Biomedical Engineering Reference
In-Depth Information
Returning to the time domain gives
10
6
The characteristic equation for the previous differential equation is
v
C
þ
10
3
v
C
þ
10
6
10
:
417
20
:
83
v
C
¼
62
:
5
2
10
3
10
6
s
þ
10
:
417
s
þ
20
:
833
¼
0
10
3
and
10
3
and the natural solution
with roots
7
:
718
2
:
7
v
C
n
ð
t
Þ¼
K
1
e
7:71810
3
þ
K
2
e
2:710
3
t
t
V
Next, we solve for the forced response, assuming that
v
C
f
ðÞ¼
K
3
. After substituting into the
differential equation, this gives
10
6
10
6
20
:
833
K
3
¼
62
:
5
or
K
3
¼
3. Thus, our solution is now
e
7:71810
3
e
2:710
3
t
t
v
C
ðÞ¼
v
C
n
ðÞþ
v
C
f
ðÞ¼
K
þ
K
þ
3V
1
2
0
,
the capacitor is replaced by an open circuit and the inductor by a short circuit as shown in the
following circuit.
v
C
0ðÞ
and
v
C
0ðÞ
Initial conditions for
are necessary to solve for
K
1
and
K
2
:
For
t
¼
i
R
1
i
R
2
i
R
3
400
Ω
100
Ω
+
i
C
i
L
(
0
−
)
+
10 V
v
C
(
0
−
)
v
L
−
−
v
L
0ðÞ¼
Notice
0 V because the inductor is a short circuit. Also note that the 500
O
resistor is
i
R
3
0ðÞ¼
not shown in the circuit, since it is shorted out by the inductor, and so
0A
:
Using the
voltage divider rule, we have
100
400
v
C
0ðÞ¼
10
100
¼
2V
þ
and by Ohm's law
10
i
L
0ðÞ¼
400
¼
0
:
02 A
100
þ
i
R
1
0ðÞ¼
i
R
2
0ðÞ¼
i
L
0ðÞ¼
It follows that
0
:
02 A
:
Because voltage across a capacitor and
0
to
0
þ
current
through an inductor are not allowed to change from
t
¼
t
¼
we have
v
C
0ðÞ¼
v
C
0ðÞ¼
i
L
0ðÞ¼
i
L
0ðÞ¼
2V and
0
:
02 A
:
