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Poles and Zeros

In the simple RC-filter example of §E.4.3, the transfer function is

$\displaystyle H(s) = \frac{1}{s+1/\tau} = \frac{RC}{RCs+1}.
$

Thus, there is a single pole at $ s=-1/\tau=-RC$, and we can say there is one zero at infinity as well. Since resistors and capacitors always have positive values, the time constant $ \tau = RC$ is always non-negative. This means the impulse response is always an exponential decay--never a growth. Since the pole is at $ s=-1/\tau$, we find that it is always in the left-half $ s$ plane. This turns out to be the case also for any complex analog one-pole filter. By consideration of the partial fraction expansion of any $ H(s)$, it is clear that, for stability of an analog filter, all poles must lie in the left half of the complex $ s$ plane. This is the analog counterpart of the requirement for digital filters that all poles lie inside the unit circle.


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Previous: The Continuous-Time Impulse
Next: RLC Filter Analysis

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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