F0O@psuvm.psu.edu (02/15/91)
I was reading an article that states the Nyquist theorm as:
"The sample frequency must be at least twice the highest frequency
component within the analog signal for an accurate representation of the
analog signal".
I'd guess here he is talking about complex signals. But what do you
do with a pure sine wave? There is only one frequency component in a sine
wave(the fundamental), and if you sample at twice that, you're not going
to get a good representation of the signal.
i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're
only going to get two points per cycle.
I'm sure I must not be understanding something here, or does the
Nyquist equation only apply to complex signals?
[Tim]north@manta.NOSC.MIL (Mark H. North) (02/16/91)
In article <91046.095459F0O@psuvm.psu.edu> F0O@psuvm.psu.edu writes: > > I was reading an article that states the Nyquist theorm as: > "The sample frequency must be at least twice the highest frequency >component within the analog signal for an accurate representation of the >analog signal". This is an incorrect statement of the Nyquist theorem. The sample freq must be *greater* than twice the highest freq component... > I'd guess here he is talking about complex signals. But what do you >do with a pure sine wave? There is only one frequency component in a sine >wave(the fundamental), and if you sample at twice that, you're not going >to get a good representation of the signal. A pure sine wave is fine. As long as you sample at greater than twice its freq. Even though it may appear that you are not getting a good represen- tation of the signal it can be shown with Fourier analysis that the sample set is unique to this component and hence the exact signal can be recovered from the sample set. > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're >only going to get two points per cycle. And imagine that those two points are phased such that they land at the zero crossing of the 60Hz signal. All your samples are zero! This is why you must sample at greater than 2nu. A good reference is "Digital Signal Analysis" by Samuel D Stearns. It is no longer in print but is available in most engr. libraries. Also there is a new edition of this book published by Printice Hall. Mark
ertas@athena.mit.edu (Mehmet D Ertas) (02/16/91)
In article <1751@manta.NOSC.MIL>, north@manta.NOSC.MIL (Mark H. North) writes: |> In article <91046.095459F0O@psuvm.psu.edu> F0O@psuvm.psu.edu writes: |> > I'd guess here he is talking about complex signals. But what do you |> >do with a pure sine wave? There is only one frequency component in a sine |> >wave(the fundamental), and if you sample at twice that, you're not going |> >to get a good representation of the signal. |> |> A pure sine wave is fine. As long as you sample at greater than twice its |> freq. Even though it may appear that you are not getting a good represen- |> tation of the signal it can be shown with Fourier analysis that the |> sample set is unique to this component and hence the exact signal can |> be recovered from the sample set. And just for ther sake of completeness, here's how you recover your original signal: Take your samples, pass them through an A-D converter and LPF the outcoming signal with cutoff freq. 0.5 times the sampling frequency. There you go! |> |> > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're |> >only going to get two points per cycle. |> |> |> |> And imagine that those two points are phased such that they land at the |> zero crossing of the 60Hz signal. All your samples are zero! This is |> why you must sample at greater than 2nu. |> That's correct; you cannot recover the amplitude of a frequency comp. exactly half the sampling frequency. |> |> A good reference is "Digital Signal Analysis" by Samuel D Stearns. It is |> no longer in print but is available in most engr. libraries. Also there |> is a new edition of this book published by Printice Hall. |> |> Mark Another useful reference may be "Digital Signal Processing" by Oppenheim & Schafer. M. Deniz Ertas
R_Tim_Coslet@cup.portal.com (02/17/91)
In Article: <1751@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes: >In article <91046.095459F0O@psuvm.psu.edu> F0O@psuvm.psu.edu writes: >> >> I was reading an article that states the Nyquist theorm as: >> "The sample frequency must be at least twice the highest frequency >>component within the analog signal for an accurate representation of the >>analog signal". > >This is an incorrect statement of the Nyquist theorem. The sample freq >must be *greater* than twice the highest freq component... > Looks accurate to me, it says "at least twice" and you say "*greater* than twice". Both wordings mean the same to me (although the second is probably clearer due to the emphasis). While the Nyquist criteria sets a minimum theoretical sampling rate, practical sampling rates are generally at least 5 times max frequency component (and more samples may be required if better reproduction is required, 5 times is just a "rule of thumb"). R. Tim Coslet Usenet: R_Tim_Coslet@cup.portal.com BIX: r.tim_coslet Free Kuwait. Disarm Iraq. Stop Soviet repression in the Baltic.
terryb.bbs@shark.cs.fau.edu (terry bohning) (02/17/91)
north@manta.NOSC.MIL (Mark H. North) writes: > > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're > >only going to get two points per cycle. > > And imagine that those two points are phased such that they land at the > zero crossing of the 60Hz signal. All your samples are zero! This is > why you must sample at greater than 2nu. > The catch is that you *know* you're sampling the highest input freq at 2 points per cycle. That is, the input signal is bandlimited. So if someone gives you a set of all zero samples and you know the sample rate is 120 Hz, the only frequency it can be is 60 Hz. The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, "Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. In reality, of course, since ideal filters are unavailable for band-limiting, the rate must be higher.
ruck@sphere.UUCP (John R Ruckstuhl Jr) (02/17/91)
In article <Abj8CR200WBN04RPUr@andrew.cmu.edu>, kr0u+@andrew.cmu.edu (Kevin William Ryan) writes: > F0O@psuvm.psu.edu > > I was reading an article that states the Nyquist theorm as: > > "The sample frequency must be at least twice the highest frequency > >component within the analog signal for an accurate representation of the > >analog signal". I think this should be "GREATER than twice the highest frequency component". > > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're > >only going to get two points per cycle. If you sample ABOVE 120Hz, you will have enough information to reconstruct your original signal as described. If you sample at precisely 120 Hz, you will be unable to reconstruct accurately. Without loss of generality, consider samples at t = kT, where k is an integer, and T is (1/120)s: x(t) = cos(120*pi*t) is indistinguishable from y(t) = 2 * cos(120*pi*t + (pi/3)) when comparing the sampled data. Therefore, there is a reconstruction ambiguity. > No, it applies to all signals, as the _minimum_ possible sampling > rate that will without error give you the signal back. Sampling a sine > wave at the Nyquist rate will give you, in the best case: > > | | | | | | | | | > ----------------- > | | | | | | | | > > where alternate samples hit the plus and minus peaks of the sine wave. > Practically, to ensure that you aren't just hitting the nodes rather > than the peaks of the waves, it's best to sample much faster. However, > it is possible to reconstruct it accurately with Nyquist rate sampling, > and impossible to reconstruct it accurately if sampled any slower. The > Nyquist criterion gives the minimum rate at which you can sample and > retain information. I believe this is misleading... One must sample ABOVE the Nyquist rate (not AT the Nyquist rate) for reconstruction to be possible. I think this is a fairly common misconception, though. Sorry kwr -- I don't mean to be impolite. Best regards, ruck -- John R Ruckstuhl, Jr ruck%sphere@cis.ufl.edu, sphere!ruck University of Florida ruck@cis.ufl.edu, uflorida!ruck
kahhan@bnr.ca (02/17/91)
In article <D04gX2w163w@shark.cs.fau.edu> terryb.bbs@shark.cs.fau.edu (terry bohning) writes: >north@manta.NOSC.MIL (Mark H. North) writes: > >> > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're >> >only going to get two points per cycle. >> >> And imagine that those two points are phased such that they land at the >> zero crossing of the 60Hz signal. All your samples are zero! This is >> why you must sample at greater than 2nu. >> >The catch is that you *know* you're sampling the highest input freq at >2 points per cycle. That is, the input signal is bandlimited. So if >someone gives you a set of all zero samples and you know the sample >rate is 120 Hz, the only frequency it can be is 60 Hz. Not quite. Another signal that will yield an all zero sample set to the 120 Hz sampling rate is DC. In general, you must sample at greater than twice the maximum frequency, not exactly twice. However, there are techniques that can be used to sample at exactly twice the frequency, under certain conditions (like looking for a single frequency, where you kick the phase of your sampler periodically to avoid sampling the input waveform only at zero crossings). A -- ---------------------------------------------------------------------------------- Larry Kahhan - NRA, NRA-ILA, CSG, GOA, GSSA | The opinions expressed here do | not necessarily represent the | views of the management. ----------------------------------------------------------------------------------
north@manta.NOSC.MIL (Mark H. North) (02/18/91)
In article <39342@cup.portal.com> R_Tim_Coslet@cup.portal.com writes: >In Article: <1751@manta.NOSC.MIL> > north@manta.NOSC.MIL (Mark H. North) writes: >> >>This is an incorrect statement of the Nyquist theorem. The sample freq >>must be *greater* than twice the highest freq component... >> >Looks accurate to me, it says "at least twice" and you say "*greater* than >twice". Both wordings mean the same to me (although the second is probably >clearer due to the emphasis). > I'm sorry I must be losing it. At least twice implies twice is good enough which it isn't. No? >While the Nyquist criteria sets a minimum theoretical sampling rate, practical >sampling rates are generally at least 5 times max frequency component (and >more samples may be required if better reproduction is required, 5 times is >just a "rule of thumb"). > The reason for requiring a much higher sampling rate than the minimum theoretical is because no filter is perfect and your nominal 'highest' frequency is not really the highest. If you knew for sure in advance that your signal had *no* components greater than nu, say, then you could do no better a reproduction with a faster sampling rate than 2nu + epsilon. Mark
north@manta.NOSC.MIL (Mark H. North) (02/18/91)
In article <D04gX2w163w@shark.cs.fau.edu> terryb.bbs@shark.cs.fau.edu (terry bohning) writes: >north@manta.NOSC.MIL (Mark H. North) writes: > >> > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're >> >only going to get two points per cycle. >> >> And imagine that those two points are phased such that they land at the >> zero crossing of the 60Hz signal. All your samples are zero! This is >> why you must sample at greater than 2nu. >> >The catch is that you *know* you're sampling the highest input freq at >2 points per cycle. That is, the input signal is bandlimited. So if >someone gives you a set of all zero samples and you know the sample >rate is 120 Hz, the only frequency it can be is 60 Hz. >The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, >"Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. >In reality, of course, since ideal filters are unavailable for >band-limiting, the rate must be higher. I knew that 8^). Actually, I got to thinking about it since this discussion came up and I believe I emailed someone mentioning the above possibility but this would be as they say a trivial (and useless) case wouldn't you agree? Mark
north@manta.NOSC.MIL (Mark H. North) (02/18/91)
In article <1758@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes: >In article <D04gX2w163w@shark.cs.fau.edu> terryb.bbs@shark.cs.fau.edu (terry bohning) writes: >>north@manta.NOSC.MIL (Mark H. North) writes: >> >>> > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're >>> >only going to get two points per cycle. >>> >>> And imagine that those two points are phased such that they land at the >>> zero crossing of the 60Hz signal. All your samples are zero! This is >>> why you must sample at greater than 2nu. >>> >>The catch is that you *know* you're sampling the highest input freq at >>2 points per cycle. That is, the input signal is bandlimited. So if >>someone gives you a set of all zero samples and you know the sample >>rate is 120 Hz, the only frequency it can be is 60 Hz. >>The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, >>"Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. >>In reality, of course, since ideal filters are unavailable for >>band-limiting, the rate must be higher. > >I knew that 8^). Actually, I got to thinking about it since this discussion >came up and I believe I emailed someone mentioning the above possibility but >this would be as they say a trivial (and useless) case wouldn't you agree? > >Mark > Sorry to answer my own post but I take that last paragraph back. I think you are wrong after all. Look at it this way -- suppose I tell you I'm going to send you one of two signals, either 1 volt 60 Hz or a DC voltage between -1 and 1 volt. You may sample at 120 Hz. You get all identical samples at 0.5 volts. Which signal did I send? Mark
ruck@sphere.UUCP (John R Ruckstuhl Jr) (02/18/91)
In article <D04gX2w163w@shark.cs.fau.edu>, terryb.bbs@shark.cs.fau.edu (terry bohning) writes: > north@manta.NOSC.MIL (Mark H. North) writes: > > > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're > > >only going to get two points per cycle. > > And imagine that those two points are phased such that they land at the > > zero crossing of the 60Hz signal. All your samples are zero! This is > > why you must sample at greater than 2nu. > The catch is that you *know* you're sampling the highest input freq at > 2 points per cycle. That is, the input signal is bandlimited. So if > someone gives you a set of all zero samples and you know the sample > rate is 120 Hz, the only frequency it can be is 60 Hz. Or 0 Hz. And supposing the signal you sampled *was* 60 Hz. You have no magnitude information. You cannot reconstruct. > The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, > "Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. Yes. They say "at least twice the highest frequency". But the equation they give is not ambiguous: Wmax < pi/Tsample (or, 2*Wmax < Wsample) (same page). Terry, please be very careful of misinformation. Best regards, ruck. -- John R Ruckstuhl, Jr ruck%sphere@cis.ufl.edu, sphere!ruck University of Florida ruck@cis.ufl.edu, uflorida!ruck
kr0u+@andrew.cmu.edu (Kevin William Ryan) (02/19/91)
amichiel@rodan.acs.syr.edu (Allen J Michielsen) >In article <D04gX2w163w@> terryb.bbs@shark.cs.fau.edu (terry bohning) writes: >>north@manta.NOSC.MIL (Mark H. North) writes: >>> > i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're >>> >only going to get two points per cycle. >>> And imagine that those two points are phased such that they land at the >>> zero crossing of the 60Hz signal. All your samples are zero! This is >>The catch is that you *know* you're sampling the highest input freq at >>2 points per cycle. That is, the input signal is bandlimited. So if >>someone gives you a set of all zero samples and you know the sample >>rate is 120 Hz, the only frequency it can be is 60 Hz. > >And what theory are you using to eliminate all other even multiples of 60 >like 120.... Then you've violated the Nyquist criterion (shame, shame) which states that the _maximum_ frequency must be less than one half the sampling rate. If the signal contains frequencies above this, good luck reconstructing it. Because you can't. You don't have the information, and in fact your reconstructed signal will contain frequencies not in the original - frequencies aliased in from the signal freqencies that were greater than 1/2 the sampling rate. This is usually enforced with some sort of prefiltering of the input frequency. The bandlimiting requirement is _very_ important. kwr Internet: kr0u+@andrew.cmu.edu
kr0u+@andrew.cmu.edu (Kevin William Ryan) (02/19/91)
Mea culpa. The requirement that you must sample _above_ 2w is
correct: sampling _at_ twice the highest frequency produces indefinite
results, as phase and therefore magnitude information is lost. Thus the
proper definition is:
max freq signal < 1/2 sampling rate
in order to preserve information.
It's been too damn long since that signals course... :-)
kwr
Internet: kr0u+@andrew.cmu.edurobf@mcs213f.cs.umr.edu (Rob Fugina) (02/19/91)
In article <1759@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes: >Sorry to answer my own post but I take that last paragraph back. I think >you are wrong after all. Look at it this way -- suppose I tell you I'm >going to send you one of two signals, either 1 volt 60 Hz or a DC voltage >between -1 and 1 volt. You may sample at 120 Hz. You get all identical >samples at 0.5 volts. Which signal did I send? >Mark You sent a DC signal of 0.5 volts. If it were AC, you the samples would be alternating positive and negative of the same magnitude. Rob robf@cs.umr.edu
kdq@demott.com (Kevin D. Quitt) (02/19/91)
In article <1759@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes: > >Sorry to answer my own post but I take that last paragraph back. I think >you are wrong after all. Look at it this way -- suppose I tell you I'm >going to send you one of two signals, either 1 volt 60 Hz or a DC voltage >between -1 and 1 volt. You may sample at 120 Hz. You get all identical >samples at 0.5 volts. Which signal did I send? You've reduced this past absurdity. If I know it *must* be one or the other, a single measure will almost always be sufficient. The discussion revolves around reconstructing *any* waveform (requires >2x sampling). -- _ Kevin D. Quitt demott!kdq kdq@demott.com DeMott Electronics Co. 14707 Keswick St. Van Nuys, CA 91405-1266 VOICE (818) 988-4975 FAX (818) 997-1190 MODEM (818) 997-4496 PEP last
terryb.bbs@shark.cs.fau.edu (terry bohning) (02/19/91)
ruck@sphere.UUCP (John R Ruckstuhl Jr) writes: > > The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, > > "Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. > > Yes. They say "at least twice the highest frequency". But the equation > they give is not ambiguous: Wmax < pi/Tsample (or, 2*Wmax < Wsample) > (same page). > > Terry, please be very careful of misinformation. > OK, OK. I'll say it "I M A D E A M I S T A K E". I'll say 5 Hail Mary's and build 10 linear phase anti-aliasing filters! It's too bad you can't make one on this board without getting hate mail in your box (not you John, I'm referring to the type of people who probably wonder why they're never invited to the meetings).
grayt@Software.Mitel.COM (Tom Gray) (02/19/91)
In article <1751@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes: }In article <91046.095459F0O@psuvm.psu.edu> F0O@psuvm.psu.edu writes: }> }> I was reading an article that states the Nyquist theorm as: }> "The sample frequency must be at least twice the highest frequency }>component within the analog signal for an accurate representation of the }>analog signal". } }This is an incorrect statement of the Nyquist theorem. The sample freq }must be *greater* than twice the highest freq component... } This is an incorrect correction. The original satement is accurate. Sampling at twice or greater than the highest freqeuncy n a band limited signal is all that is required for Nyquist sampling. }> I'd guess here he is talking about complex signals. But what do you }>do with a pure sine wave? There is only one frequency component in a sine }>wave(the fundamental), and if you sample at twice that, you're not going }>to get a good representation of the signal. } }A pure sine wave is fine. As long as you sample at greater than twice its }freq. Even though it may appear that you are not getting a good represen- }tation of the signal it can be shown with Fourier analysis that the }sample set is unique to this component and hence the exact signal can }be recovered from the sample set. } }> i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're }>only going to get two points per cycle. } }And imagine that those two points are phased such that they land at the }zero crossing of the 60Hz signal. All your samples are zero! This is }why you must sample at greater than 2nu. } This is a common misconception. The sampling pulses are of finite widht. The shape of the wave is preserved within the sampling pulse. This information allows representation of a signal at exactly 1/2 the Nyquist freqency. The origin of this misconcetion is a confusion about the sampling methods assumed for the Nyquist theorem. Nyquist assumed natural sampling in which the shape of the signal is preseved by multiplication with the sampling pulse. This is a simple multilication of the two signals in the time domain. Digital sample storage cannot do this, Only one value of the signal can be obtained per sample (not the continuous representation through out the sampling period which is obtained for natural). The digital method of sampling is called commonly flat top sampling. Flat top sampling cannot represent signals at the half sampling frequency. it is a limitation of flat top sampling and not of sampling in general ( including Nyquist sampling) which makes this limitation. If you have text books proving Nyquist by multyplying with instabtaneous pulses and referring the Dirac delat functions, I have text books which properly prove Nyquist with pulses of any width. The instantaneous pulse case is only a special case and is not true in general since it implies limitations which do not occur for pulses of finite width (ie all REAL sampling pulses). } } A good reference is "Digital Signal Analysis" by Samuel D Stearns. It is } no longer in print but is available in most engr. libraries. Also there } is a new edition of this book published by Printice Hall. } Most text books play fast and lose with Nyquist.
jfa0522@hertz.njit.edu (john f andrews ece) (02/19/91)
an added practical note: after you get through all of the theory,
sample at about 2.5 or more times the Nyquist frequency to get everything
in your signal (you will have noise and such at up to there or so).
Of course better is to seriously prefilter your analog signal to well
below the Nyquist. This will ensure that all the nasties above your
*theoretical* Nyquist will be more than 78dB or so down and thus less than
1/2 LSB of you AD converter (assuming you are using one). Then the noise will
be below the input threshold of the ADC.
-----------------------------------------------------------------------------
john f andrews SYSOP The Biomedical Engineering BBS
24 hrs 300/1200/2400 (201) 596-5679
-----------------------------------------------------------------------------
INTERNET jfa0522@hertz.njit.edu LabRat@faraday.njit.edu CIS 73710,2600
-----------------------------------------------------------------------------mac@idacrd.UUCP (Robert McGwier) (02/21/91)
From article <6607@healey>, by grayt@Software.Mitel.COM (Tom Gray): > This is a common misconception. The sampling pulses are of finite widht. > The shape of the wave is preserved within the sampling pulse. This > information allows representation of a signal at exactly 1/2 > the Nyquist freqency. > > Most text books play fast and lose with Nyquist. I pose the following question. Suppose you are sampling at rate N samples per second, and you see a constant value V for your A/D sample. Is the frequency of the signal which produced those samples 0 or N/2? Since I obviously posed this question because I know you CANNOT discriminate between these two cases, what exactly is it then that Nyquist IS telling us and am I asking about apples and oranges? Bob -- ____________________________________________________________________________ My opinions are my own no matter | Robert W. McGwier, N4HY who I work for! ;-) | CCR, AMSAT, etc. ----------------------------------------------------------------------------
gsteckel@vergil.East.Sun.COM (Geoff Steckel - Sun BOS Hardware CONTRACTOR) (02/21/91)
In article <883@idacrd.UUCP> mac@idacrd.UUCP (Robert McGwier) writes: From article <6607@healey>, by grayt@Software.Mitel.COM (Tom Gray): > The shape of the wave is preserved within the sampling pulse. This > information allows representation of a signal at exactly 1/2 > the Nyquist freqency. I pose the following question. Suppose you are sampling at rate N samples per second, and you see a constant value V for your A/D sample. Is the frequency of the signal which produced those samples 0 or N/2? Since AARRGGGHHH! The Nyquist criterion requires that sampling be GREATER THAN the highest frequency of interest. Note also that the amplitude response near Fs/2 rolls off towards 0 (sin X / X response). geoff steckel (gwes@wjh12.harvard.EDU) (...!husc6!wjh12!omnivore!gws) Disclaimer: I am not affiliated with Sun Microsystems, despite the From: line. This posting is entirely the author's responsibility.
marshall@elric.dec.com (Hunting the Snark) (02/21/91)
In article <4402@eastapps.East.Sun.COM>, gsteckel@vergil.East.Sun.COM (Geoff Steckel - Sun BOS Hardware CONTRACTOR) writes... >In article <883@idacrd.UUCP> mac@idacrd.UUCP (Robert McGwier) writes: > From article <6607@healey>, by grayt@Software.Mitel.COM (Tom Gray): > > The shape of the wave is preserved within the sampling pulse. This > > information allows representation of a signal at exactly 1/2 > > the Nyquist freqency. > > I pose the following question. Suppose you are sampling at rate N samples > per second, and you see a constant value V for your A/D sample. Is the > frequency of the signal which produced those samples 0 or N/2? Since > >AARRGGGHHH! The Nyquist criterion requires that sampling be GREATER >THAN the highest frequency of interest. Wrong answer. The answer to the question posed is that you are sampling a DC signal. if it was N/2 then the sign of the samples would alternate. The only time you can't distinguish DC and N/2 is when you just happen to sample at the zero crossings of the N/2 frequency signal. The way to get around this degenerate case is to random sample with an _average_ sample rate of N samples per second. Actually, you do not even need to be randomly sampling, you could for instance instead of sampling with a 50% duty cycle you could use a 25% duty cycle and still meet the Nyquist criteria. /> ( //------------------------------------------------------------( (*)OXOXOXOXO(*>=S=T=O=R=M=B=R=I=N=G=E=R-------- \ ( \\--------------------------------------------------------------) \> Steven Marshall "Hard to say Ma'am. I think my cerebellum just fused" -- Calvin