ECE634 Signals and Systems II, Spring 2009 - Lecture 11, February 16
Daniel S. Brogan
1
4.4 Analysis of Electrical Networks in the Frequency Domain:
•
Example 7.28 (G. E. Carlson,
Signal and Linear System Analysis
) – Circuit to equation in the
time domain
Consider the electrical circuit shown below.
The switch is closed for all
0
t
<
, so the circuit
is in equilibrium at
0
t
=
.
Find
( )
o
v
t
for
0
t
≥
Solution:
Since the voltage supply is disconnected for
0
t
≥
, there is no input signal and
( )
0
X
s
=
.
Some definitions to recall about circuit component equations:
L
L
di
v
L
dt
=
and
C
C
dv
i
C
dt
=
.
In the differential equation, any inductor will require an initial condition
(
)
0
L
i
−
and any
capacitor will require an initial condition
(
)
0
C
v
−
.
Steady-state definitions at time
0
t
−
=
specify that a capacitor is an open, and an inductor is a short.
Thus, at time
0
t
≥
, the circuit
is a simple resistor divider network such that
(
)
8
0
2
1
3
L
V
i
A
−
=
=
Ω + Ω
and
(
)
3
0
8
6
1
3
C
v
V
V
−
Ω
=
=
Ω + Ω
Now for some circuit equations…
( )
( )
2
L
o
L
di
v
t
v
t
dt
=
=
( )
( )
( )
( )
3
3
2
L
C
L
L
L
di
v
t
i
t
v
t
i
t
dt
=
+
=
+
( )
( )
C
C
L
C
dv
dv
i
t
i
t
C
dt
dt
= −
= −
= −
Converting with Laplace transforms…

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