Biomedical Engineering Reference
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which simplifies to
1
R 1 þ
1
R 2
1
R 2 v 2 ¼ C 1 v s
C 1 v 1 þ
v 1
Summing the currents leaving node 2 gives
Z t
v 2 v 1
R 2 þ C 2 v 2 þ
1
L 1
ð
v 2 v s
Þ d l ¼
0
0
Typically we eliminate integrals in the node equations by differentiating. When applied to the
previous expression, this gives
1
R 2 v 2
1
R 2 v 1 þ C 2 v 2 þ
1
L 1 v 2
1
L 1 v s ¼
0
and after rearranging yields
1
C 2 R 2 v 2 þ
1
C 2 L 1 v 2
1
C 2 R 2 v 1 ¼
1
C 2 L 1 v s
v 2 þ
When applying the node-voltage method, we generate one equation for each essential node.
To write a single differential equation involving just one node voltage and the inputs, we use
the other node equations and substitute into the node equation of the desired node voltage.
Sometimes this involves differentiation as well as substitution. The easiest case involves a node
equation containing an undesired node voltage without its derivatives. Another method for
creating a single differential equation is to use the
operator or the Laplace transform.
Consider the node equations for Example Problem 9.15, and assume that we are interested
in obtaining a single differential equation involving node voltage
D
v 1 and its derivatives, and
the input. For ease in analysis, let us assume that the values for the circuit elements are
R
¼ R
¼
1
O
,
C
¼ C
¼
1 F, and
L
¼
1 H, giving us
1
2
1
2
1
v
¼ v s
þ
2
v
v
1
1
2
and
v
þ v
þ v
v
¼ v s
2
2
2
1
Using the first equation, we solve for
v 2 , calculate
v
2 and
v
2 , and then substitute into the
second equation as follows.
v
¼ v
þ
2
v
v s
2
1
1
v 2 ¼ v 1 þ
2
v 1 v s
v 1 __ v s
After substituting into the second node equation, we have
__ v 1 þ
v 2 ¼ __ v 1 þ
2
2
v 1 __ v s þ v 1 þ
2
v 1 v s þ v 1 þ
2
v 1 v s v 1 ¼ v s
and after simplifying
v 1 þ
v 1 þ
v 1 ¼ __ v s þ v s þ v s v s
__ v 1 þ
3
2
2
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