Page 30 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

0
0
v0 = k e + k e = 5
1
2
or
k + k = 5 (1.51)
2
1
The second equation that is needed for the computation of the values of k 1 and k 2 is found from
dv dv
C
other initial condition, that is, i 0 = 2A . Since i t = C-------- = C------ , we differentiate (1.50),
L
C
dt
dt
evaluate it at t = 0 + , and we equate it with this initial condition.Then,
dv – 4t – 16t dv
------= – 4k e – 16k e and ------ = – 4k 16k (1.52)
–
dt 1 2 dt 1 2
t = 0 +
Also, at t = 0 +
1 + + dv
---v0 i 0 + + C------ = 10
L
R dt +
t = 0
dv
and solving for ------ we get
dt
t = 0 +
–
–
dv 10 5 32 2
e
------ = ------------------------------- = 502 (1.53)
dt 1640
e
t = 0 +
Next, equating (1.52) with (1.53) we get
–
– 4k 16k = 502
1
2
or
– 2k 8– k = 251 (1.54)
2
1
Simultaneous solution of (1.51) and (1.54) yields k = 291 6 , k = – 261 6 , and by substitution
e
e
1
2
into (1.50) we get the total response as
vt = v t = 291 – 4t 261 – 16t = 1 - – 4t 261e – 16t V (1.55)
---------e –
---------e
-- 291e –
n
6
6
6
Check with MATLAB:
syms t % Define symbolic variable t
y0=291*exp( 4*t)/6 261*exp( 16*t)/6; % Let solution v(t) = y0
y1=diff(y0) % Compute and display first derivative
y1 =
-194*exp(-4*t)+696*exp(-16*t)

1-18 Circuit Analysis II with MATLAB Applications
Orchard Publications