Page 19 - Circuit Analysis II with MATLAB Applications

Response of Series RLC Circuits with DC Excitation

di
and solving for ----- we get
dt
t = 0 +
di 15 0.5 u – 2.5
–
5
----- = --------------------------------------- = 10000 (1.15)
dt + 10 – 3
t = 0
Next, equating (1.14) with (1.15) we get:

–
– 200k 300k = 10000
1
2
k – – 1.5k = 50 (1.16)
2
1
Simultaneous solution of (1.13) and (1.16) yields k = 115 and k = – 110 . By substitution into (1.12)
1
2
we find the total response as
it = i t = 115e – 200t – 110e – 300t (1.17)
n
Check with MATLAB:

syms t; % Define symbolic variable t
R=0.5; L=10^( 3); C=100*10^( 3)/6;% Circuit constants
y0=115*exp( 200*t) 110*exp(-300*t); % Let solution i(t)=y0
y1=diff(y0); % Compute the first derivative of y0, i.e., di/dt
y2=diff(y0,2); % Compute the second derivative of y0, i.e, di2/dt2
% Substitute the solution i(t), i.e., equ (1.17)
% into differential equation of (1.11) to verify
% that correct solution was obtained.
% We must also verify that the initial
% conditions are satisfied
y=y2+500*y1+60000*y0;
i0=115*exp( 200*0) 110*exp( 300*0);
vC0= R*i0 L*( 23000*exp( 200*0)+33000*exp( 300*0))+15;
fprintf(' \n');...
disp('Solution was entered as y0 = '); disp(y0);...
disp('1st derivative of solution is y1 = '); disp(y1);...
disp('2nd derivative of solution is y2 = '); disp(y2);...
disp('Differential equation is satisfied since y = y2+y1+y0 = '); disp(y);...
disp('1st initial condition is satisfied since at t = 0, i0 = '); disp(i0);...
disp('2nd initial condition is also satisfied since vC+vL+vR=15 and vC0 = ');...
disp(vC0);...
fprintf(' \n')
Solution was entered as y0 =
115*exp(-200*t)-110*exp(-300*t)

1-7 Circuit Analysis II with MATLAB Applications
Orchard Publications