Biomedical Engineering Reference
In-Depth Information
Summing the currents leaving node 1 gives
Z
t
ð
v
1
v
s
Þ
þ
v
1
þ
ð
v
1
v
2
Þ
d
l þ
8
¼
0
10
0
where
:
Summing the currents leaving node 2 gives
Z
t
i
L
1
0
ðÞ¼
8A
Z
t
1
2
ð
v
2
v
1
Þ
d
l
8
þ
2
v
2
þ
ð
v
2
v
3
Þ
d
l
4
¼
0
0
0
where
Notice that the sign for the initial inductor current is negative because the
direction is from right to left and the current is defined on the circuit diagram in the opposite
direction for the node 2 equation.
Summing the currents leaving node 3 gives
i
L
2
0
ðÞ¼
4A
:
Z
t
1
2
3
v
3
¼
ð
v
3
v
2
Þ
d
l þ
4
þ
0
0
In this example, the node equations were not simplified by differentiating to remove the integral,
which would have eliminated the initial inductor currents from the node equations. If we were to
write a single differential equation involving just one node voltage and the input, a fifth-order
differential equation would result because there are five energy storing elements in the circuit. To
solve the differential equation, we would need five initial conditions, the initial node voltage for
the variable selected, and the first through fourth derivatives at time zero.
9.10.1 Discontinuities and Initial Conditions in a Circuit
Discontinuities in voltage and current occur when an input such as a unit step is applied or
a switch is thrown in a circuit. As we have seen, when solving an
n
th order differential equa-
tion, one must know
deriva-
tives at the time the input is applied or the switch thrown. As we will see, if the inputs to a
circuit are known for all time, we can solve for initial conditions directly based on energy
considerations and not have to depend on being provided with them in the problem state-
ment. Almost all of our problems involve the input applied at time zero, so our discussion
here is focused on time zero, but it may be easily extended to any time an input is applied.
Energy cannot change instantaneously for elements that store energy. Thus, there are no
discontinuities allowed in current through an inductor or voltage across a capacitor at any
time—specifically, the value of the variable remains the same at
n
initial conditions, typically the output variable and its
(n -
1
)
0
and
0
þ
:
In the
previous problem when we were given initial conditions for the inductors and capacitors,
this implied,
t
¼
t
¼
i
L
1
0ðÞ¼
i
L
1
0ðÞ
i
L
2
0ðÞ¼
i
L
2
0ðÞ
1
0ðÞ¼
v
0
þ
Þ
2
0ðÞ¼
v
0
þ
Þ
and
, and
v
ð
,
v
ð
,
1
2
3
0ðÞ¼
v
0
þ
Þ:
and
With the exception of variables associated with current through an
inductor and voltage across a capacitor, other variables can have discontinuities, especially
at a time when a unit step is applied or when a switch is thrown; however, these variables
must obey KVL and KCL.
v
ð
3
