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I know this question has been asked here numerous times. But I am still unclear here :(

My book says,

A linear time-invariant system is stable if its impulse response is absolutely summable. (G. Proakis)

So that should mean that, for a stable system, the impulse response h[n] goes to zero as n approaches infinity, right? But that filter is FIR filter. So FIR filters are always stable.

So, for unstable filters, the impulse response is not absolutely summable. In another way, the impulse response never approaches zero. Again, for IIR filter, h[h] continues to go on with n i.e. never goes to zero. So, IIR filters are supposed to be unstable.

But from Z-transform, IIR filters have poles and FIR filters don't, correct? And if the poles are inside unit circle, i.e. if their value is below one, then that filter is stable.

So, what's happening here? If IIR filters have impulse response which is not absolutely summable, how come they are stable when their poles are within unit circle? Their impulse response goes on like forever, right? Their output gets feedback. So why they can be stable depending on pole's location or value?

Thanks in advance :)

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    $\begingroup$ You've fallen for Zeno's paradox. But An infinitely long signal can have a finite sum. $\endgroup$ – hotpaw2 Dec 30 '15 at 20:17
  • $\begingroup$ I don't know about Zeno's paradox...but I got the idea. Thanks. $\endgroup$ – benjamin Dec 30 '15 at 21:10
  • $\begingroup$ A series that is not absolutely summable may have a $0$ limit. Think about the alternating series $a_n = \frac{(-1)^n}{n+1}$ for instance. $\endgroup$ – Laurent Duval Dec 30 '15 at 21:29
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Note that for stable IIR filters, the impulse response does approach zero as $n$ goes to infinity. It just never becomes exactly zero. However, the sum of the absolute values is finite. Just as an example, take the exponential impulse response

$$h[n]=a^nu[n],\qquad |a|<1\tag{1}$$

where $u[n]$ is the unit step function. The sum

$$\sum_{n=-\infty}^{\infty}|h[n]|=\sum_{n=0}^{\infty}|a|^n=\frac{1}{1-|a|}\tag{2}$$

is obviously finite (for $|a|<1$).

I agree with Hilmar's answer, but I would like to add that everything he said about poles is only valid for causal filters. In the general case it is not necessary for a stable filter to have all its poles inside the unit circle. For stability it is necessary and sufficient that the unit circle is inside the region of convergence (ROC) of the filter's $\mathcal{Z}$-transform (transfer function). So a stable filter can have poles inside as well as outside the unit circle, but in that case it can't be causal. So it's only the additional requirement of causality that forces the poles of a stable IIR filter inside the unit circle.

A similar thing holds for FIR filters. Causal FIR filters indeed have all their poles at the origin, but non-causal FIR filters also have poles at infinity. A simple example is this non-causal moving average filter

$$H(z)=\frac15\left(z^{-2}+z^{-1}+1+z+z^2\right)\tag{3}$$

which has poles at $z=0$ as well as poles at $z=\infty$, but being an FIR filter, it is obviously stable.

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IIR filters are stable when the poles are inside the unit circle. The fact the that impulse response is infinite in time doesn't mean it's not absolute summable.

Consider a simple filter like $y[n]= x[n] + \frac{1}{2} \cdot y[n-1]$. The impulse response is 1, 0.5, 0.25 ..., i.e. $h[n] = 2^{-n}$. The absolute sum of the impulse response is simply 2, even though it's an infinite series it has a finite absolute sum.

FIR filters have poles too, there just all at z = 0. Poles are fine, as long as they are inside the unit circle.

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  • $\begingroup$ I think you forgot to add a factor of $\frac12$ before $y[n-1]$, otherwise you actually have an unstable (marginally stable) filter. $\endgroup$ – Matt L. Dec 30 '15 at 20:54
  • $\begingroup$ Thanks Hilmar. However, about the example you mentioned, y[n]=x[n]+y[n−1], the pole here lies at Z=1, isn't it. So, shouldn't it be oscillatory instead of being stable? I think @Matt L. noted that already. $\endgroup$ – benjamin Dec 30 '15 at 21:09
  • $\begingroup$ Sorry, I forgot the 1/2 in the difference equation (as Matt has pointed out). But you are right: without the 0.5 the filter is only marginally stable. Fixed now. $\endgroup$ – Hilmar Dec 31 '15 at 18:01

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