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Introduction to Digital Filters
    Transfer Function Analysis
       Partial Fraction Expansion
          Repeated Poles
             So What's Up with Repeated Poles?

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So What's Up with Repeated Poles?

In the previous section, we found that repeated poles give rise to polynomial amplitude-envelopes multiplying the exponential decay due to the pole. On the other hand, two different poles can only yield a convolution (or sum) of two different exponential decays, with no polynomial envelope allowed. This is true no matter how closely the poles come together; the polynomial envelope can occur only when the poles merge exactly. This might violate one's intuitive expectation of a continuous change when passing from two closely spaced poles to a repeated pole.

To study this phenomenon further, consider the convolution of two one-pole impulse-responses $ h_1(n) = p_1^n$ and $ h_2(n) = p_2^n$:

$\displaystyle h(n) \isdef (h_1\ast h_2)(n) = \sum_{m=0}^n h_1(m)h_2(n-m) = \sum... ...^n p_1^{m}p_2^{n-m} = p_2^n\sum_{m=0}^n \left(\frac{p_1}{p_2}\right)^m \protect$ (7.14)

The finite limits on the summation result from the fact that both $ h_1$ and $ h_2$ are causal. Recall the closed-form sum of a truncated geometric series:

$\displaystyle \sum_{m=0}^n r^m = \frac{1-r^{n+1}}{1-r} $

Applying this to Eq.$ \,$(6.14) yields

$\displaystyle h(n) = p_2^n \frac{1-(p_1/p_2)^{n+1}}{1-(p_1/p_2)} = \frac{p_2^{n+1}-p_1^{n+1}}{p_2-p_1} = \frac{p_1^{n+1}-p_2^{n+1}}{p_1-p_2}. $

Note that the result is symmetric in $ p_1$ and $ p_2$. If $ \left\vert p_1\right\vert>\left\vert p_2\right\vert$, then $ h(n)$ becomes proportional to $ p_1^n$ for large $ n$, while if $ \left\vert p_2\right\vert>\left\vert p_1\right\vert$, it becomes instead proportional to $ p_2^n$.

Going back to Eq.$ \,$(6.14), we have

$\displaystyle h(n) = p_2^n\sum_{m=0}^n \left(\frac{p_1}{p_2}\right)^m = p_1^n\sum_{m=0}^n \left(\frac{p_2}{p_1}\right)^m.$ (7.15)

Setting $ p_1=p_2=p$ yields

$\displaystyle h(n) = (n+1)p^n$ (7.16)

which is the first-order polynomial amplitude-envelope case for a repeated pole. We can see that the transition from ``two convolved exponentials'' to ``single exponential with a polynomial amplitude envelope'' is perfectly continuous, as we would expect.

We also see that the polynomial amplitude-envelopes fundamentally arise from iterated convolutions. This corresponds to the repeated poles being arranged in series, rather than in parallel. The simplest case is when the repeated pole is at $ p=1$, in which case its impulse response is a constant:

$\displaystyle \frac{1}{1-z^{-1}} \eqsp 1 + z^{-1}+ z^{-2}+ \cdots \;\longleftrightarrow\; [1,1,1,\ldots] $

The convolution of a constant with itself is a ramp:

$\displaystyle h_1(n)\eqsp \sum_{m=0}^n 1\cdot 1 \eqsp n+1 $

The convolution of a constant and a ramp is a quadratic, and so on:7.9

\begin{eqnarray*} h_2(n)&=&\sum_{m=0}^n (m+1)\cdot 1 \eqsp \frac{(n+1)(n+2)}{2}\... ...+1)(m+2)}{2}\cdot 1\eqsp \frac{(n+1)(n+2)(n+3)}{3!}\\ &\cdots& \end{eqnarray*}

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Previous: Impulse Response of Repeated Poles
Next: Alternate Stability Criterion
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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