Finding the frequency response of a series RLC circuit The series RLC circuit in Fig. B.40 has R = 50 Ω, L = 4 mH, and C = 0.15 µF.
(a) Determine the series resonant frequency f_{ n } , the damping ratio δ, the quality factor Q_{ s } , the cutoff frequencies, and the bandwidth BW_s.
(b) Use PSpice/SPICE to plot the magnitude and phase angle of the output voltage for R = 50 Ω, 100 Ω, and 200 Ω. The frequency f varies from 100 Hz to 1 MHz. Assume V_{ m }=1 V \text { peak } AC .
(a) R=50 \Omega, L=4 mH \text {, and } C=0.15 \mu F , so
\omega_{ n }=\frac{1}{\sqrt{L C}}=\frac{10^{5}}{\sqrt{4 \times 1.5}}=40,825 rad / s
The series resonant frequency is
f_{ n }=\frac{\omega_{ n }}{2 \pi}=\frac{40,825}{2 \pi}=6497.5 Hz
Since \alpha=R /(2 L)=50 /\left(2 \times 4 \times 10^{-3}\right)=6250 , the damping ratio is
\delta=\frac{\alpha}{\omega_{ n }}=\frac{6250}{40,825}=0.1531
From Eq. (B.55),
\frac{I^{2} X_{ L }}{I^{2} R}=\frac{X_{ L }}{R}=\frac{2 \pi f_{ n } L}{R} (for an inductive reactance) (B.55)
Q_{ s }=\omega_{ n } \frac{L}{R}=40,825 \times 4 \times \frac{10^{-3}}{50}=3.266
For the lower cutoff frequency, Eqs. (B.85) and (B.86) give
u_{1}=-\delta+\sqrt{1+\delta^{2}} (B.85)
\omega_{1}=u_{1} \omega_{n}=\omega_{ n }\left(-\delta+\sqrt{1+\delta^{2}}\right) (B.86)
\begin{aligned}&u_{1}=-\delta+\sqrt{1+\delta^{2}}=-0.1531+\sqrt{1+0.1531^{2}}=0.85855 \\&\omega_{1}=u_{1} \omega_{ n }=0.85855 \times 40,825=35,050.4 rad / s\end{aligned}
Thus, f_{1}=35,050.4 / 2 \pi=5578 Hz For the upper cutoff frequency, Eqs. (B.83) and (B.84) give
u_{2}=\delta+\sqrt{1+\delta^{2}} (B.83)
\omega_{2}=u_{2} \omega_{ n }=\omega_{ n }\left(\delta+\sqrt{1+\delta^{2}}\right) (B.84)
\begin{aligned}&u_{2}=\delta+\sqrt{1+\delta^{2}}=0.1531+\sqrt{1+0.1531^{2}}=1.16475 \\&\omega_{2}=u_{2} \omega_{ n }=1.16475 \times 40,825=47,551 rad / s\end{aligned}
Thus, f_{2}=47,551 / 2 \pi=7568 Hz From Eq. (B.89), the bandwidth is
BW _{ s }=\frac{1}{2 \pi} \frac{R}{L}=\frac{1}{2 \pi} \frac{2 \pi f_{ n }}{Q_{ s }}=\frac{f_{ n }}{Q_{ s }} (B.89)
BW _{ s }=f_{2}-f_{1}=\frac{f_{ n }}{Q_{ s }}=\frac{6497.5}{3.266}=1989.4 Hz
(b) The series RLC circuit for PSpice simulation is shown in Fig. B.42. The list of the circuit file is as follows.
Example B.16 Frequency Response of Series RLC Circuit
.PARAM RVAL = 50
.STEP PARAM RVAL LIST 50 100 200
The PSpice plots of the magnitude and phase angle (using EXB-16.SCH) are shown in Fig. B.43. The plot for R=50 \Omega \text { gives } f_{1}=5578 Hz , f_{2}=7568 Hz , f_{ n }=6457 Hz \text {, and } BW _{ s }=f_{2}-f_{1}=1990 Hz .
