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I have two questions regarding systems with no poles:

  • Why does a system with no poles have a finite support?

  • Why if the system has a finite support it means that it is BIBO stable?

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Why does a system with no poles have a finite support?

If a system doesn't have finite poles, then its transfer function is of the form:

$$H(z) = \frac{Y(z)}{X(z)} = a_Nz^N+a_{N-1}z^{N-1}+...+a_1z+a_0$$

So if you go back to the time domain:

$$y[n]=a_Nx[n+N]+a_{N-1}x[n+N-1]+...+a_1x[n+1]+a_0x[n]$$

Note that if $x[n]=\delta[n]$, then $y[n]=h[n]$, the impulse response of the filter.

$$h[n]=a_N\delta[n+N]+a_{N-1}\delta[n+N-1]+...+a_1\delta[n+1]+a_0\delta[n]$$

It's easy to see that the equation above will return a $h[n]$ that is non-zero for a limited range of values of $n$.

This means that the filter is FIR. Therefore, it has a finite support and it is BIBO stable, leading to your second question.

Why if the system has a finite support it means that it is BIBO stable?

For BIBO stability of an LTI discrete-time system, one must check that the following condition is satisfied:

$$\sum_{n=-\infty}^{\infty}|h[n]|<\infty$$

If $h[n]$ is the impulse response of a FIR filter, then it has some length $L$:

$$\sum_{n=-\infty}^{\infty}|h[n]|=\sum_{n=K}^{L+K-1}|h[n]|$$

Due to this summation being finite, it converges. Ergo, any FIR filter is BIBO stable.

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  • $\begingroup$ Referring to the first question, for example this system: H(z)=z−2 It's really system with no poles, right? $\endgroup$ – Sama Assi Jul 14 '18 at 16:48
  • $\begingroup$ @SamaAssi Sorry, for some reason I understood some other thing. I've edited the answer. $\endgroup$ – Tendero Jul 14 '18 at 16:49
  • $\begingroup$ The system $H(z)=z-2$ has no finite poles, but it has a pole at infinity. $\endgroup$ – Matt L. Jul 14 '18 at 18:45

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