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Poles and Zeros

From the quadratic formula, the two poles are located at

$\displaystyle s =
-\eta \pm \sqrt{\eta^2 - \omega_0^2}
\;\isdef \;
-\frac{1}{2RC} \pm \sqrt{\left(\frac{1}{2RC}\right)^2 - \frac{1}{LC}}
$

and there is a zero at $ s=0$ and another at $ s=\infty$. If the damping $ R$ is sufficienly small so that $ \eta^2 < \omega_0^2$, then the poles form a complex-conjugate pair:

$\displaystyle s = -\eta \pm j\sqrt{\omega_0^2 - \eta^2}
$

Since $ \eta = 1/(2RC) > 0$, the poles are always in the left-half plane, and hence the analog RLC filter is always stable. When the damping is zero, the poles go to the $ j\omega$ axis:

$\displaystyle s = \pm j\omega_0
$


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Next: Impulse Response

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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