RLC Filter Analysis
Referring now to Fig.E.2, let's perform an impedance analysis of that RLC network.
Driving Point Impedance
By inspection, we can write
![$\displaystyle R_d(s) = R + Ls \left\Vert \frac{1}{Cs}\right.
= R +\frac{L/C}{L...
...}{Cs}}
= R + \frac{Ls}{1+LCs^2} = R + \frac{1}{C}
\frac{s}{s^2+\frac{1}{LC}}.
$](http://www.dsprelated.com/josimages_new/filters/img1826.png)
![$ \Vert$](http://www.dsprelated.com/josimages_new/filters/img1827.png)
![$\displaystyle \zbox {R_1 \Vert R_2 = \frac{R_1 R_2}{R_1 + R_2}.}
$](http://www.dsprelated.com/josimages_new/filters/img1828.png)
![$ R_1$](http://www.dsprelated.com/josimages_new/filters/img1829.png)
![$ R_2$](http://www.dsprelated.com/josimages_new/filters/img1830.png)
Transfer Function
The transfer function in this example can similarly be found using voltage divider rule:
![$\displaystyle H(s) = \frac{V_C(s)}{V_e(s)}
= \frac{\left(Ls\left\Vert\frac{1}{...
...1}{RC} s + \frac{1}{LC}}
\isdef 2\eta\cdot\frac{s}{s^2 + 2\eta s + \omega_0^2}
$](http://www.dsprelated.com/josimages_new/filters/img1831.png)
Poles and Zeros
From the quadratic formula, the two poles are located at
![$\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}}
$](http://www.dsprelated.com/josimages_new/filters/img1832.png)
![$ s=0$](http://www.dsprelated.com/josimages_new/filters/img1468.png)
![$ s=\infty$](http://www.dsprelated.com/josimages_new/filters/img1469.png)
![$ R$](http://www.dsprelated.com/josimages_new/filters/img61.png)
![$ \eta^2 < \omega_0^2$](http://www.dsprelated.com/josimages_new/filters/img1833.png)
![$\displaystyle s = -\eta \pm j\sqrt{\omega_0^2 - \eta^2}
$](http://www.dsprelated.com/josimages_new/filters/img1834.png)
![$ \eta = 1/(2RC) > 0$](http://www.dsprelated.com/josimages_new/filters/img1835.png)
![$ j\omega$](http://www.dsprelated.com/josimages_new/filters/img1488.png)
![$\displaystyle s = \pm j\omega_0
$](http://www.dsprelated.com/josimages_new/filters/img1836.png)
Impulse Response
The impulse response is again the inverse Laplace transform of the
transfer function. Expanding into a sum of complex one-pole
sections,
![$\displaystyle H(s) = 2\eta\cdot\frac{s}{s^2 + 2\eta s + \omega_0^2}
= \frac{r_...
...r_2}{s-p_2}
= \frac{(r_1+r_2)s - (r_1p_2 + r_2p_1)}{s^2-(p_1 + p_2) + p_1p_2},
$](http://www.dsprelated.com/josimages_new/filters/img1837.png)
![$ p_{1,2}\isdef -\eta \pm \sqrt{\eta^2 - \omega_0^2}$](http://www.dsprelated.com/josimages_new/filters/img1838.png)
![\begin{eqnarray*}
r_1+r_2 &=& 2\eta \;\mathrel{=}\; \frac{1}{RC}\\
r_1p_2 + r_2p_1 &=& 0.
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1839.png)
This pair of equations in two unknowns may be solved for and
.
The impulse response is then
![$\displaystyle h(t) = r_1 e^{p_1 t} u(t) + r_2 e^{p_2 t} u(t).
$](http://www.dsprelated.com/josimages_new/filters/img1842.png)
Next Section:
Relating Pole Radius to Bandwidth
Previous Section:
RC Filter Analysis