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.

The transfer function from input to output voltage is:

$$G(s)=\frac{s/(RC)}{s^2+s/(RC)+1/(LC)}$$

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 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.

Plot the frequency response of the circuit.

bodeplot(G)
grid on

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)]);

bodeplot(G,"b",G1,"r",G2,"g")
grid on
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;
subplot(3,1,1)
lp1 = lsimplot(G2,sin(t),t);
lp1.Title.FontSize = 8;
lp1.XLabel.FontSize = 8;
lp1.YLabel.FontSize = 8;
title("w = 1")
subplot(3,1,2)
lp2 = lsimplot(G2,sin(0.9*t),t);
lp2.Title.FontSize = 8;
lp2.XLabel.FontSize = 8;
lp2.YLabel.FontSize = 8;
title("w = 0.9")
subplot(3,1,3)
lp3 = lsimplot(G2,sin(1.1*t),t);
lp3.Title.FontSize = 8;
lp3.XLabel.FontSize = 8;
lp3.YLabel.FontSize = 8;
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  
                                       (rad/TimeUnit)     (TimeUnit)    
                                                                        
 -2.50e-02 + 1.00e+00i     2.50e-02       1.00e+00          4.00e+01    
 -2.50e-02 - 1.00e+00i     2.50e-02       1.00e+00          4.00e+01