Doug Kerr
Well-known member
In what follows, I have taken some liberties with rigor in the interest of trying to make clear the principles involved.
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When we first study the concept of representing a continuous variable by sampling, the vision is that the samples are just the instantaneous values of the variable at the sampling "instants".
For various reasons, we may in fact take samples that are, or are something like, the average of the variable value over a small interval centered on the sampling instant. This is in fact what happens in our digital sensors, where each detector gathers the light from a finite-sized region surrounding the theoretical "sampling point". This is said to be sampling "through a finite sampling aperture".
When we do that, we encounter an interesting result. The amplitude of any frequency component of the original signal, as represented in the train of samples, compared to its amplitude in the original signal itself, declines with increasing frequency. In effect, the use of a finite sampling window introduces a type of low-pass filter into the chain.
Why should that be so? We can get some insight through a certain look at the sampling process.
We can think of the sampling process this way: we take the original "waveform" (varying continuously) and multiply it by a time function that is mostly zero but at every sampling instant becomes one for a little while. This is just a description of a switch, operated by the sampling function, which lets the waveform "through" for a while at each sampling occurrence but gives an output of zero otherwise.
In almost the "ideal" sampling model, the length of time that the sampling function is one is "very short" (approaching zero). But for a finite sampling window, the length of time that the sampling function is one is "not so short".
Because of this multiplication of the waveform by the sampling function to give the train of samples, the frequency composition of the train of samples will be the product (over frequency) of the frequency "spectrum" of the waveform and the frequency spectrum of the sampling function (as if it were itself a "signal").
Now we recognize that the original "signal" has a spectrum (the distribution of frequencies in it), even though we might not have at hand a plot of what that is (or even care just what it is for our purposes here). Its spectrum is in fact the Fourier transform of the waveform in time.
But what about the spectrum of the sampling function? Well, it is just the Fourier transform of the sampling function in time.
If the sampling function is as described above (always either zero or one), its Fourier transform is of the form sin(kf)/kf, where f is the frequency, and k depends on the length of time that the function is one (compared to the sampling interval).
The function sin(kf)/kf starts out (when x=0) with the value 1, and then rapidly declines. (It is hard to guess this intuitively!)
Thus multiplying the spectrum of the signal with the spectrum of the sampling process is tantamount to subjecting the signal to a filter whose frequency response is sin(kf)/kf - a low-pass filter.
And this is the cause of the decline of the "representation" of signal frequency components in the sampled signal with increasing frequency - the "sampling aperture rolloff".
Best regards,
Doug
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When we first study the concept of representing a continuous variable by sampling, the vision is that the samples are just the instantaneous values of the variable at the sampling "instants".
Note that we need not convert the sample values to digital form as part of this concept, although of course we do so in most cases of interest. But this discussion will not assume that is done - the sample values are just "values" (perhaps physically the voltages of some pulses.)
For various reasons, we may in fact take samples that are, or are something like, the average of the variable value over a small interval centered on the sampling instant. This is in fact what happens in our digital sensors, where each detector gathers the light from a finite-sized region surrounding the theoretical "sampling point". This is said to be sampling "through a finite sampling aperture".
When we do that, we encounter an interesting result. The amplitude of any frequency component of the original signal, as represented in the train of samples, compared to its amplitude in the original signal itself, declines with increasing frequency. In effect, the use of a finite sampling window introduces a type of low-pass filter into the chain.
Why should that be so? We can get some insight through a certain look at the sampling process.
We can think of the sampling process this way: we take the original "waveform" (varying continuously) and multiply it by a time function that is mostly zero but at every sampling instant becomes one for a little while. This is just a description of a switch, operated by the sampling function, which lets the waveform "through" for a while at each sampling occurrence but gives an output of zero otherwise.
This is in fact the amplitude modulation of the signal by the sampling function.
In almost the "ideal" sampling model, the length of time that the sampling function is one is "very short" (approaching zero). But for a finite sampling window, the length of time that the sampling function is one is "not so short".
Because of this multiplication of the waveform by the sampling function to give the train of samples, the frequency composition of the train of samples will be the product (over frequency) of the frequency "spectrum" of the waveform and the frequency spectrum of the sampling function (as if it were itself a "signal").
Now we recognize that the original "signal" has a spectrum (the distribution of frequencies in it), even though we might not have at hand a plot of what that is (or even care just what it is for our purposes here). Its spectrum is in fact the Fourier transform of the waveform in time.
But what about the spectrum of the sampling function? Well, it is just the Fourier transform of the sampling function in time.
If the sampling function is as described above (always either zero or one), its Fourier transform is of the form sin(kf)/kf, where f is the frequency, and k depends on the length of time that the function is one (compared to the sampling interval).
The function sin(kf)/kf starts out (when x=0) with the value 1, and then rapidly declines. (It is hard to guess this intuitively!)
Thus multiplying the spectrum of the signal with the spectrum of the sampling process is tantamount to subjecting the signal to a filter whose frequency response is sin(kf)/kf - a low-pass filter.
And this is the cause of the decline of the "representation" of signal frequency components in the sampled signal with increasing frequency - the "sampling aperture rolloff".
Best regards,
Doug