Bài giải phần giải mạch P16
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Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
I(s)
+
−
1
1/s
1/s
s
22
2
)23()21s(
1
1ss
1
s1s1
s1
)s(I
++
=
++
=
++
=
= t
2
3
sine
3
2
)t(i
2t-
=)t(i
A)t866.0(sine155.1
-0.5t
Chapter 16, Solution 2.
8/s
s
s
4
2
+
−
+
V
x
−
4
V)t(u)e2e24(v
3
8
j
3
4
s
125.0
3
8
j
3
4
s
125.0
s
25.0
16
)8s8s3(s
2s
16V
s
32s16
)8s8s3(V
0VsV)s4s2(
s
)32s16(
)8s4(V
0
s
8
4
0V
2
0V
s
s
4
V
t)9428.0j3333.1(t)9428.0j3333.1(
x
2
x
2
x
x
2
x
2
x
xx
x
−−+−
++−=
−+
−
+
++
−
+−=
++
+
−=
+
=++
=+++
+
−+
=
+
−
+
−
+
−
v
x
= Vt
3
22
sine
2
6
t
3
22
cose)t(u4
3/t43/t4
−
−
−−
Chapter 16, Solution 3.
s
5/s
1/2
+
V
o
−
1/8
Current division leads to:
)625.0s(16
5
s1610
5
s
8
1
2
1
2
1
s
5
8
1
V
o
+
=
+
=
++
=
v
o
(t) =
( )
V)t(ue13125.
t625.0−
−0
Chapter 16, Solution 4.
The s-domain form of the circuit is shown below.
6 s
10/s
1/(s + 1)
+
−
+
V
o
(s)
−
Using voltage division,
+++
=
+++
=
1s
1
10s6s
10
1s
1
s106s
s10
)s(V
2
o
10s6s
CBs
1s
A
)10s6s)(1s(
10
)s(V
22
o
++
+
+
+
=
+++
=
)1s(C)ss(B)10s6s(A10
22
++++++=
Equating coefficients :
2
s:
-ABBA0 =→+=
1
s:
A5-CCA5CBA60 =→+=++=
0
s
: -10C-2,B,2AA5CA1010 ===→=+=
22222
o
1)3s(
4
1)3s(
)3s(2
1s
2
10s6s
10s2
1s
2
)s(V
++
−
++
+
−
+
=
++
+
−
+
=
=)t(v
o
V)tsin(e4)tcos(e2e2
-3t-3t-t
−−
Chapter 16, Solution 5.
s
2
2s
1
+
2
I
o
s
()
()
A)t(ut3229.1sin7559.0e
or
A)t(ueee3779.0eee3779.0e)t(i
3229.1j5.0s
)646.2j)(3229.1j5.1(
)3229.1j5.0(
3229.1j5.0s
)646.2j)(3229.1j5.1(
)3229.1j5.0(
2s
1
)3229.1j5.0s)(3229.1j5.0s)(2s(
s
2
Vs
I
)3229.1j5.0s)(3229.1j5.0s)(2s(
s2
2ss
s2
2s
1
2
s
2
1
s
1
1
2s
1
V
t2
t3229.1j2/t90t3229.1j2/t90t2
o
22
2
o
2
−=
++=
−+
++
+−
+
++
−−
−−
+
+
=
−++++
==
−++++
=
++
+
=
++
+
=
−
−°−−°−−
Chapter 16, Solution 6.
2
2s
5
+
I
o
10/s
s
Use current division.
t3sine
3
5
t3cose5)t(i
3)1s(
5
3)1s(
)1s(5
10s2s
s5
2s
5
s
10
2s
2s
I
tt
o
22222
o
−−
−=
++
−
++
+
=
++
=
+
++
+
=
Chapter 16, Solution 7.
The s-domain version of the circuit is shown below.
1/s
1
I
x
+ 2s
1
2
+s
–
Z
2
2
2
s21
1s2s2
s21
s2
1
s2
s
1
)s2(
s
1
1s2//
s
1
1Z
+
++
=
+
+=
+
+=+=
)5.0ss(
CBs
)1s(
A
)5.0ss)(1s(
1s2
1s2s2
s21
x
1s
2
Z
V
I
22
2
2
2
x
++
+
+
+
=
+++
+
=
++
+
+
==
)1s(C)ss(B)5.0ss(A1s2
222
++++++=+
BA2:s
2
+=
2CC2CBA0:s −=→+=++=
-4B ,6A3 0.5A or CA5.01:constant ==→=+=
222
x
866.0)5.0s(
)5.0s(4
1s
6
75.0)5.0s(
2s4
1s
6
I
++
+
−
+
=
++
+
−
+
=
[]
A)t(ut866.0cose46)t(i
t5.0
x
−
−=
Chapter 16, Solution 8.
(a)
)1s(s
1s5.1s
s22
)s21(
s
1
)s21//(1
s
1
Z
2
+
++
=
+
+
+=++=
(b)
)1s(s2
2s3s3
s
1
1
1
s
1
2
1
Z
1
2
+
++
=
+
++=
2s3s3
)1s(s2
Z
2
++
+
=
Chapter 16, Solution 9.
(a)
The s-domain form of the circuit is shown in Fig. (a).
=
++
+
=+=
s1s2
)s1s(2
)s1s(||2Z
in
1s2s
)1s(2
2
2
++
+
1
1
2
s2/s
1/s
s
2
(a) (b)
(b)
The s-domain equivalent circuit is shown in Fig. (b).
2s3
)2s(2
s23
)s21(2
)s21(||2
+
+
=
+
+
=+
2s3
6s5
)s21(||21
+
+
=++
=
+
+
+
+
+
⋅
=
+
+
=
2s3
6s5
s
2s3
6s5
s
2s3
6s5
||sZ
in
6s7s3
)6s5(s
2
++
+
Chapter 16, Solution 10.
To find Z
Th
, consider the circuit below.
1/s V
x
+
1V
2 V
o
2V
o
-
Applying KCL gives
s/12
V
V21
x
o
+
=+
But
xo
V
s
/12
2
V
+
=
. Hence
s3
)1s2(
V
s
/12
V
s/12
V4
1
x
xx
+
−=→
+
=
+
+
s3
)1s2(
1
V
Z
x
Th
+
−==
To find V
Th
, consider the circuit below.
1/s V
y
+
1
2
+
s
2 V
o
2V
o
-
Applying KCL gives
)1s(3
4
V
2
V
V2
1s
2
o
o
o
+
−=→=+
+
But 0V
s
1
V2V
ooy
=+•+−
)1s(s3
)2s(4
s
2s
)1s(3
4
)
s
2
1(VVV
oyTh
+
+−
=
+
+
−=+==
Chapter 16, Solution 11.
The s-domain form of the circuit is shown below.
4/s s
I
2
I
1
+
−
+
−
2
4/(s + 2)
1/s
Write the mesh equations.
21
I2I
s
4
2
s
1
−
+= (1)
21
I)2s(I-2
2s
4-
++=
+
(2)
Put equations (1) and (2) into matrix form.
+
+
=
+
2
1
I
I
2s2-
2-s42
2)(s4-
s1
)4s2s(
s
2
2
++=∆ ,
)2s(s
4s4s
2
1
+
+−
=∆
,
s
6-
2
=∆
4s2s
CBs
2s
A
)4s2s)(2s(
)4s4s(21
I
22
2
1
1
++
+
+
+
=
+++
+−⋅
=
∆
∆
=
)2s(C)s2s(B)4s2s(A)4s4s(21
222
++++++=+−⋅
Equating coefficients :
2
s:
BA21 +=
1
s:
CB2A22- ++=
0
s:
C2A42 +=
Solving these equations leads to
A 2=
,
23-B =
,
-3C =
22
1
)3()1s(
3s23-
2s
2
I
++
−
+
+
=
2222
1
)3()1s(
3
32
3-
)3()1s(
)1s(
2
3-
2s
2
I
++
⋅+
++
+
⋅+
+
=
=)t(i
1
[]
A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2
-t-2t
−−
22
2
2
2
)3()1s(
3-
)4s2s(2
s
s
6-
I
++
=
++
⋅=
∆
∆
=
== )t3sin(e
3
3-
)t(i
t-
2
A)t(u)t732.1sin(e1.732-
-t
Chapter 16, Solution 12.
We apply nodal analysis to the s-domain form of the circuit below.
V
o
10/(s + 1)
+
−
1/(2s)
4
s
3/s
o
o
o
sV2
4
V
s
3
s
V
1s
10
+=+
−
+
1s
15s1510
15
1s
10
V)ss25.01(
o
2
+
++
=+
+
=++
1s25.0s
CBs
1s
A
)1s25.0s)(1s(
25s15
V
22
o
++
+
+
+
=
+++
+
=
7
40
V)1s(A
1-so
=+=
=
)1s(C)ss(B)1s25.0s(A25s15
22
++++++=+
Equating coefficients :
2
s: -ABBA0 =→+=
1
s:
C-0.75ACBA25.015 +=++=
0
s:
CA25 +=
740A =
,
740-B =
,
7135C =
4
3
2
1
s
2
3
3
2
7
155
4
3
2
1
s
2
1
s
7
40
1s
1
7
40
4
3
2
1
s
7
135
s
7
40-
1s
7
40
V
222
o
+
+
⋅+
+
+
+
−
+
=
+
+
+
+
+
=
+
−= t
2
3
sine
)3)(7(
)2)(155(
t
2
3
cose
7
40
e
7
40
)t(v
2t-2t-t-
o
=)t(v
o
V)t866.0sin(e57.25)t866.0cos(e714.5e714.5
2-t2-t-t
+−
Chapter 16, Solution 13.
Consider the following circuit.
V
o
1/(s + 2)
1/s 2s
I
o
2
1
Applying KCL at node o,
o
oo
V
1s2
1s
s12
V
1s2
V
2s
1
+
+
=
+
+
+
=
+
)2s)(1s(
1s2
V
o
++
+
=
2s
B
1s
A
)2s)(1s(
1
1s2
V
I
o
o
+
+
+
=
++
=
+
=
1A =
,
-1B =
2s
1
1s
1
I
o
+
−
+
=
=)t(i
o
()
A)t(uee
-2t-t
−
Chapter 16, Solution 14.
We first find the initial conditions from the circuit in Fig. (a).
1
Ω
4
Ω
i
o
+
v
c
(0)
−
+
−
5 V
(a)
A5)0(i
o
=
−
, V0)0(v
c
=
−
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
2s 5/s
I
o
V
o
+
−
1 4
15/s 4/s
(b)
At node o,
0
s44
0V
s
5
s2
V
1
s15V
ooo
=
+
−
+++
−
o
V
)1s(4
s
s2
1
1
s
5
s
15
+
++=−
o
2
o
22
V
)1s(s4
2s6s5
V
)1s(s4
s2s2s4s4
s
10
+
++
=
+
++++
=
2s6s5
)1s(40
V
2
o
++
+
=
s
5
)4.0s2.1s(s
)1s(4
s
5
s2
V
I
2
o
o
+
++
+
=+=
4.0s2.1s
CBs
s
A
s
5
I
2
o
++
+
++=
sCsB)4.0s2.1s(A)1s(4
s2
++++=+
Equating coefficients :
0
s: 10AA4.04 =→=
1
s
: -84-1.2ACCA2.14 =+=→+=
2
s : -10-ABBA0 ==→+=
4.0s2.1s
8s10
s
10
s
5
I
2
o
++
+
−+=
2222
o
2.0)6.0s(
)2.0(10
2.0)6.0s(
)6.0s(10
s
15
I
++
−
++
+
−=
=)t(i
o
()[]
A)t(u)t2.0sin()t2.0cos(e1015
0.6t-
−−
Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
2s
5
+
V
o
+
−
+
−
5/s
s/4
+ V
x
−
10
3V
x
2s
5
VV
2s
5
VV,But
2s
s5
V120V)40ss2(0
2s
s5
sVVs2V120V40
0
10
2s
5
V
s/5
0V
4/s
V3V
xoox
xo
2
oo
2
xo
o
oxo
+
+=→
+
−=
+
−−++==
+
−++−
=
+
−
+
−
+
−
We can now solve for V
x
.
)40s5.0s)(2s(
)20s(
5V
2s
)20s(
10V)40s5.0s(2
0
2s
s5
V120
2s
5
V)40ss2(
2
2
x
2
x
2
xx
2
−++
+
−=
+
+
−=−+
=
+
−−
+
+++
Chapter 16, Solution 16.
We first need to find the initial conditions. For
0t <
, the circuit is shown in Fig. (a).
To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.
1
Ω
+
−
V
o
1 F
V
o
/2
+
−
1 H
i
o
+
−
2
Ω
3 V
(a)
Hence,
A1-
3
3-
i)0(i
oL
=== , V1-v
o
=
V5.2
2
1-
)1-)(2(-)0(v
c
=
−=
We now incorporate the initial conditions for
as shown in Fig. (b).
0t >
I
2
I
1
+
−
s
2.5/s
+
−
1
+
−
V
o
1/s
V
o
/2
+
−
I
o
−
+
2
-1 V
5/(s + 2)
(b)
For mesh 1,
0
2
V
s
5.2
I
s
1
I
s
1
2
2s
5-
o
21
=++−
++
+
But,
2oo
IIV ==
s
5.2
2s
5
I
s
1
2
1
I
s
1
2
21
−
+
=
−+
+
(1)
For mesh 2,
0
s
5.2
2
V
1I
s
1
I
s
1
s1
o
12
=−−+−
++
1
s
5.2
I
s
1
s
2
1
I
s
1
-
21
−=
+++ (2)
Put (1) and (2) in matrix form.
−
−
+
=
++
−+
1
s
5.2
s
5.2
2s
5
I
I
s
1
s
2
1
s
1-
s
1
2
1
s
1
2
2
1
s
3
2s2 ++=∆ ,
)2s(s
5
s
4
2-
2
+
++=∆
3s2s2
CBs
2s
A
)3s2s2)(2s(
132s-
II
22
2
2
2o
++
+
+
+
=
+++
+
=
∆
∆
==
)2s(C)s2s(B)3s2s2(A132s-
222
++++++=+
Equating coefficients :
2
s: B A22- +=
1
s
:
CB2A20 ++=
0
s:
C2A313 +=
Solving these equations leads to
7143.0A =
, ,
-3.429B = 429.5C =
5.1ss
714.2s7145.1
2s
7143.0
3s2s2
429.5s429.3
2s
7143.0
I
22
o
++
−
−
+
=
++
−
−
+
=
25.1)5.0s(
)25.1)(194.3(
25.1)5.0s(
)5.0s(7145.1
2s
7143.0
I
22
o
++
+
++
+
−
+
=
=)t(i
o
[ ]
A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0
-0.5t-0.5t-2t
+−
Chapter 16, Solution 17.
We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
I
3
+ −
1/s
I
2
I
1
4
s
1
1
For mesh 3,
0IsI
s
1
I
s
1
s
1s
2
213
=−−
++
+
(1)
For the supermesh,
0Is
s
1
I)s1(I
s
1
1
321
=
+−++
+ (2)
But
(3)
4II
21
−=
Substituting (3) into (1) and (2) leads to
+=
+−
++
s
1
14I
s
1
sI
s
1
s2
32
(4)
1s
2
s
4-
I
s
1
sI
s
1
s-
32
+
−=
++
+ (5)
Adding (4) and (5) gives
1s
2
4I2
2
+
−=
1s
1
2I
2
+
−=
==
)t(i)t(i
2o
A)t(u)e2(
-t
−
Chapter 16, Solution 18.
v
s
(t) = 3u(t) – 3u(t–1) or V
s
=
)e1(
s
3
s
e
s
3
s
s
−
−
−=−
1 Ω
+
V
o
−
+
−
1/s
V
s
2 Ω
V)]1t(u)e22()t(u)e22[()t(v
)e1(
5.1s
2
s
2
)e1(
)5.1s(s
3
V
VV)5.1s(0
2
V
sV
1
VV
)1t(5.1t5.1
o
ss
o
so
o
o
so
−−−−=
−
+
−=−
+
=
=→=++
−
−−−
−−
+
Chapter 16, Solution 19.
We incorporate the initial conditions in the s-domain circuit as shown below.
I
1/s
−
+
2 I
V
1
V
o
1/s
s
+
−
2
2
4/(s + 2)
At the supernode,
o
11
sV
s
1
s
V
2
2
V)2s(4
++=+
−+
o1
Vs
s
1
V
s
1
2
1
2
2s
2
++
+=+
+
(1)
But
and
I2VV
1o
+=
s
1V
I
1
+
=
2s
2Vs
s)2s(
s2V
V
s
)1V(2
VV
oo
1
1
1o
+
−
=
+
−
=→
+
+=
(2)
Substituting (2) into (1)
oo
Vs
2s
2
V
2s
s
s
1s2
s
1
2
2s
2
+
+
−
+
+
=−+
+
o
Vs
2s
1s2
)2s(s
)1s2(2
s
1
2
2s
2
+
+
+
=
+
+
+−+
+
o
22
V
2s
1s4s
2s
9s2
)2s(s
s9s2
+
++
=
+
+
=
+
+
732.3s
B
2679.0s
A
1s4s
9s2
V
2
o
+
+
+
=
++
+
=
443.2A =
,
4434.0-B =
732.3s
4434.0
2679.0s
443.2
V
o
+
−
+
=
Therefore,
=
)t(v
o
V)t(u)e4434.0e443.2(
-3.732t-0.2679t
−
Chapter 16, Solution 20.
We incorporate the initial conditions and transform the current source to a voltage source
as shown.
1 s
+ −
1/s
2/s
V
o
+
−
1
1/(s + 1) 1/s
At the main non-reference node, KCL gives
s
1
s
V
1
V
s11
Vs2)1s(1
ooo
++=
+
−−+
s
1s
V)s1s)(1s(Vs2
1s
s
oo
+
+++=−−
+
o
V)s12s2(2
s
1s
1s
s
++=−
+
−
+
)1s2s2)(1s(
1s4s2-
V
2
2
o
+++
−−
=
5.0ss
CBs
1s
A
)5.0ss)(1s(
5.0s2s-
V
22
o
++
+
+
+
=
+++
−−
=
1V)1s(A
1-so
=+=
=
)1s(C)ss(B)5.0ss(A5.0s2s-
222
++++++=−−
Equating coefficients :
2
s:
-2BBA1- =→+=
1
s:
-1CCBA2- =→++=
0
s:
-0.515.0CA5.00.5- =−=+=
222
o
)5.0()5.0s(
)5.0s(2
1s
1
5.0ss
1s2
1s
1
V
++
+
−
+
=
++
+
−
+
=
=)t(v
o
[]
V)t(u)2tcos(e2e
2-t-t
−
Chapter 16, Solution 21.
The s-domain version of the circuit is shown below.
1 s
V
1
V
o
+ 2/s 2 1/s
-
At node 1,
10/s
oo
o
V
s
VsV
s
s
VV
V
s
)1
2
()1(10
21
10
2
1
1
1
−++=→+
−
=
−
(1)
At node 2,
)1
2
(
2
2
1
1
++=→+=
−
s
s
VVsV
V
s
VV
oo
oo
(2)
Substituting (2) into (1) gives
ooo
VsssV
s
Vsss
)5.12()1
2
()12/)(1(10
2
2
2
++=−++++=
5.12)5.12(
10
22
++
+
+=
++
=
ss
CBs
s
A
sss
V
o
CsBsssA ++++=
22
)5.12(10
BAs +=0:
2
CAs += 20:
-40/3C -20/3,B ,3/205.110:constant ===→= AA
++
−
++
+
−=
++
+
−=
22222
7071.0)1(
7071.0
414.1
7071.0)1(
11
3
20
5.12
21
3
20
ss
s
s
ss
s
s
V
o
Taking the inverse Laplace tranform finally yields
[]
V)t(ut7071.0sine414.1t7071.0cose1
3
20
)t(v
tt
o
−−
−−=
Chapter 16, Solution 22.
The s-domain version of the circuit is shown below.
4s
V
1
V
2
1s +
12
1 2 3/s
At node 1,
s4
V
s4
1
1V
1s
12
s4
VV
1
V
1s
12
2
1
211
−
+=
+
→
−
+=
+
(1)
At node 2,
++=→+=
−
1s2s
3
4
VVV
3
s
2
V
s4
VV
2
212
221
(2)
Substituting (2) into (1),
2
22
2
V
2
3
s
3
7
s
3
4
s4
1
s4
1
11s2s
3
4
V
1s
12
++=
−
+
++=
+
)
8
9
s
4
7
s(
CBs
)1s(
A
)
8
9
s
4
7
s)(1s(
9
V
22
2
++
+
+
+
=
+++
=
)1s(C)ss(B)
8
9
s
4
7
s(A9
22
++++++=
Equating coefficients:
BA0:s
2
+=
A
4
3
CCA
4
3
CBA
4
7
0:s −=→+=++=
-18C -24,B ,24AA
8
3
CA
8
9
9:constant ===→=+=
64
23
)
8
7
s(
3
64
23
)
8
7
s(
)8/7s(24
)1s(
24
)
8
9
s
4
7
s(
18s24
)1s(
24
V
222
2
++
+
++
+
−
+
=
++
+
−
+
=
Taking the inverse of this produces:
[ ]
)t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v
t875.0t875.0t
2
−−−
+−=
Similarly,
)
8
9
s
4
7
s(
FEs
)1s(
D
)
8
9
s
4
7
s)(1s(
1s2s
3
4
9
V
22
2
1
++
+
+
+
=
+++
++
=
)1s(F)ss(E)
8
9
s
4
7
s(D1s2s
3
4
9
222
++++++=
++
Equating coefficients:
ED12:s
2
+=
D
4
3
6FFD
4
3
6or FED
4
7
18:s −=→+=++=
0F 4,E ,8DD
8
3
3or FD
8
9
9:constant ===→=+=
64
23
)
8
7
s(
2/7
64
23
)
8
7
s(
)8/7s(4
)1s(
8
)
8
9
s
4
7
s(
s4
)1s(
8
V
222
1
++
−
++
+
+
+
=
++
+
+
=
Thus,
[ ]
)t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v
t875.0t875.0t
1
−−−
−+=
Chapter 16, Solution 23.
The s-domain form of the circuit with the initial conditions is shown below.
V
I
1/sC
sL
R
-2/s
4/s 5C
At the non-reference node,
sCV
sL
V
R
V
C5
s
2
s
4
++=++
++=
+
LC
1
RC
s
s
s
CV
s
sC56
2
LC1RCss
C6s5
V
2
++
+
=
But
8
8010
1
RC
1
== , 20
804
1
LC
1
==
22222
2)4s(
)2)(230(
2)4s(
)4s(5
20s8s
480s5
V
++
+
++
+
=
++
+
=
=)t(v
V)t2sin(e230)t2cos(e5
-4t-4t
+
)20s8s(s4
480s5
sL
V
I
2
++
+
==
20s8s
CBs
s
A
)20s8s(s
120s25.1
I
22
++
+
+=
++
+
=
6A =
,
-6B =
,
-46.75C =
22222
2)4s(
)2)(375.11(
2)4s(
)4s(6
s
6
20s8s
75.46s6
s
6
I
++
−
++
+
−=
++
+
−=
=)t(i
0t),t2sin(e375.11)t2cos(e6)t(u6
-4t-4t
>−−
Chapter 16, Solution 24.
At t = 0
-
, the circuit is equivalent to that shown below.
+
9A 4
Ω
5
Ω
v
o
-
20)9(
54
4
x5)0(v
o
=
+
=
For t > 0, we have the Laplace transform of the circuit as shown below after
transforming the current source to a voltage source.
4
Ω
16
Ω
V
o
+
36V 10A 2/s 5
Ω
-
Applying KCL gives
8.12B,2.7A,
5.0s
B
s
A
)5.0s(s
s206.3
V
5
V
2
sV
10
20
V36
o
ooo
−==
+
+=
+
+
=→+=+
−
Thus,
[ ]
)t(ue8.122.7)t(v
t5.0
o
−
−=
Chapter 16, Solution 25.
For , the circuit in the s-domain is shown below.
0t >
s
6
I
Applying KVL,
+
V
−
+
−
(2s)/(s
2
+ 16)
+
−
9/s
2/s
0
s
2
I
s
9
s6
16s
s2
2
=+
+++
+
−
)16s)(9s6s(
32s4
I
22
2
+++
+
=
)16s()3s(s
288s36
s
2
s
2
I
s
9
V
22
2
++
+
+=+=
16s
EDs
)3s(
C
3s
B
s
A
s
2
22
+
+
+
+
+
+
++=
)s48s16s3s(B)144s96s25s6s(A288s36
2342342
++++++++=+
)s9s6s(E)s9s6s(D)s16s(C
232343
++++++++
Equating coefficients :
0
s
:
A144288 =
(1)
1
s:
E9C16B48A960 +++=
(2)
2
s:
E6D9B16A2536 +++=
(3)
3
s
:
ED6CB3A60 ++++=
(4)
4
s: (5)
DBA0 ++=
Solving equations (1), (2), (3), (4) and (5) gives
2A =
,
B
,
C7984.1-= 16.8-=
,
D 2016.0-=
,
765.2E =
