Biomedical Engineering Reference
In-Depth Information
While it may not seem obvious at first, a discontinuity is allowed for the derivative of the
current through an inductor and voltage across a capacitor at
0
and
0
þ
, since
t
¼
t
¼
di
L
0
ðÞ
dt
¼
v
L
0
ðÞ
L
and
dv
C
0
ðÞ
dt
¼
i
C
0
ðÞ
L
as discontinuities are allowed in
Keep in mind that the derivatives in
the previous expression are evaluated at zero after differentiation—that is,
v
L
0
ðÞ
and
i
C
0
ð :
t
¼0
þ
In calculations to determine the derivatives of variables not associated with current through
an inductor and voltage across a capacitor, the derivative of a unit step input may be
needed. Here we assume the derivative of a unit step input is zero at
t
¼0
þ
di
L
0
ðÞ
dt
¼
di
L
ðÞ
dt
and
dv
C
0
ðÞ
dt
¼
dv
C
ðÞ
dt
0
þ
.
The initial conditions for variables not associated with current through an inductor and
voltage across a capacitor at times of a discontinuity are determined only from the initial
conditions from variables associated with current through an inductor and voltage across
a capacitor and any applicable sources. The analysis is done in two steps involving KCL
and KVL or using the node-voltage method.
t
¼
0
. Recall that when a circuit is at steady state, an inductor
acts as a short circuit and a capacitor acts as an open circuit. Thus, at steady-state at
1.
First, we analyze the circuit at
t
¼
0
,
t
¼
we replace all inductors by short circuits and capacitors by open circuits in the circuit.
We then solve for the appropriate currents and voltages in the circuit to find the currents
through the inductors (actually the shorts connecting the sources and resistors) and voltages
across the capacitors (actually the open circuits among the sources and resistors).
2.
Second, we analyze the circuit at
0
þ
. Since the inductor current cannot change in
t
¼
0
þ
, we replace the inductors with current sources whose
values are the currents at
0
to
going from
t
¼
t
¼
0
:
t
¼
Moreover, since the capacitor voltage cannot change
0
to
0
þ
, we replace the capacitors with voltage sources whose
in going from
t
¼
t
¼
0
:
values are the voltages at
From this circuit we solve for all desired initial
conditions necessary to solve the differential equation.
t
¼
EXAMPLE PROBLEM 9.17
Use the node-voltage method to find
v
c
for the following circuit for
t
0.
i
R
1
i
R
2
i
R
3
400
Ω
100
Ω
+
+
i
C
i
L
+
+
−
5u(t) V
v
L
v
C
10 mH
500
Ω
5
μ
F
10 V
−
−
−
