Reply by Rune Allnor August 9, 20052005-08-09
Newbie wrote:
> Guys: > Why do the charts of the textbook representing the frequency response > of a filter always show the range from -pi to pi instead of from 0 to > 2pi? If use the range from 0 to 2pi, examples in the chapter 2 of > Understanding Digital Processing by Lyons would cause big problem. The > sampling replication would shift to the right and aliasing explaination > failed. Why "-pi to pi" not "0 to 2pi"?
Because the <-pi,pi> interval resempbles the continuous time Fourier transform. Rune
Reply by Fred Marshall August 8, 20052005-08-08
"Newbie" <EdwinYang@gmail.com> wrote in message 
news:1123513854.663337.238070@z14g2000cwz.googlegroups.com...
> Dear Fred: > Thanks for your replay. I think the representation like the chart on > the bottom of > http://www.phptr.com/content/images/chap2_0131089897/elementLinks/02fig02.gif > is not for convenient. Because if the bandwidth of the sampling > frequency were not between (-3) and (3)MHz, the replication would be > totally diferent. And the explaination of aliasing does not work out. > The sample chapter is at > http://www.phptr.com/articles/article.asp?p=345472 for your reference. >
Edward, I'm not sure how this relates to your original question. The sampling "frequency" does not have a "bandwidth" as such. I guess you mean simply the "sampling frequency" being 6MHz? Is that it? Just as shown in the figure. Well, hey, if the sampling frequency is different than one number or another then the signal frequencies that are allowed by the Nyquist criterion are different. The dispacement that we call aliasing depends on the sampling frequency and the displaced / aliased signal frequency - both. I don't see words associated with the diagram. Here's what it means: - If you sample at 6kHz and apply a signal at 7kHz, then the result will appear the same as if you had sampled a signal at 1kHz. 7kHz - 6kHz = 1kHz. - If you sample at 6kHz and apply a signal at 4kHz, then the result will appear as if you had sampled a signal at -2kHz. Of course, if the signal is real, then a signal at 7Khz also has a component at -7kHz and the sampling has terms at -6kHz and +6kHz. Sampling results in sum and difference frequencies of the input and the sampling frequency. So, 6kHz +/- 7kHz is +/-1kHz so there will be terms from 7kHz at +1kHz and =1kHz. And, 4kHz +/- 6kHz results in +/-2kHz as the aliased components. We ignore the sum terms at +/-13kHz and +/-10kHz because they are already represented by the terms at 1kHz and 4kHz. If this weren't sampled data and the spectrum continuous, then we would include terms out to infinite frequency (or at least to some practical upper limit). Fred
Reply by Jerry Avins August 8, 20052005-08-08
Newbie wrote:
> Dear Fred: > Thanks for your replay. I think the representation like the chart on > the bottom of > http://www.phptr.com/content/images/chap2_0131089897/elementLinks/02fig02.gif > is not for convenient. Because if the bandwidth of the sampling > frequency were not between (-3) and (3)MHz, the replication would be > totally diferent. And the explaination of aliasing does not work out.
What part of the explanation seems incongruous to you?
> The sample chapter is at > http://www.phptr.com/articles/article.asp?p=345472 for your reference.
Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by Newbie August 8, 20052005-08-08
Dear Fred:
 Thanks for your replay. I think the representation like the chart on
the bottom of
http://www.phptr.com/content/images/chap2_0131089897/elementLinks/02fig02.gif
is not for convenient. Because if the bandwidth of the sampling
frequency were not between (-3) and (3)MHz, the replication would be
totally diferent. And the explaination of aliasing does not work out.
The sample chapter is at
http://www.phptr.com/articles/article.asp?p=345472 for your reference.

BR,
Edward Yang

Reply by robert bristow-johnson August 7, 20052005-08-07
in article 1123432917.359421.254550@g47g2000cwa.googlegroups.com, Newbie at
EdwinYang@gmail.com wrote on 08/07/2005 12:41:

> Why do the charts of the textbook representing the frequency response > of a filter always show the range from -pi to pi instead of from 0 to > 2pi? If use the range from 0 to 2pi, examples in the chapter 2 of > Understanding Digital Processing by Lyons would cause big problem. The > sampling replication would shift to the right and aliasing explaination > failed. Why "-pi to pi" not "0 to 2pi"?
are you talking about the frequency axis or the phase response? if the former, that can represent the frequency response of a *real* signal (if symmetry about DC is present). the 0 to 2pi frequency response almost has to be a complex signal. it's kinda hard for me to find an oscilloscope that i can connect a probe up to a complex voltage. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by Fred Marshall August 7, 20052005-08-07
"Newbie" <EdwinYang@gmail.com> wrote in message 
news:1123432917.359421.254550@g47g2000cwa.googlegroups.com...
> Guys: > Why do the charts of the textbook representing the frequency response > of a filter always show the range from -pi to pi instead of from 0 to > 2pi? If use the range from 0 to 2pi, examples in the chapter 2 of > Understanding Digital Processing by Lyons would cause big problem. The > sampling replication would shift to the right and aliasing explaination > failed. Why "-pi to pi" not "0 to 2pi"? > > Repling to my email is welcome! > > BR, > Edward Yang >
Edward, Because it repeats in periods of 2pi then any defined range of 2pi gives all the information. The information from -pi to zero is identical to the information from pi to 2pi. Give from -pi to pi makes drawing pictures and visualization easier I believe. For example, if you have a weighted cosine in time with it's frequency an integer multiple of 1/NT - say K/NT - where K is an integer, where N is the number of samples and T is the temporal sampling interval / the reciprocal of the sampling frequency - then there will be two weighted samples in frequency - one at -K/NT and one at K/NT. If you plot this from zero to 2pi, then you will have one of the samples at K/NT and one at 1/T-K/NT. Since there is no sample at 1/T, this is harder to both visualize and do the index math for the sample locations. If the index system is like in Matlab / Scilab where zero frequency has index 1, then it's just that much harder. Fortran (older Fortran?) is similar to those. C starts with a zero index value in arrays making the index math easier to relate to physical values. Said another way: The frequency domain is periodic and "wraps around" - the wrap-around frequency is convenient to be zero, 1/T, 2/T .... i.e. zero, 2pi,4pi, etc. This is the way most FFT programs do it. But, the wrap-around frequency can just as easily be -pi,pi,3pi.... and may be a better choice for visualization purposes. Fred
Reply by Jerry Avins August 7, 20052005-08-07
Newbie wrote:
> Guys: > Why do the charts of the textbook representing the frequency response > of a filter always show the range from -pi to pi instead of from 0 to > 2pi? If use the range from 0 to 2pi, examples in the chapter 2 of > Understanding Digital Processing by Lyons would cause big problem. The > sampling replication would shift to the right and aliasing explaination > failed. Why "-pi to pi" not "0 to 2pi"?
Because, as you already pointed out, zero to 2pi makes big problems. Why? That's a good question. An answer is that there is really enough information (at baseband) to go from zero to pi with trigonometry, but the simpler way we do the math (using complex exponentials) requires negative frequencies. (Take that point of view with a grain of salt. Many will dispute it.) Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by Newbie August 7, 20052005-08-07
Guys:
  Why do the charts of the textbook representing the frequency response
of a filter always show the range from -pi to pi instead of from 0 to
2pi? If use the range from 0 to 2pi, examples in the chapter 2 of
Understanding Digital Processing by Lyons would cause big problem. The
sampling replication would shift to the right and aliasing explaination
failed. Why "-pi to pi" not "0 to 2pi"?

Repling to my email is welcome!

BR,
Edward Yang