Page 322 - Solutions Manual to accompany Electric Machinery Fundamentals

d v  1 v  1 v
dt C RC C RC 1

d v  1 v  V M sin t 
dt C RC C RC
The solution can be divided into two parts, a natural response and a forced response. The natural
response is the solution to the differential equation
d v  1 v  0
dt C RC C

The solution to the natural response differential equation is
t

v Cn  t  A e RC
,
where the constant A must be determined from the initial conditions in the system. The forced response is
the steady-state solution to the equation
d v  1 v  V M sin t 
dt C RC C RC
It must have a form similar to the forcing function, so the solution will be of the form



v Cf  t  B 1 sin t B 2 cos t
,
where the constants B and B must be determined by substitution into the original equation. Solving for
1
2
B and B yields:
1
2
d  sin  B cos t B 1  t  sin t   B cos t B  V M sin t
dt 1 2 RC 1 2 RC
1
  1 cos   B s in t  B  RC  t  1 sin t B  2 cos  t  B V M sin t
2
RC
cosine equation:

1
 B  1 B  2 0  B   RC B
2
1
RC
sine equation:
1 V
  B  B  M
2 1
RC RC
1 V
 2 RC B  B  M
1 1
RC RC
 2 1  V M
   RC  RC   B  1 RC

 1  2 R C   2 2 V M
  RC   B  RC
1

Finally,
V   RCV
B  1  2 M R C 2 and B  1  2 R C M 2

1

2
2
2
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