1
$\begingroup$

I have the following equation:

$$H(z) = \frac{z^3+5z^2+3z+1}{10z^3}$$

I want to find out whether if its IIR or FIR. Here is the steps I have gone so far: $$H(z) = \frac{1}{10}\frac{z^3+5z^2+3z+1}{z^3}$$ Dividing nominator with denominator to find reminder gave me(I think its wrong): $$z^3+5z^2+3z$$ After this step I have a $1$ as remainder and I do not know what to do with it.

Can someone please tell me the right procedure step by step?

Improve this question
$\endgroup$
  • $\begingroup$ Why don't you just divide each term in the numerator by $z^3$? Don't worry about negative powers of $z$; they occur because you have a causal filter. $\endgroup$ – Matt L. Jul 29 '15 at 19:43
  • $\begingroup$ @MattL. In this case it would work but what if the denum is something like $z^2(1-z^-2)$ $\endgroup$ – Saeid Yazdani Jul 30 '15 at 8:14
  • $\begingroup$ I added an answer discussing the general case. $\endgroup$ – Matt L. Jul 30 '15 at 9:26
6
$\begingroup$

In the general case you have

$$H(z)=\frac{P(z)}{Q(z)}$$

where $P(z)$ and $Q(z)$ are polynomials in $z$. If - as is the case in your example - $Q(z)$ just has one single term, $H(z)$ is definitely FIR, because you can simply divide each term of $P(z)$ by that respective power of $z$, and the number of terms of $H(z)$ equals the number of terms of $P(z)$.

If $Q(z)$ is a polynomial with finite zeros away from $z=0$, then $H(z)$ is generally IIR, except if all zeros of $Q(z)$ are cancelled by zeros of $P(z)$. One example of such a case is a recursive moving average (where $N$ is the window length):

$$H(z)=\frac{1}{N}\frac{1-z^{-N}}{1-z^{-1}}\tag{1}$$

The zero of the denominator is cancelled by one of the zeros of the numerator, so the transfer function $(1)$ can equialenty be written as

$$H(z)=\frac{1}{N}(1+z^{-1}+\ldots+z^{-N+1})\tag{2}$$

which is obviously FIR.

Improve this answer
$\endgroup$
  • $\begingroup$ +1: That pole-zero cancellation issue confuses many people into thinking that the set of recursive filters is the same as the set infinite impulse response filters... which is not true. $\endgroup$ – Peter K.♦ Jul 30 '15 at 14:19
1
$\begingroup$

If you can express it exactly with a polynomial in $z$ or $z^{-1}$ (or both), then it will be FIR. Here $H(z)= \frac{1}{10}(1+5z^{-1}+3z^{-2}+z^{-3})$ with a simple term-by-term division. So you have a FIR filter here.

Additionnally, a generic IIR $z$-transform $H(z) = \sum_{k = -\infty}^{\infty} h_k z^{-k}$ cannot generally be written as $P(z)/Q(z)$ or $R(z)$, with $P$, $Q$, $R$ polynomials in $z$ or $z^{-1}$. One of the important tasks in filter design is to find close approximations to $H(z)$ as $P(z)/Q(z)$ or $R(z)$, generally with low-degree polynomials, constraints on zeroes and poles, and different types of "closeness" measures (in time, in frequency, etc.)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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