Biomedical Engineering Reference
In-Depth Information
0
þ
is drawn by replacing the inductors in the original circuit with current
sources whose values equal the inductor currents at
The circuit for
t
¼
0
t
¼
and the capacitors with voltage
0
, as shown in the following figure with
nodes C and L and reference. Note also that the input is now 10
sources whose values equal the capacitor voltages at
t
¼
þ
5
u
ð
t
Þ¼
15 V.
i
R
1
i
R
2
i
R
3
C
L
+
+
400
100
Ω
Ω
i
C
i
L
+
−
+
−
v
C
15 V
2V
0.02 A
500
Ω
v
L
−
−
v
L
0ðÞ
To find
, we sum the currents leaving node L, yielding
v
L
2
100
þ
þ
v
L
0
:
02
500
¼
0
i
R
3
0ðÞ¼
v
L
0
ðÞ
500
v
L
0ðÞ¼
i
R
2
0ðÞ¼
þ
i
R
3
0ðÞ¼
which gives
0V
:
Now
¼
0A,
0
:
02
0
:
02 A, and
15
2
i
R
1
0ðÞ¼
¼
0
:
0325 A
:
400
i
C
0ðÞ
To find
, we write KCL at node C, giving
i
R
1
0ðÞþ
i
C
0ðÞþ
i
R
2
0ðÞ¼
0
or
i
C
0ðÞ¼
i
R
1
0ðÞ
i
R
2
0ðÞ¼
0
:
0325
0
:
02
¼
0
:
125 A
v
C
0ðÞ
i
C
0ðÞ¼
Cv
C
0ðÞ
To find
, note that
or
v
C
0ðÞ¼
i
C
0
ðÞ
C
0
:
0125
10
3
V
¼
10
6
¼
2
:
5
s
:
5
With the initial conditions, the constants
K
1
and
K
2
are solved as
v
C
0
ðÞ¼
2
¼
K
1
þ
K
2
þ
3
Next,
K
1
e
7:71810
3
K
2
e
2:710
3
10
3
t
10
3
t
v
C
ðÞ¼
7
:
718
2
:
7
Continued
