Biomedical Engineering Reference
In-Depth Information
or
Z
t
1
C
v
ð
t
Þ¼
idt
þ
v
ð
t
o
Þ
ð
9
:
26
Þ
t
o
For
t
0
¼
0, Eq. (9.26) reduces to
Z
t
1
C
v
ð
t
Þ¼
idt
þ
v
ð
0
Þ
ð
9
:
27
Þ
0
and for
t
0
¼1
, Eq. (9.27) reduces to
Z
t
1
C
v
ð
t
Þ¼
i
ðÞ
d
l
ð
9
:
28
Þ
1
The initial voltage in Eq. (9.26),
v
ð
t
Þ
, is usually defined with the same polarity as
v
,
0
which means
v
ð
t
Þ
is a positive quantity. If the polarity of
v
ð
t
Þ
is in the opposite direction
0
0
of
v
, then
v
ð
t
Þ
is negative.
0
EXAMPLE PROBLEM 9.14
Find
v
for the circuit that follows.
i
s
(A)
+
2
i
s
v
2 F
-
0
2
t (s)
Solution
The current waveform is described with three different functions: for the interval
t
0, for the
interval 0
<
t
2, and for
t
>
2. To find the voltage, we apply Eq. (9.28) for each interval as
follows.
For t < 0
Z
t
Z
0
1
C
1
2
v
ð
t
Þ¼
idt
¼
0
dt
¼
0V
1
1
