Re: Digital Watermarks for copy protection in recent Billbo (fwd)
Hi all, The Sampling Theorem in operation is a little more complicated than the model that is being discussed. Forwarded message:
From: mpd@netcom.com (Mike Duvos) Subject: Re: Digital Watermarks for copy protection in recent Billbo Date: Wed, 24 Jul 1996 21:38:36 -0700 (PDT) "Perry E. Metzger" <perry@piermont.com> writes:
The Nyquist Theorem states you need exactly twice the samples, not over twice. The magic number isn't something like 2.2, its exactly 2.
The Sampling Theorem states that equally spaced instantaneous samples must be taken at a rate GREATER THAN twice the highest frequency present in the analog signal being sampled. If this is done, the samples contain all the information in the signal, and faithful reconstruction is possible.
Actualy the sampling theorem states that if you want to reproduce a signal reliably it must be sampled at a frequency AT LEAST TWICE that of the highest frequency of interest in the FFT of the signal. This means the signal must be deformable into sine waves, not all signals qualify for this particular limitation and in general are not good signals for sampling. An example is a step change from v1 to v2. since there is no change in voltage except for a brief moment the output of the FFT is pretty much flat. What comes out the other end looks like a spike (similar to what happens when you feed a square wave to a transformer and look at the output). If you reconstruct this via the sampling theorem you get a sign wave of extremely low amplitude and frequency. If you think of a FFT as taking a signal and breaking it down into componant frequencies. Then think of a hair comb where the lowest frequencies are the bigger teeth (put them to the L. when looking at it). Because sign waves have zero crossings a single sample per cycle is not enough. By taking at least two samples a cycle you are guaranteed not to miss the presence or absence of a componant frequency. So this guarantees that we get a accurate count of componants and their phase relations to each other (where the arbitrary time reference comes in - really a fixed frequency clock). Back to the comb. Where a given signal has a componant leave the teeth. Where there is no componant break them off. You are left with a ratty looking comb. Now to each of the remaining teeth assign an amplitude for that specific frequency that is consistent with the FFT you calculated. The way this FFT is implimented in practice is called a 'Comb Filter' where it samples the signal (wide bandwidth) over a set of very small bandwidth filters in parallel. Scanning the output of the filters at a fixed rate you get a phase relation as well. Digitize the signals in a particular pattern and you are ready to cut your CD or whatever. It does NOT guarantee faithful representation of the original signal but rather a signal with the same energy spectrum and phase characteristics. One of the basic ideas of Algebra is that any given curve can be explained by a arbitrary set of equations. The Sampling Theorem just gives you a rationale for picking from that set.
Exactly twice the highest frequency won't do, and it should be obvious that sampling a sine wave at twice its frequency yields samples of constant magnitude and alternating sign which convey nothing about its phase and little useful about its amplitude either. (Drawing a little picture might be helpful here.)
Exactly what the theorem is supposed to produce. You take your original signal and run it through a FFT. You look at the bandwidth you desire. The highest frequency of interest is 1/2 or less your sampling frequency. With this information you can build a set of sine waves whose amplitude is given by the FFT along with phase relations. Since you are breaking the signal into sine waves (which happen to be well defined) all you realy need is to know the maximum and minimum amplitudes as well as their phase to some arbitrary but constant time reference. Sum them back together on the other side and what you got? A reasonable useable copy of your original signal. This is why DSP's are optimized for multiplication (multiply the amplitude of that componant by its presence in the FFT) and summing (add them together to get the target signal). Generaly because of noise and similar phenomena it is commen to multiply and add windows of samples (ie averaging).
Although anything over twice the highest frequency will work in a theoretical sense, a small fudge factor does wonders for digital signal processing, if only to reduce to a reasonable value the width of the window into the sample stream needed for various signal manipulations.
Actualy I believe the decision was made by Philips (the inventor of the CD) to settle on the 44kHz sample rate because of some design option it simplified. I unfortunately don't remember anything more specific than that. Anyone got the CD Rom bible? Jim Choate
Jim Choate <ravage@einstein.ssz.com> writes:
The Sampling Theorem states that equally spaced instantaneous samples must be taken at a rate GREATER THAN twice the highest frequency present in the analog signal being sampled. If this is done, the samples contain all the information in the signal, and faithful reconstruction is possible.
Actualy the sampling theorem states that if you want to reproduce a signal reliably it must be sampled at a frequency AT LEAST TWICE that of the highest frequency of interest in the FFT of the signal. This means the signal must be deformable into sine waves, not all signals qualify for this particular limitation and in general are not good signals for sampling.
You want a continuous Fourier transform, not a discrete one, to determine the frequency spectrum of the waveform being sampled. The FFT is simply an algorithm for computing the DFT without redundant computation. In general, any Lebesgue integrable complex function will have a Fourier transform, even one with a finite number of discontinuities. The reverse transform will faithfully reproduce the function, modulo the usual caveats about function spaces and sets of measure zero. There is no meaningful difference between speaking of the highest frequency in a signal, and the highest frequency present in its Fourier transform.
An example is a step change from v1 to v2. since there is no change in voltage except for a brief moment the output of the FFT is pretty much flat.
The Fourier transforms of step functions, square wave functions, delta functions, and other oddities are perfectly well defined. Again, the FFT is not relevant here. A step function does not have a upper cutoff in the frequency domain, so you can never reproduce it perfectly from its samples, although the faster you sample, the sharper the edges of the reconstruction will become.
What comes out the other end looks like a spike (similar to what happens when you feed a square wave to a transformer and look at the output). If you reconstruct this via the sampling theorem you get a sign wave of extremely low amplitude and frequency.
I don't think so. [Incomprehensible Deletia]
Back to the comb. Where a given signal has a componant leave the teeth. Where there is no componant break them off. You are left with a ratty looking comb. Now to each of the remaining teeth assign an amplitude for that specific frequency that is consistent with the FFT you calculated. The way this FFT is implimented in practice is called a 'Comb Filter' where it samples the signal (wide bandwidth) over a set of very small bandwidth filters in parallel.
I must have been absent on "tooth day" in Functional Analysis class.
One of the basic ideas of Algebra is that any given curve can be explained by a arbitrary set of equations. The Sampling Theorem just gives you a rationale for picking from that set.
AIIIIIIIIEEEEEEEEEEEEE! (I feel much better now :) -- Mike Duvos $ PGP 2.6 Public Key available $ mpd@netcom.com $ via Finger. $
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