I have plot some of the IIR filter's impulse responses and some of them are exponentially increasing, increasing, decreasing etc but some IIR's impulse response approaches 0. If for example I have an IIR filter defined by its lccde:

$$ 6 y[n] + y[n-1] - 2y[n-2] = x[n] - 2x[n-1] $$


$$ y[n] = \frac{1}{6}x[n] - \frac{1}{3}x[n-1] - \frac{1}{6}y[n-1] + \frac{1}{3}y[n-2] $$

when I plot its impulse response in matlab:

[h,n] = impz([1 -2],[6 1 -2]);

its impulse response would approach 0, and this zero value will bestarting at n = 23. Does this mean that the IIR is an FIR since I can cut off the impulse response at n = 23?

  • 2
    $\begingroup$ $y[23]$ is really zero? or just very tiny? $\endgroup$ Nov 29 '16 at 17:12
  • $\begingroup$ Plot the amplitude on a log scale $\endgroup$
    – Hilmar
    Nov 29 '16 at 17:21
  • $\begingroup$ Wait so this lccde does not approach 0? how do I plot this properly/ $\endgroup$ Nov 29 '16 at 17:28
  • $\begingroup$ It's an IIR. infinite impulse response, as explained in comments and answers to your two other questions, already (1, 2). you must stop tackling IIRs with the tools only applicable to FIRs! $\endgroup$ Nov 29 '16 at 17:41
  • $\begingroup$ @LeBlancLord: plot(20*log10(h)); $\endgroup$
    – Hilmar
    Nov 29 '16 at 18:51

As per the exact solution of a discrete LTI system:

$$y(t)=C A^k x_0+C\sum_{k=0}^{t-1}A^{t-k-1} B u(k)+D u(t)$$

and considering that $u(t)=\delta(t)$ and $x_0=0$ for the impulse response, we have:

$$h(t)=C A^{t-1} B+D \delta(t)$$

where $A$,$B$,$C$ and $D$ are the state space discrete matrices:


Which in this case are obtained as:

b=[1/6 -1/3 0];
a=[1 1/6 -1/3];

Hence, aside numerical issues, both methods gave the same results:

[h,t] = impz([1 -2],[6 1 -2]);

for k=2:length(t)

So the $t$=23th term on the $h(t)$ signal is: $$h(23)=C A^{t-1} B|_{t=23}=C A^{22} B=-3.39702260269318e^{-05} $$

Though small, the $t$th term never vanishes, hence, transfer functions with denominator different than the identity vector, will always be an IIR.


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