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# Documentation

## Analyzing the Response of an RLC Circuit

This example shows how to analyze the time and frequency responses of common RLC circuits as a function of their physical parameters using Control System Toolbox™ functions.

Bandpass RLC Network

The following figure shows the parallel form of a bandpass RLC circuit:

Figure 1: Bandpass RLC Network.

The transfer function from input to output voltage is:

The product LC controls the bandpass frequency while RC controls how narrow the passing band is. To build a bandpass filter tuned to the frequency 1 rad/s, set L=C=1 and use R to tune the filter band.

Analyzing the Frequency Response of the Circuit

The Bode plot is a convenient tool for investigating the bandpass characteristics of the RLC network. Use tf to specify the circuit's transfer function for the values

%|R=L=C=1|:
R = 1; L = 1; C = 1;
G = tf([1/(R*C) 0],[1 1/(R*C) 1/(L*C)])

G =

s
-----------
s^2 + s + 1

Continuous-time transfer function.



Next, use bode to plot the frequency response of the circuit:

bode(G), grid


As expected, the RLC filter has maximum gain at the frequency 1 rad/s. However, the attenuation is only -10dB half a decade away from this frequency. To get a narrower passing band, try increasing values of R as follows:

R1 = 5;   G1 = tf([1/(R1*C) 0],[1 1/(R1*C) 1/(L*C)]);
R2 = 20;  G2 = tf([1/(R2*C) 0],[1 1/(R2*C) 1/(L*C)]);
bode(G,'b',G1,'r',G2,'g'), grid
legend('R = 1','R = 5','R = 20')


The resistor value R=20 gives a filter narrowly tuned around the target frequency of 1 rad/s.

Analyzing the Time Response of the Circuit

We can confirm the attenuation properties of the circuit G2 (R=20) by simulating how this filter transforms sine waves with frequency 0.9, 1, and 1.1 rad/s:

t = 0:0.05:250;
opt = timeoptions;
opt.Title.FontWeight = 'Bold';
subplot(311), lsim(G2,sin(t),t,opt), title('w = 1')
subplot(312), lsim(G2,sin(0.9*t),t,opt), title('w = 0.9')
subplot(313), lsim(G2,sin(1.1*t),t,opt), title('w = 1.1')


The waves at 0.9 and 1.1 rad/s are considerably attenuated. The wave at 1 rad/s comes out unchanged once the transients have died off. The long transient results from the poorly damped poles of the filters, which unfortunately are required for a narrow passing band:

damp(pole(G2))


Pole              Damping       Frequency       Time Constant

rlc_gui