W5ALT CW Info
Is CW Dead
What is CW?
Is CW Digital?
Why CW Effciency?
Slowing Down a Bug
Bunnell Double Speed
TAC Hole-In-The-Wall Bug
Lionel J-36 Bug
by Walt Fair, Jr., W5ALT
We all know that CW is a narrow bandwidth mode, which is one of the reasons it is so effective under poor band conditions and low power. In fact I've heard some people think that CW has no bandwidth, because it is simply a carrier turned on and off and a carrier signal has no bandwidth. Well, it's not quite that simple! In this section we'll look at the subject of CW bandwidth and figure out what it usually is and what it might be.
CW is AM!
Although it is usually simplest to describe CW as turning on and off a carrier frequency, that isn't really a very good description from a signal point of view. Actually CW is 100% modulated AM. When the signal is turned off, its amplitude is exactly zero, and when it is turned on its amplitude is the carrier signal amplitude. So that means we can analyze a CW signal using the same techniques that are used to analyze AM signals.
Our first thought might be to evaluate a CW signal as if it were an AM signal modulated by a square wave with an amplitude of 1. That is the same as switching the carrier instantaneously on and off. As with any AM signal, we can write the resulting signal as a product of a carrier frequency and a modulating signal, in this case a square wave. But a square wave can also be represented in terms of sine functions as is shown in most basic signal processing books. In fact, a square wave is composed of an infinite series of sine waves all added together. The formula is the sum from zero to infinity of all of the odd harmonics of the modulating frequency, with a decreasing amplitude as the harmonic order increases. Now that may sound complicated, but what it means is that if we instantaneously turn on and off a carrier wave to generate CW, we end up with a signal that has an enormous bandwidth. If you listen to a signal actually generated that way, what we hear are terrible key-clicks, that represent all of the higher frequency components of the modulating square wave.
Surely that doesn't sound so good! We want a narrow band signal, but we find that simply turning on and off a carrier generates a wideband signal instead. In fact, the normal technical definition of bandwidth is the frequency range where the signal is above 30 dB of its peak value. Since the coefficients are equal to 4/Πk where k is an odd integer, the peak value is when k=1 (the fundamental modulating frequency) and the coefficients decrease with increasing values of k, the drop in magnitude of the harmonics is 30 dB < 20 log(k) or log(k) > 1.5, which implies k > 31. Recall in another section we found that the time interval for a CW element was about 1.2/W (W in words per minute) or the frequency of the symbols was about (0.8333 W). Considering that the wavelength of a modulating square wave would be equivalent to a dit and a space element, that means that the equivalent modulating square wave would have frequency of about (0.4167 W), so that the 31st harmonic would be (12.92 W) on each side of the carrier for a total bandwidth of about (26 W) Hz. That means that at 5 WPM, the bandwidth would be about 130 Hz and at 20 WPM the bandwidth would be over 500 Hz! Clearly that isn't acceptable and our experience says that it isn't usually true in practice, either.
The problem is that real electronic equipment, like radios, don't generate square waves precisely. Instead the signal rises over some time period controlled by the electronic components used. But the lesson is simple: Don't use square waves to modulate a CW signal!
Don't Be a Square
So if we don't want a square wave modulating a carrier to generate CW, what would work better? The answer can be found in trigonometry and signal processing theory. To cut past lots of theory, there is a fundamental principle of digital signal analysis that states that any periodic waveform can be described by the sum of a bunch of sine waves. In fact we don't need all possible sine waves, just a fundamental and all of its harmonics to describe any periodic waveform we want. In addition there is another principle, based on fundamental trigonometry, that says that modulating a sine wave with another sine wave (which is actually multiplying the two waveforms) yields 2 signals, one at the sum of the frequencies and the other at the difference of the frequencies. These two principles are at the heart of signal processing and apply to CW signals as well as any other.
We have already seen that modulating a carrier with a square wave is a disaster, because the square wave is composed of an infinite number of harmonics and probably at least the first 30 or so are significant. So the obvious way to get to a minimum bandwidth would be to use just 2 sine waves - one of which is our carrier. How can we determine what the other one should be?
Taking a look at the figure, it is fairly easy to see that we can create a sine wave that has a period equal to that of a dit plus the element space following the dit. If we tried to use a longer sine wave, the dit would end too soon and an abrupt change in signal level would result, which would get back to something like a square wave. (Don't worry about the second half of the sine wave - we don't need to send it, since the signal is 0 during that time.)
So if you change the shape of the signal so it has the shape of a sine wave and the period of the sine wave is exactly equal to 2 times the length of a dit, you have the lowest frequency modulating wave possible and it will have a frequency that is 1/2 the dit frequency that we already figured out: (0.8333 W)/2 = (0.4167 W) Hz. But since the resulting signal contains both the sum and difference frequencies, the bandwidth would be twice the signal frequency or (0.8333 W) Hz.
Back to the Real World
Now, you say, that's all very neat, tidy and theoretical, but what about the real world of ham transmitters? Well, as always, again it's not so simple.
The problem is that we have shown that a CW signal could have a bandwidth anywhere from (0.833 W) to somewhere around (26 W) Hz, depending on the shape of the waveform, etc. And we've also shown that for optimum (i.e. minimum) bandwidth the signal envelope should have a sinusoidal shape where the period of the sine wave varies with the CW speed. But typical electronic components do not change their values dependent on CW speed. Instead most manufacturers give specifications for their CW transmitters in terms of "rise time" and "fall time". Without getting into the details, that essentially means that the shaping is independent of the CW speed and since the shape determines the bandwidth, most ham transmitters and transceivers will have a signal bandwidth that doesn't depend on the speed. So there is no real advantage, in terms of bandwidth, in sending higher or lower speed CW.
With the newer rigs, especially those that generate CW using DSP techniques, it is indeed possible to generate optimum CW. In fact the soundcard modes are quite capable of generating "clickless" CW as are the software-defined radios. Interestingly, many hams find the sound of perfectly "clickless" CW to be somewhat "mushy" and unpleasing to the ear. I guess there's no way to satisfy everyone.
So now you know the real story of CW bandwidth. It offers the potential for extremely narrow bandwidth signals, but most often never actually lives up to its full potential. However, perhaps the most interesting observation is that applying the latest and greatest digital processing techniques to the oldest radio communications mode, yields a measureable technical improvement in terms of bandwidth. In other words, even this old CW can be updated and made new again.
OK, so now you're sitting there and thinking that all the ham rigs must be pretty darn good, because if you listen to CW on the bands, you don't hear any key-clicks, right? But wait a minute. Let's take another look.
Most CW receivers have filters in either the IF or audio stages, or both. That means that whatever you hear has been filtered through those audio stages. So what's the difference if a CW signal gets filtered at the transmitter or at the receiver? How can you tell the difference? Well, the answer is you can't tell the difference.
Remember, if there are key clicks present on a CW signal, they will extend at odd harmonics of the CW keying speed away from the carrier frequency. So if someone is racing along at 50 WPM, the keying frequency is 50/1.2 = 41.667 Hz, in other words 41.667 dits per second - that's fast! The 3rd harmonic would be at 125 Hz, the 5th harmonic would be at 208 Hz, etc. Since many CW operator use fairly narrow receiving filters, but the time you get a few odd harmonics out, the signal is outside the receiver pass band, so you don't hear them and the signal sounds fine. Remember, it is the higher frequencies that cause the clicking and take up so much bandwidth.
So if you want to know if someone is generating key clicks or not, you normally can't just listen to their signal using normal filters. You have to tune away from their carrier frequency to see if there are harmonics of the keying frequency outside your receiver passband. Without doing that you can't tell, but the people having a QSO near by sure might notice!