dar@euler.eedsp.gatech.edu (Doug Reynolds) (03/03/91)
In a pervious post, someone brought up the theoretical problem that you
sample a sine wave with frequency f at exactly 2*f, but your first sample
is perfectly lined up at t=0. Thus it would seem that your sample stream would
then be all zeros makeing it impossible to reconstruct the original sine wave.
I have tried to reconcile this, but to no ava as of yet. Does anyone else
have any ideas on this?
Doug Reynolds
.
--
Doug Reynolds [GRA M. Smith]
Georgia Tech, School of Electrical Engineering, Atlanta, GA 30332
USENET: ...!{allegra,hplabs,ihnp4,ulysses}!gatech!gt-eedsp!$me
INTERNET: $me@gteedsp.gatech.educas@media-lab.MEDIA.MIT.EDU (Scud) (03/04/91)
In article <1991Mar2.230829.21497@eedsp.gatech.edu> dar@euler.eedsp.gatech.edu (Doug Reynolds) writes: >In a pervious post, someone brought up the theoretical problem that you >sample a sine wave with frequency f at exactly 2*f, but your first sample >is perfectly lined up at t=0. Thus it would seem that your sample stream would >then be all zeros makeing it impossible to reconstruct the original sine wave. > >I have tried to reconcile this, but to no ava as of yet. Does anyone else >have any ideas on this? It's just a matter of "greater than" versus "greater than or equal to." Nyquist's theorem claims that the sampling frequency should be strictly greater than twice the frequency of the highest component of your signal. Usually this distiction isn't a big deal, but there is indeed aliasing if you sample at exactly the Nyquist rate. In the sine wave example of the previous posting, you'll get impulses at +/- f with magnitudes +/- 1/2j. When you periodically replicate the sine spectrum, the positive impulses cancel the negative impulses and you get a DTFT of zero, as predicted. In the case of a cosine you get lucky and the impulses add, so the aliasing isn't destructive and you can reconstruct your signal (although you will be off by a constant factor). Casimir Wierzynski
shmulevi@tukkasotka.tut.fi (Ilya Shmulevich) (03/04/91)
> In a pervious post, someone brought up the theoretical problem that you > sample a sine wave with frequency f at exactly 2*f, but your first sample > is perfectly lined up at t=0. Thus it would seem that your sample stream > would then be all zeros makeing it impossible to reconstruct the original > sine wave. > > I have tried to reconcile this, but to no ava as of yet. Does anyone else > have any ideas on this? > > Doug Reynolds I think the problem seems to stem from the fact that the Nyquist rate must be GREATER than (and not greater than or equal to) twice the highest frequency component of your signal. Try to imagine a bandlimited signal x(n) with a non-zero component at the highest frequency. Also, assume that the signal x(n) is real, so its Fourier Transform is conjugate symmetric. When sampled at exactly twice the highest frequency, the replicas of the spectrum WILL overlap at odd multiples of the highest frequency. This will obviously introduce distortion. Although this is not the same problem, it again demonstrates that the sampling rate must be greater than the highest frequency component. -- Ilya Shmulevich shmulevi@tut.fi Signal Processing Laboratory Tampere University of Technology, Finland
campbell@churchy.ai.mit.edu (Paul Campbell) (03/05/91)
The Nyquist rate is the rate needed to sample a given bandwidth from zero up to that frequency but not including that frequency since the frequency of the sampling introduces harmonics which will be lost, including the top of the band, and any multiple of the Nyquist rate. This is not as bad as the frequencies outside of the band, which will be aliased. This is the whole basis of the Nyquist sampling theorem, to be able to determine what sampling rate to use for a given bandwidth. Realistically, for any decent resolution of the wave, you should sample at four times the maximum, but the absolute minimum (the limit, which will obviously not work at all) is the Nyquist rate, or twice the maximum frequency.
paul@mtnmath.UUCP (Paul Budnik) (03/06/91)
In article <13687@life.ai.mit.edu>, campbell@churchy.ai.mit.edu (Paul Campbell) writes: > The Nyquist rate is the rate needed to sample a given bandwidth from zero > up to that frequency but not including that frequency ... Correct so far. > This is not as bad > as the frequencies outside of the band, which will be aliased. A frequency that is an exact multiple of the sampling rate gets aliased as a DC signal. The amplitude of this aliased signal depends on the phase where sampling occurs. > > ... Realistically, for any decent > resolution of the wave, you should sample at four times the maximum, but > the absolute minimum (the limit, which will obviously not work at all) > is the Nyquist rate, or twice the maximum frequency. When you are sampling at slightly more than twice the minimum frequency you can completely reconstruct the original signal *provided your sampling sequence is infinitely long*. Of course, it never is. You can oversample to increase frequency resolution, but you can also accomplish this by increasing your sample length. One point that may be confusing to some people is that an FFT allows you to determine the spectrum from a finite sample series. However, it assumes that the infinite series from which this finite series is taken keeps repeating the same sequence of samples out to infinity. To the degree this assumption is incorrect, the FFT does not give the exact spectrum of the infinite sequence. Thus a longer sampling length or increased sample rate can provide better resolution in an FFT when this assumption is false. Paul Budnik
bryanh@hplsla.HP.COM (Bryan Hoog) (03/06/91)
> >In a pervious post, someone brought up the theoretical problem that you >sample a sine wave with frequency f at exactly 2*f, but your first sample >is perfectly lined up at t=0. Thus it would seem that your sample stream would >then be all zeros makeing it impossible to reconstruct the original sine wave. > >I have tried to reconcile this, but to no ava as of yet. Does anyone else >have any ideas on this? > Since this is a theoretical discussion, I'll go out on a limb and claim that although a sine wave with frequency Fs/2 could not be sampled without ambiguity, a complex exponential [e^j(.5Fst+@)] could be, for any phase value @. This is only possible, though, if there is no component at [e^-j(.5Fst+@)]. In other words, you could accurately sample a "tone" at .5 Hz with a 1 Hz A/D converter, provided there was no "tone" at -.5 Hz. Of course, you'd need a complex (Re+jIm) A/D converter, since the signal to be sampled is not purely real anymore. As I recall the "nyquist sampling theorem", the sampling rate dictates the maximum bandwidth that can be sampled unambiguously with a given sampling rate, and this bandwidth interval is closed on one end and open on the other, (-.5,.5] in the above example. It is interesting to note that the nyquist bandwidth in this example violates the 2X criteria, with almost a 1 Hz bandwidth for a 1 Hz sample rate. It's been a long time since my college DSP class, so feel free to correct me if I got something wrong! Bryan Hoog