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Let $N$ be the signal graph representation network of the system function (in rationale form) of a discrete-time LTI system. For a network $N$ with no loops, the impulse response is no longer than the total number of delay elements in the network. From this, we conclude that if a network has no loops, then the system function has only zeros (except for poles at $z = 0$), and the number of zeros can be no more than the number of delay elements in the network.

Q: Why can the number of zeros be no more than the number of delay elements in the network?

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Because for a polynomial of degree $N$, the number of zeros (roots) is no more than $N$.

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There is nothing preventing you from writing down an expression that has more zeros than poles. In fact, a typical FIR filter may be designed as something like $$\sum_{n = -\frac{N-1}{2}}^{\frac{N-1}{2}} f(n) z^n$$ giving you a filter that's nice and symmetrical around zero delay.

The problem is, that in the physical world, $z$ is a lead operator -- it fetches samples from the future. Since physicists stubbornly refuse to invent time machines for us, we have to make do with filters that only operate on past values of the input.

Thus, we only allow $z^{-n}$, and accept $z^0$ because we have an unspoken agreement that it means "the latest sample, with just a bit of delay".

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  • $\begingroup$ +1; Just for the sake of completeness: transfer functions always have the same numbers of poles and zeros IF we also count poles and zeros at infinity and at the origin. $\endgroup$
    – Matt L.
    Jan 14, 2021 at 18:49

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