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I'm reading Richard Hamming's book, The Art of Doing Science and Engineering, and in the chapter Digital Filters IV he says

"These recursive filters are often called 'infinite impulse response filters' (IIR) because a single disturbance will echo around the feedback loop, which even if the filter is stable will die out only like a geometric progression. Being me, of course I asked myself if all recursive filters had to have this property, and soon found a counterexample. True, it is not the kind of filter you would normally design, but it showed their claim was superficial."

He then doesn't give the counterexample! Off the top of my head I can only think of the trivial case, which would kill all echoes immediately but also isn't really a filter so much as a cushion that muffles all incoming signals perfectly. Can anyone come up with what he may have been talking about?

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The author may be referring to "pole cancellation". A simple example would be a "moving sum" filter.

That can be implemented as an FIR filter of length $N$ as

$$y_1[n] = \sum_{k=0}^{N-1}x[n-k] \tag{1}$$

But it's more efficient to do it as IIR filter

$$y_2[n] = y[n-1]+x[n]-x[n-N] \tag{2}$$

This IIR filter has a pole at $z = 1$ but it also has zero at $z = 1$ as well so the two cancel and you are left with with a FIR.

We can take a quick look at the Z transform to ensure these are indeed identical.

$$Y_1[z] = \sum_{k=0}^{N-1}z^k \tag{3}$$

$$Y_2[z] = \frac{1-z^N}{1-z} = Y_1(z)\tag{4} $$

where we use the identity of a finite geometric series $\sum_{n=0}^{N-1} x^n = \frac{x^N-1}{x-1} $

Typically a "moving average" is more interesting than a "moving sum". We can implement this with an exponential decay as

$$y[n] = g\cdot y[n-1] + \frac{1-g}{1-g^N}\left(x[n]-g^N \cdot x[n-N]\right) \tag{5} $$

This creates an exponentially decaying window that's truncated at $N$ with a decay rate of $g$. For $g=1$ we get a rectangular moving average. Again, the truncation here happens since the single pole is cancelled by a matching zero.

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  • $\begingroup$ Uncanceled poles imply IIR, if I recall correctly? $\endgroup$ Commented Sep 4 at 19:06

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