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fundamental question

Started by bulk September 12, 2006
A very basic question:

Take two properly sampled signals (more than nyquist). Now I mulitply
the two sample streams. From a continous time view point it is easy to
see that the product could have frequencies for which the initial
sampling rate wouldnt be enough.

So in any DSP system is one not supposed to multiply two signals? Has
this something to do with whether multiplication with a variable value
is a linear operation or not? Is the solution that one has to up-sample
before doing such an operation to a rate that nyquist for the product
signal is met before multiplication?

- b

bulk wrote:

> Is the solution that one has to up-sample > before doing such an operation to a rate that nyquist for the product > signal is met before multiplication? > > - b
Correct.
> Take two properly sampled signals (more than nyquist). Now I mulitply > the two sample streams. From a continous time view point it is easy to > see that the product could have frequencies for which the initial > sampling rate wouldnt be enough. > > So in any DSP system is one not supposed to multiply two signals? Has > this something to do with whether multiplication with a variable value > is a linear operation or not? Is the solution that one has to up-sample > before doing such an operation to a rate that nyquist for the product > signal is met before multiplication?
multiplication in the time domain is the same as convolution in the frequency domain. anyway....... try multipling sin(2*pi*f1*t)*sin(2*pi*f2*t) and you get 1/2*cos(2*pi*[f1-f2]*t)-1/2*cos(2*pi*[f1+f2]*t) Lets say that f2=fs/2 and f1=fs/2 where fs is the sampling frequency. Then: Muliplying the two sine-components yields a DC-component (0Hz) and another frequency component at fs Hz.

Joe wrote:
> > > Take two properly sampled signals (more than nyquist). Now I mulitply > > the two sample streams. From a continous time view point it is easy to > > see that the product could have frequencies for which the initial > > sampling rate wouldnt be enough. > > > > So in any DSP system is one not supposed to multiply two signals? Has > > this something to do with whether multiplication with a variable value > > is a linear operation or not? Is the solution that one has to up-sample > > before doing such an operation to a rate that nyquist for the product > > signal is met before multiplication? > > multiplication in the time domain is the same as convolution in the > frequency domain. > > anyway....... > > try multipling sin(2*pi*f1*t)*sin(2*pi*f2*t) and you get > > 1/2*cos(2*pi*[f1-f2]*t)-1/2*cos(2*pi*[f1+f2]*t) > > Lets say that f2=fs/2 and f1=fs/2 where fs is the sampling frequency. > > Then: > > Muliplying the two sine-components yields a DC-component (0Hz) and > another frequency component at fs Hz.
By my calculation that gives you a lot of zeroes for an output. What was your point? -jim ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
> > By my calculation that gives you a lot of zeroes for an output. What was > your point?
My point was to illustrate what happens in the frequency domain when you multiply two signals with frequency f1 and f2, respectively.
"bulk" <unsettledinside@yahoo.com> wrote in message 
news:1158054370.069145.114760@b28g2000cwb.googlegroups.com...
>A very basic question: > > Take two properly sampled signals (more than nyquist). Now I multiply > the two sample streams. From a continous time view point it is easy to > see that the product could have frequencies for which the initial > sampling rate wouldn't be enough. > > So in any DSP system is one not supposed to multiply two signals?
You can do whatever you want.
>Has > this something to do with whether multiplication with a variable value > is a linear operation or not?
I don't think so : If it was, how would that help you?
>Is the solution that one has to up-sample > before doing such an operation to a rate that nyquist for the product > signal is met before multiplication?
That should work in all cases. Best of Luck - Mike

bulk wrote:

> A very basic question: > > Take two properly sampled signals (more than nyquist). Now I mulitply > the two sample streams. From a continous time view point it is easy to > see that the product could have frequencies for which the initial > sampling rate wouldnt be enough.
This is correct.
> So in any DSP system is one not supposed to multiply two signals?
Care should be taken to avoid the problems due to the possible aliasing.
> Has > this something to do with whether multiplication with a variable value > is a linear operation or not?
Multiplication of two variables is bilinear.
> Is the solution that one has to up-sample > before doing such an operation to a rate that nyquist for the product > signal is met before multiplication?
It depends on what are you trying to accomplish. Usually, when you are multiplying two signals, your desired result is the "difference" product. The "sum" product can be aliased, however it does not matter if it is suppresed in the algorithm. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com

Joe wrote:

>>By my calculation that gives you a lot of zeroes for an output. What was >>your point? > > > > My point was to illustrate what happens in the frequency domain when you > multiply two signals > with frequency f1 and f2, respectively. >
When you multiply F1 and F2, you get F1 - F2 and F1 + F2. Usually your goal is to obtain F1 - F2. In this case, the F1 + F2 can be aliased. As long as the alias of F1 + F2 is outside the bandwidth of interest, it can be filtered out. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com

Joe wrote:
> > > > > By my calculation that gives you a lot of zeroes for an output. What was > > your point? > > My point was to illustrate what happens in the frequency domain when you > multiply two signals > with frequency f1 and f2, respectively.
Well yes.. er.. maybe.... But given that you chose the one particular example that completely side steps the question the OP asked, I was just wondering how you expected him to draw any useful conclusions from that. I mean, multiplying by a sin function at fs/2 is probably not something the OP needs help with. -jim ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
bulk wrote:

> A very basic question: > > Take two properly sampled signals (more than nyquist). Now I mulitply > the two sample streams. From a continous time view point it is easy to > see that the product could have frequencies for which the initial > sampling rate wouldnt be enough. > > So in any DSP system is one not supposed to multiply two signals?
It's not nearly that restrictive. But before you multiply two signals (or run one through a nonlinear function) you should think about what you're doing.
> Has > this something to do with whether multiplication with a variable value > is a linear operation or not?
No, because multiplying your signal by a time varying quantity is a linear operation (it's just not time invariant).
> Is the solution that one has to up-sample > before doing such an operation to a rate that nyquist for the product > signal is met before multiplication?
If you've determined that it's going to be an issue, yes. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html