Forums

Nyquist????

Started by krish_dsp October 26, 2008
On 26 Okt, 04:26, "krish_dsp" <nmkr...@gmail.com> wrote:
> Hi everyone, > What is the dirrerence between nyquist frequency, nyquist rate and nyquist > bandwidth? I am sure that they dont point to the same concept. I know what > is a nyquist frequency. But I am not able to understand the concept of the > other two.
'Nyquist frequency' and 'Nyquist rate' are the same thing. 'Nyquist period' is the time period T = 1/2Fs where Fs is the sampling frequeny. 'Nyqust bandwidth' relates to the fact that the signal needs not be baseband. In this case, the Nyquist bandwidth relates to the Nyquist frequency which would apply if the signal was mixed down to baseband before it was sampled. Rune
Rune Allnor wrote:
(snip)

> 'Nyqust bandwidth' relates to the fact that the signal > needs not be baseband. In this case, the Nyquist bandwidth > relates to the Nyquist frequency which would apply if the > signal was mixed down to baseband before it was sampled.
That is an interesting way to say it. I think I agree. In the case of an AM-DSB signal, it should be mixed down with the original carrier at DC. (Same for AM-DSB-SC) The Nyquist bandwidth is then half the signal bandwidth. For AM-VSB it should again be mixed down with the carrier at DC, such that the Nyquist bandwidth will be somewhat less than the signal bandwidth, but more than half. Other signals with redundancy will have to be processed in other ways to extract the appropriate signal. -- glen
On Oct 28, 9:53&#2013266080;am, Rune Allnor <all...@tele.ntnu.no> wrote:
> > > 'Nyquist frequency' and 'Nyquist rate' are the same thing.
no they're not. i don't think there is a single textbook reference that would support that, Rune. some people (like O&S) might say that the Nyquist rate is always twice the Nyquist frequency (and i don't, i believe the consensus convention is that the "Nyquist rate" is a property of the signal to be sampled and the "Nyquist frequency" is a property of the system doing the sampling), but i don't know of a single reference that says they're the same. it's just a convention of definition, but it's important for efficient communication of ideas. i think that poor conventions die away due to disuse and better conventions survive. r b-j

Greg Berchin wrote:
> > On Mon, 27 Oct 2008 17:31:08 -0700 (PDT), robert bristow-johnson > <rbj@audioimagination.com> wrote: > > >the good guys say it's always 1/2 of the sampling rate, > > The "good guys"? > > I guess basing the Nyquist Frequency on characteristics of the signal > to be sampled must therefore be "elitist". :-)
Well if you are just in the planning stages of sampling a signal the definition of what may be the Nyquist Frequency might be a little murky. If you have a digital signal and you do a DFT and somebody comes along and observes that there is no energy in your signal at the Nyquist frequency - Is there any doubt as to what that means? -jim ----== Posted via Pronews.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.pronews.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= - Total Privacy via Encryption =---
On Tue, 28 Oct 2008 10:33:12 -0800, Glen Herrmannsfeldt
<gah@ugcs.caltech.edu> wrote:

>In the case of an AM-DSB signal, it should be mixed down >with the original carrier at DC. (Same for AM-DSB-SC) >The Nyquist bandwidth is then half the signal bandwidth.
I would argue that the Nyquist Bandwidth equals the full (double-sided, if baseband) signal bandwidth, because of the generalization that I presented in my earlier post. When viewed that way, the Nyquist Rate is always equal to the Nyquist Bandwidth, whether the signal is baseband or modulated, real or complex. Then the only time that the concept of Nyquist Frequency has meaning is when, for example, one wants to sample a bandpass signal or an analytic signal WITHOUT shifting the spectrum to baseband in the process. In such cases the Nyquist Frequency is equal to the highest (absolute value of) frequency of significant amplitude contained within the signal, and the Nyquist Rate is twice the Nyquist Frequency. Note that all of these entities are based upon characteristics of the signal to be sampled, not of the sampling itself. -- Greg
Greg Berchin  <gberchin@comicast.net.invalid> wrote:

>Then the only time that the concept of Nyquist Frequency has meaning >is when, for example, one wants to sample a bandpass signal or an >analytic signal WITHOUT shifting the spectrum to baseband in the >process. In such cases the Nyquist Frequency is equal to the highest >(absolute value of) frequency of significant amplitude contained >within the signal, and the Nyquist Rate is twice the Nyquist >Frequency.
I would think the Nyquist rate is necessary sample rate to sample a signal without aliasing, and in the case of a bandpass signal it is related to signal bandwidth and not absolute frequencies in the signal. Steve
On Tue, 28 Oct 2008 22:24:31 +0000 (UTC), spope33@speedymail.org
(Steve Pope) wrote:

>I would think the Nyquist rate is necessary sample rate to >sample a signal without aliasing, and in the case of a bandpass >signal it is related to signal bandwidth and not absolute >frequencies in the signal.
And that is exactly what I said in the first part of my post. But what happens if you have a strictly bandlimited bandpass signal that you want to sample without ambiguity? If you set your sampling frequency according to bandwidth alone (Nyquist Bandwidth = bandwidth; Nyquist Rate = Nyquist Bandwidth; sampling rate > Nyquist Rate), then you also need to know the center of the passband in order to reconstruct the signal. But if you set your sampling rate according to the highest |frequency| in the signal (Nyquist Frequency = |highest frequency|; sampling frequency > 2 x Nyquist Frequency), then you no longer need to know the center of the passband because it is constrained to be zero. -- Greg
Greg Berchin wrote:
> On Tue, 28 Oct 2008 22:24:31 +0000 (UTC), spope33@speedymail.org > (Steve Pope) wrote: > >> I would think the Nyquist rate is necessary sample rate to >> sample a signal without aliasing, and in the case of a bandpass >> signal it is related to signal bandwidth and not absolute >> frequencies in the signal. > > And that is exactly what I said in the first part of my post. But > what happens if you have a strictly bandlimited bandpass signal that > you want to sample without ambiguity? If you set your sampling > frequency according to bandwidth alone (Nyquist Bandwidth = bandwidth; > Nyquist Rate = Nyquist Bandwidth; sampling rate > Nyquist Rate), then > you also need to know the center of the passband in order to > reconstruct the signal. But if you set your sampling rate according > to the highest |frequency| in the signal (Nyquist Frequency = |highest > frequency|; sampling frequency > 2 x Nyquist Frequency), then you no > longer need to know the center of the passband because it is > constrained to be zero.
You don't need to know the center of the band with much precision. You need only enough information to select the right image. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
On Tue, 28 Oct 2008 21:26:26 -0400, Jerry Avins <jya@ieee.org> wrote:

>You don't need to know the center of the band with much precision. You >need only enough information to select the right image.
Doesn't knowing which image is the right image imply that you know where the images were located, which in turn implies that you know the center of the band exactly?
Greg Berchin wrote:
> On Tue, 28 Oct 2008 21:26:26 -0400, Jerry Avins <jya@ieee.org> wrote: > >> You don't need to know the center of the band with much precision. You >> need only enough information to select the right image. > > Doesn't knowing which image is the right image imply that you know > where the images were located, which in turn implies that you know the > center of the band exactly?
You can figure it out exactly, but the exact center is quantized, so choosing the correct image needs only an approximate value. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;