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I know that if a system is causal then the system function H(z) must have : a) a ROC that spans from the exterior of the most distant pole and b) the number of zeros must not be greater than the number of poles

I have found this exercise: enter image description here and the solution: enter image description here Why the number of poles and zeros must be equal?

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Regardless of causality and stability, if you count poles and zeros at the origin and at infinity, the total number of poles always equals the total number of zeros. I'll show a few examples to make this obvious.

First, take a polynomial

$$P(z)=(z-z_0)(z-z_1)\ldots (z-z_M)\tag{1}$$

It has $M+1$ zeros and no (finite) poles. However, it must have $M+1$ poles at $z=\infty$ (because of the term $z^{M+1}$).

Next consider an "all-pole" function

$$A(z)=\frac{1}{(z-z_0)(z-z_1)\ldots (z-z_N)}\tag{2}$$

with $N+1$ poles and no (finite) zeros. Clearly, due to the $N+1$ poles, the function $A(z)$ in $(2)$ has $N+1$ zeros at infinity, because of the term $z^{N+1}$ in the denominator.

In general, if you have

$$H(z)=\frac{(z-z_{0,0})(z-z_{0,1})\ldots (z-z_{0,M})}{(z-z_{\infty,0})(z-z_{\infty,1})\ldots (z-z_{\infty,N})}\tag{3}$$

and if $M>N$, you get $M-N$ poles at infinity. For $M<N$, you have $N-M$ zeros at infinity. Consequently, in all cases the number of poles equals the number of zeros if poles and zeros at infinity are included.

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  • $\begingroup$ I agree but that happens if your H(z) has this fractional form. Some LTI have this kind of form , but not all of them and the exercise does not specify which form the LTI system function. I believe that the exercise should clarify this.. $\endgroup$ – manpad May 7 at 9:00
  • $\begingroup$ @manpad: A pole-zero plot is usually understood as a graphical representation of a rational (transfer) function. $\endgroup$ – Matt L. May 7 at 10:19
  • $\begingroup$ "However, it must have M+1 poles at z=∞" why? $\endgroup$ – Filipe Pinto May 7 at 13:06
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    $\begingroup$ @FilipePinto: Check $\lim_{z\to\infty}P(z)$. If you write out $(1)$ in powers of $z$ you get a leading term $z^{M+1}$, which is a pole with multiplicity $M+1$ at $\infty$. $\endgroup$ – Matt L. May 7 at 16:37
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If a discrete-time LTI system is causal (that is that the impulse response doesn't respond before the driving impulse), then the number of zeros must not exceed the number of poles.

Consider this general transfer function:

$$\begin{align} H(z) &= A\frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\ \\ &= A\frac{z^M + b_1z^{M-1} + ... + b_{M-2}z^2 + b_{M-1}z + b_M}{z^N + a_1z^{N-1} + ... + a_{N-2}z^2 + a_{N-1}z + a_N} \end{align}$$

where $p_i$ are the poles and $q_i$ are the zeros. Now if $M=N$, then using long division, the denominator can be divided into the numerator and you'll get

$$ H(z) = A + \tfrac{(b_1-Aa_1)z^{M-1} + ... + (b_{M-2}-Aa_{N-2})z^2 + (b_{M-1}-Aa_{N-1})z + (b_M-Aa_N)}{z^N + a_1z^{N-1} + ... + a_{N-2}z^2 + a_{N-1}z + a_N} $$

and that's fine because that stand-alone $A$ term corresponds to $A\delta[n]$ in the impulse response and it's still causal. But now let's say that $M=N+1$, then you get

$$ H(z) = Az + A(b_1-1) + \tfrac{(b_1-Aa_2)z^{M-1} + ... + (b_{M-2}-Aa_{N-1})z^2 + (b_{M-1}-Aa_N)z + b_M}{z^N + a_1z^{N-1} + ... + a_{N-2}z^2 + a_{N-1}z + a_N} $$

and now you have an acausal term in the impulse response corresponding to $Az$.

So for a causal LTI system, the number of poles must be at least as many as the number of zeros. Even FIR filters have poles at the origin.

But the inverse filter has the reciprocal, $1/H(z)$ as a transfer function. If the $N>M$ for the original filter, that would be okay, but for the inverse filter you've swapped the roles of poles and zeros, now $N$ is the number of zeros and $M$ is the number of poles. So in order for both the original and inverse filters to be causal, then $N=M$, the number of poles must equal the number of zeros.

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  • $\begingroup$ That's all correct, but in the given example, the inverse system is not causal, but still they claim that the number of poles must equal the number of zeros, so there is another reason for that claim, having nothing to do with causality. $\endgroup$ – Matt L. May 7 at 8:31

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