *those* products that you hear khz away from a CW signal, not the odd harmonics of a shaped square wave"

Nope. The non-linearity just changes the rise and fall SHAPES of the signal in the time domain, and hence, modifies the spectrum characteristics, but is not wholly responsible for them.

Tim, AB0WR said:
"Those sidebands don't have power only during the leading and trailing edge of the time domain envelope but during the *entire* duration of the *entire* time domain envelope -- i.e. the keying pulse."

Again: In fact, they DO only appear during the leading and falling edges... during the "on" period between the edges, the only signal you'll find is the carrier.

Tim, AB0WR said:
"That modulating tone for the AM signal doesn't appear at the start of the envelope, disappear in the middle of the envelope, and then reappear at the end of the
envelope. It is there the whole time."

It is because you are continuously changing the amplitude of the signal, that you have continuous sidebands.


Tim, frequency domain sidebands (which are found in the spectrum, aka frequency domain phenomena) are generated by changes to the carrier during the changes only.
When a person using AM stops talking, the sidebands disappear. When a cw operator presses and holds his key down, the are momentary sidebands generated, then just the carrier remains.

(Okay, okay, I know. I said the same thing about 3 times... ;)


Tom W8JI said:

"But this is where we part in agreement. The FCC does NOT want us to use the accumulated energy over long periods of time; they are interested in PEAK envelope power on adjacent channels. If we look at peak envelope power it does not change with speed for a fixed shape and length of rise and fall no matter how fast or slow we send."

Hey readers, the key here is "on adjacent channels", meaning modulation products that are well away from the carrier frequency (say, more than 500 Hz) that have high power peaks. These are caused by rise and fall shapes, as he states. They don't change (other than how often they appear) with the cw rate because the rise and fall shape doesn't change with cw rate. (Exceptions noted)

Tom W8JI said:
"The frequency spacing of that sideband depends ONLY on how fast we slope the level of the carrier. Nothing else. If we sloped it up and down at a 1mS rise and fall rate, the complete cycle would take 2ms. The sidebands, if it was a raised sine wave shape, would appear 500Hz above and below the carrier."

Well, strictly speaking, there will be more than just the one sideband peak since they are the outcome of the *complete* character of the modulating signal; not just how fast we slope the level but also how OFTEN we slope the level. The quick way to see what products result is to simple look at the spectrum of the modulating signal alone before the modulating stage of transmitter. I'll give you this, the sidebands of interest, those within the adjacent channel(s), are governed by the rise and fall shape.

Oh, and a raised sine wave will create more than just peaks at +/-500Hz since the modulating sinusoid is discontinuous.


Folks, in a nutshell, if you don't want to produce key clicks, use rise and fall shapes which don't produce high instantaneous power outside of the bandwidth you wish to transmit within! Now, what shape is that? Well, just pass the modulated I.F. signal through a tight filter and the rise and fall shapes you desire appear. If you want 500 Hz 99% power bandwidth, then select a steep skirt filter that is 500Hz wide at its -20 dB points. Even if your power bandwidth before the filter is 1Khz wide, the filter will guarantee you meet your desired 500 Hz power bandwidth (with a few hertz error, the magnitude of which is dependent on the filter shape).


Bill W2WO said:

"W2WO:Too many people use "power" and "energy" loosely. That seems to be the heart of this disagreement. There is power in the key-click sidebands, but only for a few milliseconds. The total energy in the key clicks is small when averaged over time. Nevertheless, the power in the clicks affects receiver circuits (including AGC) and the listener.

"It seems to me that both sides are correct, but are talking about different things: power vs energy.

"Or did I get it backwards?"

Bill, I see this discussion the same way you do. You stated it very well, in my opinion.



For the record, I think Tom's method, is generally more useful. He measures the peak power over a large bandwidth, which takes key clicks into account. Measuring average power over integer number of cycles does not, so is therefore lacking. But it is also important to ensure that your average power is appropriately confined, so maybe both these measurement methods are necessary to fully qualify a CW signal.


WB1AIW
 
Bandwidth versus Keying Speed Revisited  
by AC7GO on August 19, 2007 Mail this to a friend!
This is a really interesting discussion, not because there is anything at all novel about the mathematical analysis or measurements K5MC has presented. Nor is there anything new about the non-mathematical observations offered by others. They are both correct.

How far from the carrier frequency key clicks can be heard indeed has nothing to do with the keying rate. That is because the mind/ear detector does not respond as if it were a spectrum analyzer. A spectrum analyzer must take in and process information for a very long period of time to produce an accurate result. On the other hand, the ear/mind detector is set up to respond to short bursts of information, such as the high frequency bursts in "t", "p", and "s" sounds in speech. It is a different detection system than the spectrum analyzer.

The human ear cannot tell you much about the spectrum of the code sequence, even though it can interpret the code sequence. Interestingly, the spectrum analyzer cannot give you any readily discernible interpretation of the content of the code sequence. The coded message is contained in the spectrum (well, maybe not - it typically shows only magnitude and no phase), but not in a form that can be readily interpreted.

I have encountered this circumstance in the design of radio receivers and ultrasound receivers. Frequency-domain analysis can tell you things like how wide your bandwidth needs to be to receive a certain signal. But how the signal is ultimately processed and interpreted depends upon the time-domain response of the detector, such as an envelope detector or a peak detector.

Analyses in both domains give correct results, they just differ.

 
RE: Bandwidth versus Keying Speed Revisited  
by AB7E on August 19, 2007 Mail this to a friend!
AB0WR: "... a square wave is generated by combining multiple odd harmonics. Those odd harmonics *do* exist during the ENTIRE duration of the square wave. Therefore they modulate the carrier just exactly like the tone I mentioned - they exist during the rise time of the envelope, during the constant portion of the envelope, and during the fall time of the envelope. Therefore the sidebands exist for the rise time, the constant time, and the fall time of the envelope and those sidebands are made up of all of the odd harmonics making up the keying wave."

AB0WR: "... the odd harmonics making up a square wave do NOT only exist during rise and fall time of the square wave but for the total duration of the square wave. In fact, the frequencies of the square wave are determined by the duration of the pulses in the square wave, i.e. the period of each cycle in the wave."


I was going to stay out of this round but that's the most convincing evidence of misapplied or misunderstood math from this whole thread.

Let's assume I can gate a sensitive wideband (ideal) receiver to receive only during a portion of the constant time of the envelope ... let's say a 100 msec wide window triggered 400 msec after the start of a 1000 msec pulse of a transmitted 10 MHz carrier. If you are correct, not only should I be able to hear sidebands during that 100 msec, but I would hear DIFFERENT sidebands as the pulse of RF is reduced from 1000 msec to, say, 700 msec (still keeping the position of the 100 msec gate at 400 msec into the RF pulse). Do you honestly believe that to be the case? I presume you do, but please explain how the 100 msec gate knows how to hear different sidebands due to the time shift of an event that hasn't happened yet?

 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 20, 2007 Mail this to a friend!
AC7GO - This is a really interesting discussion, not because there is anything at all novel about the mathematical analysis or measurements K5MC has presented. Nor is there anything new about the non-mathematical observations offered by others. They are both correct.

K5MC - I really appreciate AC7GO's comments. He restores a good balance to this entire discussion.

The occupied bandwidth measurement concept is clearly spelled out in the Agilent N9020A application guide and agrees with the FCC/ITU definition of occupied bandwidth. The occupied bandwidth is based on the average (mean) power of the signal and, therefore, the Fourier series approach I used in my earlier eHam article gives the correct results for "well-behaved" CW transmitters. As I've said several times now, perhaps W8JI or some other folks should derive a quantitative definition of "key-click" bandwidth and present it to the FCC for possible adoption into the Part 97 rules.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 20, 2007 Mail this to a friend!
KE3HO-The harmonics that make up the square wave do not modulate the carrier during the middle portion between the rise and fall. Take any keying waveform that you like and calculate its spectrum using the Fourier Series or the Fourier Transform. All of the frequency components calculated do indeed exist for all time (from t = -infinity to t = +infinity). However, those frequency components *ADD* in the time domain to create the keying waveshape. During the middle portion of the keying waveform, they add to one (or some constant value).

AB7E-was going to stay out of this round but that's the most convincing evidence of misapplied or misunderstood math from this whole thread.

AB7E-Let's assume I can gate a sensitive wideband (ideal) receiver to receive only during a portion of the constant time of the envelope ... let's say a 100 msec wide window triggered 400 msec after the start of a 1000 msec pulse of a transmitted 10 MHz carrier. If you are correct, not only should I be able to hear sidebands during that 100 msec, but I would hear DIFFERENT sidebands as the pulse of RF is reduced from 1000 msec to, say, 700 msec (still keeping the position of the 100 msec gate at 400 msec into the RF pulse). Do you honestly believe that to be the case? I presume you do, but please explain how the 100 msec gate knows how to hear different sidebands due to the time shift of an event that hasn't happened yet?

WB1AIW-Actually, the sidebands are created by the rise/fall of the signal so they are only there *during* the rise and fall of the signal. They DO go away during the middle of the pulse, that's why you hear only the dreaded key click each time the cw element is started and/or stopped, and not for the full duration of the pulse.

WB1AIW-Again: In fact, they DO only appear during the leading and falling edges... during the "on" period between the edges, the only signal you'll find is the carrier.

WB1AIW-Tim, frequency domain sidebands (which are found in the spectrum, aka frequency domain phenomena) are generated by changes to the carrier during the changes only. When a person using AM stops talking, the sidebands disappear. When a cw operator presses and holds his key down, the are momentary sidebands generated, then just the carrier remains.

AB0WR-I don't wish to be blunt and disrespectful about this but folks, these statements are just plain wrong.

They confuse time domain with frequency domain.

to ke3ho: It doesn't matter if all of the harmonics add up to one. That is a time domain phenomenon. It doesn't change the frequency domain in any way. In the frequency domain they exist for the whole envelope period.

to ab7e: if you change your sample time you will still get all the frequencies in the sideband since all the frequencies exist during the entire portion of the pulse. Lengthening and shortening the sample interval will *not* change the frequencies you record. If you whistle into your mic and then record that sound for 1ms or 100ms will you not record the same whistle just for different periods of time? That's because what you are recording has a frequency domain that doesn't change with time.

to wb1aiw: Exactly what do you think creates the sidebands during the rise time and fall times of the envelope? Exactly what mathematical process can you give us that models such a physical world -- that is accepted by the engineering profession and by physicists the world over?

everyone - sidebands are created by mixing a carrier signal with a modulating signal. If that modulating signal has several frequency components then all of those frequency components appear in the sideband -- they have to. The mixing process can't distinguish between frequency components and mix some and not others.

If that modulating signal is a repeating square wave, then an infinite series of odd harmonics make up that square wave. These harmonics exist during the entire period of the square wave.

Following are a number of web pages that describe and show this phenomenon quite well. Some are even animated.

http://en.wikipedia.org/wiki/Square_wave
http://slack.net/~ant/bl-synth/4.harmonics.html
http://www.allaboutcircuits.com/vol_2/chpt_7/2.html (very good site)
http://hyperphysics.phy-astr.gsu.edu/Hbase/audio/geowv.html
http://www.mathworks.com/products/matlab/demos.html?file=/products/demos/shipping/matlab/xfourier.html
http://www.freelists.org/archives/si-list/11-2001/msg00139.html
http://www.jhu.edu/~signals/phasorlecture2/indexphasorlect2.htm
http://www.tpub.com/content/neets/14181/css/14181_190.htm
http://mathworld.wolfram.com/FourierSeriesSquareWave.html
http://safari.oreilly.com/0130669466/ch02lev1sec7

If that modulating square wave has frequency components that exist during the entire square wave then they will also exist in the sidebands of the modulated wave for the entire duration of the envelope. The mixing process can't say "I'm only going to mix the carrier and the square wave harmonics during the rise and fall times of the envelope". It can only say "I'm going to mix the carrier and the square wave harmonics".

If you'll look carefully at these web pages you can see that what determines the rise and fall times are the number of harmonics that are used to generate the square wave. The fundamental and the 3rd harmonic produce a very long rise time. The fundamental, 3rd, 5th, 7th, and 9th harmonics produce a much shorter rise time but still far from zero. Those harmonics exist during the entire time of the repeating square wave.

That tells you that the way to produce a "shaped" keying waveform is to filter it through a low-pass filter. As the cutoff frequency of that low pass filter gets lower and lower you remove the higher numbered harmonics. First the 21st and 23rd, then down to the 13th and 15th, and you can finally get down to having only the 3rd and 5th harmonics. At each step the bandwidth of the signal gets less and the rise time and fall times of the waveform get longer and longer.

As the bandwidth of harmonics making up the square wave get less and less so does the bandwidth of the resulting sidebands from mixing the square wave keying waveform and the carrier.

Even here I am being imprecise. Since the filters we use do not have infinite stop bands, the harmonics of the square wave will still exist, just with reduced levels. That means that more and more of the power in the bandwidth gets concentrated in the lower level harmonics. That is what causes the 99% bandwidth to get narrower as we filter out the higher level harmonics. You may *still* be able to hear these higher level harmonics from a station running a lot of power or which has very good propagation to your station.

I urge you to study the web pages I listed above. Only by understanding how a modulating square wave (or filtered square wave) is made up of odd harmonic frequency components can we understand how a CW signal has sidebands and how the slope of the keyed envelope is generated.

Once we understand this then we can begin to address how key clicks that are heard far outside the bandwidth of the generated signal, i.e. several khz away, are generated.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 20, 2007 Mail this to a friend!
AC7GO - have encountered this circumstance in the design of radio receivers and ultrasound receivers. Frequency-domain analysis can tell you things like how wide your bandwidth needs to be to receive a certain signal. But how the signal is ultimately processed and interpreted depends upon the time-domain response of the detector, such as an envelope detector or a peak detector.

AB0WR - This may be why people say they hear key "clicks". If your operating frequency is just far enough away from another signal so that only the high frequency harmonics in the sidebands of the keying waveform are heard but not the lower frequency harmonics in the sideband, these may very well be interpreted as "clicks".

This may or may not indicate any "problems" in the transmitted signal that is partly outside of your passband but is, instead, just an artifact of how you "hear".

On the other hand, key clicks that are heard from signals several khz away are probably due to something else, either a problem in the transmitted signal or in the receiver being used.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on August 20, 2007 Mail this to a friend!
I think an application of the bandwidth theorem will put this whole discussion to rest -- especially when one tries to do an accurate measure of the spectral width of the keying action.

The bandwidth theorem is sort of like the Heisenberg Uncertainty Principle as it is applied to wave phenomina (and it is true for both quantum and macro). You can't beat the bandwidth theorem is another way of saying that you can not beat mother nature.

Of course, I may not be entirely serious about this because I have not looked into the mathematical development closely enough. I think it would be cool though if the bandwidth theorem was the controlling factor determining the end game of this discussion.
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 20, 2007 Mail this to a friend!
AB0WR: “If that modulating square wave has frequency components that exist during the entire square wave then they will also exist in the sidebands of the modulated wave for the entire duration of the envelope. The mixing process can't say "I'm only going to mix the carrier and the square wave harmonics during the rise and fall times of the envelope". It can only say "I'm going to mix the carrier and the square wave harmonics".”

Let me make sure I understand this theory correctly. A CW transmitter is fundamentally an AM transmitter in which the carrier is modulated by the keying waveform. The keying waveform, whatever it may be, has a spectrum that can be calculated using Fourier Theory. Fourier Theory says that the calculated spectral components exist for all time at their fixed calculated amplitude. Since these spectral components exist for all time at fixed amplitude, the sidebands that are produced by mixing these spectral components with the carrier also exist at constant frequency and amplitude for the entire duration of the transmission.

Is this correct?

73 - Jim
 
Bandwidth versus Keying Speed Revisited  
by AC7GO on August 20, 2007 Mail this to a friend!
K5MC: "As I've said several times now, perhaps W8JI or some other folks should derive a quantitative definition of "key-click" bandwidth and present it to the FCC for possible adoption into the Part 97 rules."

Mickey, Thanks for the interesting discussion. Do you know if there is anything presently in the rules that addresses the annoying key-click issue?


KE3HO: "Let me make sure I understand this theory correctly. A CW transmitter is fundamentally an AM transmitter in which the carrier is modulated by the keying waveform. The keying waveform, whatever it may be, has a spectrum that can be calculated using Fourier Theory. Fourier Theory says that the calculated spectral components exist for all time at their fixed calculated amplitude. Since these spectral components exist for all time at fixed amplitude, the sidebands that are produced by mixing these spectral components with the carrier also exist at constant frequency and amplitude for the entire duration of the transmission.

Is this correct?"

Jim, that is almost correct. If you were to observe the signal for an infininte length of time, and pick any of its spectal components, you would "see" that spectral component wiggling in its lazy, stationary manner ad infinitum in either direction of time, not just for the duration of the transmission.

That's not how we humans typically operate as observers, is it? We hear the effects of all of the components, at any given particular time, as processed by our rig and by our ear/mind. So we hear dits and dah's, and key-clicks ;o)

But you basically have the correct understanding of the theory. Fourier analysis is a mathematical construct, and as a mathematical construct it doesn't have to observe limitations of space and time. In fact, the formula for the Fourier components requires looking at the signal in both directions of time ad infinitum.

It is a very interesting paradoxical thing. The time domain and the frequency domain are wildly different ways of looking at exactly the same information. Stimulates the mind, doesn't it?

73 Max
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 21, 2007 Mail this to a friend!
W8JI (August 17, 2007) - Now when we change the level of that carrier sidebands are generated. No matter how slow or how fast we do that, we generate at least two sidebands. One sideband is upper, and one is lower.

The frequency spacing of that sideband depends ONLY on how fast we slope the level of the carrier. Nothing else. If we sloped it up and down at a 1mS rise and fall rate, the complete cycle would take 2ms. The sidebands, if it was a raised sine wave shape, would appear 500Hz above and below the carrier.

This is true on the leading edge, and it is true on the falling edge. If the shape of level change is not a sin shape, there would be additional sidebands generated that would be wider.

I'm sure we all agree to this point, or we had better, because that's how an amplitude modulated carrier works.

K5MC - The sinusoidal keying waveform I assumed in my earlier eHam article has a rise time and a fall time of exactly 5 milliseconds each because I used a 50-Hz sine wave as the shaping waveform for both of the "edges" for that specific keying waveform. Since the period of a 50-Hz sinusoid is 20 milliseconds, I used the 0-to-90 degrees portion for the leading edge of the keying waveform to achieve the 5-millisecond rise time. (Note that my definition of rise time here is the time required for the waveform to go from zero amplitude to 100 percent amplitude rather than the more commonly used 10% to 90% definition.) Similarly, I used the 90-to-180 degrees portion of the 50-Hz sine wave for the trailing edge of the keying waveform.

Now according to W8JI, the first pair of sidebands I should see for the CW signal using my sinusoidal keying waveform is 100 Hz above and below the carrier, regardless of the keying speed, because the sum of the rise and fall times is 10 milliseconds. However, classical Fourier analysis and the Agilent N9020A analyzer don't agree with W8JI.

Now let's think about the case of square-wave keying. According to W8JI, as the rise/fall times approach zero, the first pair of sidebands would approach an infinite frequency above and below the carrier. Wow, let's shoot for square-wave keying so that the keying sidebands are so far away in frequency (for example, at essentially DC on one end and up in the microwave bands on the other) that nobody will be bothered at all by key clicks! After all, W8JI clearly says that the frequency spacing of the first pair of sidebands is governed ONLY by the rise/fall times - as the slope increases, the key clicks move further away from the carrier (which occupies essentially zero bandwidth in W8JI's view). No doubt, this is the ultimate answer to key clicks and bandwidth for CW signals!!

Returning to reality now, the classical Fourier series concept demonstrates that the frequencies of the keying sidebands are a function of speed, not the rise/fall times. I have provided "proof" of this fact through both mathematical analysis and through actual measurements on my Kenwood 940. The spectrum plots from the Telefunken literature shown in Figure 1 provide additional proof. (See Figures 1, 16, and 17 at http://www.arrl.org/sections/?sect=LA) W8JI, however, has provided neither detailed mathematical analysis nor actual spectrum analyzer data from a "well-behaved" CW transmitter that corroborates his views.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 21, 2007 Mail this to a friend!
AC7GO - Mickey, Thanks for the interesting discussion. Do you know if there is anything presently in the rules that addresses the annoying key-click issue?

K5MC - Max, the emission standards in 97.307 are about it as far as Part 97 is concerned. Plus, there's the language in Part 2 that I've quoted a few times here on eHam. As far as I know, there's nothing currently under consideration by the FCC regarding more rules to address key clicks and splatter.

BTW, based on a few comments here and there, I have the impression that some hams believe the limitations on spurious emissions listed in 97.307 apply to key clicks. However, the FCC views key clicks as "out-of-band' emissions rather than "spurious" emissions.

I've actually thought about working on a formal definition of "key-click" bandwidth myself because it would probably be an interesting technical problem. However, at this late date I doubt that the FCC would seriously consider another definition of bandwidth, particularly if that definition applies primarily or exclusively to CW signals.

Thanks again for your comments.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by VK3GJZ on August 21, 2007 Mail this to a friend!
K5MC: Now let's think about the case of square-wave keying. According to W8JI, as the rise/fall times approach zero, the first pair of sidebands would approach an infinite frequency above and below the carrier....... No doubt, this is the ultimate answer to key clicks and bandwidth for CW signals!!

VK3GJZ: No. A perfect square wave does have "clicks" extending to infinity. But it also has clicks at all points in between.


K5MC: Returning to reality now, the classical Fourier series concept demonstrates that the frequencies of the keying sidebands are a function of speed, not the rise/fall times.

VK3GJZ: Surely the two are interchangeable. An example is to A.M. modulate a carrier with a low frequency sine wave. This exactly represents a "minimum bandwidth" CW signal sending continuous dits. As you raise the modulating frequency, the two sidebands move out. You could argue that the rise and fall times of the modulating signal are getting steeper, or that the modulating frequency is increasing. Both are correct.

Of course the usual exponential RC keying shape generates a much wider signal. As does a Delta keying shape.

Regards ............... Zim

 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 21, 2007 Mail this to a friend!
KE3HO:
Let me make sure I understand this theory correctly. A CW transmitter is fundamentally an AM transmitter in which the carrier is modulated by the keying waveform. The keying waveform, whatever it may be, has a spectrum that can be calculated using Fourier Theory. Fourier Theory says that the calculated spectral components exist for all time at their fixed calculated amplitude. Since these spectral components exist for all time at fixed amplitude, the sidebands that are produced by mixing these spectral components with the carrier also exist at constant frequency and amplitude for the entire duration of the transmission.

Is this correct?

73 - Jim
====================================
I think you have a slight misunderstanding of Fourier. Implicit in using this method is the assumption that the waveform is a periodic waveform existing from -/+ infinity. Since your result is composed of sin's and cos's you WILL end up with a mathematical description of a repeating waveform from -/+ infinity.

However, in the real world you must break your waveform up into pieces where you can, lets say PRETEND, that the portions of the waveform you are examining exists for -/+ infinity.

In other words, for one pulse from t=0 to t=1 you need to do a Fourier on the portion of the waveform from -infinity to t<0 and pretend that it will continue that way until +infinity. The result will be zero . A Fourier from t=0 to t=1 and pretend it has repeated from the beginning of time and will repeat until the end of time. Then another from t>1 to +infinity and pretend it started at -infinity. The result will be zero.

By combining these three results you will end up with a description of the real world event.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 21, 2007 Mail this to a friend!
KE3HO-Let me make sure I understand this theory correctly. A CW transmitter is fundamentally an AM transmitter in which the carrier is modulated by the keying waveform. The keying waveform, whatever it may be, has a spectrum that can be calculated using Fourier Theory. Fourier Theory says that the calculated spectral components exist for all time at their fixed calculated amplitude. Since these spectral components exist for all time at fixed amplitude, the sidebands that are produced by mixing these spectral components with the carrier also exist at constant frequency and amplitude for the entire duration of the transmission.

AB0WR-A CW transmitter is a mixer just as an AM transmitter is a mixer also. Modulation is nothing more than a mixing product -- i.e. a multiplication of two signals in the time domain resulting in a convolution in the frequency domain.

"Existing for all time" is a tricky thing to answer.

Strictly speaking, the Fourier series decomposition of a waveform only requires that the waveform be periodic, it doesn't require it to "exist for all time". The decomposition is of the form:

x(t) = (a0/2) + SUM[(ak)cos(wk)t + (bk)sin(wk)t ]
for k=1 to k=infinity.

ak and bk are coefficient values of the kth term
wk is k(w0) -- the fundamental multiplied by k

This decomposition doesn't require the periodic wave to exist for all time, just for one "period".

The Fourier *transform* on the other hand is an integral defined from t equal to negative infinity to t equal to positive infinity.

x(t) = integral[ f(t)e**-jwt] from t=-inf to t=+inf

Go to this web page:
http://mathworld.wolfram.com/FourierSeries.html
and take a look starting at equation 20 for some insight as to the relationahip between the Fourier Series and the Fourier Transform.

I think you will find after reading this web page that the Fourier Series components making up the keyed waveforms can be said to exist only so long as the periodic waveform they apply to exists also. Strictly speaking, if you send exactly one dit you can break it down using the Fourier Series to get the freq components making it up. The "periodic" waveform before the dit and after the did has, in essence, either an infinite period (making w0 equal to zero) or is undefined (since I believe the Fourier integral has to be less than +infinity to be defined).

A fundamental frequency of Zero is intuitively appealing to me since periods a little less than infinity will have very, very low frequency values but *will* exist so in the limit a frequency of zero just *seems* right. Perhaps Mickey would have some insight as to this interpretation.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 21, 2007 Mail this to a friend!
K5MC: Returning to reality now, the classical Fourier series concept demonstrates that the frequencies of the keying sidebands are a function of speed, not the rise/fall times.

VK3GJZ: Surely the two are interchangeable. An example is to A.M. modulate a carrier with a low frequency sine wave. This exactly represents a "minimum bandwidth" CW signal sending continuous dits. As you raise the modulating frequency, the two sidebands move out. You could argue that the rise and fall times of the modulating signal are getting steeper, or that the modulating frequency is increasing. Both are correct.

VK3GJZ:Of course the usual exponential RC keying shape generates a much wider signal. As does a Delta keying shape.

This is pretty close. It isn't *quite* the same as merely raising a single modulation frequency. The rise times and fall times are indicators of the *bandwidth* of the modulating signal. That bandwidth will contain multiple harmonics, ie modulating frequencies. The slope will tell you what the highest frequency harmonic, i.e. modulating frequency can be.

For example, a 10hz square wave will have harmonics at 10, 30, 50, 70, 90, 110, 130, 150, 170, 190, 210, 230hz, etc. A 20hz square wave will have harmonics at 20, 60, 100, 140, 180, 220, 260hz, etc. An RC filter with a 200hz knee will impact the 21st, 23rd, etc harmonics of the first keying wave and the 11th, 13th, etc harmonics of the second keying wave.

So the 10hz keying wave will retain more harmonics then the 20hz keying wave even though both are limited to 200hz bandwidth and will have exactly the same rise and fall times.

The RC filter will only give a 6db per octave rolloff. I'm not sure what the delta filter shape is. A gaussian filter shape has a slope that is much steeper.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 21, 2007 Mail this to a friend!
"Existing for all time" is a tricky thing to answer.

This bothered me also at first. But then I realized what this really means.

If you have a periodic waveform that exists from t = -infinity to t = +infinity, you can calculate its spectrum using the Fourier Theory and everything is neat and tidy (if you ignore the fact that in the real world you cannot have a waveform, periodic or otherwise, that exists from -infinity to +infinity). It is a theoretical mathematical construct.

What if you have a waveform that is of finite duration? What if you have a waveform that is not periodic? This is where the Fourier Transform comes into play. What the Fourier Transform does is allow us to calculate the spectrum of such waveforms. Being calculated using Fourier Theory, these frequency components exist for all time - meaning from -infinity to +infinity. At first this really upset my sensibilities. If I have a waveform that is non-zero only from, say t = 0 to t = 3 seconds, all of the required frequency components had to have existed at the proper frequency, amplitude, and phase since t = -infinity AND even more remarkably, these frequency components somehow knew from the very beginning (at t = -infinity) that my waveform was going to exist at t = 0 (which was in the infinitely remote future). Then, if you come up with some explanation of how these frequency components came into being in the infinite past just so I could have my waveform now, you can start tackling the question of where they got their energy from. These are very remarkable frequency components. Omniscient and omnipotent, one might say.

The solution to this problem is childishly simple. They don't exist prior to t = 0. Or what I should really say to be technically correct according to Fourier Theory is that all of the frequency components cancel each other prior to t = 0. They add to zero in the time domain, but more importantly, the literally annihilate each other prior to t = 0. Same thing is true after t = 3 seconds. After all, what we calculated with the Fourier Transform was the frequency components necessary to make up our original signal. What was our original signal prior to t = 0? It was zero. In Fourier's theoretical construct, they "exist" for all time, but the key is that they only exist as individual, discrete frequency components in the frequency domain. In the time domain they literally annihilate each other and do not really exist when the signal is zero (or any DC value). To demonstrate this phenomenon, what would you see if you were looking at this signal with a sweeping spectrum analyzer? Prior to t = 0 you would see nothing. Zip. Zilch. There really is NOTHING there. The frequency components do more than just add to zero, they completely cancel each other out of existence. What would you see after t = 3 seconds? Same thing - nothing.

So, let's agree for the moment that when Fourier Theory says "for all time", we will take that to mean from the start of our waveform to the end of our waveform.

Now we come to the really interesting part. What would you see between t = 0 and t = 3 seconds? (and for this, lets assume that you have some really super duper ultrafast sweeping spectrum analyzer made from 12 pounds of semiconducting Unobtainium and 4 pounds of pure Nonexistium that can sweep through your spectral region of interest in some small fraction of a second and still give you the complete spectrum in nearly real time). You would see the spectrum predicted by the Fourier Transform, right? No, not quite. Let's look at one example. What if my waveform was a pure 10 Hz sine wave from t = 0 to t = 1.5 seconds and was a pure 20 Hz sine wave from t = 1.5 seconds to t = 3 seconds. If we sampled this waveform at some reasonable sampling rate and calculate the Fourier Transform we would get the following (and I actually calculated this and plotted it. I really wish that I could include graphs here, but I can't). The spectrum has a large peak at 10 Hz and another large peak at 20 Hz. Then there are a LOT of much smaller spectral components from near 0 Hz all the way out to 5 kHz, which was the upper limit of my FFT calculation. If this was my waveform that I was watching on my imaginary spectrum analyzer, I would see a single peak at 10 Hz from t = 0 to t = 1.5 seconds. That peak would go away and a new peak would appear at 20 Hz from t = 1.5 seconds to t = 3 seconds. NONE of the other peaks would appear at any time. How can that be? Cancellation again. What would happen if you took the spectrum calculated with the FFT and you set the 10 Hz and 20 Hz components to zero, then took the inverse Fourier Transform? All that would be left would be all of the low level peaks from near DC out to 5 kHz. What would they look like in the time domain? After taking the inverse Fourier Transform, prior to t = 0 these frequency components would create a time domain waveform that was -sin(10t)-sin(20t). From t = 0 to t = 1.5 seconds the waveform would be -sin(20t). From t = 1.5 seconds to t = 3 seconds the waveform would be -sin(10t). For t > 3 seconds the waveform would again be -sin(10t)-sin(20t). In other words, all of those other frequency components cancel both the 10 Hz and 20 Hz peaks prior to t = 0, they cancel the 20 Hz component from t = 0 to t = 1.5 seconds. They cancel the 10 Hz component from t = 1.5 seconds to t = 3 seconds. Finally, they once again cancel both the 10 Hz and 20 Hz components after t = 3 seconds. And these cancellations are more than just adding to zero, they completely annihilate each other.

If you understand my explanation, then you see why I said that the spectrum analyzer has to be a sweeping analyzer and not an FFT analyzer. After all, the FFT analyzer is going to show us the Fourier spectrum with all of the UN-ADDED frequency components. I think of it this way: An "old fashioned" sweeping spectrum analyzer shows us the frequency domain from the VANTAGE POINT of the time domain - it shows us the frequency components POST-CANCELLATION. An FFT spectrum analyzer shows us the Frequency Domain from the VANTAGE POINT of the frequency domain - it shows us all of the frequency components PRE-CANCELLATION.

73 - Jim
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 21, 2007 Mail this to a friend!
KE3HO:If you understand my explanation, then you see why I said that the spectrum analyzer has to be a sweeping analyzer and not an FFT analyzer. After all, the FFT analyzer is going to show us the Fourier spectrum with all of the UN-ADDED frequency components. I think of it this way: An "old fashioned" sweeping spectrum analyzer shows us the frequency domain from the VANTAGE POINT of the time domain - it shows us the frequency components POST-CANCELLATION. An FFT spectrum analyzer shows us the Frequency Domain from the VANTAGE POINT of the frequency domain - it shows us all of the frequency components PRE-CANCELLATION.

AB0WR-I would add for consideration, the fact that the Fourier Transform is of the form

int[ f(t)(e**-jwt)dt] from t=-infinty to t=+infinity

You must be careful of how you write f(t). As far as I remember, the only restiction on f(t) is that the integral of f(t)dt not be infinite. In other words it can be a piecewise continuous function. From -infinity to -T/2, f(t)=0, from -T/2 to T/2, f(t)=1, from T/2 to +infinity, f(t) = 0.

When you break this down, you get three integrals, two of which evaluate to Zero. This is how you can break a single pulse in the time domain down into a Sinc function in the frequency domain. It is extremely important to note that the Sinc representation in the frequency domain has NO time information associated with it. If you sample the single pulse at any time using either a swept spectrum analyzer or an FFT analyzer you will find the frequencies associated with the pulse present with the amplitudes calculated by the Fourier Series coefficients. You won't find some of the frequencies existing at some time during the pulse, t0, and other frequencies existing at another time during the pulse, t1. If you sample the signal before the pulse exists or after it has ended, neither spectrum analyzer will show anything since the Fourier integral evaluates to Zero in these regions.

If you have a truly non-time variant, periodic waveform the sample time of the spectrum analyzer, i.e. the sweep time, should be irrelevant. If you have a time-variant waveform then the sample time becomes very, very important.

Consider a keyer whose frequency varies from 9hz to 11 hz over a 90sec period of time. If you run your spectrum analyzer over that 90sec of time you will find harmonic frequencies of 9, 27, 45, 63, 10, 30, 50, 70, 11, 33, 55, 77hz and everything in between. You will never get a good picture of what is actually happening with the system.

See this web page for a better explanation that I can give here:
http://cnx.org/content/m0046/latest/

This is a very useful page. See also what they show concerning power calculated in the frequency domain is exactly the same result as for power calculated in the time domain (Parseval's Theorem).

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 21, 2007 Mail this to a friend!
KE3HO - If you understand my explanation, then you see why I said that the spectrum analyzer has to be a sweeping analyzer and not an FFT analyzer. After all, the FFT analyzer is going to show us the Fourier spectrum with all of the UN-ADDED frequency components. I think of it this way: An "old fashioned" sweeping spectrum analyzer shows us the frequency domain from the VANTAGE POINT of the time domain - it shows us the frequency components POST-CANCELLATION. An FFT spectrum analyzer shows us the Frequency Domain from the VANTAGE POINT of the frequency domain - it shows us all of the frequency components PRE-CANCELLATION.

K5MC - KE3HO really should take a close look at the spectrum plots shown in Figure 1 that are posted at http://www.arrl.org/sections/?sect=LA There's no question that the spectrum analyzer used in the old Telefunken literature is a "sweeping" (i.e., swept mode) analyzer rather than an "FFT" (i.e., FFT mode) analyzer that KE3HO keeps harping on.

KE3HO's comments above are nonsense! The spectrum plots from the Telefunken literature (Figure 1) are consistent with the spectrum plots shown by the Agilent N9020A analyzer (Figures 16 and 17). I truly hope that very few other hams are under this particular delusion expressed by KE3HO.

Since I'm sure that some folks will dismiss my comments on this topic, here's some information directly quoted from page 9 of the "8 Hints for Better Spectrum Analysis" (Application Note 1286-1) that is posted by Agilent Technologies at http://cp.literature.agilent.com/litweb/pdf/5965-7009E.pdf

Agilent Technologies - A good balance between time and sensitivity is to use fast Fourier transform (FFT) that is available in the modern high-performance spectrum analyzers. By using FFT, the analyzer is able to capture the entire span in one measurement cycle. When using FFT analysis, sweep time is dictated by the frequency span instead of the RBW (resolution bandwidth) setting. Therefore, FFT mode proves shorter sweep times than the swept mode. The difference in speed is more pronounced when the RBW filter is narrow when measuring low-level signals.

K5MC - Perhaps the design engineers at Agilent should hear from KE3HO. I think they will get a very good laugh from "learning" that their newer analyzers yield completely different results than their older ones!

Just in case KE3HO or anyone else wants to pursue this particular "theory" with Agilent, here's the email address for Agilent technical support: http://contact.tm.agilent.com/Agilent/tmo/Format/K2Pages/ContactUsUSCanEmail.html?cmpid=20096

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by AB7E on August 21, 2007 Mail this to a friend!

To AB0WR:

I didn't say that I would vary the sampling interval. I said I would hold the sample time at 100 msec but vary the RF pulse width. Here is what I said:

AB7E: "Let's assume I can gate a sensitive wideband (ideal) receiver to receive only during a portion of the constant time of the envelope ... let's say a 100 msec wide window triggered 400 msec after the start of a 1000 msec pulse of a transmitted 10 MHz carrier. If you are correct, not only should I be able to hear sidebands during that 100 msec, but I would hear DIFFERENT sidebands as the pulse of RF is reduced from 1000 msec to, say, 700 msec (still keeping the position of the 100 msec gate at 400 msec into the RF pulse). Do you honestly believe that to be the case? I presume you do, but please explain how the 100 msec gate knows how to hear different sidebands due to the time shift of an event that hasn't happened yet?

You replied: "to ab7e: if you change your sample time you will still get all the frequencies in the sideband since all the frequencies exist during the entire portion of the pulse. Lengthening and shortening the sample interval will *not* change the frequencies you record. If you whistle into your mic and then record that sound for 1ms or 100ms will you not record the same whistle just for different periods of time? That's because what you are recording has a frequency domain that doesn't change with time."

I hypothesized a varying RF pulse width, not a varying sample interval.

You previously stated that the sidebands exist not only during the rise and fall times, but also during the entire intermediate constant amplitude portion. You further stated that the character of the sidebands would be directly affected by the width of the RF pulse (the constant amplitude portion) even if all other elements (rise/fall times, etc) were held constant.

So, if you could avoid being dismissive, please tell me how the sidebands I'd see/hear/measure during the fixed sampling interval can change as a function of an event that hasn't happened yet, i.e., the turnoff of the RF pulse as I change it from (in my example) 1000 msec to 700 msec.

Dave AB7E

 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 21, 2007 Mail this to a friend!
K5MC - Perhaps the design engineers at Agilent should hear from KE3HO. I think they will get a very good laugh from "learning" that their newer analyzers yield completely different results than their older ones!

Mickey,

You are correct. I am sure that they would get a good laugh from this statement. However, the comparison that I was making was between an FFT spectrum analyzer and my "make believe" ultra sweeping spectrum analyzer that could scan the entire frequency range of interest in a fraction of a second without any frequency or amplitude error, etc. It was not clear that that was the comparison that I was making.

73 - Jim
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 21, 2007 Mail this to a friend!

I have a brand new Agilent Analyzer here, and it does not read occupied bandwidth or TOI nearly as reliably as a slightly older (but still currently calibrated) unit.

It seems to have more "wobble" in readings.

All that aside it reads reasonably close to the much more expensive slightly older unit for OBW.

Now I can alter settings make it read to agree with Mickey if I change the setting to read average power over a long period of time, which happens to be what Mickey is using for his math also.

However, that is NOT what a CW receiver responds to and it is not what the FCC typically expects. They typically want PEP readings rather than long term power measurements.

This entire thing will remain unresolved because IMO Mickey is reading the wrong thing. He is calculating and measuring a long term average power rather than the PEP in a given small bandwidth as the receiver moves off frequency.

It's quite simple and easy to see why bandwidth changes when we look at long term power in sidebands compared to the carrier, but unfortunately that isn't what affects us when we are up or down frequency.

This is also why the people who use actual receivers report disagreement with the method Mickey uses.

It will be interesting to see what someone else like Ed Hare says....rather than an equipment salesman. Rather than speculate let's see how this works out long term.

Personally I think this matter won't reach resolution if people want to use a measurement or analysis method that does not fit the actual SYSTEM we are talking about. Unless it fits the system, it becomes a matter of religion or faith rather than describing what we can observe in the real world.

I prefer a measurement or analysis that agrees with how the receiver behaves under conditions of QRM (PEP), not how it might raise the temperature of water (long term accumulated power).


The long term power is what gives those asinine useless answers that “prove” when we talk slower our SSB bandwidth is less, or if we operate one hour and stay off one hour our CW bandwidth is less.

73 Tom






 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 21, 2007 Mail this to a friend!
AB7E-I hypothesized a varying RF pulse width, not a varying sample interval.

AB7E-You previously stated that the sidebands exist not only during the rise and fall times, but also during the entire intermediate constant amplitude portion. You further stated that the character of the sidebands would be directly affected by the width of the RF pulse (the constant amplitude portion) even if all other elements (rise/fall times, etc) were held constant.

AB7E-So, if you could avoid being dismissive, please tell me how the sidebands I'd see/hear/measure during the fixed sampling interval can change as a function of an event that hasn't happened yet, i.e., the turnoff of the RF pulse as I change it from (in my example) 1000 msec to 700 msec.

AB0WRThe Fourier Series analysis works on periodic signals -- i.e. something you can define as a function of time which repeats. The period of a signal *is* dependent on how long it lasts.

Therefore, events *are* known ahead of time for periodic signals.

What you are trying to do is introduce aperiodic waveforms into consideration -- i.e. change the pulse width arbitrarily. When you change that pulse width arbitrarily, then you have changed the spectrum needed to build that pulse.

The time contradiction you are attempting to introduce works both ways. With a system having no memory if you start your sample interval after the pulse has begun how do then see *any* sidebands at *any* time? How do you see anything except the carrier?

The issue that has to be grasped here is that the frequency domain does not know time. Once you do the Fourier analysis of a periodic wave the time variable disappears.

The spectrum analyzer displays that are shown show that there *are* discrete frequency components making up the sidebands -- for both swept analyzers as well as FFT analyzers. Those discrete sideband frequencies match exactly what the Fourier Series analysis predict for a carrier wave modulated by a square wave.

A Fourier Series analysis for a sawtooth wave, which is what you postulate when you say the sidebands exist only during the rise time of the modulating square wave, requires the existence of both even and odd harmonic frequencies. Yet we don't see both odd and even harmonic frequencies in the sidebands.

That means that you need to provide a mathematical basis (which I haven't seen so far) for how a sawtooth wave can produce only odd harmonics or show empirical evidence that both even and odd harmonics of the sawtooth wave exist in the sidebands of the CW pulse.

When I look at the sidebands produced by keying a CW signal with a 10hz square wave and see only 10hz, 30hz, 50hz, 70hz, 90hz, etc harmonic components I am led to believe that the math I learned is correct.

Folks, I would willingly change my position on this subject if someone could give me a textbook used in an engineering university that gives a mathematical treatise on how to analyze a periodic waveform that shows the Fourier Series only applies during part of a periodic waveform and not the whole waveform. I have been through my 40 year old textbooks and my son's 2 year old textbooks and they are pretty much the same. So are the latest Shaum's outlines on signal analysis.

I'll keep researching but don't expect me to accept "intuition says" as an explanation. I've seen too much "intuition" that proves out wrong when it comes to signal analysis.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 21, 2007 Mail this to a friend!
W8JI:However, that is NOT what a CW receiver responds to and it is not what the FCC typically expects. They typically want PEP readings rather than long term power measurements.

ab0wr-PEP is not bandwidth.



W8JI-This entire thing will remain unresolved because IMO Mickey is reading the wrong thing. He is calculating and measuring a long term average power rather than the PEP in a given small bandwidth as the receiver moves off frequency.

ab0wr:This doesn't appear to be the case. What he is measuring is the contribution to total power by each of the spectral components in the signal -- exactly what the FCC definition of occupied bandwidth is.


W8JI-It's quite simple and easy to see why bandwidth changes when we look at long term power in sidebands compared to the carrier, but unfortunately that isn't what affects us when we are up or down frequency.

ab0wr:It doesn't matter what the long term power is. What matters is the contribution to the total power by each spectral component -- which is what Mickey's traces show.

Consider a SSB transmitter with the bass response cranked up. You will see more power in the low audio frequencies than with a transmitter whose bass response is completely filtered out. This will be the same whether you input white noise and measure power long term or whether you take short term samples of actual voice.

The same thing applies to a CW keying waveform. A 10hz keyed wave whose 21st harmonic (210hz) is -30db is perceived to be more narrow than a 20hz keyed wave whose 21st harmonic (420hz) is -30db by ANYONE'S recevier.

W8JI-I prefer a measurement or analysis that agrees with how the receiver behaves under conditions of QRM (PEP), not how it might raise the temperature of water (long term accumulated power).

ab0wr-Parseval's theoremn states that the integral of the power in the time domain from -infinity to +infinity equals the integral of the power in the frequency domain summed across all frequency components from -infinity to +infinity.

You are trying to dismiss the freqency domain power analysis for purely a time domain power analysis. Parseval says they are the same. That means you can't just look at the time domain PEP at a specfic point in time without also considering the frequency components that contribute to that PEP -- especially when you are trying to define the bandwidth of a signal based on the power relationships of the frequency components.

W8JI-The long term power is what gives those asinine useless answers that “prove” when we talk slower our SSB bandwidth is less, or if we operate one hour and stay off one hour our CW bandwidth is less.

ab0wr:ROFL!! Making up strawmen to argue with are we?

I don't believe *anyone* on this thread or the prior one has ever said this. Sending slower on CW *will* lessen your bandwidth. Filtering highs and lows on SSB *will* lower your bandwidth -- which is exactly the same thing as filtering your CW keying waveform. Filters are filters whether they are RC keying waveform filters or Butterworth SSB filters.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 22, 2007 Mail this to a friend!
I'm not inventing strawmen Tim.

The fact is there are many ways to measure or calculate things, and if we calculate or measure something in a way that does not fit the actual system or effect we want to describe we can get a "correct" but totally useless answer.


An Agilent sales rep came to one of our radio club meeting and proceeded to use a spectrum analyzer incorrectly to measure spectral purity of radios. He connected the analyzer to a short whip and placed radios around the whip. Some poor guy who built his own transceiver was all in dispair about not meeting FCC requirements when the entire problem was the measurement setup.

There was also a person in the Elmers Forum who used a spectrum analyzer to incorrectly measure harmonics, and came to the conclucion his Icom only had 20-30dB harmonic suppression.


In this specific case Mickey (and you) insist on using something that is meaningless to the actual system we are trying to describe.

As long as that is done, you and a few others will continue to snipe at others or ponder why others are so dumb while you continue to focus on the wrong answer and wonder why others won't agree.

If an answer tells you the bandwidth is less when you talk slower on phone or send slower on CW, especially when we test the answer by going to an extreme or compare the answer to what we want to know in a real system and find disagreement, we know the answer has no real meaning.

Have a nice day.

73 Tom
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 22, 2007 Mail this to a friend!
I suspect Agilent has worked with the FCC to make sure what the settings on the SA should be to obtain what measurement the FCC requires. This SA will be used for measuring transmitters in other services too, probably far more than in the amateur service so one would expect that they have definite setup values to obtain this measurement.

What do they specify for the settings to obtain the FCC definition of OBW?

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 22, 2007 Mail this to a friend!
W8JI-I'm not inventing strawmen Tim.

AB0WR:When you claim someone has said that talking slower lowers your bandwidth then you are inventing a strawman.

When you claim someone has said: "if we operate one hour and stay off one hour our CW bandwidth is less" then you are inventing a strawman.

W8JI-The fact is there are many ways to measure or calculate things, and if we calculate or measure something in a way that does not fit the actual system or effect we want to describe we can get a "correct" but totally useless answer.

AB0WR: Yet we have actually seen only one way to measure or calculate occupied bandwidth presented. Talking about PEP isn't talking about bandwidth yet you seem to keep wanting to talk about measuring PEP.

W8JI-An Agilent sales rep came to one of our radio club meeting and proceeded to use a spectrum analyzer incorrectly to measure spectral purity of radios. He connected the analyzer to a short whip and placed radios around the whip. Some poor guy who built his own transceiver was all in dispair about not meeting FCC requirements when the entire problem was the measurement setup.

AB0WR: This is a red herring meant only to divert attention. It has nothing to do with how Mickey set up his measurement.

W8JI-There was also a person in the Elmers Forum who used a spectrum analyzer to incorrectly measure harmonics, and came to the conclucion his Icom only had 20-30dB harmonic suppression.

AB0WR:Another red herring. It has nothing to do with how Mickey set up his measurement.

W8JI - In this specific case Mickey (and you) insist on using something that is meaningless to the actual system we are trying to describe.

AB0WR: Frankly, you have not posted anything that has to do with showing how occupied bandwidth based on 99% power is meaningless. You just continuing to say it is is what is actually meaningless.

W8JI-As long as that is done, you and a few others will continue to snipe at others or ponder why others are so dumb while you continue to focus on the wrong answer and wonder why others won't agree.

AB0WR: No one is "sniping" here. Mickey has posted actual empirical evidence. We have both posted analyses using mathematical techniques which have been accepted for over 100years and which match with the actual empirical evidence. You have yet to show where Agilent is wrong in their design, you have yet to show where Fourier is wrong in his analysis technique of periodic waveforms, you have yet to show where Parseval's theorem is wrong, and you have yet to show where Mickey's measurement technique is wrong.

AB0WR: I would say the only one "sniping" here is you.

AB0WR: It really doesn't matter who agrees with us and who doesn't in the end. All we can do is present the math and the empirical evidence and let the readers judge the results. Fourier's method wasn't initially accepted by some of the greatest mathematicians of his time, either. That didn't keep him from presenting it anyway. No one has proved it wrong as far as I know.

W8JI-If an answer tells you the bandwidth is less when you talk slower on phone or send slower on CW, especially when we test the answer by going to an extreme or compare the answer to what we want to know in a real system and find disagreement, we know the answer has no real meaning.

AB0WR: There has been no claim and no answer that says bandwidth is less when you talk slower. I don't know where you got this or why you continue to use it. If you frequency shift someones voice, say by recording it at 45rpm and playing it back at 33 1/3rpm you *will* get a narrower speech bandwidth. This doesn't seem to be what you are saying, however.

Fourier analysis says the 99% power bandwidth *does* get less when sending a repeating waveform with longer and longer periods, i.e. sending slower. Parseval's theorem says power in the time domain equals power in the frequency domain. So analyzing bandwidth based on the power in the spectral components of a signal *is* a valid method of evaluating bandwidth. Mickey's empirical evidence matches this and is rather convincing as confiming evidence.

There are no extremes here. And the empirical evidence is from a *real* system, different ones, as a matter of fact.

W8JI-Have a nice day.

AB0WR: Stay dry.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 22, 2007 Mail this to a friend!
Tim,

You have convinced me that sidebands are indeed produced during the middle portion of a CW pulse when the carrier is constant. My earlier comment that “if the spectral components add to a DC level, then no sidebands are produced” was sheer stupidity.

73 - Jim
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 22, 2007 Mail this to a friend!
Jim,

If you have been convinced an unchanging carrier has bandwidth other than the phase or noise or other normally very small defects, you have been misled.

It is only the changing carrier that causes bandwidth.

We have to look at the energy over a period of time long enough to include a change in order to have bandwidth caused by that change. We can do that mathematically or by some system that samples and stores or accumulates energy or information about that energy, but not in a real-time system like a CW receiver.

Let me give you an example of when Mickey and Tim are correct.

Let's say we look at the spectrum of the time-varying signal level from a transmitter sending a long string of dots. Let's assume it has a 50% duty cycle on the carrier. With 100 watts envelope power, if we average that power over time, we see it is 50 watts.

Now if we sent the dots at 100 WPM or 1 WPM, the carrier power would average 50 watts as long as we covered enough dots that the stopping or starting point was insignificant in the length of the overall string.

Now the sidebands are generated only when the carrier level shifts. They have the same peak power and bandwidth for the same shape and duration no matter how often they repeat, but something odd happens when we look at them over a long window of time.

The sidebands are not, if we looked at power long enough, constant in average power level. If we increase the speed there are more of these same peak power level sidebands generated in a given window of time. This makes the POWER bandwidth of the sidebands become higher when the speed is increased even though the average power of the carrier remains the same (for the same duty cycle).

What this means is if we look at the signal over a long enough time to include the slowest changes in level, we see the power of the sidebands decrease compared to the carrier as we decrease speed.

This is actually the core of where I (and many others) disagree with Mickey and Tim.

In order to define bandwidth in a useful manner, the method we use has to match the effect on the system we are talking about. This case is much like a pulse transmitter, rather than a voice transmitter.

The receiver cannot have a long storage time, either through selectivity or recording and memory of levels, or it will not reproduce the off and on tone. This means so far as the receiver goes (or the operator), we cannot integrate power in any parts of the system beyond the length of the response time.

This is why when you tune off frequency with a CW receiver you hear the clicks, which are what the FCC defines in the rules about interference as "harmful interference", exactly the same distance away regardless of speed. For a given rise and fall if I send one dot, the peak envelope power of the signal level at a certain spacing is exactly the same as if that dot was sent 100 times a second or once a year.

Only the average power changes, and the average power is not what moves the S meter (once AGC attack delays are overcome) and it is certainly not what mutilates the CW we are trying t copy up the band. A click is interference, and the only thing more clicks per second do is increase the rate that interference occurs.

Now if I take an analyzer of any type, including a receiver, a spectrum analyzer, or a pencil and paper and integrate the power over a long time, I can prove that even very slow speeds generate sidebands. I can also prove the power in sidebands increases with speed (because the clicks repeat more often in a given time window), but what I have really do is prove something that has little to no meaning in the actual system.

So we have no disagreement at all in what Mickey or Tim proposes, except what they describe has no effect on the real system and does not define the bandwidth actually occupied by the CW signal.

You can prove this yourself by sending a series of dots at a modest rate and tuning off to see how far away those dots would bother another station, and then doubling or tripling the speed. If you listen and if the ALC or other characteristics do not change the shape and duration of rise and falls, you will see absolutely no change in bandwidth. If you use a method that is out of context of the system we are talking about, like an analyzer set to read average power rather than peak power, you will see the instrument shows at higher speeds the energy in the sidebands is proportionally higher at higher speeds.

Now if we compare that energy to the carrier we find the average POWER bandwidth changes, but in the real time system it does not.

What we really have here is a case where people are insisting on using a valid analysis for other applications in an application where the result is totally meaningless. It cannot be used to define bandwidth because the system by definition cannot have energy storage over time, and is not subject to interference on that basis.

In other words Mickey and Tim are misapplying something that is correct in other applications to a system where the analysis is largely meaningless.

Because of this, this thread has become more of a religious argument than anything meaningful or useful to CW ops.

If you want to know how far away from a carrier the sidebands that cause clicks extend, then we need to use PEP. If you want to use power bandwidth you will get an incorrect answer.

When I punch the button to measure occupied BW on my current calibrated instrument, it uses PEP in any small window. It does not average or accumulate power, because with off and on keying the FCC does not normally want average power.

Now I do have two other pieces of gear that can be set to average power, and they will indeed produce a result that agrees with Tim and Mickey, but that result is meaningless for the CW system. The actual CW system shows the only thing that happens for a given envelope shape is that the clicks happen more often as speed is increased, but the distance away from the carrier remains the same.

It's up to everyone to decide for themselves what is correct. If we feel that we should define interference bandwidth by the distance off frequency one CW signal bothers another one, then a PEP measurement is the way to go.

If we feel that something only shown on paper or on an instrument that integrates or stores energy over a long period of time gives a more valuable indication than the actual system, then we should use that system.

I'd just add one word of caution. If we never set a reasonable limit on what is on and what is off times and we use the analysis as presented, we can also show that bandwidth is less if we transmit fewer hours in a day. We can also prove bandwidth is less when we transmit fewer hours in a year, or fewer years in a decade.

Myself, I'm going to stick with how far away the clicks cause problems. That's what the FCC uses in the rule on bandwidth, and that's what bothers everyone on the air.

But then I'm a CW operator, not someone trying to prove something with a method that doesn't describe the things we really want to know.

73 Tom
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 22, 2007 Mail this to a friend!
Tom,

After long consideration, debating the various theories, and looking back over my own analysis, I have come to the conclusion that I was wrong about one point. That point being the business of sidebands during the constant portion of the keying waveform when the carrier is at constant magnitude. Tim's arguments were persuasive, although I resisted them initially. It was actually our discussion of the Fourier Theory assertion that the spectral components exist "for all time" and just what that means in the real world. I initially said that if the spectral components add to a DC level, then no sidebands are generated. I believe now that I was wrong. Even a pure sine wave requires bandwidth if it exists for a finite period of time. If it had no bandwidth, it would either not exist at all, or would exist from t = -infinity to t = +infinity.

That said, I am still convinced that the bandwidth of a CW signal is dominated by the shape of the rising and falling edge of the keying waveform. The keying speed has some effect on bandwidth, but that effect is small, IMO, compared to the effect of the rising and falling shape of the keying waveform.

If someone were to design a rig with mathematically optimized keying waveform to generate the smallest possible bandwidth, then the keying speed might become the dominant effect. I don't think that has happened yet. In real world radios, the rising and falling edge is far from optimal, and in some cases (as you have documented) is closer to worst-case than best-case.

Don't count me as a "lost soul" yet :-)

73 - Jim
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 22, 2007 Mail this to a friend!
W8JI - So we have no disagreement at all in what Mickey or Tim proposes, except what they describe has no effect on the real system and does not define the bandwidth actually occupied by the CW signal.

K5MC - The "we" in W8JI's comment is, I believe, getting to be a small crowd. There are still some amateur radio operators, no doubt, who continue to subscribe to the theory of signal analysis "published" by W8JI. However, you will find very few, if any, communication engineers who actually graduated from an accredited electrical engineering program who will reject "traditional" Fourier analysis in favor of W8JI's theory.

As I pointed out earlier, the fact that the occupied bandwidth (as defined by the FCC in Part 2) of a CW/ASK signal varies with both the keying speed and the rise/fall characteristics of the keying waveform is well recognized in the professional world. There is no controversy in the professional literature (that written and reviewed by electrical engineers and others holding similar degrees who have studied such topics as Fourier series, Fourier transforms, information theory, etc.) on this subject.

W8JI continues to misrepresent the occupied bandwidth as defined by the FCC and other entities such as the ITU and Agilent Technologies (which went by the name of Hewlett-Packard until about 1999). The words "peak envelope power" do not appear at all in the definition. The FCC talks about the bandwidth "occupying" 99 percent of the total mean power radiated by a given emission. The word "mean" is a fancy word for average. The given "emission" is the particular signal being transmitted. Since we have been assuming a periodic string of dits as our signal, we can use a Fourier series to model that signal.

No matter how much W8JI argues otherwise, he is the misleading one on this topic. The occupied bandwidth (as defined by the FCC) of a "well-behaved" CW transmitter is a function of both the keying speed and the rise/fall characteristics of the keying waveform.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 22, 2007 Mail this to a friend!
W8JI - It will be interesting to see what someone else like Ed Hare says....rather than an equipment salesman.

K5MC - W8JI is apparently referring to the technical support engineer at Agilent Technologies who I mentioned in one of my earlier posts. I've had the pleasure of communicating with this particular engineer (who also happens to be a ham) by both telephone and email. Because he is a representative of Agilent, this engineer will not make public statements on this "controversy" here on eHam and I certainly can't fault him for taking that position.

Concerning Ed Hare, I had the pleasure of meeting him in January 2006 when I visited ARRL Headquarters in my official role as the LA Section Manager to discuss "lessons learned" during Katrina. I made a special point to visit Ed in the ARRL Lab to give him a copy of a paper I co-authored on the coupling between HF communication systems and BPL systems installed on overhead distribution lines. (BTW, this paper was published in the October 2006 issue of the IEEE Transactions on Power Delivery.) As I mentioned earlier, I emailed Ed several weeks ago about whether the ARRL Lab had any data on the occupied bandwidth (the FCC's definition rather than W8JI's "definition") for typical ham transmitters as a function of keying speed. Ed acknowledged my email, but I have not heard anything else from him on this matter.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 23, 2007 Mail this to a friend!
Very entertaining discussion.

To me, CW signals are NOT periodic. I generally tune right past the lids who send long strings of dots to tune up or mess with a pileup or whatever.

Difficult as it may be, try to turn this discussion to real-world CW signals with continuously changing spectral content. For non-periodic signal it is appropriate to use a moving window and analyze the portion of the signal captured in the window by Fast Fourier Transform. The data could be presented in form of spectrogram which would indicate how spectrum is changing with time.

Choose small enough window and spectrogram will show broad spectrum at time of transitions, and narrow spectrum during steady carrier. Choose long window and you loose information about how spectrum varies with time.

I have no objection to the analysis presented for periodic strings of dots, but everyone should be aware of the limited application to CW signals that are not periodic.

73, Chris VA3NR.
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 23, 2007 Mail this to a friend!
Chris,

The FCC defines CW bandwidth as the bandwidth that does not cause keyclicks to appear and cause interference to other communications, not the power bandwidth or whatever Mickey wants to use.

When people insist on using a model that does not behave like the real world it will always become a long unresolved disagreement.

By definition the receiver isn't affected by average power so a snapshot over a long period of time is useless.

Try a real receiver, you will see what I mean.

73 Tom

 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 23, 2007 Mail this to a friend!
W8JI:The FCC defines CW bandwidth as the bandwidth that does not cause keyclicks to appear and cause interference to other communications, not the power bandwidth or whatever Mickey wants to use.

When people insist on using a model that does not behave like the real world it will always become a long unresolved disagreement.

By definition the receiver isn't affected by average power so a snapshot over a long period of time is useless.

======================================================
W8JI is correct in what the FCC defines for CW necessary bandwidth in Part 97. However, using this as an argument that their Part 2 definition of occupied bandwidth doesn't have any applicability to determining the actual bandwidth is an obvious misfire. I suspect that the FCC, nor anyone else, would have a problem with a cw transmitter that meets this definition. Consequently, it IS something to shoot for.

Better yet, this discussion is important in determining just exactly what the processes are in a cw transmitter that result in key clicks. If one can show that an ideal and properly operating cw transmitter does have an output spectrum as defined by a Fourier series, then we have a starting point and can determine what other processes are occurring that change this.

However, as long as the view is held that the "rise time" is the causal factor rather than the result of these other processes then there is no way to address the actual cause.

For example, in a perfect world with an ideal transmitter one should be able to modulate a carrier with a series of dits consisting of square pulses with near infinite rise times and the output spectrum should look like that predicted by a fourier series. Decreasing the rise time to 5 ms should not have an adverse effect on the output spectrum of this ideal transmitter. In other words, "rise time" is not the causal factor in increasing the output spectrum.

Unless there can be agreement on this, there is no common starting point.

I fear the folks that are pursuing "rise time" as a causal factor are wondering in the dark and will never find the light!

Here is something to think about. In the icom 751a (and the 745, 751, and 761 at least) the cw oscillator is keyed off and on in cw regardless of the mode, i.e. full breakin, semi-breakin, or full on. Yet this same oscillator is also used for rtty. The cw and rtty mark uses exactly the same components. The rtty space is generated by switching in a fixed and trimmer capacitor. 800 Hz shift uses an extra fixed capacitor. The only difference - the oscillator stays on the whole time you're in rtty transmit.

Have you ever heard anyone complain about key-clicks when operating rtty? I wonder why that is. Same IF chain. Same RF amp chain. The rtty space is nothing more than a cw signal that goes on and off with a pretty fast rise time! Yet why no key clicks?

Could spinning up and spinning down the cw oscillator be generating a number of intermod spectrum? Is this one of those processes that we should be looking at?

K5MC might help by letting us know if his cw oscillator are keyed in his transmitters.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 23, 2007 Mail this to a friend!
VA3NR-Difficult as it may be, try to turn this discussion to real-world CW signals with continuously changing spectral content. For non-periodic signal it is appropriate to use a moving window and analyze the portion of the signal captured in the window by Fast Fourier Transform. The data could be presented in form of spectrogram which would indicate how spectrum is changing with time.

VA3NR-Choose small enough window and spectrogram will show broad spectrum at time of transitions, and narrow spectrum during steady carrier. Choose long window and you loose information about how spectrum varies with time.
----------------------------------------

AB0WR: I don't think this is what will happen. It's very noisy tonight but I recorded part of a CW conversation on 80m using my Softrock SDR and KGKSDR software as a wave file. I then used CoolEdit Pro to select out just one dit from the sending train.

I then used ScopeDSP to analyze the dit. No matter what section of the dit I select, or even if I select the whole dit, the FFT in ScopeDSP shows a frequency spectrum of about 450hz to 900hz.

Some of this is an artifact of the 500hz filter I was using. I'm not sure what the total bandwidth would have been if I had done the recording using a 2.1khz filter. I will try that and see what it shows.

In any case, you do NOT get a broad spectrum at the start and end of the pulse with a narrow spectrum in the middle. You get the same spectrum throughout the pulse.

I will try recording some signals when the noise is less, do the analysis using the software above, and post it somewhere for everyone to see.

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 23, 2007 Mail this to a friend!
VA3NR - Difficult as it may be, try to turn this discussion to real-world CW signals with continuously changing spectral content. For non-periodic signal it is appropriate to use a moving window and analyze the portion of the signal captured in the window by Fast Fourier Transform. The data could be presented in form of spectrogram which would indicate how spectrum is changing with time.

K5MC - Here are some additional occupied bandwidth (99% power bandwidth) measurements for my Kenwood 940:

1. About 30 Hz bandwidth when sending a string of dahs (dashes) at 6 wpm and about 60 Hz bandwidth when sending a string of dits (dots) at 6 wpm.

2. About 150 Hz bandwidth when sending a string of dahs at 53 wpm and about 260 Hz bandwidth when sending a string of dits at 53 wpm.

3. About 45 Hz bandwidth when sending the word PARIS repeatedly at 6 wpm and about 220 Hz bandwidth when sending the word PARIS repeatedly at 53 wpm.

My old Autek memory keyer was used to key my Kenwood 940 and Ten Tec Corsair. The slowest and fastest speeds for the Autek keyer are 6 wpm and 53 wpm. I used the internal keyers when I measured my Ten Tec Orion II and Kenwood 2000 rigs. Both of those internal keyers have a minimum speed of 10 wpm.

The occupied bandwidths that are generated by a periodic string of dits and a periodic string of dahs do a good job in establishing the upper and lower bounds on the occupied bandwidth of a practical CW signal. The periodic string of dits is a good test signal exactly because it does yield the "worst-case" occupied bandwidth for a specific transmitter at a specific speed. (Of course, the rise/fall times will also have to be specified for the newer rigs that have adjustable rise/fall times.)

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 23, 2007 Mail this to a friend!
WA0LYK - K5MC might help by letting us know if his cw oscillator are keyed in his transmitters.

K5MC - Jim, I would be extremely surprised to learn that the oscillators are directly keyed on any of the rigs I tested. All of my rigs are completely unmodified as far as their keying circuits are concerned, but I have not looked at the design details of any of them to know with absolute certainty that the oscillators are not being keyed.

The only rig that I've ever owned that directly keyed the oscillator (to my knowledge) was my old Heathkit DX-60B novice transmitter that I built in 1968. Even the matching outboard VFO for the DX-60B gave the operator the option to either key or not key the oscillator. When I started checking into CW traffic nets in 1970, I remember "hearing" my oscillator on 80 meters during key-up times because the DX-60 didn't use frequency multiplication on that band. Since I wanted to run QSK on the traffic nets, I started keying the oscillator, but the resulting chirp was really bad. Fortunately, I was able to upgrade to a Drake T-4XB transmitter shortly thereafter and I "retired" my DX-60!

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 23, 2007 Mail this to a friend!
Brief intro to Short-time Fourier transform:
http://en.wikipedia.org/wiki/Short-time_Fourier_transform

If experimenting with STFT, take care to apply a window function that does not impact the spectral analysis. For example if you simply truncate a recorded signal you are effectively applying a rectangular window function. The FFT result will reflect the instantaneous step changes at the boundaries of the window. Better choice might be the Gaussian window that is available in various spectrogram packages.

If you select correct window function, and choose window length that captures only the constant amplitude portion of a dot, result will be narrow spike at sinusoid frequency. As window moves and captures a transition, spike amplitude will drop and spectrum width will broaden. Window function, window length, # of values in FFT, all will affect the time and frequency resolution. Choice of window is critical. (I think that length of window is cause for much of the disagreement on this topic.)

It might be educational to experiment with one of the various spectrogram packages:

Spectrum Lab (http://freenet-homepage.de/dl4yhf/spectra1.html)
or

Spectran
http://digilander.libero.it/i2phd/spectran.html

or maybe praat
http://www.fon.hum.uva.nl/praat/

Interestingly, there is sample screen print on the Spectran homepage where you can see broad spectrum spikes at the start and end of each code element on one of the CW signals. Look for the '73' on the right screen at about 600 Hz.

73, Chris VA3NR.
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 24, 2007 Mail this to a friend!
I wanted to make a comment on my recent reply in this thread. I do understand the difference between AM and FSK. I was not trying to equate these modulation schemes but only trying to emphasize that different process can have different results. In one case a sudden frequency change can occur with little or no intermod while with a sudden amplitude change it appears there is lots of intermod while using the same components.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by W8JI on August 24, 2007 Mail this to a friend!
Jim,

There are many things that can happen in a transmitter and cause problems, and like anything in life there are multiple ways to analyze a problem.

CW is exactly like an AM mode so far as sidebands produced, but besides the bandwidth of the modulating signal (which is set by the carrier rise and fall slope) there can be other defects that introduce problems. One common problem is ALC. You can see the effects of ALC at:

http://www.w8ji.com/occupied_bw_of_cw.htm

If you look at the FT1000 OBW, you see a "wobble" in bandwidth as dot speed is progressively increased. This is caused by the ALC in the FT1000. It modifies the envelope rise and fall as speed is changed.

If I set the TX IF gain in the FT1000MP MKV so the ALC is barely active, that wobble goes completely away and the FT1000 has a constant OBW with dot speed change.

My measurements and descriptions will disagree with people who insist on using a meaningless description like "power bandwidth" where accumulated or average power over long periods of time are used.

I'm using a response that acts like a typical CW receiver, since we are trying to describe the effects on a CW receiver and not a tub of water being heated.

You can see analysis of CW bandwidth using the same method, where the analysis matches what a receiver or operator hears, at the following links:

By myself and several others at:
http://www.w8ji.com/keyclicks.htm

By SM5BSZ of Linrad and moonbounce fame at:
http://www.nitehawk.com/sm5bsz/dynrange/dubus204/dubus204.htm

By Kevin Schmidt at this link:
http://fermi.la.asu.edu/w9cf/articles/click/index.html

By Doug Smith, who did the CW system in the Ten Tec Orion at this link:
http://www.nitehawk.com/sm5bsz/others/occbw.htm

There are dozens more of analysis that actually use methods that duplicate or replicate the actual CW system, so this really isn't a matter of W8JI vs. two or three other people.

It is really a few people who want to use an analysis or method that does not fit the system being discussed vs. the people who do.


73 Tom


 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 24, 2007 Mail this to a friend!
VA3NR:Interestingly, there is sample screen print on the Spectran homepage where you can see broad spectrum spikes at the start and end of each code element on one of the CW signals. Look for the '73' on the right screen at about 600 Hz.

73, Chris VA3NR.

===================================================

Yet not all of the signals show this. What is your conclusion from these screen prints?

 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 24, 2007 Mail this to a friend!
VA3NR-If experimenting with STFT, take care to apply a window function that does not impact the spectral analysis. For example if you simply truncate a recorded signal you are effectively applying a rectangular window function. The FFT result will reflect the instantaneous step changes at the boundaries of the window. Better choice might be the Gaussian window that is available in various spectrogram packages.

VA3NR-If you select correct window function, and choose window length that captures only the constant amplitude portion of a dot, result will be narrow spike at sinusoid frequency. As window moves and captures a transition, spike amplitude will drop and spectrum width will broaden. Window function, window length, # of values in FFT, all will affect the time and frequency resolution. Choice of window is critical. (I think that length of window is cause for much of the disagreement on this topic.)
------------------------------------------------

AB0WR:Yes, you must be *very* careful of the window you use.

I have never used an STFT to analyze a cw envelope. I'll have to do some reading on the subject.

Knowing, however, that a CW envelope is produced from the multiplication of a square wave and a sinusoid which is a direct convolution in the frequency domain, and knowing that the harmonics of a square wave exist for the entire square wave, I would question an STFT that shows the sidebands of a CW envelope only existing during a portion of the envelope.

Even if you consider a CW dit as a "gating function", you wind up with a convolution in the frequency domain of a continuous spectrum of the form (sin x)/x, i.e. the sinc function, with the carrier sinusoid so you still wind up with the sidebands being there all the time.

I'm not sure very short interval windows such as the STFT uses will work with very low frequencies like you will get at baseband, e.g. an audio envelope with a carrier of 600hz to 700hz. Won't you run into a resolution problem of trying to separate 600hz from 610hz from 630hz from etc?

To get the result that sidebands only exist during the rise time of the CW envelope, I think it will have to somehow be shown that the harmonis of a square wave exist only during the rise time of the square wave. Either that or show that CW envelope generation is not a case of linear mixing in the time domain but, instead, follows some other mathematical description of the process other than a multiplication of cw(t) with carrier(t).

I've not seen this kind of treatise anywhere, have you?

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 24, 2007 Mail this to a friend!
W8JI (June 4, 2007) - Now if I view this signal with an instrument approved and accepted by the FCC to measure occupied bandwidth, it also detects the peak powers. It samples the peak powers and stores them. It doesn't care in the least if the carrier peak power is on for 5 days or 5 milliseconds, it is whatever it is. It might be 100 watts in a typical radio. The sideband, the level being determined by the change in amplitude and the frequency being determined by the slope of that change, repeats exactly so long as nothing alters that rise and fall. Every single time the analyzer sweeps by a frequency it reads the peak power at that instant, and it records that peak power. One peak an hour, one peak a millisecond, if it is 10 watts peak power the level indicated on a particular frequency will be constant unless we somehow alter the slope or level change of the rising edge.

This is why a spectrum analyzer agrees with the ear, and why a receiver agrees with the ear and the analyzer.

W8JI (June 8, 2007) - You keep picking on my test gear, but I have current expensive Agilent equipment that has a resoltion bandwith of only 10 Hz.

Forst you wrongly publically claimed I was syncing the dots to the trace, now you imply my rigs or my test gear has some odd abnormal problem.

The fact is ALL of my rigs are common rigs and all of the "abnormal problems" with them are common to all equipment.

K5MC - Perhaps W8JI will enlighten the "equipment salesman" at Agilent and the rest of us with the EXACT DETAILS of his measurement setup using his spectrum analyzer. Now that I have access to a very good (and expensive!) analyzer, which is giving me results at odds with W8JI, I need to know whether I should send the analyzer back to Agilent for a refund.

W8JI needs to be sure to not leave anything out in his setup description. As a courtesy to W8JI, http://cp.literature.agilent.com/litweb/pdf/N9060-90010.pdf is the online link to the "User's and Programmer's Reference Measurement Application" manual for my N9020A analyzer. According to the "equipment salesman" at Agilent, W8JI's posted spectrum plots indicate one of the 8590E series analyzers. The online link for that manual is http://cp.literature.agilent.com/litweb/pdf/08590-90301.pdf

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by WB1AIW on August 24, 2007 Mail this to a friend!
I see a lot of mis-information continuing to circulate here, jeez Louise!


Readers, please don't be swayed. Sideband power exists only when carrier amplitude changes! This is the only way it can be in the real world. Since nothing is known about the signal before it happens; systems can only react to their input, they can't predict it. Suppose someone sends a sequence of dits, then decides to stop just before the next rising edge. If the sidebands were continuous, and non-fading, how would they know that the operator has decided to stop? The input is zero volts either way, so these sidebands have no way of being informed that the next edge isn't going to happen. So do they instantly disappear at the exact moment that the edge would have occurred? What if the edge came 10us late, do these sidebands instantly disappear, then re-appear when the edge occurs??? And what if the edge came early? Now, Tim (AB0WR) says they add together to create the signal, so if the operator is late with the edge, wouldn't these sidebands all add in phase at the time the edge should have happened and create the edge anyway??? How can an operator control them once they've started? What piece of information from the past tells them how long to exist for?


A spectrum analysis over a full cycle of transmitted dits shows the average power of those sidebands but gives NO information about their nature in the time domain. So concluding they are there continuously is faulty.

Readers, I ask that you give this your thoughtful consideration.


>VA3NR:Interestingly, there is sample screen print on the Spectran homepage where you can see broad spectrum spikes at the start and end of each code element on one of the CW signals. Look for the '73' on the right screen at about 600 Hz.
>
>73, Chris VA3NR.
>
>===================================================
>
>WA0LYK: Yet not all of the signals show this. What is your conclusion from these screen prints?
>

That screen print (http://digilander.libero.it/i2phd/spectran.html) nicely shows when the sideband is present; during the amplitude changes only.

Tim, AB0WR, you claim that for a signal made from odd harmonics of sine waves, the sidebands exist for the whole period. I agree 100%! And you can tell that from the fact that the high (dot) and low parts of the resulting waveform have some nasty looking ripple. Note that even the "off" part of the transmission shows ripple, which would cause the transmitter to emit a bit of power between elements. I contend that a CW transmitter doesn't create its waveforms this way, so the output has a nice flat top and is fully off between elements. I looked very closely at some CW signals on the air using Spectrum Lab, and I'll be darned if I could see any power emitted between those elements. So I conclude that the model your using for a CW waveform doesn't match reality. Chen and others tell the story well.

Let's look at some other statements:

AB0WR - That fact that you have rise times on the outputs indicates that you have a bandwidth limited system response. If the input bandwidth is wider than the system response bandwidth you get exactly what Tom is showing in his displays. You have extrapolated that into saying that the system response will *ALWAYS* be that wide no matter what the input happens to be. Yet your displays only include inputs that are *wider* than the system response. Therefore you have not provided a full system test.



Controlling the rise and fall times IS applying bandwidth control. So long as the rise and fall times don't change, the bandwidth envelope away from the carrier doesn't change, as shown nicely in Tom's SA plots.

If you want your bandwidth envelope to change in proportion to your keying speed, then change the rising and falling edge rate proportionally with your keying speed.


AB0WR - If key clicks are generated 5khz away from a square wave with a 1hz frequency then you are hearing the 5000th harmonic. For a keying waveform with a 2hz frequency you would be hearing the 2500th harmonic. For a keying waveform of 4hz you would be hearing the 1250th harmonic. The 1250th harmonic would be something like -62db down from the fundamental.
Question 4. Just how strong of a signal are you putting out that you can hear the 1250th harmonic of a 4hz keying wave that is 62db down from the fundamental?
Question 5. If you can hear the 1250th harmonic of a 4hz keying wave how sure are you that you aren't nearing artifacts from an overloaded receiver?

Here's another nail in the coffin for the continuous sideband power model.
The average power may be -62dBc, but the peak power doesn't change; the click you hear has the same amplitude and duration, regardless of the repetition rate yet completely characterized by the edge shape.


AB0WR - Look at it this way - an RC low pass filter with a cutoff of 600 hz would eliminate all of the harmonics of a 100hz square wave that are above 600hz.

Not quite, my friend. An RC filter would be a huge disappointment to you if you expected it to eliminate everything above its 600Hz cutoff.


AB0WR - The 3rd harmonic would be at 300hz, the 5th harmonic would be at 500hz, and the 7th harmonic would be at 700hz. So you would wind up with a square wave consisting of only the fundamental and the 3rd and 5th harmonics. It's not going to look like a very good square wave.
If you apply a 5hz signal to that same filter, you will get odd harmonics all the way out to the 119th. A much better square wave will result.


If you believe this, then you'll be surprised to discover that the rise and fall shapes will be the same whether the square wave frequency is 100 Hz or 5 Hz. The falling edge will have this shape A(t) = Ao*e^(-t/RC), and the rising edge will have this shape A(t) = Ao - Ao*e^(-t/RC). Go ahead, look at it with an o'scope.


Lin
WB1AIW


 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 24, 2007 Mail this to a friend!
W8JI - There are dozens more of analysis that actually use methods that duplicate or replicate the actual CW system, so this really isn't a matter of W8JI vs. two or three other people.

It is really a few people who want to use an analysis or method that does not fit the system being discussed vs. the people who do.

K5MC - The various links that W8JI provides are studies/reports/discussions relating to the key-click or interference "bandwidth" generated during the rise/fall times of CW waveforms. I have no argument with these studies. As a matter of fact, I briefly discussed the use of "short-time" Fourier transforms (STFTs) a few days after my original eHam article appeared:

K5MC (May 28, 2007) - Fourier analysis can be misleading when used to study signal transients. As I pointed out earlier, the frequency information of a signal calculated by the classical Fourier transform is an average over the entire time duration of the signal. If there is a local transient over some small interval of time in the lifetime of the signal (such as the rise/fall times of a single keying pulse having a relatively long total time duration), the transient will contribute to the Fourier transform but its location on the time axis will be lost.

The "short-time" Fourier transform is one attempt to overcome this limitation, but a more recent mathematical approach to studying signal transients is "wavelet" analysis. Rather than using everlasting sinusoids, the wavelet functions are "local" both in time and frequency. There are many different families of wavelet functions, however, and choosing the most appropriate one to use for a specific signal isn't usually a trivial matter. Perhaps some day a manufacturer will develop a "wavelet" analyzer to study signal transients, but as far as I know, no such instrument is readily available and wavelet analysis remains more of a research tool. (Studying the transients generated by a variety of CW keying waveforms using wavelets would probably make an interesting research topic for my next graduate student!)

K5MC (August 24) - One of the research areas I've pursued as an electrical engineering professor (with a particular emphasis in electric power engineering) has been the transient analysis of time-varying loads (such as arc furnaces) using such tools as STFTs and wavelets. Power engineers study the spectra of these transients, but they rarely, if ever, compute the "bandwidths" of these transients. Similarly, in communication systems the "bandwidths" of the signal transients by themselves are rarely defined or calculated. The transients can certainly cause problems if they are not sufficiently limited, but the transients themselves convey little or no useful "information" to the receiver. On page 37 of the seventh edition of Couch's book (the textbook I referenced in my original eHam article) you will read the following:

Couch - In communication systems, if the received average signal power is sufficiently large compared to the average noise power, information may be recovered. This concept was demonstrated by the Shannon channel capacity formula. Consequently, AVERAGE power is an important concept that needs to be clearly understood.

K5MC - You can find similar statements to Couch's in dozens of communication textbooks. It's the average (mean) powers of the individual pulses (the dits and the dahs) of our CW signals that are important in terms of conveying information, not the "peak envelope powers" of the turn-on and turn-off transients of the pulses!

If someone answers my CQ on my calling frequency, I don't know if his signal has significant clicks or not. If his signal is weak, however, the sending operator can slow down and I can go to a sharper filter in my receiver because the 99% "power" bandwidth (which is equivalent to the FCC's definition of occupied bandwidth) of the signal decreases when the sending speed decreases. By doing these things, the signal-to-noise ratio will go up and we will be more likely to have an enjoyable QSO.

The fact that the occupied bandwidth (as defined by the FCC, the ITU, Agilent Technologies, etc.) does vary with the sending speed is unquestioned in professional circles. W8JI is apparently so consumed by key clicks to the exclusion of everything else that he still can't acknowledge this basic fact.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by KE3HO on August 24, 2007 Mail this to a friend!
Lin,

You bring up some valid points. However, I think that you are misquoting Tim’s position. Nobody here, as far as I can tell, has suggested that the sidebands exist between CW dits and dahs. Even if the keying waveform had some ripple between elements, the transmitter is off anyway, so it is certainly not transmitting sidebands. In addition, if you took a square or rectangular wave and decided to allow all harmonics up to some number N and to completely remove all harmonics beyond N, then the results would be a signal with ringing and ripple. However, in real transmitters, this doesn’t happen. Instead, harmonics are rolled off with something like an RC filter. This creates a situation where all harmonics up to N are unattenuated, but harmonics beyond N are progressively rolled off as you go up in frequency. The result is a keying waveform with a smooth, rounded, sloped transition between levels, not ringing and ripple.

Tim has said that the sidebands produced during a CW pulse are constant throughout the entire pulse. I don’t agree with that, although Tim has convinced me that sidebands are produced during the constant “un-modulated” part of the pulse. If you look at the link that Tom posted here, http://fermi.la.asu.edu/w9cf/articles/click/index.html , and you scroll down to figures 2, 3, and 4, these show the sidebands produced by a CW keying function that has identical rise and fall time and shape, with the only difference being the width of the pulse. If sidebands are ONLY produced during the rising and falling edges, then figure 2 and figure 3 would be identical. After all, the rising and falling edges in these two plots are identical. There is also some information on Tom’s website showing some Fourier analysis that I did several months ago that shows the very same thing. In my analysis, as in W9CF’s, I was looking not at a string of pulses, but at a single pulse.

While I now agree with Tim and Mickey that sidebands are produced during the central portion of the CW pulse, I do not agree that the sidebands are constant during the entire pulse. I don’t have a good mathematical basis for this yet, but I would contend that the sidebands are very different during the rising and falling edges of the keying waveform compared to the sidebands during the central portion of the pulse.

Now, someone could (and probably will) argue that you can calculate the spectrum of the keying waveform with Fourier theory, and Fourier theory says that all of the calculated spectral components exist “for all time”. This is true, whether you wish for that to mean from –infinity < t < infinity, or if you wish for that to mean for the entire duration of the pulse. My personal opinion is that this is one of the pitfalls of Fourier analysis. You can see the spectral elements in the frequency domain, but you can’t see how they add in the time domain (other than seeing the original keying waveform).

Let’s look at this argument that the spectral components of the keying waveform can be calculated using Fourier Theory, and as such they exist for the entire duration of the pulse, therefore the sidebands produced must be constant for the entire duration of the pulse.

A CW transmitter is fundamentally an AM transmitter where the carrier is modulated by the keying waveform. If I sit down at my AM transmitter and I say, “CW is fun.” into the microphone, the sidebands produced vary over time, following the fluctuating amplitude of my voice. Yet I could say that same sentence into a microphone connected to the input of my sound card, digitize it at some reasonable sampling rate, and run the resulting data through an FFT algorithm and calculate the spectrum of my voice speaking this sentence using Fourier theory. These calculated spectral components would also exist “for all time”, just like for the CW keying waveform. Should I conclude then that the sidebands produced are constant throughout my entire transmission? Clearly not.

It is my belief that the following is true, and again I don’t have a good mathematical basis for this yet.
• If the calculated spectral components add to zero during some period in the time domain, no sidebands are produced.
• If the calculated spectral components add to a constant (DC) non-zero level during some period in the time domain, then the sidebands produced during that period will be constant and will be determined by the width of this time period.
• If the calculated spectral components add to a time-varying signal during some period in the time domain, then time-varying sidebands are produced.

It is, therefore, my belief that during the rising and falling edges of the CW keying waveform, time-varying sidebands are produced, and these sidebands are determined by the shape of the rising and falling edges of the keying waveform. During the central, flat-top portion of the CW keying waveform, constant sidebands are produced that are determined by the width of the central flat-top portion of the keying waveform.

As can be seen in the analysis done by W9CF as well as in the analysis that I did which W8JI posted to his web site, you can see that the SHAPE or ENVELOPE of the spectrum is determined almost exclusively by the shape of the rising and falling edges of the keying waveform. And, what is more, this same thing can be seen in the spectral plots that Mickey posted. For example, in Mickey’s figures 9, 10, and 11 (Kenwood TS-2000), you can look at any frequency difference from the carrierer and look at how far down that particular frequency component is from the carrier, and they all match regardless of keying speed. Same things for figures 12 and 13 (TenTec Corsair). Same thing for figures 15 and 16 (TenTec Orion II). One interesting thing to note is that this does NOT hold true for the Kenwood TS-940 (figures 6, 7, and 8). The shape of the spectrum changes significantly with keying speed. Mickey stated that he observed the CW envelope on a scope to make sure that that rise and fall times were the same for all keying speeds. I have no reason not to believe him. However, looking at these plots, the conclusion that I would draw is that if the rise and fall times were indeed the same, then the SHAPE of the rise and fall must have been different in some way. Even a small difference in shape could account for the difference in spectral envelope shape. I know nothing about the TS-940, so I don’t know how they generate the keying waveform. I am only drawing a conclusion from the spectral plots.

For any of these rigs other than the TS-940, if you were listening with your CW receiver at, say, 250 Hz away from the carrier, you would not notice any difference in the level of interference that you received with changes in keying speed.

Finally, going back to Mickey’s original assertion that the bandwidth changes with keying speed, I would draw your attention once again to W9CF’s page, specifically figure 4 and the 4 lines of text that follow it. While the overall shape of the spectrum is constant regardless of pulse width at frequencies removed from the carrier, there is a difference in the near-carrier spectrum. W9CF says that the energy near the carrier is 2.5 times higher for the wider pulse width. If you were to integrate the power under the curve for each spectrum in figure 2 and figure 3, then began at the carrier integrating out in both directions looking for the point where you reached 99% of the total area, they you would get a different bandwidth if 99% power was your definition of bandwidth, just because there is so much more power near the carrier in the narrower pulse. This is Mickey’s argument, and I cannot argue with it. However, I have to agree with Tom that this definition of bandwidth does not reflect on what the guy listening 250 Hz away would hear. Or the guy at 500 Hz away. Or the guy at 1 kHz away (assuming the transmitter is not using narrow filters in the TX chain). The 99% power bandwidth is not a good measure of CW bandwidth, as transmitters can have significantly different 99% power bandwidths and yet have virtually identical power at 100 Hz from the carrier, at 200 Hz from the carrier, at 300 Hz from the carrier, etc.

At this point, I would say that I agree with Mickey’s mathematical analysis. I agree that the 99% power bandwidth does indeed change with keying speed as he claims. I just don’t agree that 99% power bandwidth is a useful measure of CW bandwidth. A better measure, IMO, is to look at sideband attenuation (relative to carrier) versus frequency. In this latter case, the bandwidth will be dominated by the shape of the rising and falling edges of the keying waveform and will be almost independent of keying speed.

73 - Jim
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 25, 2007 Mail this to a friend!
KE3HO:The 99% power bandwidth is not a good measure of CW bandwidth, as transmitters can have significantly different 99% power bandwidths and yet have virtually identical power at 100 Hz from the carrier, at 200 Hz from the carrier, at 300 Hz from the carrier, etc.

==============================================

I may be wrong, but I don't think this can hold up if you do the math. If the 99% bandwidth shrinks, then how can the power in 200 or 300 Hz remain the same? You only have 1% of the power to spread out amongst the remaining spectra which by definition is larger since the 99% part has shrunk.

Also, don't mix time and frequency domains, you can't mix the two. Part of what we are discussing here is due to causality. How do you know what bandwidth is required until the "pulse" is complete? You don't and can't know. This may be a less than accurate analogy but I'll use it anyway. Suppose you gave someone a lead ball and told him to climb a ladder and drop it. You can't calculate the force with which it will hit until you know the height to which he climbed. You don't need to know the time he takes to climb or if he holds it for a length of time at a given height before dropping, just the height(and the acceleration I guess, grin). But the fact remains, you can't know the force until the event happens. Seeing the bandwidth of a pulse is kind of like that, you can't predict the future, therefore you can't know the bandwidth until the pulse happens, completely.

Some less than intuitive concepts here I know, but they are good to discuss.

As to whether the freqs in a square wave exist throughout, look at it this way. If I took 30 oscillators and set them up at the right frequencies with 1/n power ratios then summed them what would I get if you looked at the output of the summer? Are the tones there during the middle, flat part of the result? Can you see the tones if you used a spectrum analyzer set to trigger/measure only during the middle of the result? Do you know if I turned the oscillators on or off or simply gated the output (i.e. do they exist for all time or just for the length of the pulse? Does it matter? If I use the result to modulate a carrier are the original tones still there?

I think part of the problem is that folks are having a hard time believing there are spectra actually there during the middle of a dit or dah especially when you don't REALLY know when the pulse is going to end. Yet all I can tell you is that they are. There are too many pictures of even old analog spectrum analyzers examining cw pulses and that measure individual frequencies and the power in each and really show they are there.

Another problem folks have is that how can a physical switch used to turn on/off a carrier "generate" sideband frequenies. Are the sidebands created when I close the switch or when I open it? The switch isn't made up of tones so where do the "sidebands" come from. You just have to know that when you modulate/mix what are basically two signals, that is what you end up with and that they can be seen on a frequency measuring device. They are real!

Jim
WA0LYK
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 25, 2007 Mail this to a friend!
KE3HO - At this point, I would say that I agree with Mickey’s mathematical analysis. I agree that the 99% power bandwidth does indeed change with keying speed as he claims. I just don’t agree that 99% power bandwidth is a useful measure of CW bandwidth. A better measure, IMO, is to look at sideband attenuation (relative to carrier) versus frequency. In this latter case, the bandwidth will be dominated by the shape of the rising and falling edges of the keying waveform and will be almost independent of keying speed.

K5MC - Jim, if you decide to approach the FCC with some sort of "key-click" or "interference" bandwidth definition, I suggest you also readily acknowledge to the FCC the fact that the current definition of "occupied" bandwidth in Part 2 is equivalent to the 99% power bandwidth.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 25, 2007 Mail this to a friend!
WA0LYK - Some less than intuitive concepts here I know, but they are good to discuss.

K5MC - Jim makes some very good comments here. I've been down this road before and you can really tie yourself up into knots if you try to rely too much on your "intuition" to "figure out" exactly what's happening in the frequency domain from one instant of time to the next.

The 99.1% power bandwidth calculations I carried out for the two keying waveforms in my earlier eHam article were all done in the time domain. Since the occupied/power bandwidth definition uses mean (average) powers, it was a fairly simple matter to calculate the average powers of the individual frequency components and use the superposition theorem from circuit theory to arrive at the final answers. I didn't assume that the frequency components responsible for the various sidebands were only there during the rise/fall times, etc. The various frequency components making up the keying waveform are always there mathematically! Again, since I was assuming a string of dits at 50% duty cycle, I was able to stay in the "time domain" with my calculations and use algebra and trigonometry to do the calculations (once I had calculated the coefficients of the Fourier series terms by cranking out the integrals!). I did not calculate any Fourier transform (frequency domain) integrals in my calculations.

The purpose of my earlier eHam article was to point out to hams that there are many definitions of "bandwidth" and one must understand the particular type of bandwidth under discussion. Otherwise, there will be endless debates. We've had a very long debate here despite my best efforts to provide clear bandwidth definitions from "official" sources. At this point in this discussion, however, I can't help but believe that the vast majority of hams who have been following my two articles will now agree with the following statements:

1. For CW signals, the 99% power bandwidth is equivalent to the "occupied" bandwidth as defined by the FCC in Part 2. (I provided the link to Part 2 in my earlier article.)

2. The 99% power bandwidth/occupied bandwidth does vary with the keying speed as I've now demonstrated both mathematically and experimentally.

I can understand that many hams think that the emphasis on the "bandwidth" of our signals should be on their potential to cause interference to stations on adjacent frequencies. However, hams should also, in my opinion, be able to appreciate the relationships between such items as rate of information (keying speed), bandwidth, and signal-to-noise ratio! As amateur radio moves more toward digital communications in the future, these relationships (as exemplified by such concepts as Shannon's equation, the exchange between signal power and signal bandwidth, etc.) will hopefully become more obvious to most hams.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 25, 2007 Mail this to a friend!
I had a short, off-line, conversation with WA0LYK about the short time fourier transfer (STFT) process.

Our conclusion is that the STFT works well for sampling a waveform in order to rebuild it but that all that one sample will give you is the sine and cosine components necessary to rebuild that one little, bitty, piece of the overall waveform that you sample. It doesn't really tell you anything about the overall bandwidth of the total pulse.

If you have a waveform whose frequency varies over time the STFT will give you a method of tracking that change in the fequency domain that would be impossible to do in any other manner. But with a periodic, non-frequency varying waveform, one sample isn't going to tell you much about the overall picture.

It would be like pulling one brick out of a building. That small sample will tell you how to rebuild that one little, bitty piece of the building. It will tell you if that brick has a face that is concave, convex, carved, cracked, or whatever. But it won't tell you how tall the building is, how wide it is, or what its footprint on the ground is.

If you want to know how much space that building takes up you better look at the whole picture, not just a piece of it.

The same thing applies to a keying envelope. If your window is the equal to or less than the period of the carrier signal, what the STFT will give you is the frequency components necessary to rebuild that little, bitty piece of the overall envelope. It *will* be dominated by the frequency of the carrier. It can't be anything else. But that little, bitty piece of the overall waveform won't tell you much about the overall waveform.

Personally, I can't find any references on the internet that talk about using the STFT to find the overall bandwidth of a periodic signal, just the components involved in that one small sample. If you want to analyze an entire waveform you have to combine many, many small samples which, I suspect, will wind up giving you exactly what a Fourier analysis of the entire waveform will give you.

So, do we have the concept correct or not?

tim ab0wr

 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 25, 2007 Mail this to a friend!
KE3HO-Now, someone could (and probably will) argue that you can calculate the spectrum of the keying waveform with Fourier theory, and Fourier theory says that all of the calculated spectral components exist “for all time”. This is true, whether you wish for that to mean from –infinity < t < infinity, or if you wish for that to mean for the entire duration of the pulse. My personal opinion is that this is one of the pitfalls of Fourier analysis. You can see the spectral elements in the frequency domain, but you can’t see how they add in the time domain (other than seeing the original keying waveform).

Jim, your last sentence says a mouthful.

I suspect the important concept to take away from this is that determining the bandwidth of a signal is not possible in the time domain, it is a frequency domain concept.

When it comes right down to it, it really isn't even important when in time the bandwidth occurs, it is just important that it *does* occur. And that bandwidth can be calculated. That bandwidth doesn't spread out willy-nilly in a manner that can't be determined.

The second important concept is that rise time is a result and not a cause. Rise time is a result from inputs to a system being acted upon by the system, the inputs to the system and the transfer function of the esystem are not a result of the rise time.

If you have a brick wall low-pass filter the highest component seen in a keying waveform will vary litle. It will be that odd harmonic that is closest to the brickwall. That does not mean that the power bandwidth of the signal resulting from the system with that brickwall filter will be constant. Slower speeds will have more harmonics pass the filter and the 99% bandwidth of those harmonics will be closer to the fundamental.

The rise times of all these waveforms will be similar because the rise time is determined by the highest harmonic and not the number of the harmonics in the signal.

This is where the casual stuff just seems to get lost. If you bandwidth limit a signal then you bandwidth limit the signal. If the output of the system then has a bandwidth larger than this, it must be an indication of something else happening in the transmitter.

It is this "something else" that you seem to be searching for and which no one seems to have the answer to. It can't be the rise time -- not if a 5ms rise time indicates a system with a 70hz bandwidth.

I have yet to see anyone refute mathematically that the bandwidth of a 5ms rise time is other than 70hz. I have yet to see anyone refute mathematically that a square wave is made up of continuous, odd harmonics. I have yet to see anyone refute mathematically that a linear mixing process of continuous tones and a carrier can result in the disappearance of those tones for a period of time during the mixing process.

That tells me that if you are seeing or hearing keyclicks kilohertz away from the fundamental frequency during the rise and fall of a CW pulse, something else is going on to generate them.

Now, I will also admit that even in a bandwidth limited system like we are talking about that higher numbered harmonics will pass through a simple RC filter, but they *will* be attenuated. For a 20hz keying wave, a harmonic at 500hz (25th harmonic) will already be down to 1/25th of the fundamental. Add in another 18db of loss (70 x 2 = 140, 140 x 2= 280, 280 x 2 = 560, about 3 octaves above 70hz) and that harmonic is going to be down a long ways (about 40db or so?). A 40db drop would put the power at about 10mw or so from a 100watt transmitter. I sincerely doubt that this level of output is going to cause objectionable keyclicks to very many people.

My only conclusion can be that something else is causing these key clicks and we'll never find the answer till we begin to look beyond "bandwidth" of a CW keying envelope. None of the math indicates that a well-formed keying envelope at the speeds commonly used on the ham bands are going to produce objectionable keyclicks very often, let alone from almost all transmitters.

There has to be something else at play.

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 25, 2007 Mail this to a friend!
AB0WR - The same thing applies to a keying envelope. If your window is the equal to or less than the period of the carrier signal, what the STFT will give you is the frequency components necessary to rebuild that little, bitty piece of the overall envelope. It *will* be dominated by the frequency of the carrier. It can't be anything else. But that little, bitty piece of the overall waveform won't tell you much about the overall waveform.

K5MC - In one sense, the folks who are using STFTs to examine the spectra during the rise/fall times are simply demonstrating what is shown in the signal analysis textbooks: the reciprocity of "signal" duration and "bandwidth." Fourier transform theory "proves" that as the width of a square-wave pulse (also referred to as a rectangular pulse or gate pulse in the literature) increases in the time domain, its spectrum decreases in width. Of course, the folks are looking at well-shaped pulses (e.g., raised-cosine, sinusoidal, or whatever) that have finite rise/fall times. These folks also have to decide what type of "window" function (Hamming, Hanning, flat top, uniform, exponential, etc.) to use when their FFT software packages calculate the spectra of these pulses during the rise/fall times.

One of the main problems here is exactly what do we mean by the "signal" or "emission" duration? If one's focus is on the spectra of the key clicks, then it certainly makes sense to use STFTs over the rise/fall times of the keying envelope. If we are interested in the occupied bandwidth (as defined by the FCC) of our CW signals, however, then we obviously need to look over at least one period of our signal if we are sending periodic signals such as a string of dits or a string of dahs. The occupied bandwidths for these two specific periodic signals (which can be calculated by a Fourier series approach or measured by a suitable spectrum analyzer) establish the upper and lower bounds for the occupied bandwidth when sending a random message at the same speed.

73, K5MC
 
Bandwidth versus Keying Speed Revisited  
by K5MC on August 25, 2007 Mail this to a friend!
K5MC - Fourier transform theory "proves" that as the width of a square-wave pulse (also referred to as a rectangular pulse or gate pulse in the literature) increases in the time domain, its spectrum decreases in width.

In my sentence above, I meant to say that it is relatively easy to "prove" using Fourier theory that as the width of a square-wave pulse increases in the time domain, its spectrum decreases in width. The same basic relationship holds for any given pulse shape, but the mathematical details become messier!

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by AB7E on August 25, 2007 Mail this to a friend!
AB0WR:"It would be like pulling one brick out of a building. That small sample will tell you how to rebuild that one little, bitty piece of the building. It will tell you if that brick has a face that is concave, convex, carved, cracked, or whatever. But it won't tell you how tall the building is, how wide it is, or what its footprint on the ground is.

If you want to know how much space that building takes up you better look at the whole picture, not just a piece of it.

The same thing applies to a keying envelope. If your window is the equal to or less than the period of the carrier signal, what the STFT will give you is the frequency components necessary to rebuild that little, bitty piece of the overall envelope. It *will* be dominated by the frequency of the carrier. It can't be anything else. But that little, bitty piece of the overall waveform won't tell you much about the overall waveform. "


AB7E: I guess this, as I believe W8JI has suggested, is where the differences in perspective come from. My ear responds exactly to those little slices of time ... in real time. Even at moderately high code speeds my ear and mind separate the effects of the rise and fall times from that of the steady carrier in between. My mind only integrates the pieces when the code speed gets fast enough that either my receiver or my mental neurons can't separate them. What I experience in terms of real time interference has little practical relevance to a mathematical analysis that tries to "capture the whole picture" by integrating over a longer period of time than I am able to easily distinguish.
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 25, 2007 Mail this to a friend!
We don't send CW messages by the linear multiplication of a carrier with a series of tones making a square wave. We send CW by making and breaking a key contact and transmitting the carrier at key down. The step-like change from zero to carrier amplitude requires bandwidth. A true unit step function would require infinite bandwidth. Think about that - it wouldn't matter how often we sent a unit step, just sending one would require infinite bandwidth.

Instead of unit step function, you can choose to model rise & fall times as first-order (RC) system if you want. The spectrum of the unit step put through a first order RC filter is still infinite, and it is continuous. There are no 'spikes' at certain harmonics.

We can choose to key transmitter at regular intervals. The repetition of step changes at periodic intervals creates re-occuring spectra that reinforce at the repetition rate and its harmonics. Location of harmonics is predicted by Fourier. I'm sure the 99% average power bandwidth of such a periodic signal will be exactly as Mickey presents.

That doesn't change fact that at the transitions, there was a large spectrum created by the (filtered) step change. If we are monitoring spectrum with, for example, a collection of receivers with notch filters, we would hear the transitions at a range of frequencies centered at the carrier frequency. I guess if receivers had infinite sensitivity and zero noise we would hear it all frequencies, but with finite sensitivity the range of frequencies where we hear the transition is dictated by path gain, noise level, transmitter power, and *** how fast the spectrum of the filtered step change rolls off ***.

The range of frequencies away from the carrier where we could hear the transitions would be same regardless of whether there was 100 transitions per second or just one transition and then none ever again!
Assuming path gain, tx power, and noise level are fixed, the only adjustment we can make to our transmitter to decrease the 'audible transition bandwidth' is to roll off the spectrum of the step change at faster rate. For example - change waveshape away from first-order exponential or else choose larger time constant in the exponential. The time constant in the exponential is directly related to rise time. Longer rise time = faster roll off of the step change's spectrum = narrower range of frequencies where transition is audible.

73, Chris VA3NR.
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 25, 2007 Mail this to a friend!
VA3NR-A true unit step function would require infinite bandwidth. Think about that - it wouldn't matter how often we sent a unit step, just sending one would require infinite bandwidth.

AB0WR: Even though infinite bandwidth is needed I seem to remember the amplitude falling off as (1/w) where w is the radian frequency. That drops the power at the higher frequencies pretty quickly. At 100hz you will be down something like -56db. A 6db per octave rolloff filter (i.e. an RC filter) with a 70hz cutoff would knock it down even more. At 300hz you would be over 80db down with a simple RC filter.

VA3NR- That doesn't change fact that at the transitions, there was a large spectrum created by the (filtered) step change. If we are monitoring spectrum with, for example, a collection of receivers with notch filters, we would hear the transitions at a range of frequencies centered at the carrier frequency. I guess if receivers had infinite sensitivity and zero noise we would hear it all frequencies, but with finite sensitivity the range of frequencies where we hear the transition is dictated by path gain, noise level, transmitter power, and *** how fast the spectrum of the filtered step change rolls off ***.

AB0WR: Since the spectrum rolls off at a 1/w rate already, even a simple filter will put the spectrum down significantly.

VA3NR-The range of frequencies away from the carrier where we could hear the transitions would be same regardless of whether there was 100 transitions per second or just one transition and then none ever again!

AB0WR: You cannot have a series of "on" transitions with0ut a corresponding series of "off" transistions, i.e. a negative step function. When this happens you have generated at least a pulse that can be analyzed as a gate pulse if nothing else. If you are using the step functions to generate a periodic waveform, i.e. a string of dits, the the Fourier of the periodic waveform becomes the controlling process. You can't take part of a waveform, arbitrarily terminate it and analyze that. I think it was *you* that pointed this out in an earlier message.

VA3NR-Longer rise time = faster roll off of the step change's spectrum = narrower range of frequencies where transition is audible.

AB0WR: I agree with this. However the audibility of two different rates of keying may be different at the same distance away from the carrier. A slower speed will decay to a specfiic audibility level at a frequency that is closer to the carrier than a faster keying speed will.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 25, 2007 Mail this to a friend!
VA3NR-Longer rise time = faster roll off of the step change's spectrum = narrower range of frequencies where transition is audible.

AB0WR: I agree with this. However the audibility of two different rates of keying may be different at the same distance away from the carrier. A slower speed will decay to a specfiic audibility level at a frequency that is closer to the carrier than a faster keying speed will.
_____________________

VA3NR - I'm glad you agree the transition is audible over range of frequencies. Now if we've already heard a transition, how would the time delay until the next transition make any difference at all? Is the negative step somehow going to remove our the memory of the first transition? or, if we hear two clicks closer together are we going to interpret that as one louder click? Are repeated clicks at certain rate somehow going to make our ears or receiver more sensitive? No need to answer really - just give it some thought.

de VA3NR nw qrt.
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 26, 2007 Mail this to a friend!
VA3NR-Longer rise time = faster roll off of the step change's spectrum = narrower range of frequencies where transition is audible.

AB0WR: I agree with this. However the audibility of two different rates of keying may be different at the same distance away from the carrier. A slower speed will decay to a specfiic audibility level at a frequency that is closer to the carrier than a faster keying speed will.
_____________________

VA3NR - I'm glad you agree the transition is audible over range of frequencies. Now if we've already heard a transition, how would the time delay until the next transition make any difference at all? Is the negative step somehow going to remove our the memory of the first transition? or, if we hear two clicks closer together are we going to interpret that as one louder click? Are repeated clicks at certain rate somehow going to make our ears or receiver more sensitive? No need to answer really - just give it some thought.

AB0WR: I didn't agree that we hear the "transition". Don't quote me out of context!

AB0WR:Did you not read the part where I said "You cannot have a series of "on" transitions with0ut a corresponding series of "off" transistions, i.e. a negative step function. When this happens you have generated at least a pulse that can be analyzed as a gate pulse if nothing else. If you are using the step functions to generate a periodic waveform, i.e. a string of dits, the the Fourier of the periodic waveform becomes the controlling process. You can't take part of a waveform, arbitrarily terminate it and analyze that. I think it was *you* that pointed this out in an earlier message."

AB0WR:Do you have any empirical evidence showing that the leading edge of a square wave generates a spectrum similar to that of a step function? I have yet to see a spectrum analyzer display that shows such exists.

AB0WR:You cannot simply truncate a waveform and do an analysis on that piece. The spectrum associated with the waveform is a function of the entire waveform.

AB0WR:If you apply a rectangular window to the start of a square wave pulse that has a rise time then you have, in effect, multiplied the waveform within that window *by* that window. If it is a rectangular window then the result you are analyzing becomes equivalent to a single sawtooth wave with a sloped frontal edge and a vertical back edge.

AB0WR:What you come up with will let you rebuild that piece of the original waveform you sampled but it will *not* represent the spectrum involved in the overall waveform.

AB0WR: A square wave *is* made up of the sum of numerous oscillators consisting of the fundamental and the odd harmonics.

AB0WR: What happens to these frquencies when applied to a linear mixer? How do they disappear either at the beginning of a keying envelope, during the middle of a keying envelope, or at the end of a keying envelope? If you have some mathematical model that will answer this *then* we may have something to begin discussing.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 26, 2007 Mail this to a friend!
VA3NR-We don't send CW messages by the linear multiplication of a carrier with a series of tones making a square wave. We send CW by making and breaking a key contact and transmitting the carrier at key down. The step-like change from zero to carrier amplitude requires bandwidth. A true unit step function would require infinite bandwidth. Think about that - it wouldn't matter how often we sent a unit step, just sending one would require infinite bandwidth.

AB0WR - I suspect this is part of our disagreement.

AB0WR - A square wave *is* certainly made up of a series of frequencies. I really don't know of anyone who seriously doubts this or questions it. Are you actually questioning this or is this just for purposes of discussion?

AB0WR - Sending a CW signal, particularly one made up of a string of dits or dahs *is* doing a multiplication of the two signals in the time domain. if c(t) is the carrier signal and k(t) is the keying waveform then the resulting time domain signal is c(t) x k(t). Since k(t) alternates between 0 and 1, the product of the two has the carrier alternating between being on and off. This multiplication in the time domain is a convolution of the two signals in the frequency domain. Since k(t) is a square wave made up of a distinct frequency spectrum depending on its period, that spectrum gets convolved with the carrier to form a signal consisting of an Upper Sideband, a Lower Sideband, and the Carrier.

AB0WR- This really *is* Linear Systems and Signals 101.

AB0WR - A square wave actually is an infinite sum of odd harmonics so it has the same "infinite" bandwidth you speak of. When applied to the real world, however, the amplitude of those harmonics fall off by a factor of 1/n and soon reach a level where they are below the noise level we work with due to atmospheric noise, receiver noise, path loss, etc. At this point they become inaudible. Use of a keying filter only hastens how quickly the harmonics become inaudible. That doesn't mean they don't exist, we just can't detect them.

AB0WR - It also needs to be recognized that a rise time indicates a system bandwidth. If the rise time is 5ms that indicates a system bandwidth of 70hz using the old formula of BW x Ristime = 0.35. This means even if your transition analysis is valid that you will not have a wider perceived bandwidth at transistions than you will have for repetitive waveforms unless the spectrum components for the transition have much higher amplitudes than the spectrum components of the repetitive waveform. Since the spectrum for an exponential rise time goes down by 1/w as opposed to 1/n for a square wave, the frequency components of a transition will be less at any specific frequency than for a square wave. This means perceiving a "click" of a transistion would be highly unlikely at any distance from the carrier. At 100hz (=628 rad/sec) you would be down 20log(1/628)=-56db. At 200hz (=1256rad/sec) you would be down 20log(1/1256)=-62db. Someone listening to another signal centered in a 500hz filter would hear your transition signal, if it even exists, at something around a -65db. If your fundamental has a power output of 100watts, the clicks you would be sending would be at 50mw or less. Yes, that could be heard in many cases. However it is doubtful that such adjacent channel interference would be considered as harmful interference or even as "darned keyclicks" except perhaps around the QRP calling frequency.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on August 26, 2007 Mail this to a friend!
AB0WR - Since the spectrum for an exponential rise time goes down by 1/w as opposed to 1/n for a square wave, the frequency components of a transition will be less at any specific frequency than for a square wave.

VA3NR: Repeated step changes at regular intervals create re-occuring spectra that re-inforce at frequencies harmonically related to repetition rate. (think of it as sum of positive and negative unit steps time-shifted by integer multiples of dot length.) For square wave the harmonics will add up such that amplitude rolls off 1/n as you say. Problem applying that to key clicks is that when copying code real-time our receiver and ears don't add re-occuring spectrum. We hear and recognize the click associated with the spectrum of each individual transition. Yes, the spectrum should roll off 1/f and more with exponential filter with well chosen rise time. But when you're chasing weak DX on 160m and some local starts CQing with a 40 db over S9 signal, putting the key clicks down 60 or even 70dB down sometimes doesn't seem to be enough.

____
Thanks discussion. Time for me to spin the VFO.
73, Chris VA3NR.
 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on August 26, 2007 Mail this to a friend!
Just some minor corrections and explanations....

AB0WR: I agree with this. However the audibility of two different rates of keying may be different at the same distance away from the carrier. A slower speed will decay to a specfiic audibility level at a frequency that is closer to the carrier than a faster keying speed will.


No, that isn't right. The average power goes down with slower keying, but the instantaneous power stays the same, and is only dependent on the edge shape. So the ear will hear identical click sounds, just not as often. (By edge I mean either edge. It's direction is irrelevant in terms of amplitude; however, the phases are opposite.)


AB0WR: You cannot have a series of "on" transitions with0ut a corresponding series of "off" transistions, i.e. a negative step function. When this happens you have generated at least a pulse that can be analyzed as a gate pulse if nothing else. If you are using the step functions to generate a periodic waveform, i.e. a string of dots, the the Fourier of the periodic waveform becomes the controlling process. You can't take part of a waveform, arbitrarily terminate it and analyze that. I think it was *you* that pointed this out in an earlier message.

AB0WR - A square wave actually is an infinite sum of odd harmonics so it has the same "infinite" bandwidth you speak of. When applied to the real world, however, the amplitude of those harmonics fall off by a factor of 1/n and soon reach a level where they are below the noise level we work with due to atmospheric noise, receiver noise, path loss, etc. At this point they become inaudible. Use of a keying filter only hastens how quickly the harmonics become inaudible. That doesn't mean they don't exist, we just can't detect them.


When and only when the bandwidth of your receiving system is narrowed to the point where it approaches the keying frequency, will you be able to begin to discern the peaks and nulls associated with the keying rate that are predicted by Fourier analysis as you tune away from Fc (carrier frequency). This makes sense doesn't it? To find the transform over one cycle, the time interval is T, so the bandwidth is 1/T. For a 24 wpm, T is 100ms, so the receiving bandwidth needs to be <= 10Hz before sidebands manifest themselves and therefore can be distinguished.

Another way to say it is this. When receiving a series of 24 WPM dots with a bandwidth of <= 10 Hz, the receiver system response time is slow enough to still be reacting to the previous edge when the new edge arrives. Only when this situation occurs, do you start getting cancellations and reinforcements due to the edges that start looking like the Fourier results.
A good RX bandwidth for CW might be 500Hz, so the RX system response time would be roughly 2ms. So 2ms after seeing an edge (or a piece of it, if Tr/Tf = 5ms) the system is pretty much done responding to it. The previous edge was soooo long ago, from the system's perspective, that it doesn't play any part in how the system is responding to the current edge. Hence, when you tune off freq a bit, where the carrier freq is cut off by your RX system's filter, the sideband power from the edges are heard as ~7ms clicks because they only exist in the RX system for ~7ms (5ms of edge and ~2ms of system response time).


My Orion II minimum BW is 100 Hz. Using this BW, what equivalent CW WPM rate I would I need to receive before the system response to two edges start to interact?? Well, the dot rate needs to be 100 Hz, so using the formula 2.4 WPM = 1 dot/sec, the sending speed would need to be greater than 240 wpm. Then my receiver could start to reveal any waviness to the sideband envelope.
In other words, the slope of the key click amplitude rolls off dependent only on the edge shape as I tune away from the carrier frequency, unless the sending speed approaches 240 wpm.


AB0WR - It also needs to be recognized that a rise time indicates a system bandwidth. If the rise time is 5ms that indicates a system bandwidth of 70hz using the old formula of BW x Ristime = 0.35. This means even if your transition analysis is valid that you will not have a wider perceived bandwidth at transistions than you will have for repetitive waveforms unless the spectrum components for the transition have much higher amplitudes than the spectrum components of the repetitive waveform. Since the spectrum for an exponential rise time goes down by 1/w as opposed to 1/n for a square wave, the frequency components of a transition will be less at any specific frequency than for a square wave. This means perceiving a "click" of a transistion would be highly unlikely at any distance from the carrier. At 100hz (=628 rad/sec) you would be down 20log(1/628)=-56db. At 200hz (=1256rad/sec) you would be down 20log(1/1256)=-62db. Someone listening to another signal centered in a 500hz filter would hear your transition signal, if it even exists, at something around a -65db. If your fundamental has a power output of 100watts, the clicks you would be sending would be at 50mw or less. Yes, that could be heard in many cases. However it is doubtful that such adjacent channel interference would be considered as harmful interference or even as "darned keyclicks" except perhaps around the QRP calling frequency.



Adding to what AB0WR said here, I just want to point out that the spectral power envelope from the signal edges fall off in accordance to the edge shape, not with the a square wave envelope which is wider. An exponential edge has a slower fall off slope than a gaussian edge, and so will be audible above the noise farther away from the Fc.(See http://fermi.la.asu.edu/w9cf/articles/click/index.html.)

More importantly, shortening up an edge duration by half (but maintaining it's shape) doubles its average sideband energy. What's more, this doubled energy is expended in half the time, so the instantaneous power actually quadruples. This difference of 6 dB is quite noticeable and one would have to tune twice as far from the carrier frequency to hear the click as before!

From W9CF's article, if the edges were of the exponential decay type (RC filter), the average power of the sidebands 500 Hz away are about -52dBc. If the carrier power is 100W or 50dBm, the average power is thus 0.6mW, or -2dBm. With a reciever bw of ~500Hz, the sideband power within this bandwidth (250 to 750 Hz) is additive. I'll use a factor of 10 (conservative for sure) for this summing. So now the average is 6mW, or 8dBm. If the rise time is ~4ms, and the pulse width is 50ms, then the instantaneous power is 6mW * 50ms/~4ms ~= 75mW, or 18.8dBm. This is only ~30 dB down from the carrier power, so would be an S-9 signal if the carrier was +30 over S-9, or an S-4 signal if the carrier was S-9.


Lin
WB1AIW

 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on August 26, 2007 Mail this to a friend!
Correction,

I previously wrote:

"When and only when the bandwidth of your receiving system is narrowed to the point where it approaches the keying frequency, will you be able to begin to discern the peaks and nulls associated with the keying rate that are predicted by Fourier analysis as you tune away from Fc (carrier frequency). This makes sense doesn't it? To find the transform over one cycle, the time interval is T, so the bandwidth is 1/T. For a 24 wpm, T is 100ms, so the receiving bandwidth needs to be <= 10Hz before sidebands manifest themselves and therefore can be distinguished."



Please replace the last sentence with:

For a 24 wpm, T is 100ms so the receiving bandwidth needs to be <= 10Hz before the sideband power starts appearing as frequency-dependent lobes on either side of the carrier.



Lin
WB1AIW


 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 26, 2007 Mail this to a friend!
VA3NR:We don't send CW messages by the linear multiplication of a carrier with a series of tones making a square wave. We send CW by making and breaking a key contact and transmitting the carrier at key down. The step-like change from zero to carrier amplitude requires bandwidth. A true unit step function would require infinite bandwidth. Think about that - it wouldn't matter how often we sent a unit step, just sending one would require infinite bandwidth.

Instead of unit step function, you can choose to model rise & fall times as first-order (RC) system if you want. The spectrum of the unit step put through a first order RC filter is still infinite, and it is continuous. There are no 'spikes' at certain harmonics.

We can choose to key transmitter at regular intervals. The repetition of step changes at periodic intervals creates re-occuring spectra that reinforce at the repetition rate and its harmonics. Location of harmonics is predicted by Fourier. I'm sure the 99% average power bandwidth of such a periodic signal will be exactly as Mickey presents.

===============================================

I'm sorry but the "turning on/off" of a carrier wave is a multiplication by "on/off" waveform. It doesn't matter if you use a mechanical switch or an electronic switch. I sincerely doubt that many transmitters have the mechanical key in the output of the transmitter which is what your statement implies. Otherwise, you ARE using "electronic switches". In either case, this is modulation.


-----------------------------------------

VA3NR:That doesn't change fact that at the transitions, there was a large spectrum created by the (filtered) step change. If we are monitoring spectrum with, for example, a collection of receivers with notch filters, we would hear the transitions at a range of frequencies centered at the carrier frequency. I guess if receivers had infinite sensitivity and zero noise we would hear it all frequencies, but with finite sensitivity the range of frequencies where we hear the transition is dictated by path gain, noise level, transmitter power, and *** how fast the spectrum of the filtered step change rolls off ***.

The range of frequencies away from the carrier where we could hear the transitions would be same regardless of whether there was 100 transitions per second or just one transition and then none ever again!
Assuming path gain, tx power, and noise level are fixed, the only adjustment we can make to our transmitter to decrease the 'audible transition bandwidth' is to roll off the spectrum of the step change at faster rate. For example - change waveshape away from first-order exponential or else choose larger time constant in the exponential. The time constant in the exponential is directly related to rise time. Longer rise time = faster roll off of the step change's spectrum = narrower range of frequencies where transition is audible.

=============================================

On the other hand, when you use a spectrum analyzer whose sweep scans all the frequencies around the carrier, you would expect to see the spectra you mention if it had any power in it. Granted, at any given time, the analyzer may not be sitting on a frequency when it is actually generated. However, I'll bet you can let your dits run at 50 wpm for an hour and the analyzer will never, even with a free running sweep speed several times (or even many times) the period of the keying waveform, "see" any of the frequencies you discuss.

This can only happen because they either aren't there or they are of such lower power, spectrum analyzers can't detect them. Either way, key click interference from them are unlikely.

Now maybe I'm wrong, I would like to see a spectrum analyzer picture from a controlled lab environment that shows otherwise. By the way, I'm not sure that a transmitter with a wide spectrum is a good candidate for such a "lab" experiment. It should be able to be duplicated with a "clean" signal from a lab device.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 26, 2007 Mail this to a friend!
WB1AIW-No, that isn't right. The average power goes down with slower keying, but the instantaneous power stays the same, and is only dependent on the edge shape. So the ear will hear identical click sounds, just not as often. (By edge I mean either edge. It's direction is irrelevant in terms of amplitude; however, the phases are opposite.)

AB0WR: Lin, all I can do is tell you to go look up Parseval's theorem. The power in the time domain is the same as the power in the frequency domain. That means that the power in each individual frequency in the frequency domain must be considered, whether it is in calculating average power or instantaneous power.

AB0WR: This isn't my theorem. It is an accepted engineering theorem.
----------------------------------


WB1AIW-When and only when the bandwidth of your receiving system is narrowed to the point where it approaches the keying frequency, will you be able to begin to discern the peaks and nulls associated with the keying rate that are predicted by Fourier analysis as you tune away from Fc (carrier frequency). This makes sense doesn't it? To find the transform over one cycle, the time interval is T, so the bandwidth is 1/T. For a 24 wpm, T is 100ms, so the receiving bandwidth needs to be <= 10Hz before sidebands manifest themselves and therefore can be distinguished.

AB0WR - Both of my sons played violin in the high school orchestra. I think any musician will tell you that it doesn't matter if you can distinguish each individual harmonic that is being played. What you hear is the sum of all of the harmonics.

The fundamental frequency will be given by 1/T. The harmonics are the odd harmonics, however. 30hz, 50hz, etc. So you only need a 20hz bandwidth to resolve individual frequencies for a 10hz keying rate. BTW, I think 24wpm is actually a keying speed of about 20hz. So your resolution bandwidth will be about 40hz.
------------------------------------


WB1AIW-Another way to say it is this. When receiving a series of 24 WPM dots with a bandwidth of <= 10 Hz, the receiver system response time is slow enough to still be reacting to the previous edge when the new edge arrives. Only when this situation occurs, do you start getting cancellations and reinforcements due to the edges that start looking like the Fourier results.

AB0WR: Huh? Resolution isn't the same thing as response time. I'm not sure exactly where you are coming from with this.

High frequency rolloff is not the same thing as "response time" either, at least how you seem to be defining it. It only means that higher frequencies don't appear in the output, it doesn't mean that the receiver has a "reflex" time like a driver in an automobile.

Does a SSB filter with a 2.4khz high frequency limit prevent the receiver from properly responding to a voice spectrum? Do syllables run together because the high frequencies in the voice are being cut off?
---------------------------------------


WB1AIW-The previous edge was soooo long ago, from the system's perspective, that it doesn't play any part in how the system is responding to the current edge. Hence, when you tune off freq a bit, where the carrier freq is cut off by your RX system's filter, the sideband power from the edges are heard as ~7ms clicks because they only exist in the RX system for ~7ms (5ms of edge and ~2ms of system response time).

AB0WR:Certainly removing part of the frequency spectrum associated with a periodic wave will change the way it "sounds". Listen to someone with a a 4.5khz ESSB signal using a 1.8khz SSB filter and then listen with a 2.4khz filter. Their voice will "sound" different. It will sound even more different when using a 4.5khz filter.

This hasn't got anything to do with the "response" time. It has to do with the spectrum being received.

If the bandwidth of your receiver is greater than the bandwidth of the signal being received (i.e. a 2ms rise time receiver bandwidth versus a 7m rise time on the received signal) there won't be any additive factors encountered. You will hear the full spectrum of the waveform being received.

You may, in fact, intrepret what you hear as a "click". I've never said any differently. But that doesn't mean that the frequencies you hear are from an "edge" of the transmitting waveform. What you hear is based on what is transmitted. The received frequencies are not generated by the receiver they are generated by the transmitter.

Just as you cannot tell what bandwidth of voice a transmitter is sending by listening to the received signal with a narrower filter than the transmitter is using you can't tell what frequency spectrum a CW transmitter is sending by listening to only part of the spectrum by tuning off to the side of the CW signal.

Just because you might hear clicks from an adjacent signal when using a 500hz CW filter in your receiver doesn't mean that adjacent signal is generating key clicks as a spurious product. If the transmitted spectrum of that CW signal is 400hz (plus and minus 200hz) and you are listening with the bottom of your filter 100hz away from the CW signals carrier frequency you will hear only a part of the spectrum being transmitted. What you hear may sound like a click. That doesn't mean that the adjacent signal has any problems whatsoever. Those clicks you would hear would be perfectly legitimate.
---------------------------------------



WB2AIW-My Orion II minimum BW is 100 Hz. Using this BW, what equivalent CW WPM rate I would I need to receive before the system response to two edges start to interact?? Well, the dot rate needs to be 100 Hz, so using the formula 2.4 WPM = 1 dot/sec, the sending speed would need to be greater than 240 wpm. Then my receiver could start to reveal any waviness to the sideband envelope.
In other words, the slope of the key click amplitude rolls off dependent only on the edge shape as I tune away from the carrier frequency, unless the sending speed approaches 240 wpm.

AB0WR: You are lost in the casual forest again. Your receiver doesn't determine the bandwidth of the transmitted signal. Only the transmitter can do that. All you can control at the receiver is the bandwidth of your receiver.

If the transmitter has a 5msec rise time (or fall time) then your receiver would have to have a narrower bandwidth before it starts impacting what you hear. A 100hz bandwidth indicates a rise time of about 0.35/100 = 3.5msec. Since your rise time is LESS than the rise time of the received signal, your receiver should faithfully reproduce what you receive.

Again, I'm not sure where you are coming from with the term "response". If you don't expect to hear syllables in words get slurred together when you use a filter narrower than the filter the transmitter used why would you expect the elements of a CW signal to get slurred together?

If you narrow your received filter down enough so that you could not hear either the fundamental of the keying waveform or any of its odd harmonics, i.e. could only "hear" the carrier, the worst that would happen is that you wouldn't hear *anything*. It would be like using a 100hz CW filter to listen to a 4.5khz SSB signal. You wouldn't hear anything you could make out as intelligence.

-------------------------------------

WB1AIW-More importantly, shortening up an edge duration by half (but maintaining it's shape) doubles its average sideband energy. What's more, this doubled energy is expended in half the time, so the instantaneous power actually quadruples. This difference of 6 dB is quite noticeable and one would have to tune twice as far from the carrier frequency to hear the click as before!

AB0WR: What's your point? That going from a 5ms to a 2.5ms rise time in the keyed waveform is *bad*?

Have you seen anyone on here argue otherwise?

I would also point out, however, that most amateur transmitters are peak power limited. The theory you quote would be appropriate for a non-peak limited system. For a peak limited system, the power available actually becomes less in the higher frequency spectrum as you increase bandwidth since the available limited peak power gets spread over a larger spectrum. Whether you hear an increase, no change, or actually a decrease in adjacent channels will depend greatly on how close you are to the peak power limit of your transmitter when you change the bandwidth.

Btw, I think you are mixing up power and work (Power x Time = Work, it doesn't equal power), I'll cover it more later.
------------------------------------------

WB1AIW-I'll use a factor of 10 (conservative for sure) for this summing. So now the average is 6mW, or 8dBm. If the rise time is ~4ms, and the pulse width is 50ms, then the instantaneous power is 6mW * 50ms/~4ms ~= 75mW, or 18.8dBm.

AB0WR: ROFL. Lin, you are trying to have it both ways now. I thought you were arguing that the sidebands only exists during the rise and fall times? Yet you use the entire pulse width as a multiplying factor? That would have to indicate that the sidebands exist for the entire pulse. If they exist for the entire pulse then W9CF's analysis is incorrect as Mickey has shown with his spectrum analyzer pictures.

Which way is it? I think you are at a point where you are going to have to choose if you want your calculations to make any sense.

Oh, btw, your calculations don't make any sense to me. When you sum instantaneous power over several freqency components you still have instantaneous power, not average power. Power is joules/sec. It is a rate of doing work. If the amount of work done is 6mw x 50msec and you want the amount of work done during a 4ms period in that 50ms you multiply by 4/50 or just multiply the 6mw x 4ms. That gives you joules not watts.

I think you are still confusing the time domain with the frequency domain.

Now, Tom, W8JI, is correct when he says that a receiver doesn't respond to work done, i.e. power x time, at least not directly. A receiver responds to the instantaneous power. The work is done when that power is converted to force and is applied to a speaker causing the cone to move and thus you have (Force x Distance) = Work.

You are also back in the casual forest again when you say that the rise time generates the sidebands. The rise time is associated with an output resulting from an input and a system response function. An output cannot create an input. An output cannot create a system response.

The *sidebands* that are generated by mixing a square wave with a carrier frequency determine the rise time of the output signal. The output doesn't determine the input square wave nor does it determine the carrier input frequency.

I have yet to see anyone explain mathematically how the sidebands you discuss are generated by mixing input signals (i.e. a keying waveform and a carrier frequency).

If just one person could give me a reference to a university level Linear System or Signal Analysis textbook that shows modulation of a carrier by a square wave resulting in anything other than a mixing of the square fundamental and its harmonics with the carrier frequency I will totally change my viewpoint.

Is there anyone out there with such a reference? Is there anyone out there with empirical evidence otherwise?

I hate to be blunt about it but so far I've just seen repeats of the same old argument: Outputs cause inputs. I've seen lots of subtle approaches to this like turning a CW pulse into a sawtooth and then saying the sidebands from this approach are what people hear -- with no mathematical explanation of where the sidebands generating the sawtooth output come from. Magic I guess. Or the casual impossibility of the output causing the input.

If it can't be shown mathematically where those input sidebands come from in order to mix them with the carrier to get an output then how does anyone reading this forum expect expect it to be called anything except majik?

(btw, the web page VA3?? quoted is not deterministic, not all of the CW signals show the same spectrographic evidence nor is it obvious that the one signal showing such evidence doesn't have problems such as tremendous intermod products from ALC interaction).

Doesn't anyone on here have the math to back up their "intuition"?

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on August 26, 2007 Mail this to a friend!
AB0WR:Doesn't anyone on here have the math to back up their "intuition"?

===============================================

This is what I would like to see along with some spectrum analyzer outputs confirming what is calculated.

With the proper math, you should be able to calculate where these sidebands are located and find them with a spectrum analyzer.

Heck, you can even turn off the freq sweep on the analyzer and tune to the calculated sidebands to see them, if they are there!

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by W3BC on August 27, 2007 Mail this to a friend!
For example, what's the OBW of "CQ TEST CQ TEST DE W3BC W3BC TEST"

I would think the OBW would be mostly affected by the modulation products of the rise and fall waveforms. Square keying would be a very short rise time, and contain odd-order harmonics of the corresponding frequency.

A fourier analysis of a typical CW message ("CQ TEST" ?) would be most meaningful to the real world, as it would consider

a. Dits
b. Dahs
c. Rise time
d. Fall time
e. The periods of silence between dits and dahs
f. The periods of silence between letters
g. The periods of silence between words

Some of these change with sending speed. A few are constant, and specific to a particular radio.

In other words, as with all things in nature, there are a lot more variables to consider than we initially believe.

"Beware the simple solution."

73 - Joe
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 27, 2007 Mail this to a friend!
W3BC-I would think the OBW would be mostly affected by the modulation products of the rise and fall waveforms. Square keying would be a very short rise time, and contain odd-order harmonics of the corresponding frequency.

AB0WR: Rise and fall times don't have modulation products. Modulation products have rise and fall times.

The rise and fall times is an indicator of the bandwidth of a system. A short rise time indicates a wider bandwidth. The strength of the harmonics at that bandwidth limit, however, is determined by the period of the square wave and the 1/n factor. At a 300hz bandwidth, a 10hz square wave would generate the 29th harmonic at the bandwidth limit which would be at a level of (1/29)th of the fundamental. A 20hz keying waveform would generate the 15th harmonic which would only be down to (1/15)th of the fundamental. So at 300hz away from the fundamental the 20hz square wave would be almost twice as strong as the 10hz square wave. Both would have almost the same rise time.

------------------------------------------


>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
W3BC -A fourier analysis of a typical CW message ("CQ TEST" ?) would be most meaningful to the real world, as it would consider

a. Dits
b. Dahs
c. Rise time
d. Fall time
e. The periods of silence between dits and dahs
f. The periods of silence between letters
g. The periods of silence between words

Some of these change with sending speed. A few are constant, and specific to a particular radio.

In other words, as with all things in nature, there are a lot more variables to consider than we initially believe.
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

AB0WR:The widest bandwidth of a CW signal occurs when multiple dits are sent together. This represents the shortest period waveform and therefore the highest frequency waveform. If your bandwidth is acceptable when sending multiple dits it should be acceptable when sending any other element.
------------------------------------------------

W3BC-"Beware the simple solution."

AB0WR: Occam might disagree with you.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on August 27, 2007 Mail this to a friend!
For those who want to see spectrograms of cw signals showing the bandwidths I have uploaded 4 jpg files to my account at photobucket. I took pictures from my Softrock SDR using three different programs; PowerSDR, KGKSDR, and Winrad.

None of the spectrograms I recorded showed far-off harmonics being stronger at the start and end of the cw elements, at least not that the resolution in these programs are capable of showing. They all show exactly what one would expect of a well-behaved transmitter sending the sidebands from a bandwidth-limited square wave. The strongest frequency components are close to the carrier and they fall off as you move away.

The Winrad spectrogram actually shows the spectral lines you would expect to see from a square wave modulation of a carrier. I will have to try this with a controlled signal that will give me the ability to count the spectral lines to see if they are where they should be or if they are just an artifact of the analysis algorithm the software is using.

Goto:
http://s205.photobucket.com/albums/bb178/ab0wr/
to see the four pictures.

If someone has a better way of looking at the signals, please let me know.

I really like the Softrock SDR and the various software programs you can use with it for looking at signals in ways I never had the capability of doing before. My hats are off to Tony Parks for making the Softrock available to the masses and to those software guru's who have coded the programs to use with it.

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by W1GDQ on August 29, 2007 Mail this to a friend!
All these years I thought the opposite was true. Thanks for the article.
 
Bandwidth versus Keying Speed Revisited  
by OK1RR on September 1, 2007 Mail this to a friend!
Totally false! The bandwidth depends exclusively on the shape and not on the speed, however the readability at particular speed depends also on the element shape. The relation of speed to bandwidth is therefore indirect. The human factor cannot be omitted, a well trained CW operator can copy rounded, long-sloped narrow bandwidth characters which another, less trained OP can't read. With machine, it is even worse.

An exact analysis is on the Kevin Schmidt, W9CF web site:
http://fermi.la.asu.edu/w9cf/articles/click/index.html
however you can cut down the bandwidth on both sides (TX or RX) until the code became unreadable. The 'readability limit' is very subjective.

Remember that most powerful device is the human brain which can make wonders, with most advanced math zillion times confirmed as impossible. Also, no device can pull an idiot to a genius level.
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 1, 2007 Mail this to a friend!
ok1rr - Totally false! The bandwidth depends exclusively on the shape and not on the speed,

AB0WR: Really? So the harmonics of the square wave go away when the slope of the envelope goes to zero, right?

AB0WR: What happens in a two-tone test of a SSB transmitter when the slope of the envelope goes to zero? Do the sidebands, i.e. the two tones, disappear as well? What is the amplitude of those two tones in the time domain when the envelope is at its maximum peak or minimum peak, i.e. when the slope of the envelope is zero?

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on September 1, 2007 Mail this to a friend!
OK1RR - Totally false! The bandwidth depends exclusively on the shape and not on the speed, however the readability at particular speed depends also on the element shape. . . An exact analysis is on the Kevin Schmidt, W9CF web site:
http://fermi.la.asu.edu/w9cf/articles/click/index.html

K5MC - In terms of the "occupied" bandwidth as defined by the FCC, the ITU, Agilent Technologies, etc., the occupied bandwidths of typical CW transmitters do vary with the keying speed as my two eHam articles discussed. It must be understood that the occupied bandwidth as defined by these entities is not the same concept as the so-called "key-click" or "interference" bandwidth.

As far as the analysis by W9CF is concerned, a careful comparison between Figure 2 and Figure 3 will reveal that the "essential" bandwidth of the 20-millisecond pulse shown in Figure 2 is larger than that of the 50-millisecond pulse shown in Figure 3. For example, the 20 dB bandwidth of the 20-ms pulse is about 75 Hz, whereas the 20 dB bandwidth of the 50-ms pulse is about 40 Hz. The 50 dB bandwidths are approximately 425 Hz and 300 Hz for the 20-ms and 50-ms pulses, respectively. Although he appears to not understand his results in terms of "bandwidth" in its usual communications sense, W9CF is, in fact, simply demonstrating a basic concept discussed in practically all communication/signal analysis textbooks: the inverse relationship between a pulse's time duration and its spectrum width as given by Fourier analysis.

W9CF writes, "The keying speed does not effect the overall bandwidth." If by "overall" bandwidth W9CF means the "absolute" bandwidth, then I have no argument with him. He is simply stating a well-known fact from Fourier analysis: any time-limited signal has an absolute bandwidth of infinity.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on September 1, 2007 Mail this to a friend!
Square Wave ?

OK, I must admit that I have not been following the discussion for the last couple of days (or, has it been weeks already). But...

Where is this square wave?

It could be a semantics thing but I am thinking that a square wave is a DC voltage that is started and stopped but maintained DC by the multitude of odd harmonics.

But, that has nothing to do with a CW keying wave form does it? I mean, the keying of a CW is on and off keying of a sine wave of a given SINGLE frequency. The only harmonics that come into play are during the actual on and off keying action but this is not a square wave. At best, it might be considered a very sharp, almost Gaussian shape of a highly peaked wave if you remove the middle part of constant maximum peak envelope on the sign wave.
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 2, 2007 Mail this to a friend!
K7PEH-Square Wave ?

OK, I must admit that I have not been following the discussion for the last couple of days (or, has it been weeks already). But...

Where is this square wave?

It could be a semantics thing but I am thinking that a square wave is a DC voltage that is started and stopped but maintained DC by the multitude of odd harmonics.

But, that has nothing to do with a CW keying wave form does it? I mean, the keying of a CW is on and off keying of a sine wave of a given SINGLE frequency. The only harmonics that come into play are during the actual on and off keying action but this is not a square wave. At best, it might be considered a very sharp, almost Gaussian shape of a highly peaked wave if you remove the middle part of constant maximum peak envelope on the sign wave.
-------------------------------------------------

AB0WR: How do you turn that sine wave off and on?

In an Icom 751a, it is done by applying a "square wave" bias voltage to the keyed CW oscillator.

This makes the stage not only an oscillator but a mixer.

Consider the old single-balance mixer using a dual-gate mosfet. The carrier signal to be mixed was applied to gate1. The other signal to be mixed was applied to gate2. The signal on gate2 would turn on and off the signal on gate1 as it varied from a negative bias to a positive bias and back.

If that signal on gate2 happened to be a square wave what do you suppose the output of the mixer would be? If that signal on gate2 happened to be a keyed CW square wave signal, what do you suppose the output of the mixer would be?

I go through this to point out that any time you are turning a stage on and off what you have is a linear mixer that results in an output that combines the input to the stage with the square wave off/on signal.

If your input signal is f(t) and the on/off signal is g(t) you get f(t)g(t), the two signals multiplied together. This multiplication in the time domain is a convolution in the frequency domain, commonly known as a mixing process. Since the square wave in the frequency domain consists of discrete spectra (i.e. the odd harmonics) these wind up appearing as upper and lower sidebands associated with the carrier.

Go take a look at any of the websites K5MC or myself have provided to see pictures taken by a spectrum analyzer (by K5MC) or from the output of a software defined radio (mine) that show this.

Turning a signal on and off and on and off *is* mixing a signal and a square wave. Linear Systems and Signal Analysis 101.

73,

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on September 2, 2007 Mail this to a friend!
>>Turning a signal on and off and on and off *is* mixing a signal and a square wave. Linear Systems and Signal Analysis 101.


I had to think about that for a minute since I did not take Linear Systems and Signal Analysis 101. However, after thinking about it I do see what you mean. My background is physics and math, not engineering so I missed out on the signal analysis 101 course work.

But, in thinking about a typical signal of y(t) = A sin(wt), I had not thought about the importance of separating out the amplitude as a simple example of this idea. So, I created a step function of A where A is now a function of t, A(t) with the value of zero for all t less then pi and the value of 1 for all t greater then or equal to pi to represent turning on the signal. So, then I have the product of two functions (mixing): y(t) = A(t)*sin(wt). And, of course, the square wave is merely another step function.

About linear systems 101, I checked my library and I don't have a single book on signal analysis. Actually, that surprises me because before I went on my 15 minute search I would have guessed I had at least one in the mix of several hundred physics and math books.
 
Bandwidth versus Keying Speed Revisited  
by W5VPU on September 2, 2007 Mail this to a friend!
I want to be kind to you, Mickey. I believe you are convinced this information is extremely important for every ham to know. Otherwise you would not have invested so much of yourself in it. I wish to offer you a gentle observation followed by a suggestion.

My own reaction is less enthusiastic than yours. Even after my 60 years of education, training and experience in electronic physics.

You prove your hypothesis to the wrong people. You confuse us with people who care about such things. You deserve better than we as a whole can give you.

With a shrug, we dismiss your study with the title of a Shakespearan play: "Much Ado About Nothing." But to you it is far more than "nothing."

A kind suggestion: Present your pearls to a different audience. To those who will understand the value and significance of what you do. This theory is beyond my capacity to apply to very ordinary days lived in a very ordinary life. Find those who can be your peers.
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on September 2, 2007 Mail this to a friend!
>>>
A kind suggestion: Present your pearls to a different audience. To those who will understand the value and significance of what you do. This theory is beyond my capacity to apply to very ordinary days lived in a very ordinary life. Find those who can be your peers.
<<<


My two-bits.

Although I have not read all the detailed messages, especially the various attempts at "one up-man-ship" type posts that have occurred, I do very much enjoy this kind of topic. I am not too old to learn something new. Indeed, I spend most of my day learning new things. And, since most of those new things are not in the core electronic or engineering field of this ilk, I savor these types of postings when they come.

There is a kind of battle of who's right and who's wrong in the previous messages but I ignore that and take each for what I gain from it. If this subject were beyond and above me totally and I felt incapable of learning from it, then I can easily choose not to spend time reading it. I mean, isn't it better for each of us to do our own censorship rather than censoring an author as a whole or censoring a topic as a whole.

Therefore, keep it up.

phil, K7PEH
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on September 2, 2007 Mail this to a friend!
K7PEHa:Square Wave ?

<<snip>>

Where is this square wave?

It could be a semantics thing but I am thinking that a square wave is a DC voltage that is started and stopped but maintained DC by the multitude of odd harmonics.

But, that has nothing to do with a CW keying wave form does it? I mean, the keying of a CW is on and off keying of a sine wave of a given SINGLE frequency. The only harmonics that come into play are during the actual on and off keying action but this is not a square wave. At best, it might be considered a very sharp, almost Gaussian shape of a highly peaked wave if you remove the middle part of constant maximum peak envelope on the sign wave.

=============================

This has been discussed in the previous posts. If it is nothing more than switching a carrier wave, then why would there be harmonics of a square wave that are predicted by a Fourier series? See the spectrum analyzer displays posted by Mickey, K5MC. These are not theoretical, they are real.

Look at it from a simple point of view, you mention a DC voltage that is stopped and started. Think about the plate (collector) voltage, it is DC is it not? How about the plate (collector) current? I don't know of any rigs that display AC current, only DC current. Aren't you switching that DC current on/off?

You are also making a common mistake where you see the "envelope" as a voltage impressed on the carrier wave. Keep in mind that the "envelope" is really the amplitude of the carrier's sine wave changing value. There is no "envelope" voltage really there, although it is really easy to come to this conclusion by looking at an oscilloscope.

This is why the cw keying wave "modulates" the carrier sine wave rather than simply turning it on/off. Remember, the "envelope" developed by a 10 Hz keying wave modulating the carrier is no different than a 1000 Hz sine wave modulating the carrier wave.

Lastly, if you are going to assume that you can analyze the waveform by removing the "flat" portion, then you must also assume that the keying action causing the "rising" portion of the envelope goes back to zero before the "flat" portion begins. Otherwise, it will have an effect on the "flat" portion. That means you are discussing a saw tooth wave. Saw tooth waveforms require both even and odd harmonics to form. Since the spectrum pictures show no even harmonics, this assumption would be incorrect.

Jim
WA0LYK
 
RE: Bandwidth versus Keying Speed Revisited  
by WA0LYK on September 2, 2007 Mail this to a friend!
K7PEH:My two-bits.

<<snip>>
There is a kind of battle of who's right and who's wrong in the previous messages but I ignore that and take each for what I gain from it. If this subject were beyond and above me totally and I felt incapable of learning from it, then I can easily choose not to spend time reading it. I mean, isn't it better for each of us to do our own censorship rather than censoring an author as a whole or censoring a topic as a whole.

Therefore, keep it up.

phil, K7PEH

=================================================

You make excellent points. If there is no right/wrong in the previous messages, then there is no point in the argument. Since we are discussing real, actual, physical responses, there should be a correct (rather than right/wrong) interpretation.

Lastly, telling someone to take discussions like this elsewhere is pretty egocentric, in fact, to the point of narcissism.

I would hope there is still room for "advancement of the radio art" within amateur radio. Although the discussion is very high level and CW is rather mundane, the fundamentals of signal analysis is something we as a group need more of. As we move toward a digital future these concepts are even more important for understanding the effects different signals will have. An excellent example is the discussion at
http://mysite.verizon.net/nb6z/psk31.htm
concerning psk31. Much thought was obviously given to designing this mode to be bandwidth conserving and to have little interference outside the occupied bandwidth.

I only wish more thought like this would be given to other modes we are seeing.

Jim
WA0LYK

 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on September 2, 2007 Mail this to a friend!
Phil,

You asked some great questions.

Some folks appear confused over the issue of sideband power, especially over when, during the CW waveform, the power is emitted.

I'm not going to be mathematically rigorous here--too much time to look up every detail, but does the following make sense?

---

Here's the disconnect for some folks. They've seen the Fourier Series and mis-apply it to this discussion. Fourier invented his series to solve a thermodynamics problem. He found that he could simplify the math if he modelled a square wave with a set of sine waves. It's just a model, a mathematical tool or simplification that worked for his application. It isn't applicable here. Since Fourier's time, others built on his work and came up with the transform named in his honor, which is applicable here, but one must understand how to apply it and how to interperet the results.

As far as I can tell, Tim AB0WR believes that the square-ish waves we modulate the carrier with are exactly in the form of the series, something like this;

f(t) = A*(sin(wt) + (1/3)sin(3wt) + (1/5)sin(5wt) + .....)

Which obviously they are not, else even when we lift the key, these components would still exist. That isn't even physically possible to do with our off the shelf TX equipment.

A more accurate model we could use is; (this is given in conceptual form)

0 < t <= T/2, f(t) = A*(1-exp(-t/RC))
T/2 < t <= T, f(t) = A*exp(-t/RC)
.
.
.

when our keying wave shape is formed with an RC filter. Of course, this isn't the only way to form the shape, but this is an easy example.


Back in Fourier's time, he couldn't solve this problem, but today we can.


The transform shows lobes at intervals of the odd harmonics, that is;

F(w)*(a*sin(a'w) + b*sin(3b'w) + c*sin(5c'w) + ....).

I'm guessing that these components, since they look similar to the Series, are also causing confusion for some readers. As one can see though, there is only w in the result, no t. No time information is present except the interval over which the transform was derived. Therefore, one can't say *when* during the interval the sideband energy was expended, we only know the *total* sideband energy expended.

To recontruct the sideband energy expenditure over time, we need to dismantle the waveform. By slicing up our original interval into smaller increments, and finding the transform of each slice, we find the sideband energy contribution of each slice. In this way, we can build up a power versus frequency versus time table.

Oh, but isn't that what a spectrum waterfall is? Absolutely! So we can all download, say, Spectrum Laboratory, and have it do all of the above and graph it to boot! Just adjust the sampling rate and the number of samples per calculation to control the time slice length, set the waterfall scroll speed to an appropriate value and off you go.


I have some nice JPEG images of CW waveforms gathered from this tool. If anyone is interested, I can send them along. Email me at linwood.davis@gmail.com.

Lin
WB1AIW
 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on September 2, 2007 Mail this to a friend!
I just want to add that this discourse has motivated me to brush the dust off old textbooks and delve into the math again.

It's been fun and I wish I had the time to go through it in more detail.

Also, I don't have anything against any of the contributors that I've fenced with here! I want to understand this as much as you and argue my points after careful thought is given to each response.

Take care and be well...

...but in the meantime, En Guarde!

Lin
WB1AIW
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 2, 2007 Mail this to a friend!
WB1AIW-A more accurate model we could use is; (this is given in conceptual form)

0 < t <= T/2, f(t) = A*(1-exp(-t/RC))
T/2 < t <= T, f(t) = A*exp(-t/RC)
.
.
.

when our keying wave shape is formed with an RC filter. Of course, this isn't the only way to form the shape, but this is an easy example.
-----------------------------------------------

AB0WR:Yes, but what is the input to the RC filter? It is a square wave with odd harmonics.

The RC filter will have transfer function of the form (a)/[(w**2 + a**2)**(1/2)]

The output of the RC filter will be the square wave, s(t) multiplied by the transfer function.

What you wind up with in the frequency domain is a square wave with odd harmonics whose amplitude values are modified by the transfer function.
****************************************************************





WB1AIW - Back in Fourier's time, he couldn't solve this problem, but today we can.

AB0WR: I'm not sure why you say this.
***************************************************************





WB1AIW-The transform shows lobes at intervals of the odd harmonics, that is;

F(w)*(a*sin(a'w) + b*sin(3b'w) + c*sin(5c'w) + ....).

I'm guessing that these components, since they look similar to the Series, are also causing confusion for some readers. As one can see though, there is only w in the result, no t. No time information is present except the interval over which the transform was derived. Therefore, one can't say *when* during the interval the sideband energy was expended, we only know the *total* sideband energy expended.

To recontruct the sideband energy expenditure over time, we need to dismantle the waveform. By slicing up our original interval into smaller increments, and finding the transform of each slice, we find the sideband energy contribution of each slice. In this way, we can build up a power versus frequency versus time table.
------------------------------------------------------------

AB0WR:ROFL!!

There is no confusion here. You got exactly what would be expected from a filtered square wave. What you got is similar to the Fourier Series for a square wave because a square wave is the base waveform you are using to generate your output.

The Fourier Transform will move the time domain function into the frequency domain. That's why there is no (t) in the frequency domain representation.

If you go look at any of the ten web sites I gave you, you will find that the odd harmonics of the square wave exist during the entirety of the square wave. All the RC filter does is change the relationship of the amplitudes of the frequency components. The RC filter won't change the anything else. The bandwidth of the RC filter determines what the rise time/fall time will be by limiting the high frequency components of the waveform.

You can't dismantle the waveform in the manner in which you are trying.

If you take the rise time of a square wave by itself, you have to assume subtraction of a step function at the end of rise time. You can't just end a periodic waveform out in the "air" and perform a Fourier Transform on it. The FT is for periodic waveforms so you must have zero crossings somewhere in order to calculate the period of the waveform. What you will wind up doing is calculating the FT of a sawtooth with an exponential rise time and a vertical fall time. You will wind up doing the same thing at the end of the waveform. You will have a periodic wave with a vertical rise time and an exponential fall time.

The periods of the sawtooth waveforms will be the length of the rise and fall times and not the entire peroid of the whole keying pulse. This will cause all kinds of errors.

*******************************************************************


AB0WR:You can go look at Mickey's spectrum analyzer screens and my screen captures from my SDR radio showing exactly what you are speaking of. Take a close look at the outputs from the Winrad program. It is quite easy to see the spectral components from the CW pulse.

Please email me the CW waveforms you have captured from this tool. I am always looking for additional data to add to my collection. My email is ab0wr@ab0wr.net

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 2, 2007 Mail this to a friend!
WB1AIW-As far as I can tell, Tim AB0WR believes that the square-ish waves we modulate the carrier with are exactly in the form of the series, something like this;

f(t) = A*(sin(wt) + (1/3)sin(3wt) + (1/5)sin(5wt) + .....)
-----------------------------------------------------

AB0WR: BTW, Lin, there is such a thing as an inverse Fourier Transform. You *can* take what you find in the frequency domain back to the time domain.

Do that with the frequency components you found for your waveform with the RC filtering and see what you get!

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 2, 2007 Mail this to a friend!
K7PEH-If this subject were beyond and above me totally and I felt incapable of learning from it, then I can easily choose not to spend time reading it. I mean, isn't it better for each of us to do our own censorship rather than censoring an author as a whole or censoring a topic as a whole.

Therefore, keep it up.



AB0WR:The mark of a scholar! Most of the discussion here isn't personal. That doesn't mean it isn't heated at times!

I welcome your insights.

*YOU* keep it up!

73,

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 2, 2007 Mail this to a friend!
K7PEH-But, in thinking about a typical signal of y(t) = A sin(wt), I had not thought about the importance of separating out the amplitude as a simple example of this idea. So, I created a step function of A where A is now a function of t, A(t) with the value of zero for all t less then pi and the value of 1 for all t greater then or equal to pi to represent turning on the signal. So, then I have the product of two functions (mixing): y(t) = A(t)*sin(wt). And, of course, the square wave is merely another step function.

AB0WR:I'd say you have it down pretty well. It doesn't appear you needed LS and SA 101!

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by K7PEH on September 2, 2007 Mail this to a friend!
AB0WR:I'd say you have it down pretty well. It doesn't appear you needed LS and SA 101!

Well, I am planning to go to U of W bookstore and take a look at the various linear systems and signal processing texts. I will probably pick up one for one of the engineering classes and go through that.

My math background though is fairly mature. I was pursuing a PhD in Math when I chose to quit and get a job (motivated primarily by the fact my wife was pregnant but basically my heart was not in the world of academia at that time). My undergraduate degree was Physics so my thesis subject area was Geometric Algebras but I didn't get far enough to even start on my research. However there are close connections between Physics and Geometric Algebras. However, all this was 35 years ago!
 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on September 2, 2007 Mail this to a friend!
K7PEH - My math background though is fairly mature. I was pursuing a PhD in Math when I chose to quit and get a job (motivated primarily by the fact my wife was pregnant but basically my heart was not in the world of academia at that time). My undergraduate degree was Physics so my thesis subject area was Geometric Algebras but I didn't get far enough to even start on my research. However there are close connections between Physics and Geometric Algebras. However, all this was 35 years ago!


I hear that! I graduated with a BS in Engineering Physics (EE minor) 26 years ago and have been a communications engineer for almost my entire career. I've learned so much practical stuff since then. However, to go back to the theory, knowing what I know now, and tie it all together would be fantastic!

Good luck. I hope you get back to it someday.
Lin
WB1AIW

 
Bandwidth versus Keying Speed Revisited  
by K5MC on September 2, 2007 Mail this to a friend!
WB1AIW - Here's the disconnect for some folks. They've seen the Fourier Series and mis-apply it to this discussion. Fourier invented his series to solve a thermodynamics problem. He found that he could simplify the math if he modelled a square wave with a set of sine waves. It's just a model, a mathematical tool or simplification that worked for his application. It isn't applicable here. Since Fourier's time, others built on his work and came up with the transform named in his honor, which is applicable here, but one must understand how to apply it and how to interperet the results.

K5MC - If one observes the RF waveform in the time domain on an oscilloscope, it is obvious that a typical CW transmitter sending a string of dits (a periodic signal) can be represented very well by a Fourier series as I described in my original eHam article. The keying waveform (the baseband signal) is modeled as a Fourier series; this baseband signal, in turn, multiples the high-frequency carrier to form the RF output signal. I carried out almost all of my calculations on the baseband signal, which is a very common practice in signal analysis. My calculations were all performed in the time domain as I've stated before. There was no need for me to calculate any Fourier transforms (frequency domain) since my signals of interest were periodic. Algebra and trigonometry were the main math tools I used to calculate the average powers of the individual frequency terms. (I did have to calculate the coefficients of the Fourier series by integration, however.) My experimental results (obtained from four different CW transmitters using a state of the art RF spectrum analyzer) agree quite well with my Fourier series models.

Fourier transforms can be calculated for both periodic and non-periodic signals. As I said, however, there was no need for me to use Fourier transforms to represent the keying waveforms I used in my earlier calculations.

The Fourier series approach is a powerful concept used to model periodic signals. Practically any electrical engineering textbook on signal analysis/communications will include a discussion of Fourier series! (I have taught Fourier analysis to electrical engineering students for over 20 years now.)

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on September 3, 2007 Mail this to a friend!
WB1AIW - They've seen the Fourier Series and mis-apply it to this discussion.


K5MC - If one observes the RF waveform in the time domain on an oscilloscope, it is obvious that a typical CW transmitter sending a string of dits (a periodic signal) can be represented very well by a Fourier series as I described in my original eHam article.


Mickey, I don't have any argument with someone applying the Series to solve for the *average* energy spectrum density, and thus find the power bandwidth. By "this discussion", I was referring to the off-topic discussion about the energy density spectrum as a function of time. "When are sidebands emitted" was the question I was trying to answer. Sorry for the confusion.
 
RE: Bandwidth versus Keying Speed Revisited  
by WB1AIW on September 3, 2007 Mail this to a friend!
AB0WR - The RC filter will have transfer function of the form (a)/[(w**2 + a**2)**(1/2)];

A textbook I have entitled "Signals and Linear Systems" by Gabel and Roberts has the transform of exp(-at) as 2a/(a^2 +w^2).


AB0Wr - You can't dismantle the waveform in the manner in which you are trying.

Indeed you can. Whats more is that the DFT is based on this concept.

AB0WR: BTW, Lin, there is such a thing as an inverse Fourier Transform. You *can* take what you find in the frequency domain back to the time domain.

AB0WR - Do that with the frequency components you found for your waveform with the RC filtering and see what you get!

Well, since the Fourier transform has the property of symmetry, if I apply the inverse transform to the transform, I get my original function back. It doesn't come back as a Fourier Series. Please don't dismiss out of hand. This IS a very important property!!! One of the cornerstones upon which the Fourier transform sits.

You're certainly not giving much thought and consideration to what I've written and that's distressing to me. Why is this? I'm not pulling this stuff out of my butt; I have two good textbooks (one from a college course I took) on the subject and 25 years experience with digital communications as an engineer.

And besides, the Fourier Series is just a approximation of a square wave. It converges upon the square wave, yet never fully reaches it; a point where the Mean Square Error is zero. Check out Gibbs Phenomenon.

So why would I chose to model the squarewave with the Fourier Series when I can directly transform my RC-derived waveform without it?

Very simplistically, I know that a DC level has a nice single frequency transform at w=0; the impulse function. Whenever the amplitude of this DC is disturbed, the energy density spectrum broadens while the disturbance is occuring. At the point where the amplitude becomes unchanging, the spectrum once again becomes an impulse function at w=0. I know this because I've looked at thousands of signals with the trusty spectrum analyzer, and if I were to view a signal composed of a single low-to-high transition with an SA, I would see spectral content only briefly, while the transition was occuring.

I sent to you three images of SPEC LAB waterfalls showing how the spectrum changes over time over the course of CW elements. Did you receive them?


Lin
WB1AIW
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 3, 2007 Mail this to a friend!
AB0Wr - You can't dismantle the waveform in the manner in which you are trying.

WB1AIW:Indeed you can. Whats more is that the DFT is based on this concept.

AB0WR: You can't do it the way you are doing it.

The DFT isn't based on dismantling a periodic waveform into other waveforms. The DFT is a method for determining frequency components involved in a signal based on sampling the signal.

That's not the same thing as breaking up a square wave with rise and fall times into a beginning sawtooth, a perfect square wave, and an ending sawtooth (and then trying to call the perfect square wave a DC component). And then trying to find a Fourier series for each and saying that makes up all the frequency components involved in the signal.

Just what do you think a DFT of a time domain square wave is going to give you for frequency components in the frequency domain?

The DFT and the Fast Fourier Transform (FFT) are essentially the same thing. What do you think software programs like Cool Edit, ScopeDSP, and program you referenced use? What do you think they will come up with for the FFT of a square wave?

I assure you it will be different than it will be for a sawtooth.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 3, 2007 Mail this to a friend!
WB1AIW-Well, since the Fourier transform has the property of symmetry, if I apply the inverse transform to the transform, I get my original function back. It doesn't come back as a Fourier Series. Please don't dismiss out of hand. This IS a very important property!!! One of the cornerstones upon which the Fourier transform sits.

AB0WR- You haven't looked at a single one of the web sites I referenced, have you?

If you have a frequency component (nw) in the frequency domain, what do you think that represents in the time domain?

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 3, 2007 Mail this to a friend!
WB1AIW-You're certainly not giving much thought and consideration to what I've written and that's distressing to me. Why is this? I'm not pulling this stuff out of my butt; I have two good textbooks (one from a college course I took) on the subject and 25 years experience with digital communications as an engineer.

AB0WR- I'm sorry this is distressing to you. It's apparent you have spent a lot of time on this. But it is also apparent that you are trying to come up with reasons for saying that the odd harmonics that make up a square wave don't exist throughout the entire square wave. If you would reference *any* of the ten web sites I gave earlier in the thread you would know that your quest is futile.

I have looked at square waves using different programs till I'm blue in the face, I even spent money on buying ScopeDSP. I have been through my 35 year old Signal Analysis textbook and my son's two year old Signal Analysis textbook. I even built a square wave generator, a 10khz oscillator, and a linear mixer circuit to study the result of mixing the two. Result---- Square waves are made up of a series of cosines (or sines depending on where you start) including the fundamental and all the odd harmonics. The odd harmonics fall of as the ration of 1/n. They last for the duration of the square wave.

When you mix the fundamental and the odd harmonics of a square wave with a carrier signal you wind up with upper and lower sidebands consisting of the square wave fundamental and all of its odd harmonics. They last for the duration of the square wave.

You can filter out the higher frequency harmonics to make the square wave more "rounded". This decreases the bandwidth of the sidebands. Various filters can be used but I have only used the standard RC lowpass filter.

The envelope resulting from mixing the two signals, however, is NOT a mixture of a beginning sawtooth, a DC component, and another ending sawtooth. None of the programs I have used to record the resultant 10khz signal and then analyze it using an FFT show this at all. They show the upper and lower sidebands exactly the way you would expect.

So, again, I'm sorry if you are distressed. But I can't find a single reference book that shows analyzing a square wave by analyzing the rise and fall times as if they are sawtooths. Do *you* have any that show breaking down a square wave this way?

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by K5MC on September 3, 2007 Mail this to a friend!
WB1AIW - Mickey, I don't have any argument with someone applying the Series to solve for the *average* energy spectrum density, and thus find the power bandwidth. By "this discussion", I was referring to the off-topic discussion about the energy density spectrum as a function of time. "When are sidebands emitted" was the question I was trying to answer. Sorry for the confusion.

K5MC - OK, Lin, it's hard for me to distinguish between the "main" topic and these "off-topic" discussions. I hope you also agree that the 99% power bandwidth is identical to the FCC's "occupied" bandwidth as defined in Part 2.

Your question about "when are sidebands emitted" can be a very tricky proposition. To add to Tim's earlier comments about the two-tone SSB signal, let me quote W8JI below and then make some follow-up comments.

W8JI (May 27, 2007) - 1.) Bandwidth is required only when the carrier changes level. When at full power or zero power it has no bandwidth, it has no modulation. 2.) Assuming a stable carrier the rate of power level change controls the highest sideband frequency produced. The level change at that slope (or any angle of slope during the transition) controls the amplitude of the sidebands.

K5MC - Let's apply W8JI's theory above to conventional amplitude modulation (DSB-AM) where the modulating signal is a simple sinusoidal signal. For example, let's assume the modulating signal is a 100-Hz sine wave. Let's further assume that we are talking about ideal modulation here (no splatter, etc.) up to 100% modulation.

At the crests and troughs of the RF envelope, the rate of change of amplitude is zero. Therefore, W8JI (and some others?) must also believe that the "instantaneous" bandwidth is zero at those specific times, which corresponds to the same exact times that the instantaneous values of the modulating signal itself are either maximum positive or maximum negative. On the other hand, the "instantaneous" bandwidth of the AM signal must be greatest at the exact times when the modulating signal itself has zero amplitude. If we had an "ideal" spectrum analyzer, then we would see the amplitudes and frequencies of these sidebands change from one instant to the next if we believe W8JI.

If one argues that the "bandwidth" of a CW signal suddenly goes to zero during the time interval when the envelope is constant, then one has to argue along the same lines as I discuss above for simple tone modulation. How many folks are willing to go that far?

This discussion of when/if the sidebands/bandwidth disappears is fraught with danger. I know of no communication/signal analysis textbooks that discuss the sudden appearance and disappearance of the sidebands/bandwidth as apparently described by some hams here for the test signals that I used in my two eHam articles. If you "sample" a signal as a function of time over a particular time interval, if you then study that signal in the frequency domain by calculating its Fourier transform (or its Fourier series if the signal is periodic), all information in the time domain is lost over that particular time interval. Of course, you can take the FFTs of samples over smaller time portions and piece them together to form a spectrogram (such as commonly done to study speech waveforms), which is the basic idea behind short-time Fourier transforms.

73, K5MC
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 3, 2007 Mail this to a friend!
WB1AIW-Very simplistically, I know that a DC level has a nice single frequency transform at w=0; the impulse function. Whenever the amplitude of this DC is disturbed, the energy density spectrum broadens while the disturbance is occuring. At the point where the amplitude becomes unchanging, the spectrum once again becomes an impulse function at w=0. I know this because I've looked at thousands of signals with the trusty spectrum analyzer, and if I were to view a signal composed of a single low-to-high transition with an SA, I would see spectral content only briefly, while the transition was occuring.

AB0WR-Mathematically, this isn't correct. A true DC signal would have to last from -infinity to +infinity. What you have is a step function going up. Mathematically, when you turn the signal off it becomes a gate pulse. This is described mathematically as the rect(t) function in the time domain and as the sinc function in the frequency domain. It has the form of (1/2pi)[sin(w/2pi)/(w/2pi)] in the frequency domain. Since (w/2pi) is the period, what happens when the period gets longer and longer? (hint: how low in frequency does your spectrum analyzer go?) I should also point out that the step function doesn't satisfy the usual requirement that the integral from -T/2 to +T/2 of the waveform be finite if you want to do a Fourier Series on the waveform. A step function has no period and therefore can't meet the requirement for a Fourier Series.

A step function does have an impulse at w=0. The transform is (pi)[Impulse(w)] + (1/jw). So you get a continuous freqency spectrum from -infinity to +infinity with a maximum at w=0 and that tails off as 1/jw.

I have yet to see any square wave with any rise time show such a frequency spectrum using any analysis program I have available. I can find no analog spectrum analyzer display showing this for a square wave.

If you could show me mathematically how the continuous-frequency Fourier Transform of a step up function and the continous-freqneucy step down function could add to come up with a discrete Fourier Series for a square wave I would be in your debt. You would also be a lot further along in showing how a square wave only has frequency components during the rise time and the fall time. It would be even nicer if you could show us pictures of spectrum analyzer displays showing this continous spectra tailing off as 1/w. That should be easy to do with a spectrum analyzer that can be synced to the incoming signal.

As I said earlier, it is obvious that you are trying to prove that you can take a square wave apart into a separate step up function, a separate DC component, and a separate step down function. I can't find anything in the literature showing that you can do this. I can't find any spectral displays anywhere of a square wave showing anything other than Fourier Series components.

It should be obvious that a square wave can't be a product of *both* a Fourier Series *and* a continous spectrum from w=-infinity to +infinity at the beginning and end of the square wave. These components would not add up to be a square wave.

You have also built yourself somewhat of a connundrum. If the square wave has a step-function frequency spectrum at the beginning and the end and no spectrum in the middle, then where do the odd-harmonic spectral oomponents all of the spectrum analyzers show come from?

I await your mathematical explanation.

tim ab0wr
 
Bandwidth versus Keying Speed Revisited  
by AC7GO on September 3, 2007 Mail this to a friend!
Anyone know of an easy-to-use, readily available program for calculating DFTs? It seems to me that one way people can arrive at a resolution of these arguments would be to generate time-domain modulated (keyed carrier) waveforms, then look at the DFT outputs.

Put a short window (shorter than a "dit") between carrier bursts, and the DFT will give you no spectral components whatsoever. Put one in the middle of a carrier burst, and only the carrier will show up in the spectrum (unless your sample window is other than an integral number of carrier periods). Center a window on a transition, and you will get something very broadly spread out in frequency. You could see the effect of widening the window. When you make it really wide, thousands of times wider than a "dit", you will get a spectrum that is consistent with occupied bandwidth calculations. But be aware that it will take some work to make sense of the DFT outputs, and to set up the right sample frequency, etc.

Many of the basic assertions presented here are basically correct, but they seem contradictory. Frequency-domain and time-domain representations of a signal are not contradictory. But in some cases it takes real effort to resolve the paradoxes.

It takes effort and exploration to develop an "intuition" for the relationship between time-domain and frequency-domain signals. Calculus and complex number theory helps, too. Believe me, when you put some time into it, it makes sense. Really, the principles are beautiful, too.
 
Bandwidth versus Keying Speed Revisited  
by AC7GO on September 3, 2007 Mail this to a friend!
Anyone know of an easy-to-use, readily available program for calculating DFTs? It seems to me that one way people can arrive at a resolution of these arguments would be to generate time-domain modulated (keyed carrier) waveforms, then look at the DFT outputs.

Put a short window (shorter than a "dit") between carrier bursts, and the DFT will give you no spectral components whatsoever. Put one in the middle of a carrier burst, and only the carrier will show up in the spectrum (unless your sample window is other than an integral number of carrier periods). Center a window on a transition, and you will get something very broadly spread out in frequency. You could see the effect of widening the window. When you make it really wide, thousands of times wider than a "dit", you will get a spectrum that is consistent with occupied bandwidth calculations. But be aware that it will take some work to make sense of the DFT outputs, and to set up the right sample frequency, etc.

Many of the basic assertions presented here are basically correct, but they seem contradictory. Frequency-domain and time-domain representations of a signal are not contradictory. But in some cases it takes real effort to resolve the paradoxes.

It takes effort and exploration to develop an "intuition" for the relationship between time-domain and frequency-domain signals. Calculus and complex number theory helps, too. Believe me, when you put some time into it, it makes sense. Really, the principles are beautiful, too.
 
RE: Bandwidth versus Keying Speed Revisited  
by AB0WR on September 4, 2007 Mail this to a friend!
AC7GO-Put one in the middle of a carrier burst, and only the carrier will show up in the spectrum (unless your sample window is other than an integral number of carrier periods). Center a window on a transition, and you will get something very broadly spread out in frequency.

AB0WR: Why do you say this? My copy of ScopeDSP shows frequency spectra from a window in the middle of a square wave. ScopeDSP using a Fast Fourier Transform to calculate the DFT.

There seems to be a disconnect here that a carrier "burst" doesn't have an envelope. If you crank the sweep speed up fast enough on your oscilloscope to see the individual periods of an RF carrier, certainly you lose sight of the envelope since it is at a much, much lower frequency. That doesn't mean it goes away.

This somehow gets translated into there being only the carrier frequency existing in the middle of an envelope.

If a DFT didn't pick up both the low frequencies forming the shape of the envelope as well as the high frequency of the carrier, what good would the DFT be? How could you ever use a DSP to "detect" the audio frequencies in a modulated SSB signal?

I think those who contend that a square wave or a carrier modulated by a square wave has no frequency components where the slope of the signal is Zero need to answer Mickey's and my questions about what happens to the modulating tone of an AM signal or to the 2-tone SSB test tones where the slope of the envelope is Zero. That would be at the maximum and minimum values of the envelope.

Do these tones disappear when the slope of the envelope is Zero?

The answer to these questions will go far in clearing up what happens during a string of CW dits.

tim ab0wr
 
RE: Bandwidth versus Keying Speed Revisited  
by VA3NR on September 5, 2007 Mail this to a friend!
If anyone is interested, its fairly easy to use Matlab or other plotting program to illustrate how repeating positive and negative unit steps converge to spectrum of square wave.

Use:
Fourier transform of unit step: FT{u(t)} = 1/jw+pi*dirac(w)
Linearity property.
Time shift property: FT{f(t-dot)} = exp(-j*dot*w)*FT{f(t)}
w:=omega, dot:=time shift`, j^2=-1

Sol'n hints:
One pulse = u(t)-u(t-dot)
FT{one pulse}=1/jw*(1-exp(-j*dot*w))
Two pulses = u(t)-u(t-dot)+u(t-2*dot)-u(t-3*dot)
FT{two pulses}=1/jw*(1-exp(-j*dot*w)+exp(-j*2*dot*w)-exp(-j*3*dot*w))
etc.

Spectrum of one pulse is sync function. Spectrum of two pulses shows peaks at odd harmonics. Add more pulses and peaks narrow and approach values predicted for square wave. The problem modelling CW with strings of unit steps is same as using square waves - we perceive the key click bandwidth associated with each step, not cummulative steps.

_________
Commands to plot in Matlab:
f=0.01:0.01:30;
w=2*pi*f;
dot=0.5;
ONE=abs((1./(j*w)).*(1-exp(-j*dot.*w)));
plot(f,ONE)
TWO=abs((1./(j*w)).*(1-exp(-j*dot.*w)+exp(-j*2*dot.*w)-exp(-j*3*dot.*w)));
plot(f,ONE,f,TWO)
THREE=abs((1./(j*w)).*(1-exp(-j*dot.*w)+exp(-j*2*dot.*w)-exp(-j*3*dot.*w)+exp(-j*4*dot.*w)-exp(-j*5*dot.*w)));
plot(f,ONE,f,TWO,f,THREE)

 
Bandwidth versus Keying Speed Revisited  
by K5MC on September 5, 2007 Mail this to a friend!
VA3NR -If anyone is interested, its fairly easy to use Matlab or other plotting program to illustrate how repeating positive and negative unit steps converge to spectrum of square wave.

K5MC - It's even easier to show that the Fourier transform of a periodic square wave is an infinite series of impulses. The positive and negative frequencies of the impulses are determined by the period of the square wave and the strengths (areas) of the impulses are proportional to the coefficients of the Fourier series of the square wave.

73, K5MC
 
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