Xref: utzoo sci.electronics:18076 comp.dsp:1306 Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!usc!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!att!ucbvax!pasteur!galileo.berkeley.edu!jbuck From: jbuck@galileo.berkeley.edu (Joe Buck) Newsgroups: sci.electronics,comp.dsp Subject: Re: A question about the Nyquist theorm Message-ID: <11515@pasteur.Berkeley.EDU> Date: 28 Feb 91 03:24:47 GMT References: <20408@shlump.nac.dec.com> <625@ctycal.UUCP> Sender: news@pasteur.Berkeley.EDU Reply-To: jbuck@galileo.berkeley.edu (Joe Buck) Lines: 46 In article <625@ctycal.UUCP>, ingoldsb@ctycal.UUCP (Terry Ingoldsby) writes: |> Pursuing the discussion of the Nyquist theorem, I have a question |> about practical sampling applications. If you have a sine wave at |> frequency f, which you sample at just over 2f samples per second then |> the Nyquist theorem is satisfied. I know that by performing a Fourier |> transform it is possible to recover all of the signal, i.e. deduce that |> the original wave was at frequency f. |> |> Note that this is different than just playing connect the dots with the |> samples. Most of the algorithms I've heard of used with CD players |> perform a variety of interpolation, oversampling, etc., but these all |> seem to be elaborate versions of connect the dots. I'm not aware that |> the digital signal processing customarily done will restore the wave to |> anything resembling its original. The Nyquist sampling theorem says more than just that you need to sample at a rate higher than twice the highest frequency. It also gives the formula for the reconstructed time series. If you have a signal with no frequency components higher than f = 1/2T, where T is the spacing between samples, then the original waveform x(t) may be found exactly at any point by computing the sum x(t) = sum from m=-infinity to infinity x[m] * sinc (pi (t - m*T) / T) where sinc(x) is just sin(x)/x (note: sinc(0) is 1). This is exactly what you get when you pass a series of Dirac delta functions with weights x[m] through an ideal low pass filter with cutoff frequency 1/2T; the impulse response of such a filter is sinc(pi*t/T). You can't make an ideal low pass filter; for one thing, it's noncausal. All you can do is approximate this. To know more about how this works, you need to study some digital signal processing; then you can go laugh at your CD or DAT sales critter when he attempts to tell you about why one system is better than another. Example of CD salespeak: pushing oversampling as an advanced technical feature. Oversampling is simply inserting zeros between the digital samples and thus increasing the sampling rate. It's used because then you can use cheaper, less complex analog filters; it reduces the system cost. Still, some sales critters think it's an advanced technical extra. -- Joe Buck jbuck@galileo.berkeley.edu {uunet,ucbvax}!galileo.berkeley.edu!jbuck