Problem 2.9.6.5
quit
R
i(t)
Switch closes at t = 0
e(t)
L
(c)
C
v
C
(0)
i
L
(0)
The di erential equation governing the circuit is
v(t)=i(t)R+ L
di(t)
dt
+ v
c
(0)+
1
C
Z
t
0
i(t)dt:
Taking the Laplace transform of both sides of this equation yields
V (s);
v
c
(0)
s
= I(s)R+ LsI(s);Li
L
(0)+
I(s)
Cs
;;
or
V (s);
v
c
(0)
s
+ Li
L
(0) = I(s)R+ LsI(s)+
I(s)
Cs
:
Solving this last equation for I(s)yields
I(s) =
sV (s);v
c
(0)+ sLi
L
(0)
s(R + Ls +
1
Cs
)
=
sV (s)=L;v
c
(0)=L+ si
L
(0)
s
2
+
R
L
s +
1
LC
:
=
sV (s)=L
s
2
+
R
L
s +
1
LC
:
+
si
L
(0);v
c
(0)=L
s
2
+
R
L
s +
1
LC
:
In this last form wesee that wehavearesponse due to the forcing function
and a response due to anynonzero initial conditions. The time functions
1
will be the same for both the forced portion of the response and the initial
condition portion.
The roots of
s
2
+
R
L
s +
1
LC
=0
are
s = ;
R
2L
1
2
s
R
L
2
;
4
LC
:
If
R
L
2
>
4
LC
;;
The roots are real.
If
R
L
2
=
4
LC
;;
the roots are real and repeated.
If
R
L
2
<
4
LC
;;
The roots are complex with negativereal part. Since
1
2
s
R
L
2
;
4
LC
<
R
2L
;;
in every case the roots lie in the open left half of the s-plane. This means
that the time functions associated with the roots will decaywith time, as
shown next.
Letting
V(s)=
V
s
;;
V aconstant, wehave
I(s) =
s
L
(
V
s
)
s
2
+
R
L
s +
1
LC
:
+
si
L
(0);v
c
(0)=L
s
2
+
R
L
s +
1
LC
:
=
V=L
s(s
2
+
R
L
s +
1
LC
):
+
si
L
(0);v
c
(0)=L
s
2
+
R
L
s +
1
LC
:
2
Wenow nd the explicit time response for the case of unrepeated real
roots and complex roots. Considering the forced response rst, wehave
I
f
(s) =
V=L
s(s
2
+
R
L
s +
1
LC
):
=
V=L
s(s + )(s + ):
=
A
s
+
D
1
s +
+
D
2
s +
;;
where
= ;
R
2L
+
1
2
s
R
L
2
;
4
LC
= ;
R
2L
;
1
2
s
R
L
2
;
4
LC
:
Then
A =
sV=L
s(s + )(s + ):
j
s=0
=
V=L
(s + )(s + ):
j
s=0
=
V=L
D
1
=
V=L(s + )
s(s + )(s + ):
j
s=;
=
V=L
s(s + ):
j
s=;
=
V=L
; ( ; )
=
V=L
( ; )
D
2
=
V=L(s+ )
s(s + )(s + ):
j
s=;
=
V=L
s(s + ):
j
s=;
=
V=L
; ( ; )
:
3
For the portion of the response due to the initial conditions, wehave
I
ic
(s) =
si
L
(0);v
c
(0)=L
s
2
+
R
L
s +
1
LC
:
=
si
L
(0);v
c
(0)=L
(s + )(s + ):
=
E
1
s +
+
E
2
s +
Then
E
1
=
(si
L
(0);v
c
(0)=L)(s+ )
(s + )(s + )
j
s=;
=
(si
L
(0);v
c
(0)=L)
(s + )
j
s=;
=
(; i
L
(0);v
c
(0)=L)
( ; )
E
2
=
(si
L
(0);v
c
(0)=L)(s+ )
(s + )(s + )
j
s=;
=
(si
L
(0);v
c
(0)=L)
(s + )
j
s=;
=
(; i
L
(0);v
c
(0)=L)
( ; )
The total response in the time domain is then
i(t)=
h
A +(D
1
+ E
1
)e
; t
+(D
2
+ E
2
)e
;betat
i
1(t):
Weseethat at steady state wewill haveascaled, phase shifted version
of the forcing function, the response due to the poles of the system having
decayed to zero.
For the case of repeated real roots, the system response will be
R
L
2
=
4
LC
;;
The time response is
i(t)=
h
A + C
1
e
;at
+ C
2
te
;at
i
1(t);;
The partial fraction expansions can be developed in a similar fashion to the
case of unrepeated real roots.
4
The most interesting case from an engineering standpointis
R
L
2
<
4
LC
;;
in whichcase
i(t)=
h
A +2jMje
;at
cosbt +
i
1(t):
with
a =
R
2L
b =
1
2
s
4
LC
;
R
L
2
:
If we break the response down as we did for the case of unrepeated roots we
have:
I(s) =
V=L
s(s + a;jb)(s+ a + jb)
+
si
L
(0);v
c
(0)=L
(s + a;jb)(s+ a + jb)
:
For the partial fraction expanasion we rstconsiderthe term that contains
the forcing function. Then
V=L
s(s + a;jb)(s+ a + jb)
=
A
s
+
D
s + a;jb
+
D
s + a + jb
:
Then
A =
V=L
(s + j!)(s+ a;jb)(s+ a + jb)
j
s=0
=
V=L
(a;jb)(a+ jb)
j
s=0
=
V=L
a
2
+ b
2
:
Similarly,
D =
(V=L)(s+ a;jb)
s(s + a;jb)(s+ a + jb)
j
s=;a+jb
=
V=L
s(s + a + jb)
j
s=;a+jb
=
V=L
(;a + jb)(2jb)
=
V=2bL
6
(;=2)
[
p
a
2
+ b
2
6
^
1
]
=
V
6
( ;=2)
2bL
p
a
2
+ b
2
6
( ;=2)
5
where
=tan
;1
(b=a):
Thus, the response due to the forcing function will be
i(t)=[A +2jDjcos(bt + )]1(t):
It should be clear that the contribution of the initial conditions will be
i
init
(t)=[2jEje
;at
cos(bt+
3
)]1(t);;
where
E =
(s + a;jb)
si
L
(0);v
c
(0)=L
(s+ a;jb)(s+ a + jb)
s=;a+jb
= jEje
j
:
Then the total response is
i(t)=
h
A +2jDje
;at
cos(bt+
2
)+2jEje
;at
cos(bt +
3
)
i
1(t):
Once the transient dies out weare left with a scarled version of the input
step.
6
