1
$\begingroup$

My professor recently said something along the lines of, "finite length signals can't have poles because poles correspond to recursive difference equations which correspond to infinite length signals."

I can't really wrap my head around this. Can someone explain in a different or more simple way why finite length signals can't have poles? Would much prefer a simple, hand-wavy explanation that I can remember easily rather than a rigorous explanation/proof.

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ your professor is wrong if the "finite-length signals" are causal. such as the impulse response of a causal FIR filter. in that case (a causal FIR), there are as many poles as there are zeros. but all of the poles are at the origin. however, if your "finite-length signals" are symmetrical about the origin, that is $x[-n]=x[n]$, then there are no poles. see also this old answer. $\endgroup$
    – robert bristow-johnson
    Dec 16, 2016 at 2:30
  • 1
    $\begingroup$ Sorry for being pedantic here... signals don't have poles, only systems (or their transfer function) do. $\endgroup$
    – MBaz
    Dec 16, 2016 at 3:08
  • 1
    $\begingroup$ As MBaz points out, your professor most likely talks about the poles of the transfer function. Maybe you understood a function with a pole discontinuity, like x(t) = tan(t) ? $\endgroup$
    – Maximilian Matthé
    Dec 16, 2016 at 4:17
  • $\begingroup$ Sorry yes I meant to ask why the transfer function of a finite length signal has no polls $\endgroup$
    – Austin
    Dec 16, 2016 at 4:19
  • $\begingroup$ what do you mean by a "transfer function of a finite length signal"? do you mean "impulse response" in the place of "signal"? $\endgroup$
    – robert bristow-johnson
    Dec 16, 2016 at 5:10

2 Answers 2

Reset to default
1
$\begingroup$

Hand-wavy? A rational fraction $$\frac{N(X)}{D(X)}$$ where $N(X)$ and $D(X)$ are polynomials can be written as some polynomial $P(X)$ plus a finite amount of terms of the shape:

$$\frac{a_i}{X-X_i}\,.$$

Let us say that a $0$ pole ($X_i=0$) is harmless, because it is just some $X^{-1}$. So on the one hand, you have the monomials $X^d$, $d\ge0$, from $P$ and possibly a $X^{-1}$. If you replace the $X$ by a $z^{-n}$, you will still have a finite number of $z$ to the power of some integer, which are just delays.

On the other hand, if you have one non-zero pole:

$$\frac{1}{X-X_j} = \frac{1}{X}\left(1+ \left(\frac{X_j}{X}\right)^1+\left(\frac{X_j}{X}\right)^2+\cdots\right)$$ and replace the $X$ by some $z^{-n}$, you will get an infinite quantity of $z$ to the power of some integer, and thus an infinite impulse response (IIR). The complicated part is to show that several non-zero poles do not "cancel" each other. A lot of people confuse IIR and recursive.

Be cautious though that an apparently "recursive implementation" does not always imply non-zero poles: a FIR system with $z$-transform $1-z^{-2}+z^{-4}-z^{-6}$ can be written as:

$$ \frac{1-z^{-8}}{1+z^{-2}}$$

that is, you can rewrite: $$y ( k ) = x ( k ) - x ( k - 2 ) + x ( k - 4 ) - x ( k - 6 )$$ as $$y ( k ) = x ( k ) - x ( k - 8 ) - y ( k - 2 )\,,$$

which is a recursive difference equation, yet with a FIR.

Improve this answer
$\endgroup$
0
$\begingroup$

Hand-wavy :)....

A Finite Impulse Response convoluted with any signal will never lead to infinite values!....

Because it is finite, you are just multiplying and adding a limited number of times.

With an Infinite Impulse Response, you have the danger for diverging with some kind of signals (at some frequencies). Those frequencies are... the poles!....

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.