Check if a filter is IIR or FIR

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?

• 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. – Matt L. Jul 29 '15 at 19:43
• @MattL. In this case it would work but what if the denum is something like $z^2(1-z^-2)$ – Saeid Yazdani Jul 30 '15 at 8:14
• I added an answer discussing the general case. – Matt L. Jul 30 '15 at 9:26

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.

• +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. – Peter K. Jul 30 '15 at 14:19

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.)