clear all

clear all
close all
clc
% Charging of an RC circuit
%
c = 10e-6;
r1 = 1e3;
tau1 = c*r1;
t = 0:0.002:0.05;
v1 = 10*(1-exp(-t/tau1));
r2 = 10e3;
tau2 = c*r2;
v2 = 10*(1-exp(-t/tau2));
r3 = .1e3;
tau3 = c*r3;
v3 = 10*(1-exp(-t/tau3));
plot(t,v1,'+',t,v2,'o', t,v3,'*')
axis([0 0.06 0 12])
title('Charging of a capacitor with three time constants')
xlabel('Time, s')
ylabel('Voltage across capacitor')
text(0.03, 5.0, '+ for R = 1 Kilohms')
text(0.03, 6.0, 'o for R = 10 Kilohms')
text(0.03, 7.0, '* for R = 0.1 Kilohms')

5.3 RLC CIRCUIT
For the series RLC circuit shown in Figure 5.9, we can use KVL to obtain
the Equation (5.15).

MATLAB Script
p = [1 40 1000];
lambda = roots(p)
lambda =
-20.0000 +24.4949i
-20.0000 -24.4949i

Differentiating the above expression, we get

devide by L:

The homogeneous solution can be found by making Vt S () = constant, thus

The characteristic equation is

The roots of the characteristic equation can be determined. If we assume that the roots are

then, the solution to the homogeneous solution is

where; A1 and A2 are constants. If vt S () is a constant, then the forced solution will also be
a constant and be given as :

From Circuit Analysis II with MATLAB Applications

1.3 Response of Series RLC Circuits with DC Excitation
Depending on the circuit constants R, L, and C, the total response of a series RLC circuit that is
excited by a DC source, may be overdamped, critically damped, or underdamped. In this section
we will derive the total response of series RLC circuits that are excited by DC sources.
Example 1.1

For the circuit of Figure 1.5, i L( 0) = 5 A, v C( 0) = 2.5 V, and the 0.5 ? resistor represents the
resistance of the inductor. Compute and sketch i( t) for t > 0.

Figure 1.5: Circuit for Example 1.1

Solution:
This circuit can be represented by the integrodifferential equation

Differentiating and noting that the derivatives of the constants v C( 0) and 15 are zero, we obtain
the homogeneous differential equation

or

and by substitution of the known values R, L, and C

The roots of the characteristic equation of (1.11) are s 1 = ? 200 and s 2 = ?300. The total
response is just the natural response and for this example it is overdamped. Therefore, from (1.7),

The constants k 1 and k 2 can be evaluated from the initial conditions. Thus from the first initial
condition i L( 0) = i( 0) = 5 A and (1.12) we get

%series_rlc.m
%Transient in RLC series circuit forced by unit step
%solution vC(t)=vn(t)+vf(t)=natural response+forced response
%here forcedresponse(capacitor voltage)due to DC voltage will bf(t)=Vs
Vs=1; R=400; L=1e-3; C=5e-9;
omega=1/sqrt(L*C);
alpha=R/(2*L);
zeta=alpha/omega;
s1=omega*(-zeta+sqrt(zeta^2-1));
s2=omega*(-zeta-sqrt(zeta^2-1));
%solve for the capacitor voltage vC(t):
%given the initial condition vC(0)=i1, iL(0)=i2;
%[K1; K2]=[1, 1; s1, s2]\[vC(0)-Vs; iL(0)/C]%t=0, K(1)+K(2)+Vs=vC(0)
i1=0; %vC(0)=0 (continuity)
i2=0; %iL(0)=0 (continuity)
K=[1,1; s1,s2]\[i1-Vs; i2/C];
t=0:0.5e-6:30e-6;
vC=K(1)*exp(s1*t)+K(2)*exp(s2*t)+Vs;
plot(t,vC); xlabel(’time in sec’); ylabel(’vC(t)’);
title(’Series RLC circuit forced by unit step’);
grid;
print -deps series_rlc.eps