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Author Topic: what exactly *IS* CW... and how are there sidebands?  (Read 6325 times)
AK4YA
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« on: August 26, 2012, 07:59:07 PM »

*just* got my upgrade to general.  All along I've thought morse was just carrier off, then carrier on for short period for the dots and carrier on for a longer period for the dashes.  (My Keesler AFB tech school completely skipped morse and went straight to fsk and other type of digital modulation)  But now I've discovered that apparently CW has sidebands, so I don't see how my original thinking there is correct.  I understood a sideband to be a product of nonlinear combination of two separate signals. 

The only other thing I can think is a morse transmission requires 100% duty cycle, and some tone is used to modulate the carrier and only the resulting sideband is sent?  If this is the case, then if you tuned off center freq a bit, wouldnt that change the perceived pitch of the dots/dashes?


ugh, thanks
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WX7G
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« Reply #1 on: August 26, 2012, 08:07:25 PM »

CW as used by amateurs has a 5 ms rise and a 5 ms fall time. During the rise/fall transitions sidebands are created about 70 Hz either side of the carrier.

Another way to think of it is like this: One cannot convey information with zero bandwidth.
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PA0BLAH
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« Reply #2 on: August 27, 2012, 03:06:20 AM »

CW as used by amateurs has a 5 ms rise and a 5 ms fall time. During the rise/fall transitions sidebands are created about 70 Hz either side of the carrier.

Another way to think of it is like this: One cannot convey information with zero bandwidth.

Well, when the FCC wants to hand out a ticket to some unlis, they need the information wether of not he has a carrier in the air. So the information, carrier or no carrier is always present even when that is an infinite long during situation.

5 ms is pretty long. When the bandwidth is too small CW sounds lousy, looks like the dits and dashes are not switched on and off but are transmitted with an audio volume knob turning open and close. When I look at RUFZXP, a widely used program by call sign trainers - just for fun or to exercise for contesting - and HST guys, the fade in and fade out , that is between 0 and 100 %, is 0.6 and 0.7 ms as maximum  adjustable value.

Yes, when you switch a carrier on and off with cosine-squared leading and trailing edges of 5 ms and you do that 100 times per second  the envelope is a sine wave and you have a carrier with 2 sidebands, not more.

Are you using shorter, infinite short leading and trailing edges, then you modulate your carrier not with a sinewave of 100 Hz but with a symmetrical block wave of 100 Hz fundamental frequency. Hence the carrier has then sidebands at 100, 300, 500, 700 Hz distance. A dash is not symmetric, hence also smaller components at 200, 400, 600 etc.

So when you want, in order to preserve sufficient short leading and trailing edges after IF filtering in your receiver all components up to and including the 5 th harmonic, that makes a total bandwith of 1000 Hz, when the signal is block modulated with 100 Hz.
Block modulating with 100 Hz is a dittime (dot mark time)  of 5 ms, according to PARIS standard that is 240 words per minute Morse speed. Hence  the  bandwith you need in your CW filter is at least 4 times the speed in wpm.

Normally Xtal filters are ringing on such a signal so in practice somewhat more bandwidth is desirable. In fact CW filters are sold mentioning the number of poles, and by a frequency vs attenuation graph. They should also show the envelope of a 25 Hz on off modulated carrier after filtering.

When you use not an on/off carrier but carrier shift in frequency, you actually make two AM (on/off) modulated transmitters, One modulated with the marks as on, and the other modulated with the spaces as on. So you get roughly a total bandwith that is the addition of the original CW bandwith plus the frequency shift, when that shift is limited.

When the CW is not symmetrical going through a small band filter the sidebands have phase and amplitude distortion, that affects the envelope.

Normally CW is detected by interference with a BFO. When all the components are present left and right of the carrier, that yields an audio tone with the same slopes as he filter in the IF allowed. When you cut one sideband of the CW by filtering, the pitch will be the same but the envelope is not preserved.
« Last Edit: August 27, 2012, 03:29:30 AM by PA0BLAH » Logged
STAYVERTICAL
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« Reply #3 on: August 27, 2012, 03:40:07 AM »

This question was handled very well by Claude Shannon way back in the days of slide rules when he invented information theory.
Basically as WX7G said " IF YOU WANT TO CONVEY INFORMATION YOU NEED BANDWIDTH".
That is the answer to your question.

Of course your modulation system can either convey the information in the minimum bandwidth, or take more than its share.
Key clicks, transitions and all that stuff are really secondary to the actual reason - just imperfect ways of modulating information.
Alternatively, you can devise modulation systems to increase throughput within a smaller bandwidth.
One way is the systems used in modems in the early days, where quadrature modulation was used to convey more information in the same bandwidth.

Another system is incorporated in PSK31 where cosine modulation is used to avoid rapid transitions which take more bandwidth.

BASICALLY, ANY SQUARE WAVE CAN BE REPRESENTED BY A SERIES OF SINE WAVES MATHEMATICALLY.
If you take a sine wave and add progressively higher odd order multiples to it, the waveform will become "squarer" and more true.
The steeper you want the sides, the more higher order sine waves you need to add.
If you slope the sides, then the waveform looks like a sloppy chunky sine wave, so you don't need as many sine waves 3/5/7...etc.

This is why if you reduce the rise and fall times of a keying waveform ( obviously a square wave), the bandwidth will decrease.
These multiples are not the odd harmonics of the R.F. frequency, but of the keying rate.
So the faster you send, the wider the signal.

So as you can see, morse keying has to obey the basic laws of information theory, and that is why it has bandwidth.

73 - Rob
« Last Edit: August 27, 2012, 04:33:10 AM by STAYVERTICAL » Logged
W8JI
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« Reply #4 on: August 27, 2012, 05:07:31 AM »

The only other thing I can think is a morse transmission requires 100% duty cycle, and some tone is used to modulate the carrier and only the resulting sideband is sent?  If this is the case, then if you tuned off center freq a bit, wouldnt that change the perceived pitch of the dots/dashes?

Morse code CW is 100% modulated amplitude modulation. It is amplitude modulated differently than what you might think of normally, because there is no tone added to the carrier, like we might think of with normal AM. The carrier goes from zero to full, and then back to zero level. It is off-on keying. 

Because the carrier goes to zero (for very long periods at times), and because it reaches a peak, it is amplitude modulated.

There are two modulation rates involved. One is the rate the carrier changes level at the start and stop of each dot and dash. The other rate is the information rate, just like the baud rate of any information-carrying mode.

The rise and fall rates are fast compared to the times we have information, and the rise and fall rates set the ultimate bandwidth because they are so fast. The rise and fall is normally what makes "clicks". 

5 mS rise and fall is actually not long, and sounds perfectly fine. A rise is like 1/2 of a cycle, so is a fall. With a 5 mS rise and 5 mS fall, the LOWEST frequency sidebands would be 1/(.005*2) = 100 Hz away from the carrier. This is because it takes a rise and fall to make one cycle, so a rise (or fall) is just half of a cycle. This means a CW signal with off-on keying and a sine shaped curve on rise and fall with 5 mS would have two sidebands. The sidebands would be 100Hz up and down from the carrier, and only there as the carrier is increasing or decreasing at the start and stop of each dot and dash element.

The common problem comes from the shape of the rise and fall, and the time. If the shape is sharper than a sine shaped curve, harmonics of the 100 Hz appear. There will usually be multiple harmonics, and these harmonics might extend out 500 Hz or more. This is when, on a good receiver, you hear a wide clicking signal.

The narrowest possible signal the reciprocal of the TOTAL rise and fall time up and down from the carrier if everything is perfect, which it never is.  So if we used a PERFECT 2.5 mS rise and fall the sidebands would be 200 Hz up and down. Everything made is worse than that, because nothing is perfect in symmetry and shape.

There is more explained here:

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

73 Tom
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PA0BLAH
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« Reply #5 on: August 27, 2012, 05:18:19 AM »


Basically as WX7G said " IF YOU WANT TO CONVEY INFORMATION YOU NEED BANDWIDTH".
That is the answer to your question.


OK Rob, but how much bandwidth is needed   in the practice of ham radio Morse code reception.

When I have rise and fall times that are so long, due to limited bandwidth, that the the rise and fall time spans touch each other in the time domain of a dot , the dots   become  sine wave shaped.  So we want at least up to the fifth harmonic transmitted, and passed through our CW receiver filter.  However, we can produce that higher harmonics for free AFTER detection (without taking transmission channel bandwidth) when we put the received detected signal through a limiter with a decision level at half the amplitude of the received signal. So you can reproduce with a simple non linear operation ( a limiter) the whole frequency spectrum of the transmitting key, by only receiving the first sideband component. That is because you KNOW it is a key transmitting, so when you know the first sideband you can add all the other components that determine the steep edges, not transmitted or filtered out in your receiver, with a limiter.

Bob

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STAYVERTICAL
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« Reply #6 on: August 27, 2012, 05:49:58 AM »


Basically as WX7G said " IF YOU WANT TO CONVEY INFORMATION YOU NEED BANDWIDTH".
That is the answer to your question.


OK Rob, but how much bandwidth is needed   in the practice of ham radio Morse code reception.

When I have rise and fall times that are so long, due to limited bandwidth, that the the rise and fall time spans touch each other in the time domain of a dot , the dots   become  sine wave shaped.  So we want at least up to the fifth harmonic transmitted, and passed through our CW receiver filter.  However, we can produce that higher harmonics for free AFTER detection (without taking transmission channel bandwidth) when we put the received detected signal through a limiter with a decision level at half the amplitude of the received signal. So you can reproduce with a simple non linear operation ( a limiter) the whole frequency spectrum of the transmitting key, by only receiving the first sideband component. That is because you KNOW it is a key transmitting, so when you know the first sideband you can add all the other components that determine the steep edges, not transmitted or filtered out in your receiver, with a limiter.

Bob


Sorry Bob, it does not work like that.
If it did, you would have a pretty good compression system there.
You need to have enough bandwidth to allow what would be the equivalent of the nyquist frequency.
That would work if you reduced the keying rate enough to accommodate the recovery of the information.
You have no doubt used a very narrow filter (say DSP) and noticed that it becomes very difficult to differentiate dits and dahs as you go narrower.
Some of that is due to the filter ringing but it also is caused by the lack of "sidebands" being received.

What you are suggesting is a form of "decimation" which is used in DSP when oversampled signals are passed through a low pass filter.
Then they are downsampled to make computation easier.
But this still requires the maximum frequency component to be half the nyquist frequency for this signal.
If you have frequency components greater than this frequency they will be lost and you will have aliasing errors.
This information is lost, and you will not recover it later as you suggest.

A morse signal is really just a digital signal and the same theories which apply to detection of digital signals apply to morse in the end.
I understand the concept of detecting digital signals with threshold detectors etc.
But you still need a particular bandwidth cable, otherwise what you suggest would mean you could detect Gigabit ethernet packets on rubbish cable by just detecting a narrow section of the wide bandwidth required.
Another way of looking at it is like a power supply low pass filter.
As you reduce the frequency spectrum you go from 60/120hz ripple to D.C.
The DC has lost the "information" in the ripple.

I agree that you don't always need the full bandwidth, take SSB for example, but there is always a minimum bandwidth required for a rate of information transfer.
If you have devised some way of defeating information theory, then please patent it quickly, and put it next to your perpetual motion machine.

73 - Rob

« Last Edit: August 27, 2012, 05:55:45 AM by STAYVERTICAL » Logged
PA0BLAH
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« Reply #7 on: August 27, 2012, 06:41:54 AM »

The only other thing I can think is a Morse transmission requires 100% duty cycle, and some tone is used to modulate the carrier and only the resulting sideband is sent?  If this is the case, then if you tuned off center freq a bit, wouldn't that change the perceived pitch of the dots/dashes?

Morse code CW is 100% modulated amplitude modulation. It is amplitude modulated differently than what you might think of normally, because there is no tone added to the carrier, like we might think of with normal AM. The carrier goes from zero to full, and then back to zero level. It is off-on keying.  

Sri, no difference with AM modulation, the AM-modulation is 100% depth not with a sine wave but with a block wave of limited bandwidth.
Quote
Because the carrier goes to zero (for very long periods at times), and because it reaches a peak, it is amplitude modulated.
Doesn't matter in my humble opinion,  it is still a bandwidth limited block wave  of some random changing mark space ratio.


Quote
The rise and fall rates are fast compared to the times we have information, and the rise and fall rates set the ultimate bandwidth because they are so fast. The rise and fall is normally what makes "clicks".  

5 mS rise and fall is actually not long,

5 mS is 5 millisiemens that is something you like to call in non-ISO standard 5 millimho's.
5 ms is 5 millisecond (singularis) and 5 msec is 5 meter of dry alcoholic beverages.


Quote
and sounds perfectly fine. A rise is like 1/2 of a cycle, so is a fall. With a 5 mS rise and 5 mS fall, the LOWEST frequency sidebands would be 1/(.005*2) = 100 Hz away from the carrier. This is because it takes a rise and fall to make one cycle, so a rise (or fall) is just half of a cycle. This means a CW signal with off-on keying and a sine shaped curve on rise and fall with 5 mS would have two sidebands. The sidebands would be 100Hz up and down from the carrier, and only there as the carrier is increasing or decreasing at the start and stop of each dot and dash element.

Not only then, but in the case of a periodic string of dots or dashes always and constant.

Yes except that blooper that is what I wrote, thanks for repeating the information without my language bloopers, hi.

When you want to receive the 5-th harmonic in order to get a nice detectable signal by the  ear, (your 5 ms transient)  after your product  detector,  you need an  IF filter is 200 Hz wide, and phase linear, which yields 5 ms transient time, you need 20 times the dit (=baud) rate - (that is 10 times the dot rate) as bandwith of the IF filter, you get that as I have shown with a 200 Hz ideal IF bandwith filter 20 dots/s = 24 wpm as max CW speed.
When your speed is higher you lose your fifth harmonic that gave you the 5 ms transient time and for higer speeds then 24 wpm  your transient time of a series of dots will be 15 ms instead of 5 ms. So the transient inceases dramatically and the duration of dots is simultaneously shorter with a speed > 25 wpm. That will be a ceiling for the guys trying to break the 25 wpm speed  with an ideal 200 Hz wide CW filter phase linear filter.
Quote
The common problem comes from the shape of the rise and fall, and the time. If the shape is sharper than a sine shaped curve, harmonics of the 100 Hz appear. There will usually be multiple harmonics, and these harmonics might extend out 500 Hz or more. This is when, on a good receiver, you hear a wide clicking signal.

I cannot understand how a 200 Hz wide linear filter, can pass 500 Hz components in base band.
Quote
The narrowest possible signal the reciprocal of the TOTAL rise and fall time up and down from the carrier if everything is perfect, which it never is.  So if we used a PERFECT 2.5 mS rise and fall the sidebands would be 200 Hz up and down.

that filter should be 400 Hz wide.


Are we really answering the question of the topic starter? He was talking about single side band CW with suppressed carrier.
« Last Edit: August 27, 2012, 06:46:39 AM by PA0BLAH » Logged
PA0BLAH
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« Reply #8 on: August 27, 2012, 08:47:57 AM »



Sorry Bob, it does not work like that.

First of all, Rob, thank for your reply. I wonder what is the reason of investing in appliances and Morse code, when I meet you here and we can talk nice, about the technique supporting this all.
Quote
If it did, you would have a pretty good compression system there.
Yes sure, but it even can be better when only one sideband of the CW is used as questioner asked, but everybody is playing back his own perpetual record, that is not really an answer on his question.

Quote
You need to have enough bandwidth to allow what would be the equivalent of the nyquist frequency.


Yes I agree, but Shannon shows that when you use multilevel transmission with the same baudrate and hence a higher bit rate, it is still the same bandwidth. The bandwidth is determinde by the baud rate, not the bit rate.

So in 100 Hz bandwith you can push 100 bits per second when using on off keying a carrier, but when you use not on/off but 4 transmisson amplitudes you are already on 200 bits ber second with the same baud rate (shortest element time) of 100 baud. Every time you double the number of possible levels (amplitude or phase) you can add a bit extra in that signal element. However because there is noise (and quantum effects) you can not keep adding bits in a signal element by doubling levels, because the signal noise ratio makes differing between them error prone, as Shannon shows. So Nyquist limits the number of signal elements in a certain bandwidth and Shannon limits the channel capacity that you can increase by more levels in a signal element, due to signal noise ratio.

Quote

That would work if you reduced the keying rate enough to accommodate the recovery of the information.
You have no doubt used a very narrow filter (say DSP) and noticed that it becomes very difficult to differentiate dits and dahs as you go narrower.

Yes, that is what I mean when I say that it is no longer on/off but il looks like you fade away and increase the signal strength with a potmeter. The envelope of a series of dots becomes a sinusoid, hard to detect on ear.

Quote
Some of that is due to the filter ringing but it also is caused by the lack of "sidebands" being received.[

I agree. It has no sense to sent any more spectrum components then the filter in the receiver of the QSO partner passes through. So, with the mentioned 5 ms slope after a 200 Hz IF filter, is has no sense to transmit less then 5 ms slopes, because the additional wider spectrum you transmit does not improve the readablity at the receiver side, when he has a 200 Hz wide CW filter. After (product)detection it is 5 ms slope.

However when you receive a signal and you limit it in bandwidth to the utmost, such that a dot is a half sine wave, you still can reproduce the wide original spectrum of the key at the sending station, bij putting a limiter after the detector with decison level the half amplitude of the received signal. Out of your limiter is then coming a signal with extreme steep slopes and the same width as the original dots and dashes were. That is because the limiter generates the by the IF filter cut off sidebands.

Quote
What you are suggesting is a form of "decimation" which is used in DSP when oversampled signals are passed through a low pass filter.
Then they are downsampled to make computation easier.

No not so. No sampling at all, just remaking a block that is transformed to a sinewave in transit by just limiting the spectrum to the first harmonic.  Remodelling by  a limiter, to the same block as the original block was, while the transmission bandwith is limited to the first harmonic of the block. Normally when you only give a user the first harmonic, nothing else, he cannot know the envelope of the original signal. However with Morse code we KNOW it was  a block, not a sawtooth  (ramp), Dirac, Gaussian pulse or what have you.

So take a morse signal (string of dots) , bandlimit it with cut off frequency  the first harmonic of the dot frequency, and you get a signal with slopes that are a dotmark wide. Hardly to detect by ear, however put it through a limiter with dicisionlevel half the amplitude of the incoming bandlimited signal and you get your original infinite steep sloped signal back with the same width of dits dahs and spaces. That is because the information it was a block at the tx is known, that is no information, when that was not known we cannot know the original slope, and form of pulses at the transmitting side because we have only the first harmonic available.  But because we KNOW it is a block we can recover the original Morse code from the first harmonic.


Quote
But this still requires the maximum frequency component to be half the nyquist frequency for this signal.
If you have frequency components greater than this frequency they will be lost and you will have aliasing errors.
This information is lost, and you will not recover it later as you suggest.

I recover no information, the timepoints of the original on/off points are passed by the zero crossings of the first harmonic, and that is all I need to recover the infinite wide spectrum of the original Morse code, because the fact that is was a block is no information transmitted, it is known information.

Simple: I have a 50 Hz block, It has harmonics further the 1 kHz. I wanted to transmit it. It suffices to sent a 50 Hz sinewave, because the receiver can, knowing it was a block and not a 50 Hz ramp , recover it with a limiter.
Quote
A morse signal is really just a digital signal and the same theories which apply to detection of digital signals apply to morse in the end.
I understand the concept of detecting digital signals with threshold detectors etc.
But you still need a particular bandwidth cable, otherwise what you suggest would mean you could detect Gigabit ethernet packets on rubbish cable by just detecting a narrow section of the wide bandwidth required.

Yes morse is a digital , even binary, signal, and when you limit the bandwidth and keep it 2 level, the minimum bandwidth required is half the number of dits per second according to Nyquist and Shannon for 2 level signal elements. So a 30 wpm morsesignal which has a baudrate of 25 baud, can be transmitted in 12.5 Hz bandwidth. However when you do, you cannot detect it on ear. You need faster slopes in order to do that. You can recover the fast slopes with a limiter.

Quote
Another way of looking at it is like a power supply low pass filter.
As you reduce the frequency spectrum you go from 60/120hz ripple to D.C.
The DC has lost the "information" in the ripple.

Yes you limit the filter bandwidth below the fundamental frequency of 60 Hz. When you omit the fundamental you cannot recover it.
Quote
I agree that you don't always need the full bandwidth, take SSB for example, but there is always a minimum bandwidth required for a rate of information transfer.
If you have devised some way of defeating information theory, then please patent it quickly, and put it next to your perpetual motion machine.

Haha did I claim a perpetual moving machine?
You have two kinds: of the first and second order, the first order is just moving, but you cannot distract energy of it and the second order keeps moving even when you distract energy from them.

73 - Bob


« Last Edit: August 27, 2012, 09:28:02 AM by PA0BLAH » Logged
STAYVERTICAL
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« Reply #9 on: August 27, 2012, 02:34:24 PM »

I feel sorry for 5VNN.

He must be hitting his head against a wall wondering how his simple question led to this avalanche of theoretical information and debate.
But that is good.

In previous era's debate was used in public places to find answers.
Many minds sifting facts and ideas, leads to understanding and common ground when executed well.

I think 5VNN has no doubt about why CW produces sidebands now (HI).

73 - Rob
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PA0BLAH
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« Reply #10 on: August 28, 2012, 04:14:13 AM »


The only other thing I can think is a morse transmission requires 100% duty cycle, and some tone is used to modulate the carrier and only the resulting sideband is sent?  If this is the case, then if you tuned off center freq a bit, wouldnt that change the perceived pitch of the dots/dashes?

ugh, thanks

Still not answered, I am afraid.
Do not understand completely your question.
What you mentioned here is MCW modulated CW, as an on/off tone, modulated on a carrier.


The on/off tone has itself a spectrum like a bare RF CW signal, kind of irregular block wave AM modulated on a carrier. So when the tone is 600 Hz it has itself little sidebands left and right of 600 Hz because it is an AM (on/off) modulated tone of 600 Hz. When you modulate a carrier in a hamband AM with that signal, the spectrum appears to be that carrier en left and right of it (lower en higher freq) appear at 600 Hz offset the tone carrier with his sidebands.

So with a 200 Hz wide receiver you find regular CW in the lower sideband, tuning upwards: the sole carrier, tuning upwards further the higher sideband which is a regular CW signal.

So calling CQ in MCW mode with a tone of 600 Hz can yield 2 amateurs thinking it is a regular CW signal and one calling back 600 Hz lower and the other ham 600 Hz higher then your carrier frequency.

When your receiver is 3 kHz wide, and in the AM detection mode (no local BFO) you hear 600 Hz CW because the 2 sidebands at + and - 600 Hz of the carrier are detected  with the carrier as interference source. So mistuning the receiver has no effect on the pitch because the distance between sidebands and carrier is not changed by mistuning.

Good luck, We have a scientific hobby, that is something else as buying appliances and playing CB at HF, so go on for extra, try to learn CW, in order to meet non CB appliance operators an go up first for extra.

BTW: congrats with your just acquired general!

Bob
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VA7CPC
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« Reply #11 on: August 28, 2012, 02:41:51 PM »

When I read the original post, my first reaction was:

. . . He's thinking about _receiving_ CW -- which can be done either in USB or LSB.

If that's what the OP was asking about, we don't need Claude Shannon to answer the question.

           Charles
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AK4YA
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« Reply #12 on: August 28, 2012, 06:26:27 PM »

thanks everyone, im barely hanging in on the discussion.  But I have gotten my main questions answered.  I was already vaguely familiar with Shannon and his equation for max BW, and I knew that you cant convey info with zero BW.  What I didnt get was how on/off dits and dots created the sidebands, and so now that is much clearer as far as how CW constitutes sidebands and thus occupies bandwidth.

This discussion has raised more questions though, but basically, the faster you key, the further from the carrier the sidebands are, and thus the more frequency bandwidth the signal occupies?  How much power is in the carrier, and how much are in the sidebands?  If those last 2 dont make any sense to you then obviously im confused.
« Last Edit: August 28, 2012, 06:28:36 PM by KC5VNN » Logged
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« Reply #13 on: August 28, 2012, 09:36:17 PM »

Ok, let's let Claude rest in peace.

You may already know this, since it is bread and butter stuff, so pardon me if it is redundant.
But, it looks to me, that where the difficulty lies, is in thinking about the modulation in two different domains.

In the time domain, we see square waves, pure and simple - nothing more complicated.
But if you look in the frequency domain, we see a series of sine waves of different amplitudes.

The most common way of translating between the two is using fourier analysis(or Fast Fourier Transforms for ease of use).
If you sample the Morse keying waveform, and apply an FFT to the resulting samples, you will obtain an output which is frequency versus amplitude.
There are certain issues of course with sample size, integer lengths, sample rate etc, but in the end you will see what a spectrum analyser sees.

So forgetting all the other theory, the equivalence of time and frequency domains is what can explain a series of morse pulses just as it can any other waveform.

The web is full of good tutorials and many ready-to-run programs which allow you to experiment with the FFT and other DSP techniques.
A very good starting point is "The Scientists and Engineers guide to Digital Signal Processing" at www.dspguide.com

There are links on this site to read the entire book online or you can download each chapter in PDF format - free and legal.

I would strongly suggest you look at this or other resources on DSP, as with computers being so common, you can learn and experiment easily.

73 - Rob
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PA0BLAH
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« Reply #14 on: August 29, 2012, 01:13:50 AM »

Rob

Internet is a "mere a boire", but I do not think it will be helpfull to complicate the matter and the path to understanding by adding in the explanation the sampling phenomenon. The book you mention is also mentioned by the designer of Pick 'n Star , transceiver. A design published in the RSGB (the companion ham organisation of Great Britain like the ARRL in the States) Pick n Star is really home building, not home brewing, because the design is done by the author. And no kit available in order to assemble without understanding.

KC5VNN came up with another question, how much power is in the sidebands of the CW.

Answer:

Suppose you sent continuously dots, hence mark/space ratio 1:1 on and off.
In case of 'on' the amplitude is mentioned A [volt] effective value.
The switched carrier has an amplitude of 0,5 A and all the sidebands together add up the other half A during on time and compensate the carrier during off time.
So all the sidebands together have the same power as the carrier. The carrier power is A.A/4, hence the sidebands together are A.A/4. All the power in the dotting CW signal is hence A.A/2.

That is during on time of the CW A.A and during 'off' zero, hence in the average A.A/2.

Yes, faster keying puts the sidebands wider apart, and I calculated that with a 200 Hz wide CW filter your speed may not pass 24 wpm, When you go faster the slope on the edges of the CW signal is decreased from 5 to 15 millisecond, due to the fact the 5-th harmonic of the keying signal can't pass the filter anymore.

---

What Charles VA7CPC said about upper and lower sideband CW is also a confusing subject for me.

I suppose it is about receiving CW on the same frequency as you sent your CW, but the filter of the receiver, when in SSB 2.8 kHz wide, passes a lot of QRM. The switch LSB-CW to USB-CW puts the filter passband in the bulkpart higher or lower then the receiving CW signal, so that you have the choice between receiving QRM lower or QRM higher than the received CW signal. The receiving CW signal keeps the same frequency on your freq scale, but the BFO is 600 Hz higher or 600 Hz lower, and also the  2.8 kHz filter pass band is shifted.  

Is there a flaw in the meaning of this description Charles?

When my explanation  is true, it has no sense to switch LSB/USB when you are using a 200 Hz wide CW filter. However when you turn your tuning knob slightly to a higher frequency the pitch of the received CW will change higher or lower dependent on the LSB USB switch position.
« Last Edit: August 29, 2012, 01:37:43 AM by PA0BLAH » Logged
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