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eHam Forums => CW => Topic started by: AK4YA on August 26, 2012, 07:59:07 PM



Title: what exactly *IS* CW... and how are there sidebands?
Post by: AK4YA 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: KH6AQ 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.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: W8JI 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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



Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL 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



Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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




Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: VA7CPC 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: AK4YA 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.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL 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


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH 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.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL on August 29, 2012, 03:12:55 AM
Bob,

Gaining an understanding of Fourier analysis will lead to real understanding.
This is equivalent to teaching a man to fish, rather than giving him a fish.

You can explain how a CW waveform has sidebands, but what about PSK31, or SSB, or FSK.....

This is why I always look at directing people to more general concepts and techniques.

I had a friend who worked for a company which repaired certain electronic equipment.
He was a guy who refused to learn the basics.
Day after day, he would come to me and ask me how a different circuit worked, so he could fix it.
I tried to teach him fundamental electronics, but he was not motivated to do so, and eventually lost that job.
Without those basics, every new circuit diagram was foreign to him.

My technique of learning anything is to learn the fundamentals first, and then you can understand the implementation easily.

This is why I suggest learning, or at least playing with FFT's.
It is like explaining how phase shift occurs in capacitors for example.
You could try explaining it by talking about charge building up and so on, or you could just use differential calculus.
For example,  the rate of change of voltage determines the current flow.
So if you apply a sine wave sin(theta) to a capacitor, the first derivative (rate of change), of Sin(theta) is Cos(theta) - a 90 degree change.

Not only does it simply and elegantly explain the phase difference, but it can also be applied to other waveforms.
This is why I suggest general means of understanding electronics - rather than special cases.
It is easier, more complete, and teaches the concept without making it a one off case.

We all learn differently, but I am by no means a math wizard - so if I can do it, anyone can.
Oh, and if you learn everything in that book I suggested, you will certainly be very competent in DSP techniques.

I guess, in the end, we are all different, and learn differently as well.
So whether we use empirical, mathematical or voodoo techniques to learn is not important or better - as long as we reach understanding.

73 - Rob


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 04:06:40 AM
Bob,

Gaining an understanding of Fourier analysis will lead to real understanding.

Mwah, all those integral transformations are nothing else than expanding the signal in a sum of terms of an infinite set of orthogonal functions. Why Fourier, that does it with a set of sine waves sin(nwt+fie(n)) as terms with n a natural number of zero to infinite. You can do it with an infinite orthogonal set of Walsh functions, they are binary, better for the understanding of digital signals. I handled once an Antenna problem with the expansion in an orthogonal set of Bessel functions.

When you do it by FFT you have to sample the signal first, and that makes the spectrum infinite wide, so your FFT is actually repetitive. Fancy windows like Hannah have to be introduced to mask the introduced errors.

Do it then with the Z-transform, that maps the left half plane of the complex plane to the inner of a circle, and projects the infinite number you generate by sampling  exactly over each other in the Z-plane.
Quote

This is equivalent to teaching a man to fish, rather than giving him a fish.

You have to give him a fish in order not to bring him in starvation, during your teaching

Quote
You can explain how a CW waveform has sidebands, but what about PSK31, or SSB, or FSK.....
No problem. FSK is just the sum of the AM spectra of 2 transmitters the frequency shift apart, that are keyed a "baud"time different in time. So when the one is on the other is off.

PSK31 is a 2 tone signal, with the smallest possible bandwidth. Phase modulation with 180 degree difference between mark and space, hence also the spectrum of 2 AM transmitters keyed on and off, however on the same frequency and with carriers 180 degrees out of phase.

You don't need FFT when you just take simple signals, like the carrier is cos(wt) and the audio is cos(ut) for DSB suppressed carrier,  or (1+cos(ut)) for AM. Multiply them and look at the result. cos(a)cos(b)=0.5[cos(a+b)+cos(a-b)] Just  math from school learned when I was 15, just before I was forced to drop out, because I challenged the teachers with poisoned questions.

Rob, You have to go to sleep right now when you live somewhere in Australia.

Bob


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL on August 29, 2012, 04:25:36 AM
Bob,

I am glad you finally see the utility of mathematical techniques.
If it is good enough for you, it should be good enough for others.
I feel I have achieved something there.

No, I never need much sleep.
I work on a 24 hour sliding sleep cycle.
I am a fan of the late Buckminster Fuller, who pioneered this technique.
His wife finally made him give it up, because she could not take him sleeping and being awake all around the clock.
But I don't have that problem, so expect me to post anytime around the clock.

As regards leaving school at 15 because of poisonous questioning of your teachers - I am astounded at that.
There is nothing in your numerous posts to indicate that personality trait.

On to the next joust I guess.

73 - Rob


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 05:38:39 AM
Bob,

I am glad you finally see the utility of mathematical techniques.

That is not finally Rob, I see that very well, where the brain fails to make complex logic steps, math is the tool, and where the brain is too slow, the computer is the  tool. All programmed (math AND the computer) by the slow brain formulating simple axioms and stepstoning on them .  But I think somebody with quite another job then mathematics and electronics involved, has not to get the lesson that sounds: "First your math, and after that you can enjoy your hobby".

I am glad to hear that a ham that just got his general, is questioning about how and what, so I think it is best to show him in a way he hopefully can understand, how and what.

Forums here have the severe and overwhelming  disadvantage, and are due to that in effect worthless, because everybody that want to show himself (like a rooster) tells his kukelekuu (dutch rooster) his quoquelicot (french rooster) nothing to say as : here is the sperm gland, come to me hopefully you evaluate at the long term to a PA0BLAH alike. (Passing on the route Stayvertical-alike hahaha)

Quote

If it is good enough for you, it should be good enough for others.
I feel I have achieved something there.

Pity that you are so far away, we could be closed friends.
Insulting each other at the utmost, not being hurt.

What in fact is insulting?
When I say you are a lousy snake, you will be not be insulted, because you know you are not a lousy snake.

So people can't insult me, when they try, because I know they are wrong.

However when you are a rooster with only air and feathers trying to impress the environment, I can insult you by saying you are only a rooster with nothing as air and feathers to show and impress the environment.

I saw you editted your msg, before I replied. So you don't exactly write, finally, what you thought at the moment of writing, that is a pity. But I read it as first published before the edit. So I appreciate your personality more before then after the edit.

Was it not you that opened my eyes when you  said persons learn different? You are a retired professional teacher in adult courses in Australia, I suppose. Well I hate teachers, I am at least 5 times faster learning from a book. May be infinite, because I never learned something aural from a teacher. It must be an extreme handicap when you can only learn something in conversation with a teacher. That handicapped persons were your gratefully students.

BoB



Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL on August 29, 2012, 06:10:05 AM
I am glad we could be friends Bob,
Yes, I changed my first edit, because, in the end, I did not want to be too hurtful.
It is not the first time I have done that, and frankly I misread your comments.
I learned many years ago, that not admitting a mistake is just piling error upon error.

It is sometimes difficult to infer your meaning, although I do respect your skills in english.
So sometimes it takes a few readings to get to the core of what you are saying.

People who cannot admit to being wrong, are doomed to a life of misery.
In the end, what we express here are opinions.
They are not facts which we brought into existence - they are merely facts we have learned from others who came before.

Where we can be truly creative is in explaining things to others in ways which help them to learn.
You obviously have a wide range of technical skills, and it would be a pity to waste them by clothing them in a cloak of sarcasm.

Belittling people is not helping them, and appears more like bullying than elmering.
And I will let the statement about teaching handicapped people pass, but say it is in very poor taste.
A society is judged by how it treats those who are weakest within it, and is a gauge of compassion.

You have much to offer those of us who need the skills you have gained in your lifetime.
But the price should not be ridicule and poisonous posts.

I, for one, would be very interested in seeing your knowledge used to elmer people who could use your skills.
But, perhaps you could leave the sarcasm and ridicule out of the equation, since this only turns off the lights, not lights the way.

73 Friend Bob,

- Rob




Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: AK4YA on August 29, 2012, 06:58:05 AM
<SNIP>
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.
<SNIP>
73 - Rob
EURIKA of course.  this is where my disconnect was.  I always read and knew that a square wave was the sum of a carrier and its odd harmonics, I just never put 2 and 2 together to understand the big picture.

My next goal is to understand how the sidebands and carrier are demodulated in a BFO type of receiver, specifically, wouldnt faster morse code transmissions have a higher pitch than slower ones, since they are demodulated with a product betector/BFO, just like SSB is demodulated?


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 08:21:00 AM

People who cannot admit to being wrong, are doomed to a life of misery.

The priests and ayatollahs of different tasting churches will not be happy with this statement. But they,  in name of their supreme,  forgive you when you are willful to join their only true church. It remembers me of the photoshopped away expensive watch, forgetting the accompanying glance of the mirror image in the table surface,  of the leader of some church.

Quote
In the end, what we express here are opinions.
They are not facts which we brought into existence - they are merely facts we have learned from others who came before.
Where we can be truly creative is in explaining things to others in ways which help them to learn.

Of course you will in order to be able to be proud of your skills and your former profession, and you have obviously skills, which I conclude reading the last post op topic starter, but realise that not those learned facts are the basis of our present world, but the creative brains, that created the manmade world as you see it around you, based on the scientific facts that are presented and understood.

Quote
You obviously have a wide range of technical skills, and it would be a pity to waste them by clothing them in a cloak of sarcasm.

I didn't waste them, certainly not, but my professional environment were always people of around the same or higher level, (because I am pretty stupid) so when you are retired, you fall back in a world of roosters, and that give some reaction, as noticed by you. A long lasting reaction, that is true, because I left my profession longer ago then I worked in it.


Quote
Belittling people is not helping them, and appears more like bullying than elmering.

I do not see that. When a guy asks me how much is  2^10, I say : 1024 decimal or  are your numbers hexadecimal? In that case is the answer 65536 decimal, SUCKER. Ask me again when they are octal based.

So he knows I find him a sucker, AND he knows the correct answer on his question.

It is the price he pays me for being lazy and not to find out for himself without questioning on a forum, and the answer is for free. May be it encourages the next time to find out for himself.


Quote
And I will let the statement about teaching handicapped people pass, but say it is in very poor taste.
A society is judged by how it treats those who are weakest within it, and is a gauge of compassion.

Come to Holland, There was a time that we passed the amount one million inhabitants having benefits from  the law of the too disabled to work.  And we have about 6 million people in the age able to work, so one out of 6 is disabled.

Bob



Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: K8AXW on August 29, 2012, 10:19:06 AM
STAYVERTICAL:

Quote
Where we can be truly creative is in explaining things to others in ways which help them to learn.
You obviously have a wide range of technical skills, and it would be a pity to waste them by clothing them in a cloak of sarcasm.

Belittling people is not helping them, and appears more like bullying than elmering.
And I will let the statement about teaching handicapped people pass, but say it is in very poor taste.
A society is judged by how it treats those who are weakest within it, and is a gauge of compassion.

You have much to offer those of us who need the skills you have gained in your lifetime.
But the price should not be ridicule and poisonous posts.

I, for one, would be very interested in seeing your knowledge used to elmer people who could use your skills.
But, perhaps you could leave the sarcasm and ridicule out of the equation, since this only turns off the lights, not lights the way.

73 Friend Bob,

- Rob



Very well written Rob! 

PA0BLAH:  I've followed your many posts here on this thread and see that you're a very well educated (school or self taught - Irrelevant) man with excellent English skills.  Yet you prefer to act as you do which serves no purpose except to alienate people.  There's no damned excuse for doing this and your reasoning for doing so simply doesn't fly!


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: K7KBN on August 29, 2012, 12:11:06 PM
<SNIP>
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.
<SNIP>
73 - Rob
EURIKA of course.  this is where my disconnect was.  I always read and knew that a square wave was the sum of a carrier and its odd harmonics, I just never put 2 and 2 together to understand the big picture.

My next goal is to understand how the sidebands and carrier are demodulated in a BFO type of receiver, specifically, wouldnt faster morse code transmissions have a higher pitch than slower ones, since they are demodulated with a product betector/BFO, just like SSB is demodulated?

The frequency of the RF signal being received doesn't change.  If your receiver is stable and receiving a signal at 14007.38 KHz, AND if the sending station operator doesn't bump his frequency control, AND if you don't bump your tuning or BFO controls, the tone of the dits and dahs won't change, whether the sending station is keying at 5WPM and then increases to 50WPM.

Now, if one station is in a very fast aircraft, you might be able to detect some doppler ...


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 12:39:19 PM


PA0BLAH:  I've followed your many posts here on this thread and see that you're a very well educated (school or self taught - Irrelevant) man with excellent English skills.  Yet you prefer to act as you do which serves no purpose except to alienate people.  There's no damned excuse for doing this and your reasoning for doing so simply doesn't fly!

Al,  you are without doubt a nice guy, a guy to drink a glass of beer and to talk about our past life, but when you use your computer you have to behave yourself the way the "stupid" computer likes, just touch the right keys in the right sequence and the computer does what you want him (must be a him nor a her)   to do. I am the computer and you did not touch the right keys in the right sequence. That's all. But not the slightest problem, so try to insult me or push me in heaven (with 7 times 7 virgins) , doesn't make any difference in handling your msgs.
 
Sometimes I think:"Why do I explicit and only communicate in CW in the hambands?" I am starting to understand that better since I am writing in the CW chapter of this forum.

Bob


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 01:03:41 PM

My next goal is to understand how the sidebands and carrier are demodulated in a BFO type of receiver, specifically, wouldnt faster morse code transmissions have a higher pitch than slower ones, since they are demodulated with a product betector/BFO, just like SSB is demodulated?

The BFO and the CW carrier difference in frequency give you the desired tone. Say 600 Hz. The mixing process if pure multiplying 2 signals, is linear, that means the whole CW spectrum, including his sidebands, are transformed to 600 Hz audio and the 2 sidebands look for the on off keying of the 600 Hz audio. So faster CW with more space between its sidebands yields faster 600 Hz switching because the 600 Hz has wider spread sidebands.

Stayvertical (who is right now laying horizontal) can translate it in understandable language, when you can't understand  this.

In earlier times (in Dutch: Vroegah)  they detected with a BFO and a diode detector. Then however the strongest signal is the carrier, and that may be a QRM signal. So product detection is surely better.

BoB
( You know the joke of Brigitte Bardot with on her left and right buttock a B , her initials,  tattooed?)


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: STAYVERTICAL on August 29, 2012, 01:52:32 PM
Bob,

I am sorry to see that you refuse to do anything except cause dissent.
Your gaining of knowledge is like a fine meal, prepared meticulously, and then dumping a bucket of mustard on top.

Knowledge is only useful if it is transferred to others, or used to improve your own life.

I understand that you see yourself as a liberator of those oppressed by inflated roosters.
Unfortunately, it appears that you are like the legendary ghost ship, the Flying Dutchman, doomed to sail the seas forever in your quest.

However, despite your frequent rages against various religions, I will have to take the advice of one.
Paraphrasing, it goes something like "If your teaching is not accepted, shake the dust off your shoes as you leave".

This thread has been hijacked long enough, and in fairness to the original poster, I will retire from the ring.

I am certain, you will have enough in this post to write a three page scornful post about it, so enjoy yourself.
It appears that you are on a personal crusade against shadows from your past - and I wish you well in your campaign.

73 - Rob


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 29, 2012, 03:45:14 PM
Bob,

Your gaining of knowledge is like a fine meal, prepared meticulously, and then dumping a bucket of mustard on top.

Actually this remembers me on the situation in a fine food French restaurant (don't use the toilet facilities) of an American guy eating only with a fork, left hand under the table, that drops half a bottle of tomato ketchup on his served exclusive fine meal.

Quote
Knowledge is only useful if it is transferred to others, or used to improve your own life.

I understand that you see yourself as a liberator of those oppressed by inflated roosters.
Unfortunately, it appears that you are like the legendary ghost ship, the Flying Dutchman, doomed to sail the seas forever in your quest.

Amazing that other people that know nothing about me, and write that they are  not believing what I write, such as being ham for 65 years, living in Holland, and pounding only the brass, write that they think they know more then I write.
Quote
However, despite your frequent rages against various religions,
I only reacted on your statement that people that wouldn't admit their wrong thinking are destined to live in misery, something like that. And because there are a large number of religions, and splitting offs, and all say they are the whole and only sole truth, I only can conclude with the knowledge I share now and here, and sharing that is the value of knowledge as you just said, that at least all of them minus one must be wrong. They know that, must be,  do not admit that fact, and still dont live in misery, as I adstructed with a recent example.

They call this kind of reasoning "Mathematical Logic" (in order to distinguish if from "Female Logic" I suppose)

Quote
Paraphrasing, it goes something like "If your teaching is not accepted, shake the dust off your shoes as you leave".

May be that statement is valid when you try to teach an opinion, or religion,  hence indoctrination, not when you teach the unchanging laws of nature and all her consequences.
 
Quote
This thread has been hijacked long enough, and in fairness to the original poster, I will retire from the ring.
Well Topic starter got his answers, I guarantee they are correct, and I am not going to another forum chapter, in order to meet the no code appliance Cb-ers overthere with their alfa bravo  chit chat.
Quote
I am certain, you will have enough in this post to write a three page scornful post about it, so enjoy yourself.
It appears that you are on a personal crusade against shadows from your past - and I wish you well in your campaign.
No not at all. I am just reading and answering if it suits me.
CUL Rob best 73 Bob

73 - Rob
[/quote]


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: K8AXW on August 30, 2012, 09:07:49 AM
PA0BLAH:

Yes, I am a nice guy to drink a beer with.  I've drank beer with many people, including Dutch and former enemies in Germany.  However, I would never knowingly drink a beer with you.  I socialize only with people I respect.  After seeing that you are capable of holding a serious and intelligent conversation and then go back to your insulting ways, I have no respect for you.

As for my computer ..."but when you use your computer you have to behave yourself the way the "stupid" computer likes, just touch the right keys in the right sequence and the computer does what you want him (must be a him nor a her)   to do. I am the computer and you did not touch the right keys in the right sequence."

I don't know what that means.  I can tell you that one use of my computer here on eHam is to try to help others, not insult or demean them.  As STAYVERITCAL says, it's a shame when you have such great knowledge but don't use it to help but instead to insult. (Paraphrased)

You make many nasty references to CB operators Bob but you are exactly the same! You use a "handle" (PA0BLAH) which is not a government issued callsign and probably a fake name (Bob) to hide behind so you can spread your hate and poison without anyone knowing who you really are.  Or is it because down deep you feel some shame?

I'm sure you will answer this so please when you do, give it your best shot.  This will be my last to you.  AR VA CL



Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 30, 2012, 10:05:22 AM
PA0BLAH:

Yes, I am a nice guy to drink a beer with.  I've drank beer with many people, including Dutch and former enemies in Germany.


Just as Overste Karremans (some superhigh military rank) with his shy dog-eyes and martial mustache , commander of "Dutch bat" that drank a beer with Mladic, while 8000 males were separated from their family and executed in Sebrenica?

Quote
However, I would never knowingly drink a beer with you.  I socialize only with people I respect.  After seeing that you are capable of holding a serious and intelligent conversation and then go back to your insulting ways, I have no respect for you.

That is your fundamental right Al. No problem, only for yourself, because it turns out you can not handle people that are different in the way  you expect them to be, and  above that they refuse to be bended in the way you want them to be, in order to like them. They call that different "personalities", I suppose. Not everybody in the world can be a person respected by you, but don't forget: your respect is a time variant behavior when you drink a beer with former enemies, as you said.  

It must be comfortable to have no talents, then you can't waste them anyway. So those people  have always a position to stimulate other people to be envy on them. 40 year in an 8 hour shift copying Morse number stations, retirement benefits and that's it.

Quote
As for my computer ..."but when you use your computer you have to behave yourself the way the "stupid" computer likes, just touch the right keys in the right sequence and the computer does what you want him (must be a him nor a her)   to do. I am the computer and you did not touch the right keys in the right sequence."

I don't know what that means.

Quite simple Al, When you (try to) kick me under the ass, like you tried a few times, you just get it back. No problem no bad feelings whatsoever.

I actually enjoy myself with the primitive completely forcastable  results of my injected messages on this forum. People are obviously non linear systems. When you kick them they react on the expected way,  when you kick them twice as hard the result is not double the kicking force and additional when you give a number of kicks, spread in time,  the result is not the sum of the time-shifted time functions of each kick.  So thanks Al for proofing that. Have you any objection to show up in an article, with your call and QRZ published data, handling this kind of behavior? (no reaction is understood as no objection) Stay vertical can explain you that you are  a time variant non linear system to be described as a set of non linear differential equations with non constant coefficients . (When he has understood the over simplified book he advices to read in order to become  a DSP expert, hi)

Bob


 


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: AK4YA on August 30, 2012, 03:13:32 PM
Please knock it off guys.  Some of you are acting like a bunch of little girls.  I do appreciate the answers I am getting here, 1 last topic:

I have been assuming that each odd harmonic was a sideband, but after thinking about it and the Fourier summation for a square wave, that cannot be correct.  So tell me if this is correct:

1. The rise/fall shape determines how many harmonics are created
2. Each harmonic ends up 100% AM modulating the carrier to create 2 sidebands, 1 above and 1 below the carrier.

Are those 2 correct?

And, for some reason, at the minumum, the 5th harmonic is the lowest harmonic you want.  Im not sure why.  Im sure it has something to do with the squareness of the waveform but I cannot find any math for that.


Title: RE: what exactly *IS* CW... and how are there sidebands?
Post by: PA0BLAH on August 30, 2012, 04:32:18 PM
Please knock it off guys.  Some of you are acting like a bunch of little girls.  I do appreciate the answers I am getting here, 1 last topic:

I have been assuming that each odd harmonic was a sideband, but after thinking about it and the Fourier summation for a square wave, that cannot be correct.  So tell me if this is correct:

1. The rise/fall shape determines how many harmonics are created
Correct. The blockwave consist of a fundamental also mentioned first harmonic and higher harmonics. The first harmonic is somewhat larger then the block amplitude.  The n-th harmonic has amplitude 1/n. for n odd,

All those harmonics you find left AND right of the carrier as sidebands. Not every sideband is separately modulating the carrier 100% because they are smaller, However together they do that,
Quote
2. Each harmonic ends up 100% AM modulating the carrier to create 2 sidebands, 1 above and 1 below the carrier.

Are those 2 correct?
One harmonic , the  fifth, has amplitude somewhat more then 20% of the block amplitude, so the upper and lower sideband will be both 10% , and the modulation depth of that 5 th harmonic only 20%
Quote
And, for some reason, at the minumum, the 5th harmonic is the lowest harmonic you want.  Im not sure why.  Im sure it has something to do with the squareness of the waveform but I cannot find any math for that.

Draw on a piece of graph paper 3 sinewaves, sin x ,  0.33 sin 3x and   0.2 sin 5x ; for x 0 to 360  degree, add on each time the 3 momentary values and you get your block like signal with moderate slope. You can also calculate the sum and draw the result for every x.