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Signal Processing

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  • $\begingroup$ Why do you specify that the zero is at negative infinity? A zero at infinity would mean that $H(z) \xrightarrow{z\to\infty}0$. This is the same as saying that the system has a pole at the origin. $\endgroup$ Commented Apr 5, 2016 at 0:05
  • $\begingroup$ Yes, correct SleuthEye. Thanks for catching that. $\endgroup$ Commented Apr 5, 2016 at 1:01
  • $\begingroup$ M. S. That was given in the prompt for the problem. I just e-mailed my Prof. and he said that if we were to represent $H(z)$ as $\frac{\sum_{m=0}^{M} b_m z^{-m}}{1+\sum_{k=1}^{N} a_k z^{-k}}$ = $\frac{b_0+b_1 z^{-1}+\dots +b_M z^{-M}}{1+a_1 z^{-1}+\dots+a_N z^{-N}}$, a zero at negative infinity means that $b_0=0$. $\endgroup$ Commented Apr 5, 2016 at 1:10
  • $\begingroup$ We talk about "negative infinity" when calculating limits, for example, in real calculus. When dealing with complex numbers, the "point at infinity" is unique and is neither positive or negative. If you want more information about this topic, you can investigate about the extended complex plane. Regarding the last equation you wrote, the fact that that $H(z)$ has or not a zero at infinity depends on the values of $M$ and $N$. Only if the denominator is of higher order than the numerator will there be zero(s) at infinity. $\endgroup$ Commented Apr 5, 2016 at 15:11