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Impulse Response of Repeated Poles

In the time domain, repeated poles give rise to polynomial amplitude envelopes on the decaying exponentials corresponding to the (stable) poles. For example, in the case of a single pole repeated twice, we have

$\displaystyle \zbox {\frac{1}{\left(1-pz^{-1}\right)^2}
\;\longleftrightarrow\;
(n+1) p^n, \quad n=0,1,2,\ldots\,.}
$



Proof: First note that

$\displaystyle \frac{d}{dz^{-1}}\left(\frac{1}{1-pz^{-1}}\right) = (-1)(1-pz^{-1})^{-2}(-p)
= \frac{p}{\left(1-pz^{-1}\right)^2}\;.
$

Therefore,
$\displaystyle \frac{1}{\left(1-pz^{-1}\right)^2}$ $\displaystyle =$ $\displaystyle \frac{1}{p}\, \frac{d}{dz^{-1}}\left(\frac{1}{1-pz^{-1}}\right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{p}\, \frac{d}{dz^{-1}} \left(1 + pz^{-1}+ p^2z^{-2}+ p^3 z^{-3}
+ \cdots \right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{p} \left(0 + p + 2p^2z^{-1}+ 3p^3z^{-2}+ \cdots \right)$  
  $\displaystyle =$ $\displaystyle 1 + 2pz^{-1}+ 3p^2z^{-2}+ \cdots$  
  $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}(n+1)p^n z^{-n}$  
  $\displaystyle \isdef$ $\displaystyle {\cal Z}\left\{(n+1)p^n\right\} \;\longleftrightarrow\; (n+1)p^n.$ (7.13)

Note that $ n+1$ is a first-order polynomial in $ n$. Similarly, a pole repeated three times corresponds to an impulse-response component that is an exponential decay multiplied by a quadratic polynomial in $ n$, and so on. As long as $ \vert p\vert<1$, the impulse response will eventually decay to zero, because exponential decay always overtakes polynomial growth in the limit as $ n$ goes to infinity.


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Previous: Example
Next: So What's Up with Repeated Poles?

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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