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Frequency convolution

Started by Seppo October 10, 2005
Hello everyone. I have a question I cant find an answer to:

Say you have, X(z) the z transform of a sampled time response x[n],
i.e.

Z{x[n]} = X(z)

Does a Y(z) exist s.t.

Z{x[n]^2} = Y(z)

and if so, how does it relate to X(z)? If you dont feel challenged
enough, try this other one: Does a Y2(z) exist such that:

Z{abs(x[n])} = Y2(z)?

For the first point I am aware that multiplication in the time domain
is the same as convolution in the frequency domain, and that
convolution in the z-domain is expressed as a contour integral, but I
have no idea how to interpret these.

My guess is that Y(z) will only have positive coefficients.

Seppo wrote:

> Hello everyone. I have a question I cant find an answer to: > > Say you have, X(z) the z transform of a sampled time response x[n], > i.e. > > Z{x[n]} = X(z) > > Does a Y(z) exist s.t. > > Z{x[n]^2} = Y(z)
Yes; if you can express y[n] = f(x[n]) you can find the z transform of y given f and x.
> > and if so, how does it relate to X(z)?
Not necessarily very directly, if you have X(z) in the form of a ratio of polynomials.
> If you dont feel challenged > enough, try this other one: Does a Y2(z) exist such that: > > Z{abs(x[n])} = Y2(z)?
Yes, see above.
> > For the first point I am aware that multiplication in the time domain > is the same as convolution in the frequency domain, and that > convolution in the z-domain is expressed as a contour integral, but I > have no idea how to interpret these. > > My guess is that Y(z) will only have positive coefficients. >
If you are doing Fourier analysis in discrete-time then the y[n] = x[n]^2 case is fairly easy to analyze; the y[n] = abs(x[n]) isn't really worth it. If you could find a Y(z) that was a ratio of polynomials then it wouldn't necessarily have all positive coefficients. Answering these questions yourself will give you a deeper understanding of the z transform, but it's the trip, not the destination, that does the teaching. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com