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There's no such THING, Mart. That's where you err.

207.207.243.246

Posted by jj on February 19, 2000 at 14:18:54

In Reply to: Re: You mean, "it can" posted by Mart on February 19, 2000 at 14:08:05:

You appear to be assuming that because a damped sine wave with a 20kHz carrier has a 20kHz carrier, it has only frequency content at 20kHz.

This is not true, going back to the "convolution theorem" discussed below, or somewhere around here, you must convolve the frequency shape of the damping signal (e^-at) with the sine wave. The shape of the damped sine wave, especialy if it has a very sharp attack, is quite wide, very likely more than 4khz. This means that the FREQUENCY CONTENT of a "damped 20khz sine wave" is WELL ABOVE 22.1 kHz.

So, it has out of band signals. You can, of course, filter that signal BEFORE you sample it, and prevent any aliasing, and remove the out-of-band information at the same time. This will change, somewhat, the shape (mostly the attack) of the damped sine wave. This is not wrong, this is not distortion, it is REMOVING COMPONENTS ABOVE THE NYQUIST LIMIT.

Just because the sine wave is at 20kHz does not, repeat NOT mean that the signal spectrum consists only of 20kHz, and in fact it does not consist only of 20kHz unless the sine wave is of infinite duration.

For anything other than infinite duration, you must convolve that line at 20khz with the shape of the time window, be it a naturally damped signal, a guassian window (which has the narrowest time/bandwidth tradeoff, i.e. it's the case where t*f = 1, where t is time resolution and f is frequency resolution, and we're talking about one sigma in each case, please see Morrison for a complete reduction of this case), a Hann window, a Hamming window, Blackmun, rectangular, Kaiser, ... what-have you.

All windows that are not identically a single constant over all time introduce spreading of the spectrum. This is why you can only recover something at exactly half the sampling frequency with infinite delay, btw, because that is the only time that a sine wave exactly at the Nyquist bound will have NO signal above it.

But, please, understand this, you can get arbitrarily close to that bound.

You can not, though, get closer than that tf>=1 relationship, although you can, if you must, redefine that relationship to have a different error criterion (some people have done that, but all it does is create confusion). That means that if the signal is of length 't' you never get any signal closer than fs/2 - 1/t resolved, because it will necessarily have out of band components, and they will get removed in the input antialias filter.

Is this getting through?

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