I remember being surprised to learn that a CW signal has a bandwidth (albeit small), but when I thought it over, it made sense. Essentially, we are modulating the carrier with a low frequency square wave. This would explain key clicks: a perfect square wave has infinite bandwidth, but if you smooth the rising and falling edges, the bandwidth goes down.

This got me thinking. If, theoretically, we could key up and key down precisely on the zero-crossings of the carrier, it seems to me that there would not be any distinct rising or falling edge. Then, would the signal have zero bandwidth?

(I haven't the slightest idea whether this would be practical in practice. I have to assume either a) it's not or b) I'm wrong about this theory, because otherwise I'm sure transmitters would already do this.)

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    $\begingroup$ Good question. I've always guessed phase-coherent switching has a reduced bandwidth, but I've not seen a mathematical explanation as to why it's any better than non-coherent switching. $\endgroup$ Commented Oct 25, 2017 at 18:04
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    $\begingroup$ Turning on a carrier instantaneously, at a zero crossing or not, uses an enormous amount of bandwidth. Theoretically you could modulate the carrier in this way at every cycle, so the bandwidth of the resultant signal, if not filtered later, is about half the carrier freq. Big key clicks! A narrow band signal changes slowly! Imagine a 100 Hz wide CW signal - the carrier must go from Off to On slowly over about 1/100 s, which could be 100,000 carrier wave cycles. In this thought experiment, the exact phase of the modulation, relative to the carrier, is irrelevant. $\endgroup$
    – tomnexus
    Commented Oct 25, 2017 at 19:52
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    $\begingroup$ I asked a similar question over on the DSP.SE and got some interesting (but very mathematical and honestly still a bit beyond my own current grok!) answers: dsp.stackexchange.com/questions/31044/… $\endgroup$ Commented Oct 26, 2017 at 1:30
  • $\begingroup$ @natevw-AF7TB Thanks for the link. That looks like it would explain everything if only it weren't going over my head! My rusty, basic background in complex analysis helps, but it isn't really enough. Some more Googling might help, though. $\endgroup$ Commented Oct 26, 2017 at 6:25

6 Answers 6


You raise an excellent question and your thought processes are indeed on the right track.

First some background. An ideal, uninterrupted sinusoidal carrier has zero bandwidth. Real world factors such as phase noise, amplifier distortion, etc. produce a measurable bandwidth of the carrier. When the carrier is keyed on and off as it is with Morse code, this is a special form of ASK (Amplitude Shift Keying) called OOK (On Off Keying) as it is discussed in professional literature. Normally, the bandwidth of the modulated OOK signal is minimized when the keying waveform rise and fall times take on a Gaussian or raised cosine form. Most amateur radio transceivers are using simpler RC filtering to shape the keying waveform, resulting in less than ideal bandwidth but at least avoiding the key clicks from the otherwise sharp raise and fall time of the modulating signal. In any case, it is the rise and fall shape of the keying waveform that determines the bandwidth of the modulated signal.

When the transition of the modulating signal is synchronized with the zero crossing of the carrier signal, this is known as coherent OOK modulation. This is, of course, much more difficult to implement than a simple, non-coherent RC filtering scheme so it is not commonly seen in amateur radio applications. However, when implemented properly coherent OOK does reduce, but does not eliminate, the signal bandwidth. The "residual" bandwidth is largely due to a variety of real world factors including non-instantaneous keying time which introduces a distortion of the carrier waveform. Any distortion, no matter how small, of the sinusoidal carrier will result in a non-zero bandwidth.

[EDIT] Even with ideal zero crossing switching and ideal amplification, there will be minor sidebands but these will roll off as a function of frequency at a significantly greater rate than switching at non-zero points. With zero crossing switching, however, non-linear amplification present in most CW transmitters will likely introduce greater side band components than those caused by zero crossing switching. [/EDIT]

So you could win over some of the CW aficionados with a coherent OOK transmitter but in a hobby market the commercial viability of such a transmitter would be a question of price elasticity. Perhaps in this new era of DSP based radios, the lowered cost of implementation will make this commercially viable. But given that few, such as the Elecraft K3, amateur radio transceivers implement Gaussian keying, we still have a ways to go to optimize the bandwidth of our CW signals.

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    $\begingroup$ Surely the "residual bandwidth" is not just due to real-world imperfections as your fourth paragraph claims. A zero bandwidth would violate the Shannon–Hartley theorem which says that communicating any information at all requires nonzero bandwidth. $\endgroup$
    – Kevin Reid AG6YO
    Commented Oct 25, 2017 at 17:37
  • $\begingroup$ Thanks for such a thorough answer. It's interesting to know some of the other real-world factors that come into play.I am curious about what @KevinReidAG6YO asked, too. $\endgroup$ Commented Oct 25, 2017 at 18:00
  • $\begingroup$ Accidentally hit enter and missed the edit window for my last comment, so adding here. I'm not sure if this is a straightforward question, but if so, I'm kind of curious: When coherent OOK is implemented properly, what sort of bandwidth is attainable in the real world today? $\endgroup$ Commented Oct 25, 2017 at 18:11
  • $\begingroup$ @KevinReidAG6YO SHT does not apply at all here since it describes an analog receive channel and here we are talking about the transmitter. In any case, the SHT does not describe an absolute boundary but rather a relative threshold of an arbitrary low error rate. And a receiver channel bandwidth of zero is a boundary condition of SHT It is more instructive to think of what happens as the rcv channel bandwidth approaches zero. For example, at a 1 millihertz (0.001 Hertz) channel bandwidth, a 1000 bit/second rate threshold is described with only a 60 dB SNR. I can expand in my answer if needed. $\endgroup$
    – Glenn W9IQ
    Commented Oct 26, 2017 at 6:22
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    $\begingroup$ You don't need the SHT to explain how a CW signal with zero bandwidth is impossible. As the bandwidth approaches zero, the limiting case is a frequency domain becomes a Dirac delta function. The Fourier transform of that (getting us back into the time domain) is a sinc function, which has nonzero values out to infinity and is thus impossible to realize. $\endgroup$ Commented Oct 26, 2017 at 15:48

There is no way to transmit information in a signal with zero bandwidth. Switching the carrier at the zero crossings would reduce bandwidth but not take it to zero.

There's a mathematical explanation on DSP StackExchange: Does “keying on” a sine wave at a zero-crossing reduce its bandwidth? Summary: in the worst case of switching on/off at the peak instantaneous amplitude, the sidebands fall off in proportion to 1/f. In the best case of switching at zero crossings, the proportion is 1/f2. A significant improvement, but still far from zero bandwidth!

Without delving into math, the simple explanation is that a band-limited signal must not only be continuous itself, but so too must be its derivatives. A hard switch at a zero crossing has a discontinuity in the first derivative at the switching point.

For having smooth derivatives, the Gaussian function is a limiting case because the derivative of a Gaussian function is another Gaussian. The Gaussian function comes up in the central limit theorem for similar reasons.

It's generally true that resolving something precisely in the time domain requires a wide bandwidth and vice-versa, and the Gaussian function is a "middle ground" which maximizes the rise and fall times while minimizing sidebands to the extent mathematically possible. This is because the Fourier transform of a Gaussian is a Gaussian. So in this sense, the minimal bandwidth keying envelope for CW would be a Gaussian function.


The "Coherent CW" folks use something similar.

Very slow CW is keyed at zero crossings with Raised Cosine envelope to give minimal bandwidth.

But yeah, it's the rise and fall times which gives key-clicks, not whether it's keyed at zero crossings or not

  • $\begingroup$ +1 because you are correct about the risetimes. Hopefully those are filtered out in FCC Type-accepted commercial radios. Also the vote is for the very interesting topics raised. Can you please elaborate in this answer exactly what is meant by Coherent CW and Raised Cosine envelopes? $\endgroup$
    – SDsolar
    Commented Oct 29, 2017 at 0:43

Rob Sherwood NC0B of Sherwood Engineering Inc. did a study and produced a video showing that the rise time (and presumable the fall time too) is the key to reducing bandwidth. He compared rise times of 1 ms to 10 ms. Too fast of a rise time significantly increases bandwidth. To answer Dominick's question, I looked at the CTU (Contest University) files. Rob has mentioned this in nearly every one of his presentations every year he has spoken. THey can be found in the videos here or the files here.

  • $\begingroup$ Do you have a link to the video? I can't find it on that website (there are a lot of presentations there). Also, I don't want to make any judgements without seeing the video, but are you sure it is relevant to the question at hand? Slower rise times generally mean lower bandwidth--there's no question about that--but does he discuss what happens (or doesn't happen) if we time a transition to align precisely with the zero crossing? That's what this question is about. $\endgroup$ Commented Oct 26, 2017 at 15:41
  • $\begingroup$ Dominick, I will try to find the video. Also implicit in the question is whether or not there is a significant bandwidth reduction if you transition at the zero crossing. This answer directly deals with that aspect of the question. The answer is NO, it has more to do with rise time NOT zero crossing. $\endgroup$ Commented Oct 27, 2017 at 0:14
  • $\begingroup$ Ah, I see. Perhaps you want to clarify that in your answer? It wasn't clear that was what you meant, at least not to me. $\endgroup$ Commented Oct 27, 2017 at 0:59

First off, there is no such thing as a signal with zero bandwidth. That is like the imaginary line between two imaginary points - the line with no width between points of no size.

It is like saying isotropic antenna. (An FCC favorite)

There simply are no such things except as theoretical constructs - end even then, to prevent dividing by zero, they always must be expressed as limits of variables as they approach zero.

CW transmitters always begin at zero output (what you are calling a zero-crossing, but at the boundary condition) until they are commanded to produce a signal by the closure of a contact on a key or by other means. Then they return to zero when that output command is removed.

CW transmitters do not come up to full power instantaneously, and produce waveforms like this:

enter image description here

Note that as this is displayed in the time domain relative to the keying frequency, the actual carrier is not visible.

So even if the modulation itself (the key, acting as a switch) is a square wave, the resulting output is not truly square due to the physical constraints of the components of the transmitter.

This is true for both the rise and fall of the envelope.

Here is another example:

enter image description here

In this plot, relative to the carrier frequency, it is clearly seen that in a practical sense the transmitted wave envelope does indeed begin and end at the boundary conditions of zero.

Of course, philosophers have been known to argue that it never actually reaches zero - just gets closer and closer to it. Then the quantum physicists bring up the Planck limits, and so on, ad nauseaum.

In the frequency domain it will look like this:

enter image description here

Note that this is a plot of a signal modulated by 1 KHz square-waves, which is why the bandwidth is spread so widely. At 1 KHz keying, the ratio of signal to modulation frequencies are getting to the point where it is producing sidebands. The amplitudes of the sidebands fall off in the normal way. This plot does not show the effects of input filtering.

Consider a generic mixer circuit - the output should always be filtered in such a way that only the desired modulated carrier is the resultant output.

The risetime portion is as if it has been modulated at some very high rate. That produces spurious emissions in the driver circuit which are then filtered out. They should not be within the bandpass of the final power amplifier.

But since no filter is perfect some of those spurs will end up on the output. Hence the need for guidelines regarding limitation of spurious emissions. These are specifically regulated for FCC Type-Accepted commercial transmitters.

Ultimately, you want a transmitter that exhibits higher Q as a measure of how well it applies available power to the desired output frequency for higher efficiency.

Here is an interesting article on the subject (slightly TL;DR):

Effect of Keying Waveform on CW Bandwidth


the transfer function of a real transmitter is NOT linear.

i.e., a square-wave input (where the rise and fall time is as if the frequency of modulation is at the boundary condition of the limit of it being at infinity, where the limit of the period of modulation approaches zero.

Bottom line is that since there is no carrier between the Morse elements, the output always begins at zero and ends at zero.

Hand keying is at such a slow rate as compared to the carrier frequency that it is not very relevant.

The O-scope trace at the top is taken from a 630 Meter transmitter, and even then the modulation rate is still so slow compared to even that low of carrier frequency that there are no clear artifacts introduced by keying at random time relative to the carrier oscillations.

It is the modulation frequency that is the dominant factor in producing the final signal bandwidth.

As an aside, back in the day when the FCC had monitoring stations all over, I would visit sometimes and watch as they used what they called "TXID" - Transmitter identification - the way a signal from a CW or FM transmitter wanders in frequency and amplitude as they were keyed up. They were very good at this, and often could tell who was transmitting before any identifying modulation was applied. i.e, saying or keying a callsign.

They told me that even with the same models of transmitters made on the same production line on the same day there would be measurable differences due to the tolerances of the various components that went into the final product. They considered this to be closely-held information at the time.

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    $\begingroup$ I think the question is about the transition between carrier on and carrier off, is it not? $\endgroup$ Commented Oct 26, 2017 at 17:12
  • $\begingroup$ Correct, and the linked article goes on at length about that. I once built a small two-transistor transmitter on a breadboard where I keyed the actual oscillator, which by definition meant that it began at zero. It produced very noticeable chirp. So I was advised to change my switching method such that the oscillator continued to run in between the keying of the PA. The transition from power off to power on always begins at zero, and the oscillator continues to run at a stable frequency. I did not have any output filtering so it probably produced a lot of spurious "key clicks" $\endgroup$
    – SDsolar
    Commented Oct 26, 2017 at 17:37
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    $\begingroup$ It doesn't matter where you switch the carrier: the question asks about transmitted signal, not the implementation. At the end of a dah or dit, it can do so where the instantaneous amplitude is zero, or it can do it where the amplitude is non-zero. $\endgroup$ Commented Oct 26, 2017 at 17:37
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    $\begingroup$ Interesting aside about the TXID! Was that a specific device, or just something the operators had gotten good at themselves? (I added a cross reference to here from my answer about tracking unlicensed FRS/GMRS stations: ham.stackexchange.com/a/6492/1362) $\endgroup$ Commented Oct 26, 2017 at 17:39
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    $\begingroup$ Partly answering my own question I guess: looks like there was a device called the "MoTron TxID-1" that as of 1997 according to citeseerx.ist.psu.edu/viewdoc/… was "currently the only system used by spectrum management personnel in Canada and the United States for the purpose of identifying unlawfully operated transmitters". Some more discussion at kb9mwr.blogspot.com/2008/04/transmitter-fingerprinting.html too. $\endgroup$ Commented Oct 26, 2017 at 17:46

From a mathematical point of view, any finite length (e.g. keyed) signal has infinite bandwidth. You can reduce the side-bands by choosing a keying envelope whose FT has low sidebands, which implies slowed rise/fall times of the envelope and to a diminishing degree all its derivatives.


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