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

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

Started by ●February 16, 2006

Reply by ●February 17, 20062006-02-17

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