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If a z transform has a pole on the unit circle does that signal have a fourier transform?

Started by Abhishek February 16, 2006
Hi,
If I have been given a z-transform and I know that the z transform has
a pole on the unit circle at a certain angle, does it mean that the
fourier transform does not exist at all, because I have read a paper
where the author tries to derive the fourier transform from the z
transform even when it has a poleon the unit circle. I know if all the
poles are inside the unit circle and if the signal is causal, we can
substitute exp(jw) for z and get the z transform..but what if the
signal is causal and the z transform has a pole on the unit circle? I
can post the research paper if you want.
Bye

Z in polar form z=rexp(jw).
r=|z|, w=angle.
Within, Region of convergence(ROC), substitute z,
X(Z) at z=rexp(jw) = SUM[x(n)r(power of -n) exp(-jwn) ] SUM over +inf
to -inf

X(Z) can be interpreted as fourier transform of [x(n) r(power of -n) ]
weight factor (r(power of -n))  decays with r>1

If X(Z) converges with |z|=1then it reduces to fourier transform for
x(n).

If X(Z) does not converge for |z|=1 (when unit circle is not within
ROC) fourier transform for X(Z) does not exist.

It is important to note that existance of Z transform needs [x(n)
r(power of -n) ] absolutely summable i.e it leads to condition for ROC.
Cheers,
Santosh