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Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \displaystyle\sum_{k=1}^K \alpha_k p_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $p_k$'s are free parameters.

Am I correct in understanding that

  • This is the impulse response of an IIR filter of order $K$.
  • The filter's poles are the $p_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - p_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\displaystyle\sum_{k=1}^{K} \beta_k z^{k}}{\displaystyle\prod_{k=1}^K (z - p_k)}$$

From the denominator one can infer the poles are $p_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

  • Will this hold for unstable versions where $|p_k| < 1$ is not satisfied?
  • Do I have to account for the Z-transform's ROC in this case, and if so, how?
  • Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?
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  • $\begingroup$ Turns out that: $$ H(z) = A\frac{\prod\limits_{k=1}^K (z - q_k)}{\prod\limits_{k=1}^K (z - p_k)}$$ where $A$ is an overall gain parameter, $q_k$ are zeros, and $p_k$ are poles. Some of the zeros or poles may be at the origin. $\endgroup$ Mar 7 at 23:09
  • $\begingroup$ Also, your equation is correct if you make one small adjustment: $$ H(z) = \frac{\sum\limits_{k=0}^{K} \beta_k z^{k}}{\prod\limits_{k=1}^K (z - p_k)}$$ $\endgroup$ Mar 7 at 23:14

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From the denominator one can infer the poles are pk , right?

Yes

And the order of both numerator and denominator polynomials is K , hence that's also the order of the filter?

Yes

Will this hold for unstable versions where |pk|<1 is not satisfied?

It holds inside a ROC for all poles. However, if that ROC doesn't include the unit circle, you can't do anything with the filter, so it's moot question

Do I have to account for the Z-transform's ROC in this case, and if so, how?

The ROC is a circle with radius of $\max(|p_k|)$

Is there a simple rule for αk's to make this an all-pole filter?

You basically have a sum of first order single pole filters. If you want an all-pole filter make it a cascade, i.e.

$$ A(z) = g \cdot \prod_{k=1}^K \frac{1}{1 - p_k z^{-1}} $$

You can convert cascade and parallel filter structures into each other through partial fraction expansion/decompostion and use that to derive the $\alpha_k$. Whether that's "simple" or not is in the eye of the beholder. If the poles are close to the unit circle, this can be numerically quite tricky.

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  • $\begingroup$ "... if that ROC doesn't include the unit circle, you can't ..." well, you can code it up and try it, and watch it blow up. Or you can wrap it with a feedback loop that stabilizes it. Both of these possibilities have potential for practical value, but only in very niche applications. $\endgroup$
    – TimWescott
    Mar 9 at 4:33

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