Chapter 14 The Laplace Transform

Prévia do material em texto

Problems 
 
Section 14-2: Laplace Transform 
 
P14.2-1 
( ) ( )
( ) ( ) ( ) ( )
1 1
2 2
1 1 2 2cos
A f t A F s
AsF ss sf t t F s
s
ωω ω
⎫=⎡ ⎤⎣ ⎦ ⎪ ⇒ =⎬ += ⇒ = ⎪+ ⎭
L
 
 
P14.2-2 
( ) 11 1 21! 1Fn nnt s ts s s+ +⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦L L 1
! = 
 
 
P14.2-3 
( ) ( ) ( ) ( )
( ) ( )
( ) [ ] ( )
( )
1 1 2 2 1 1 2 2
1 2
3
1 1
2 22
2
Linearity: a a
Here 1
1
 3
1
1 1so
3
t
f t a f t F s a F s
a a
f t e F s
s
f t t F s
s
F s
s s
−
+ = +⎡ ⎤⎣ ⎦
= =
⎡ ⎤= = =⎡ ⎤⎣ ⎦ ⎣ ⎦ +
= = =⎡ ⎤⎣ ⎦
= ++
L
L L
L L
 
 
 
P14.2-4 
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
1
1 2
2 3
1 2 3
1 ( )
1 1
1 1 , 
1 1
bt
bt bt
f t A e u t A f t
f t e u t u t e u t f t f t
F s F s
s s b
AbF s AF s A F s F s A
s s b s s b
−
− −
= − =
= − = − = +
−= = +
⎡ ⎤
3
∴ = = + = − =⎡ ⎤⎣ ⎦ ⎢ ⎥+ +⎣ ⎦
 
 
 
 
 
1 
Section 14-3: Impulse Function and Time Shift Property 
 
P14.3-1 
( ) ( ) ( )f t A u t u t T= − −⎡ ⎤⎣ ⎦ 
( ) ( ) ( ) ( )1 sTsT eA AeF s A u t A u t T A
s s s
−− −= − − = − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦L L 
 
 
P14.3-2 
( ) ( ) ( ) ( ) ( ) ( )at atf t u t u t T e F s e u t u t T⎡ ⎤= − − ⇒ = − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦L 
( ) ( )
( ) ( )
( ) ( )( )
1
1
sT
s a T
at
eu t u t T es F s
s ae g t G s a
−
−⎫−− − =⎡ ⎤ −⎪⎣ ⎦ ⇒ =⎬ −⎪⎡ ⎤ = −⎣ ⎦ ⎭
L
L
 
 
P14.3-3 
(a) ( ) ( )3
2
3
F s
s
= + 
 
(b) ( ) ( ) ( ) ( )sT sTf t t T F s e t eδ δ− −= − ⇒ = =⎡ ⎤⎣ ⎦L 
 
(c) ( ) ( ) ( ) ( )2 2 22
5 55
8 48 16 254 5
F s
s ss ss
= = =
1+ ++ + ++ + 
 
 
 
P14.3-4 
( ) ( ) ( ) ( )( (0.5 0.5)) 0.5 ( 0.5)0.5 0.5 0.5t t tg t e u t e u t e e u t− − + − − − −= − = − = − 
 
( ) ( ) ( ) 0.5 0.50.5 0.50.5 ( 0.5) 0.5 ( 0.5) 0.5 0.50.5 0.5
1 1
ss
t t s t ee ee e u t e e u t e e e u t
s s
−−
− − − − − − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤− = − = = =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ + +L L L 
 
 
 
P14.3-5 
( ) ( ) ( ) 2 
sT s
sT e et T tu t T e u t t u t
T T T T
− −
−−⎡ ⎤ ⎡ ⎤− − = − = − = −⎡ ⎤⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦L L L
T
s
 
 
 
1 
P14.3-6 
 
( ) ( ) ( ) (5 55 4.2
3 3
f t t u t t u t⎛ ⎞ ⎛ ⎞= − + − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ )4.2 
 
( ) ( )4.24.22 2 15 5 15 5 53 3 3
s
s
s e
F s e
s s s s
−
− + −⎛ ⎞ ⎛ ⎞= − + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 2
 
 
P14.3-7 
 
 
 
2 22
0 0
0
3 3(1 )( ) ( ) 3 
st s
st st e eF s f t e dt e dt
ss
− −∞ − − −= = = =−∫ ∫ 
 
 
P14.3-8 
 
 
( ) 5 2 0 2
0 otherwis
t t
f t
< <⎧= ⎨⎩ e 
 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) 2 2 2 22 2 2
5 5 5 5 2 2 2 2 2 ( 2)
2 2 2 2
5 1 2 5 1 1 2
2 2
s s
s s
f t t u t u t t u t t u t t u t t u t u t
e eF s f t e se
s s s s
− −
− −
= − − = − − = − − − − −⎡ ⎤ ⎡⎣ ⎦ ⎣
⎡ ⎤ ⎡ ⎤
⎤⎦
∴ = = − − = − −⎡ ⎤ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦L
 
 
 
2 
Section 14-4: Inverse Laplace Transform 
 
P14.4-1 
( ) ( )3 2 22
3 3( )
3 6 4 1 2 41 1 3
s s AF s
s s s s s ss s
+ += = = ++ + + + + +⎡ ⎤+ + +⎣ ⎦
Bs C+ 
where 
( )2 1
3 2
31 3 s
sA
s =−
+= =+ + 
Then 
( )
( ) ( ) ( ) 222
2
3 2 43 3 ( )
1 2 4 3 3 31 2 4
s Bs C s B s B C s
s s ss s s
+ + ⎛ ⎞= + ⇒ + = + + + + + +⎜ ⎟+ + ++ + + ⎝ ⎠
8 C 
 
Equating coefficient yields 
2 2 2s : 0 
3 3
4 2 1s : 1 
3 3 3
B B
C C
= + ⇒ = −
= − + ⇒ =
 
 
Then 
( ) ( )
( )
( ) ( )2 2
12 2 1 2 2 31
33 3 3 3 3
1 11 3 1 3 1 3
s s
F s
s ss s s
− + − +
= + = + ++ + 2+ + + + + + 
 
Taking the inverse Laplace transform yields 
 
( ) 2 2 1cos 3 sin 3 , 0
3 3 3
t t tf t e e t e t t− − −= − + ≥ 
 
 
 
1 
P14.4-2 
( ) ( ) ( )( )
2 2 *
3 2
2 1 2 1
3 4 2 1 1 1 1 1
s s s s a a bF s
s s s s s j s j s j s j s
− + − += = = ++ + + + + − + + + − + + +1+ 
where 
( )
( ) ( )
2
2
2
*
2 1 4
1 1 1
2 1 3 4 3 2
1 1 2 21
3 2
2
s sb
s s
s s ja j
s s j s j
a j
− += =+ + =−
− + −= =+ + + −=− +
= − −
=− + 
Then 
( )
3 32 2 42 2
1 1
j j
F s
s j s j s
− + − −
= + +
1+ − + + + 
Next 
( ) ( )22 15 23 2 2 and tan 126.932
2
m θ −
⎛ ⎞⎜ ⎟= − + = = =⎜ ⎟⎜ ⎟−⎝ ⎠
° 
From Equation 14.5-8 
( ) ( ) ( )5 cos 127 4t tf t e t e u− −⎡ ⎤= + ° +⎣ ⎦ t 
 
 
 
P14.4-3 
( ) ( ) ( )2 2
5 1( )
1 21 2 1
s A B CF s
s ss s s
−= = + ++ −+ − + 
 
where ( )21 2
5 1 5 12 and 1
2 1s s
s sB C
s s=− =
− −= = =− + = 
Then ( ) ( ) ( )
2
21
1
91 1
2s
s
dA s F s
ds s=− =−
−⎡ ⎤= + = =⎣ ⎦ − − 
 
Finally ( ) ( ) ( )
2
2
1 2 1( ) 2 + 
1 21
t t tF s f t e t e e u t
s ss
− −− ⎡ ⎤= + + ⇒ = − +⎣ ⎦+ −+ 
 
 
2 
P14.4-4 
( ) ( ) ( ) ( ) ( ) ( )222
1 1
11 2 2 1 11 1 1
A Bs CY s
ss s s ss s
+= = = ++⎡ ⎤+ + + + ++ + +⎣ ⎦
 
where 2
1
1 1
2 2 s
A
s s =−
= =+ + 
Next 
( ) ( ) ( )
( ) ( )
2
22
2
1 1 1 2 2 ( 1)
1 2 21 2 2
1 1 2
B s C s s Bs C s
s s ss s s
B s B C s C 2
+= + ⇒ = + + + + ++ + ++ + +
⇒ = + + + + + +
 
Equating coefficients: 
2s : 0 1 1
: 0 2 1
B B
s B C C
= + ⇒ = −
= + + ⇒ =− 
 
Finally ( ) ( ) ( ) ( )2
1 cos
1 1 1
t tsY s y t e e t u t
s s
− −1 + ⎡ ⎤= − ⇒ = −⎣ ⎦+ + + 
 
 
P14.4-5 
( ) ( )( ) ( ) ( )( ) ( )2 22
2 3 11 2
11 2 5 1 4 1 4
s s
F s
ss s s s s
+ − += = + +++ + + + + + + 
 
( ) ( ) ( ) ( )cos 2 sin 2t t tf t e e t e t u− − −⎡ ⎤= − +⎣ ⎦ t 
 
 
 
P14.4-6 
( ) ( )( ) ( )
2 s+3
s s+1 2 1 2
A B CF s
s s s s
= = + ++ + + 
where 
( ) ( )( ) ( ) ( ) ( )
( )
( )0 1 10
2 3 2 3
3, 1 4
1 2 2s s ss
s s
A sF s B s F s
s s s s= =− =−=
+ += = = = + = =+ + + − 
and 
( ) ( ) ( )( )2 2
2 3
2 1
1s s
s
s F s C
s s=− =−
++ = =+ = 
Finally 
( ) ( ) ( ) ( )23 4 1 3 41 2 t tF s f t e e u ts s s − −−= + + ⇒ = − ++ + 
 
 
3 
P14.4-7 
( ) ( )( )
( ) [ ] ( ) ( ) ( ) ( ) (
2 2
2 2 1
2 2
1
2
1 , tan
2
 cos sin cos cos
j j
at at at
cs ca d c jd c jdF s
s a j s a js a
me me dwhere m c d cs a j s a j
)f t e c t d t u t e c d t u t m e t u t
θ θ
ω
ω ωω
θω ω
ω ω ω θ ω θ
−
−
− − −
+ − ⎡ ⎤+ −= = +⎢ ⎥+ − + ++ + ⎣ ⎦
⎡ ⎤= + = + =⎢ ⎥+ − + +⎣ ⎦
⎡ ⎤∴ = − = + + = +⎣ ⎦
 
 
P14.4-8 
(a) ( ) ( )( )
( )( )
( ) ( )
( ) ( ) ( )
22
2 21
2
2 8 38 3 1
4 13 2 2 9
3 8 2
 2, 8, 3 & 3 7.33
3
6.33 tan 38.4 , 8 6.33 10.85
8
10.85 cos 3 42.5t
ssF s
s s s
a c ca d d
m
f t e t u t
ω ω
θ − °
−
−−= = ×+ + + +
− −∴ = = = − =− ⇒ = =−
⎛ ⎞∴ = = = + =⎜ ⎟⎝ ⎠
⇒ = +
 
 
(b) ( ) ( ) ( )( )( )
( ) ( ) ( ) ( ) ( )
1 22 2
1
1 1 1
2 33 3 1Given , first consider .
2 17 2 17 2 1 16
3 4Identify 1, 0, 4 3 3 4. Then | | 3 4, tan 90
0
So ( ) (3 4) sin 4 . Next, 1 . Fin
s
t s
eF s F s
s s s s s
a c and d d m d
f t e t u t F s e F s f t f t
ω ω θ
−
−
− −
= = = ×+ + + + + +
−⎛ ⎞= = = − = ⇒ =− = = = =−⎜ ⎟⎝ ⎠
= = ⇒ = − ally
° 
 
( ) ( ) ( )( 1)(3 4) sin 4 1 1tf t e t u t− −∴ = −⎡ ⎤⎣ ⎦ − 
 
 
 
P14.4-9 
(a) ( ) ( ) ( )
2
2 2
5
11 1
s A B CF s
s ss s s
−= = + +++ + 
where 
( ) ( ) ( )20 15 1| 5 and 1 | 41 1s sA sF s C s F s= =
5
−
− −= = = − = + = − =
6
 
 
Multiply both sides by ( )21s s +
 
( ) ( )22 5 5 1 1 4s s Bs s s B− = − + + + + ⇒ = 
4 
Then 
( ) ( )2
5 6 4
1 1
F s
s s s
−= + ++ + 
Finally 
( ) ( )5 6 4 , 0t tf t e t e− − t= − + + ≥ 
 
(b) ( ) ( ) ( ) ( ) ( )
2
3 2
4
33 3
s A B CF s
ss s
= = + +++ + 33s+ 
where 
( ) ( ) ( ) ( )2 3 32 3 31 3 4, 32 s s
d dA s F s B s F s
ds ds=− =−
⎡ ⎤ ⎡ ⎤= + = = + =⎣ ⎦ ⎣ ⎦ 24−
6
 
and 
( ) ( )3 33 3sC s F s =−= + =Then 
( ) ( ) ( ) ( )2 3
4 24 36
3 3 3
F s
s s s
−= + ++ + + 
Finally 
( ) ( )2 34 24 18 , 0tf t t t e t−= − + ≥ 
 
 
 
 
5 
Section 14-5: Initial and Final Value Theorems 
 
P14.5-1 
 
(a) ( ) ( ) 2 22 22 3 4 20 lim lim 3 2
s s sf sF s
s s ss s
− += = =+ +→∞ →∞ 2= 
 
(b) ( ) ( ) 4lim 2
20
f sF s
s
∞ = = =→
 
 
 
P14.5-2 
Initial value: ( ) ( ) ( ) 22 216 160 lim lim lim 14 12 4 12s s s
s s s sv sV s
s s s s→∞ →∞ →∞
+ += = =+ + + + = 
 
Final value: ( ) 22 20 016 16lim lim 04 12 4 12s s
s sv s
s s s s→ →
⎛ ⎞+ +∞ = = =⎜ ⎟+ + + +⎝ ⎠
s 
(Check: V(s) is stable because { }Re 0 since 2 2.828i ip p j< = − ± . We 
expect the final value to exist.) 
 
 
P14.5-3 
Initial value: 2
3 2
10(0) lim ( ) lim 0
3 2s s
s sv sV s
s s s→∞ →∞
+= = + + = 
 
Final value: ( ) ( ) ( )( )20 0
10
lim lim 10
3 2 1s s
s s
v sV s
s s s→ →
+∞ = = =+ + 
(Check: V(s) is stable because 0.333 0.471p ji = − ± . We expect the 
final value to exist.) 
 
 
P14.5-4 
Initial value: ( ) ( ) 22 2 140 lim lim 22 10ss
s sf s F s
s s→∞→∞
− −= = − + = − 
 
Final value: F(s) is not stable because { }Re 0 since 1 3i ip p> = j± . No final value of 
( )f t exists. 
 
 
 
14-1 
Section 14-6: Solution of Differential Equations Describing a Circuit 
 
P14.6-1 
 
KVL: 
42 1050 0.001 2 tdii v
dt
− ×+ + = e for t ≥ 0 
The capacitor current and voltage are related 
by 
 
( )62.5 10 dvi dt−= × 
 
 
42 10
1 = 2e V
tv − × , (0) 1 A, (0) 8 Vi v= =
 
Taking the Laplace transforms of these equations yields 
 
[ ]
( ) ( )
4
6
250 ( ) 0.001 ( ) (0) ( )
 2 10
( ) 2.5 10 ( ) 0
I s s I s i V s
s
I s sV s v−
+ − + = + ×
= × −⎡ ⎤⎣ ⎦
 
 
Solving for I(s) yields 
 
( ) ( ) ( ) ( )
2 4 8
4 44 4 4
1.4 10 1.6 10
10 2 10 4 1010 2 10 4 10
s s A BI s
s s ss s s
+ × − ×= = ++ + × + ×+ + × + × 4
C+ 
where 
( ) ( ) ( )( )
( ) ( ) ( ) ( )
( ) ( )
2 4 8 8
4
4 84 4= 10
4
= 10
2 4 8 8
4
4 84 4= 2 10
4
= 2 10
2
4
4
= 4 10
1.4 10 1.6 10 2 10 2 10
3 10 32 10 4 10
s 1.4 10 s 1.6 10 .4 10 1 2 10 
2 10 5s 10 s 4 10
s 1.4 4 10 
s
s
s
s
s
s sA s I s
s s
B s I s
C s I s
−
−
− ×
− ×
− ×
+ × − × − × −= + = = =×+ × + ×
+ × − × ×= + × = = =×+ + ×
+ ×= + × = ( ) ( )
4 8 8
84 4
4
= 4 10
10 s 1.6 10 8.8 10 22 
6 10 15s 10 s 2 10
s − ×
− × ×= =×+ + ×
 
Then 
 
( ) ( ) ( )4 4 410 t 2x10 4x104 4 4 12 3 1 5 22 15 10e 3e 22e A10 2 10 4 10 15 t tI s i t u ts s s − − −⎡ ⎤= − + + ⇒ = − + +⎣ ⎦+ + × + × 
 
 
 
 
14-1 
P14.6-2 
 
We are given ( ) 160cos 400v t t= . 
The capacitor is initially uncharged, so 
( )C 0 0 Vv = . Then 
 
( ) ( )160cos 400 0 00 160 A
1
i
× −= = 
KCL yields 
C C3 10
100
d v v
i
dt
− + = 
Apply Ohm’s law to the 1 Ω resistor to get 
C
C 1
v v
i v v i
−= ⇒ = − 
Solving yields 
( )41010 1600 cos 400 6.4 10 sin 400d i i tdt + = − × t 
 
Taking the Laplace transform yields 
( ) ( )
( ) ( )
( )
2
2 22 2
6.4 10 4001600s( ) (0) 1010 ( )
s 400 400
s I s i I s
s
×− + = −+ + 
so 
( )
7
2 2
160 1600s 2.5 10( )
1010 1010 (400)
I s
s s s
− ×= ++ ⎡ ⎤+ +⎣ ⎦
 
Next 
( )
7 *
2 2
1600 2.5 10 
1010 400 4001010 (400)
s A B
s s j s js s
− × = + ++ + −⎡ ⎤+ +⎣ ⎦
B 
where 
( )
7
22
 = 1010
1600 2.5 10 23.1
400
s
sA
s −
− ×= =+ − , 
( ) ( )
7 7
*
5
 = 400
1600 2.5 10 2.56 10 1.4 11.5 27.2 and 11.5 27.2
1010 400 8.69 10 68.4s j
s xB j
s s j
°
°
−
− × ∠= = = −+ − × ∠
B j= + 
Then 
( ) 136.9 11.5 27.2 11.5 27.2 
1010 400 400
j jI s
s s j s j
− += + ++ + − 
Finally 
( ) ( ) ( )1010
1010
 136.9 2 11.5 cos 400 2 27.2 sin 400 for 0
 136.9 23.0 cos 400 54.4sin 400 for 0
t
t
i t e t t t
e t t t
−
−
= + −
= + − >
>
 
14-2 
P14.6-3 
 
C (0) 0v = 
3
c
c
c3 c
15 10 10cos 2
2 20cos 21 10
30
v i t
d v v td v dti
dt
−
⎫+ × = ⎪ ⇒ + =⎬⎛ ⎞= ×⎜ ⎟ ⎪⎝ ⎠ ⎭
 
 
Taking the Laplace Transform yields: 
 
( ) ( ) ( ) ( ) ( )( )
*
C C C C2 2
200 2 20
4 22 4
s s AsV s v V s V s
s ss s
− + = ⇒ = = + +
2 2
B B
s j s j+ + + −+ + 
where 
( )( )2 = 2 = 2
20 40 20 5 5 5 5 5*5, and 
4 8 2 2 1 2 2 2s s j
s sA B j
s s s j j− −
−= = = − = = = + = −+ + − − 2B j 
Then 
( ) ( ) ( )2C C
5 5 5 5
5 2 2 2 2 5 5 cos 2 sin 2
2 2 2
t
j j
V s v t e t t
s s j s j
−
+ −−= + + ⇒ = − + ++ + − V 
 
 
 
 
 
P14.6-4 
L c
c L L12 2 8 and 
d i d vv i i C
dt dt
− + + = − = −
 Taking the Laplace transform yields 
 
( ) ( ) ( ) ( )L Lc L 812 2 0V s I s sI s i s− + + − = −⎡ ⎤⎣ ⎦ 
( ) ( ) ( )c cL 0I s C sV s v= − −⎡ ⎤⎣ ⎦ 
 
 
c L(0) 0, (0) 0v i= = 
 
 
 
 
14-3 
Solving yields 
( )c
2
4 
6
2
CV s
Cs s s
= ⎛ ⎞+ +⎜ ⎟⎝ ⎠
 
 
(a) 1 F
18
C = ( ) ( ) ( )2 2
72 
33 3
c
ca bV s
s ss s s
= = + +++ + 
 
( ) ( )2
248 88, 8, and 24
3 3
ca b c V s s s s
−−= = − = − ⇒ = + ++ + 
( ) 3 3 8 8 24 V, 0t tcv t e t e t− −= − − ≥ 
 
(b) 1 F
10
C = ( ) ( ) ( )
40 
1 5 1 5c
ca bV s
s s s s s s
= = ++ + + ++ 
 
( )C 28 108, 10, and 2 1 5a b c V s s s s
−= = − = ⇒ = + ++ + 
 
( ) 5 8 10 2 V, 0t tcv t e e t− −= − + ≥ 
 
 
 
P14.6-5 ( )c L(0 ) 10 V, 0 0 Av i− −= = 
 
( )6 c c5 10 and 400 1 0d v d ii idt dt−= × + + =v 
 
Taking Laplace transforms yields 
( ) ( ) ( )( )
( ) ( )( ) ( ) ( ) ( )
6
c
22 5 2
c
1 4005 10 10 10 40
400 2 10400 0 0 200 400
I s sV s
I s
s sI s s I s V s s
− ⎛ ⎞−⎜ ⎟⎫= × − −⎪ ⎝ ⎠⇒ = =⎬ + + ×+ − + = + +⎪⎭
 
so 
( ) ( ) ( )2001 sin 400 A
40
ti t e t u t−= − 
 
 
 
 
 
14-4 
P14.6-6 
After the switch opens, apply KCL and KVL to 
get 
( ) ( ) ( )1 sdR i t C v t v t Vdt
⎛ ⎞+ +⎜ ⎟⎝ ⎠ = 
 
Apply KVL to get 
 
( ) ( ) ( )2dv t L i t R i tdt= + 
 
Substituting into the first equation gives ( )v t 
 
( ) ( ) ( ) ( ) ( )1 2d d d 2 sR i t C L i t R i t L i t R i t Vdt dt dt
⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ = 
then ( ) ( ) ( ) ( ) ( )21 1 2 1 22d d sR C L i t R C R L i t R R i t Vdtdt + + + + = 
Dividing by : 1R C L
( ) ( ) ( )2 1 2 1 2 s2
1 1
R C R L R R Vd di t i t i t
1R C L dt R C L R C Ldt
⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 
 
With the given values: ( ) ( ) ( )22 25 156.25 125d di t i t i tdtdt + + = 
 
Taking the Laplace transform: 
 
( ) ( ) ( ) ( ) ( ) ( )2 1250 0 25 0 156.25ds I s i s i s I s i I s
dt s
⎡ ⎤⎛ ⎞− + + + + − + + =⎡ ⎤⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠⎣ ⎦ 
 
We need the initial conditions. For t < 0, the switch is 
closed and the circuit is at steady state. At steady state, 
the capacitor acts like an open circuit and the inductor 
acts like a short circuit. Using voltage division 
 
( ) ( )
90 20 14.754 V
9 16 || 4
v − = =+ 
 
Then, using current division 
 
 
( ) ( )040 0.328 A
16 4 9
v
i
−⎛ ⎞− = =⎜ ⎟+⎝ ⎠ 
 
The capacitor voltage and inductor current are continuous so ( ) ( )0 0v v+ = − and ( ) ( )0 0i i+ = − . 
After the switch opens 
 
( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 9 0 9 0.32814.7540 29.5080.4 0.4 0.4 0.4
v id dv t L i t R i t i
dt dt
+ += + ⇒ + = + = + = 
 
Substituting these initial conditions into the Laplace transformed differential equation gives 
 
( ) ( ) ( ) ( )2 12529.508 0.328 25 0.328 156.25s I s s s I s I s
s
⎡ ⎤− + + − + =⎡ ⎤⎣ ⎦⎣ ⎦ 
 
( ) ( ) ( )2 12525 156.25 29.508 0.328 25 0.328s s I s ss+ + = + + + ( ) 
so 
( ) ( )( )( )
( )( )
( ) ( )
2
2
2
2 2
0.328 29.508 25 0.328 125
25 156.25
0.328 29.508 25 0.328 125 0.471 23.6 0.812.512.5 12.5
s
I s
s s s
s
s ss s s
+ + += + +
+ + + −= = ++ ++ +
 
 
Taking the inverse Laplace transform 
 
( ) ( )12.50.8 23.6 0.471 A for 0ti t e t t−= + − ≥ 
so 
( ) ( )12.5
0.328 A for 0
0.8 23.6 0.471 A for 0t
t
i t
e t t−
≤⎧= ⎨ + − ≥⎩
 
 
(checked using LNAP 10/11/04) 
 
P14.6-7 
 
 
KCL: 1 67
5
tv i e−+ = 
KVL: 1 14 3 0 4
di dii v v i
dt dt
3+ − = ⇒ = + 
Then 6 6
4 3 357 2
5 4
t t
di i didt i e i e
dt
− −
+
+ = ⇒ + = 
 
Taking the Laplace transform of the differential equation: 
 
35 1 35 1( ) (0) 2 ( ) ( )
4 6 4 ( 2)( 6
s I s i I s I s
s s
− + = ⇒ =
)s+ + + 
 
Where we have used . Next, we perform partial fraction expansion. (0) 0i =
 
2 
1 1 1 where and 
( 2) ( 6) 2 6 6 4 2 4s s
A B A B
s s s s s s=− = −
= + = = = = −+ + + + + + 6
1 1 
Then 
2 635 3535 1 35 1( ) ( ) A, 0
16 2 16 6 16 16
t tI s i t e e
s s
− −= − ⇒ = −+ + t ≥ 
 
 
P14.6-8 
 
Apply KCL at node a to get 
 
1 2 1 1
1 2
1 2 2
48 24
d v v v d v
v v
dt dt
−= ⇒ + = 
 
Apply KCL at node b to get 
 
 
2 2 1 2 2 2
1 2
50cos 2 1 0 3 60
20 24 30 24
v t v v v d v d v
v v
dt dt
− −+ + + = ⇒ − + + = cos 2 t 
 
Take the Laplace transforms of these equations, using 1 2(0) 10 V and (0) 25 Vv v= = , to get 
 
( ) ( ) 21 2 1 2 225 60 1002 ( ) 2 ( ) 10 and ( ) 3 ( ) 4
s ss V s V s V s s V s
s
+ ++ − = − + + = + 
 
Solve these equations using Cramer’s rule to get 
 
 
( )
( )
( )
( )( ) (
( )( )( )
)
( )( )( )
2
2 22
2 2
3 2
2
25 60 1002 10 2 25 60 100 10 44
2 (3 ) 2 4 1 4
25 120 220 240
4 1 4
s ss s s s ss
V s
s s s s s
s s s
s s s
⎛ ⎞+ ++ +⎜ ⎟ + + + + ++⎝ ⎠= =+ + − + + +
+ + += + + +
 
Next, partial fraction expansion gives 
 
( ) *2V 2 2 1 4
A A B Cs
s j s j s s
= + + ++ − + + 
where 
( ) ( ) ( )
( ) ( )
( ) ( )
3 2
2
*
3 2
2
1
3 2
2
4
25 120 220 240 240 240 6 6
1 4 2 40
6 6
25 120 220 240 115 23 
15 34 4
25 120 220 240 320 16 
60 34 1
s j
s
s
s s s jA j
s s s j
A j
s s sB
s s
s s sC
s s
=−
=−
=−
+ + + − −= =+ + − −
= −
+ + += = =+ +
+ + + −= = =−+ +
= +
 
Then 
( )2 6 6 6 6 23 3 16 32 2 1
j jV s
s j s j s s 4
+ −= + + ++ − + + 
Finally 
4
2
23 16( ) 12cos 2 12sin 2 V 0
3 3
t tv t t t e e t− −= + + + ≥ 
 
 
 
Section 14-7: Circuit Analysis Using Impedance and Initial Conditions 
 
P14.7-1 
 
 
400
6 0.010 1.2 0.002 .003 .005( )
5 2000 ( 400) 400
2 mA 0
( )
3 5 mA 0
L
L t
ssI s
s s s s s
t
i t
e t−
− −= = = −+ +
− <⎧= ⎨ − >⎩
+
 
 
 
P14.7-2 
 
 
 
( )L LL
L
L
800
3
L
10 8( ) ( ) .015( ) 30 ( 8002000 4000 5
3
8
( ) 0.15 0.003 0.005 0.00215( ) 8008005
33
( ) 5 2 mA, 0
L
t
V s V sV ss V s
s s
V sI s
s s s ss s
i t e t
−
− − −= + + ⇒ =
+
+= = + = −⎛ ⎞ ++⎜ ⎟⎝ ⎠
= − >
)
 
 
14-1 
P14.7-3 
 
6
1000
8( )( )0.006 0
102000
.5
6000 8500 ( ) 0.5 ( ) 0
8 12000 12 4( )
( 1000) 1000
( ) 12 4 V, 0
c
c
c c
c
t
c
V sV s s
s
s
V s s V s
s s
sV s
s s s s
V t e t−
−
− + + =
⎛ ⎞− + + − =⎜ ⎟⎝ ⎠
+= = −+ +
= − >
 
 
P14.7-4 
 
 
( )
6
1500
6( ) ( ) 0.5 8( ) 0
2000 4000 10
6 8500 ( ) 250 ( ) 0.5 ( ) 0
6000 8 4 4( )
1500 1500
( ) 4 4 V, 0
c
c
c
c c c
c
t
c
V s V s ss V s
s
V s V s s V s
s s
sV s
s s s s
v t e t−
− ⎛ ⎞ ⎛ ⎞+ + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞− + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+= = ++ +
= + >
 
 
14-2 
P14.7-5 
 
 
Node equations: 
 
( ) ( ) ( ) ( ) ( )a C a a C1 66 6
V s V s V s
V s V s
s s s
− + = ⇒ = + 6
6s+ + 
( ) ( ) ( ) ( )C CC C
6 66
1 36 63 0
4 4
V s V sV s ss ss V s
s s
⎛ ⎞− +− ⎜ ⎟+ +⎝ ⎠+ + + +
2
− =
)
 
 
After quite a bite of algebra: 
( ) ( )( )(
2
C
6 56 132
2 3
s sV s
s s s
+ +=
5+ + + 
Partial fraction expansion: 
( )( )( )
2
44 1
6 56 132 93 3( )c 3 2 5 2 3
s sV s
s s s s s s
+ += = −
5
++ + + + + + 
Inverse Laplace transform: 
2 3 544 1( ) 9 V, 0c 3 3
t t tv t e e e t− − −= − + ≥ 
 
 
 
14-3 
P14.7–6 
 
 
Write a node equation in the frequency domain: 
( ) ( ) ( )
2 2
1 1o o
o
1 2
22
1010 5 10 5 10
5 01 11
1
R R
s
R C RV s V ss C V s
R R s ss sCs R CR C
R
⎡ ⎤+ −⎢ ⎥⎢ ⎥+ − + = ⇒ = − = − +⎢ ⎥⎛ ⎞ +⎢ ⎥+⎜ ⎟⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠
 
Inverse Laplace transform: 
( ) 22 2 1000o
1 1
10 5 10 10 5 V for 0t R C t
R R
v t e e t
R R
− −⎡ ⎤⎛ ⎞ ⎡ ⎤= − + − = − − >⎢ ⎥⎜ ⎟ ⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
 
14-4 
P14.7-7 
Here are the equations describing the coupled coils: 
 
( ) ( )
( ) ( )
1 2
1 1 1 1 2 1 2
2 1
2 2 2 1 2 1 2
( ) ( ) 3 ( ) 2 ( ) 3 3 ( ) ( ) 9
( ) ( ) ( ) 2 2 ( ) 3 ( ) 2 ( ) 8
di div t L M V s s I s sI s s I s sI s
dt dt
di div t L M V s s I s sI s sI s sI s
dt dt
= + ⇒ = − + − = +
= + ⇒ = − + − = +
−
−
 
 
Writing mesh equations: 
 
( ) ( ) ( ) ( )
( )
1 2 1 1 2 1 2 1 2
1 2 2 1 2 1 2 2 1 2
55 2 ( ) ( ) 2 ( ) ( ) 3 ( ) ( ) 9 3 2 2 9
( ) ( ) 1 ( ) 3 ( ) ( ) 9 ( ) 2 ( ) 8 ( ) 2 1 1
I s I s V I s I s s I s sI s s I s I
s s
V s V s I s s I s sI s sI s sI s I s s I s I
== + + + + + − ⇒ + + + = +
= + ⇒ + − = + − + ⇒ − + =
 
 
Solving the mesh equations for I2(s): 
 
( ) ( )( )2 2
15 8 3 1.6 0.64 2.36 = + 
0.26 1.54 + 0.26 + 1.545 9 2
s sI s
s s s ss s
+ += = + ++ + 
 
Taking the inverse Laplace transform: 
 
0.26 1.54
2( ) 0.64 2.36 A for 0
t ti t e e t− −= + > 
 
 
P14.7-8 
 
t<0 
 
 
 
time domain 
 
frequency domain 
 
Mesh equations in the frequency domain: 
 
( ) ( ) ( )( ) ( ) ( ) ( )1 1 2 1 1 212 2 26 6 6 0 3 3I s I s I s I s I s I ss s+ − + + = ⇒ = − 
14-5 
( ) ( ) ( )( ) ( ) ( )2 1 2 2 12 6 26 0 6 6I s I s I s I s I ss s s⎛ ⎞− − − = ⇒ + − =⎜ ⎟⎝ ⎠ 6s 
 
Solving for I2(s): 
( ) ( ) ( )2 2 2
1
2 2 2 6 26 6 13 3
2
I s I s I s
s s s s
⎛ ⎞⎛ ⎞+ − − = ⇒ =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ +
 
Calculate for Vo(s): 
( ) ( )o 2
1
1 6 1 6 22
1 12 2
2 2
V s I s
s ss s
⎛ ⎞⎜ ⎟ −= − = − =⎜ ⎟⎜ ⎟+ +⎝ ⎠
4
s
− 
Take the Inverse Laplace transform: 
 
( ) ( )/ 24 2 V for 0tov t e t−= − + > 
 
(Checked using LNAP, 12/29/02) 
 
P14.7-9 
 
t<0 
 
 
 
time domain 
 
 
frequency domain 
 
Writing a mesh equation: 
( ) ( ) ( )
26
12 3 354 5 30 0 44
55
s
s I s I s
s s ss s
⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎝ ⎠+ + + = ⇒ = = − +⎜ ⎟⎛ ⎞ ⎜ ⎟++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
 
Take the Inverse Laplace transform: 
 
( ) ( )0.83 1 A for 0ti t e t−= − + > 
 
(Checked using LNAP, 12/29/02) 
14-6 
P14.7-10 
Steady-state for t<0: 
 
 
 
From the equation for vo(t): 
 
( ) ( )o 26 12 6 Vv e− ∞∞ = + = 
 
From the circuit: 
( ) ( )o 3 183v R∞ = + 
Therefore: 
( )36 18 6
3
R
R
= ⇒ =+ Ω 
 
Steady-state for t>0: 
 
 
 
 
 
( ) ( )1 18 6 62 0 1
2
I s I s
C s s s s
C
⎛ ⎞ −+ + − = ⇒ =⎜ ⎟⎝ ⎠ +
 
( ) ( )o 1 18 1 6 18 12 12 18 121 1
2 2
V s I s
Cs s Cs s s s ss s s
C C
⎛ ⎞⎜ ⎟− −= + = + = + + =⎜ ⎟⎜ ⎟+ +⎜ ⎟⎝ ⎠
6
1
2C
+
+
 
 
Taking the inverse Laplace transform: 
 
( ) / 2o 6 12 V for 0t Cv t e t−= + > 
Comparing this to the given equation for vo(t), we see that 
12 0.25 F
2
C
C
= ⇒ = . 
 
(Checked using LNAP, 12/29/02)
14-7 
 
P14.7-11 
We will determine , the Laplace transform of the output, twice, once from the given 
equation and once from the circuit. From the given equation for the output, we have 
( )oV s
 
( )o 10 5100V s s s= + + 
 
Next, we determine fromthe circuit. For , we represent the circuit in the frequency 
domain using the Laplace transform. To do so we need to determine the initial condition for the 
capacitor. 
( )oV s 0t ≥
 
When and the circuit is at steady state, 
the capacitor acts like an open circuit. Apply 
KCL at the noninverting input of the op amp to 
get 
0t <
( ) ( )
1
3 0
0 0
v
v
R
− − = ⇒ − = 3 V 
The initial condition is 
 
( ) ( )0 0 3v v+ = − = V 
Now we can represent the circuit in the frequency domain, using Laplace transforms. 
 
 
 
Apply KCL at the noninverting input of the op 
am to get 
( ) ( )
6
1
2 3
10
V s V s
s s
R
s
− −
= 
Solving gives 
( )
6
1
6 6
1 1
103 2
2 1
10 10
s
R
V s
s
s s s
R R
+
= = +⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
Apply KCL at the inverting input of the op amp to get 
 
( ) ( ) ( ) ( ) ( )o 2o 62
1
2 11 1
1000 1000 1000 10
V s V s R RV s
V s V s
R s
s
R
⎛ ⎞⎜ ⎟− ⎛ ⎞ ⎛ ⎞⎜ ⎟= ⇒ = + = + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎛ ⎞⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
2 
The expressions for Vo(s) must be equal, so 
2
6
1
10 5 2 11
100 1000 10
R
s s s
s
R
⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟+ = + +⎜ ⎟⎜ ⎟+ ⎛ ⎞⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
Equating coefficients gives 
 
2
21 5 41000
R
R+ = ⇒ = kΩ and 
6
1
1
10 100 10 kR
R
= ⇒ = Ω 
 
(checked using LNAPTR 7/31/04) 
 
 
P14.7-12 
For t < 0, The input is constant. At steady state, 
the capacitor acts like an open circuit and the 
inductor acts like a short circuit. 
 
The circuit is at steady state at time 0t = − so 
 
( )C 0v − = 0 and ( )L 0i B− =
 
The capacitor voltage and inductor current are continuous so ( ) ( )C C0 0v v+ = −
)
and 
. ( ) (L L0 0i i+ = −
 
For t < 0, represent the circuit in the 
frequency domain using the Laplace 
transform as shown. is the node 
voltage at the top node of the circuit. 
Writing a node equation gives 
( )CV s
 
( ) ( ) ( )C C CV s V sA B B C sV ss R s L s
+ = + + + 
so 
 
( )2 CA L s R R LC s V ss R L s
+ += 
Then 
( )C 2
2 1 1
A
A R L CV s
R LC s L s R s s
RC LC
= =+ + + +
 
and 
( ) ( )CL
2 1 1
A
V s B BLCI s
L s s s
s s s
RC LC
= + = +⎛ ⎞+ +⎜ ⎟⎝ ⎠
 
a.) When 12 , 4.5 H, F, 5 mA and 2 mA
9
R L C A B= Ω = = = = − , then 
( ) ( )L 2
5 40
10 2 3 7 7
144.5 2
2
I s
s s ss s s s
−= + = + ++ + − +
 
Taking the inverse Laplace transform gives 
 
( ) 4 0.5L 5 53 mA for 07 7t ti t e e t− −= + − ≥ 
 
b.) When , then 1 , 0.4 H, 0.1 F, 1 mA and 2 mAR L C A B= Ω = = = = −
 
( ) ( ) ( ) ( )L 2 22
25 2 25 2 1 5 1
510 25 5 5
I s
s s ss s s s s s
⎛ ⎞− −= + = + = − +⎜ ⎟⎜ ⎟++ + + +⎝ ⎠s
+ 
 
Taking the inverse Laplace transform gives 
 
( ) ( )5 5L 1 5 mA for 0t ti t t e e t− −= − + − ≥ 
 
c.) When , then 1 , 0.08 H, 0.1 F, 0.2 mA and 2 mAR L C A B= Ω = = = = −
 
( ) ( ) ( ) ( ) ( )L 2 22 2 2
25 2 1.8 0.2 2 1.8 5 100.2 0.1
10 125 5 10 5 10 5 10
s sI s
s s ss s s s s s
− − − − − += + = + = − −+ + + + + + + +2 2
 
Taking the inverse Laplace transform gives 
 
( ) ( ) ( )( )5L 1.8 0.2cos 10 0.1sin 10 mA for 0ti t e t t t−= − − + ≥ 
 
P14.7-13 
For t < 0, the switch is open and the circuit is at 
steady state. and the circuit is at steady state. At 
steady state, the capacitor acts like an open 
circuit. 
( )
2
Ai t
R
= and ( )C 2
Av t =
 
Consequently, 
( )0
2
Ai
R
− = and ( )C 0 2
Av − =
 
Also 
( )C 0 0i − = 
 
The capacitor voltage and inductor current are continuous so ( ) ( )C C0 0v v+ = −
)
and 
. ( ) (L L0 0i i+ = −
 
For t > 0, the voltage source voltage is 
12 V. Represent the circuit in the 
frequency domain using the Laplace 
transform as shown. 
 
( ) ( )L C and I s I s are mesh currents. 
Writing a mesh equations gives 
 
 
( ) ( ) ( )( )L L C 02A L AL s I s R I s I sR s− + − − = 
( ) ( ) ( )( )C L C1 02AI s R I s I sC s s− + − = 
Or, in matrix form 
( )
( )LC
2
1
2
A L A
L s R R I s R s
R R I s A
C s
s
⎛ ⎞++ −⎛ ⎞ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟− + ⎜ ⎟⎜ ⎟⎝ ⎠ −⎝ ⎠ ⎜ ⎟⎝ ⎠
 
 
( )
( )
( )C 22
2 2 2
1 11
A A L A AL s R R
s R s LI s
s sL s R R R RC LCC s
⎛ ⎞ ⎛ ⎞+ − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =⎛ ⎞ + ++ + −⎜ ⎟⎝ ⎠
 
 
 
a.) When 13 , 2 H, F and 12 V
24
R L C A= Ω = = = , 
( ) ( )( )C 2
3 3
3 2 4 4
8 12 2 6 2 8
I s
s s s s s s
= = = −+ + + + + + . 
 
Taking the inverse Laplace transform gives 
 
( ) ( )2 6C 3 3 A4 4t ti t e e u t− −
⎛ ⎞= −⎜ ⎟⎝ ⎠ 
 
b.) 12 , 2 H, F and 12 V
8
R L C A= Ω = = = , 
 ( ) ( )C 22
3 3
4 4 2
I s
s s s
= =+ + + 
 
Taking the inverse Laplace transform gives 
 
( ) ( )2C 3 ti t t e u t−= A 
 
c.) 110 , 2 H, F and 12 V
40
R L C A= Ω = = = 
 
( ) ( ) ( )C 2 22
3 3 3 4
4 20 42 16 2 1
I s
s s s s
= = = ×+ + 6+ + + + 
 
Taking the inverse Laplace transform gives 
 
( ) ( ) ( )2C 3 sin 4 A4 ti t e t u t−= 
 
(checked using LNAP 4/11/01) 
 
 
 
P14.7-14 
For t < 0, The input is 12 V. At steady state, the 
capacitor acts like an open circuit. 
 
Notice that v(t) is a node voltage. Express the 
controlling voltage of the dependent source as a 
function of the node voltage: 
 
va = −v(t) 
 
Writing a node equation: 
 
( ) ( ) ( )12 3 0
8 4 4
v t v t
v t
−⎛ ⎞ ⎛ ⎞− + + −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
= 
 
( ) ( ) ( ) ( )12 2 6 0 4 Vv t v t v t v t− + + − = ⇒ = − 
 
 
 
 
( ) ( )0 0 4v v+ = − = − V 
 
For t < 0, represent the circuit in the frequency 
domain using the Laplace transform as shown. 
 
( )V s is a node voltage. Express the controlling 
voltage of the dependent source in terms of the 
node voltages 
( ) ( )aV s V s= − 
 
Writing a node equation gives 
 
( ) ( ) ( ) ( )
6
3 4 0.75
8 4 40
V s V s ss V s V s
s
− ⎛ ⎞+ + + =⎜ ⎟⎝ ⎠ 
 
Solving gives 
( ) ( ) ( ) ( )
10 10 4 2 2 4 1 15 4 2
5 5 5 5
s V s V s
s s s s s s s s
− ⎛ ⎞− = − ⇒ = − = + − = − +⎜ ⎟5s− − − − ⎝ ⎠− 
 
Taking the inverse Laplace transform gives 
 
( ) ( )52 1 V for 0tv t e t= − + ≥ 
 
This voltage becomes very large as time goes on. 
 
 
 
P14.7-15 
For t < 0, the voltage source voltage is 2 V 
and the circuit is at steady state. At steady 
state, the capacitor acts like an open circuit. 
 
( ) 3 32 00 0.010 10 40 10i
−− = =× + × 4 mA 
and 
 
( ) ( )( )3 3C 0 40 10 0.04 10 1.6 Vv −− = × × = 
The capacitor voltage is continuous so ( ) ( )C C0 0v v+ = −
o
. 
 
For t > 0, the voltage source voltage is 
12 V. Represent the circuit in the 
frequency domain using the Laplace 
transform as shown. 
 
( ) ( )C and V s V s are node voltages. 
Writing a node equation gives 
 
( ) ( ) ( ) ( ) ( ) ( )C C C C C63 3
12 1.6
12 1.60 4 0.08 0
0.5 1010 10 40 10
V s V s V ss s V s s V s V s
s s
s
− − ⎛ ⎞ ⎛ ⎞+ + = ⇒ − + − +⎜ ⎟ ⎜ ⎟×× × ⎝ ⎠ ⎝ ⎠ C = 
( )( ) ( ) ( ) ( )C C
48 80 48 1.6 600 9.6 80.08 5 0.128
0.08 5 62.5 62.5
s sV s s V s
s s s s s s s
+ + −+ = + ⇒ = = = ++ + + 
 
Taking the inverse Laplace transform gives 
 
( ) 62.5C 9.6 8 V for 0tv t e t−= − ≥ 
 
The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so 
 
( ) ( ) ( )62.5 62.5o C50 50 9.6 8 12 10 V for 040 40 t tv t v t e e t− −= − = − − = − + ≥ 
so 
( )o 62.52 V for 012 10 V for 0t
t
v t
e t−
− ≤⎧= ⎨− + ≥⎩ 
 
(checked using LNAP 10/11/04) 
 
P14.7-16 
For t < 0, the voltage source voltage is 5 V and the 
circuit is at steady state. At steady state, the capacitor 
acts like an open circuit. Using voltage division twice 
 
( ) 32 300 5 5 0
32 96 120 30
v − = − =+ + .25 V
 V
 
and 
( ) ( )0 0 0.25v v+ = −= 
 
 
 
For t > 0, the voltage source voltage is 20 V. 
Represent the circuit in the frequency domain using 
the Laplace transform as shown. 
 
We could write mesh or node equations, but finding a 
Thevenin equivalent of the part of the circuit to the left 
of terminals a-b seems promising. 
 
 
 
 
 
 
 
Using voltage division twice 
 
 
( )oc 32 20 30 20 5 4 1 V32 96 120 30V s s s s
−⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ s= 
 
 
 
 
 
 
 
 
 
( ) ( )t 96 || 32 120 || 30 24 24 48 Z = + = + = Ω 
 
After replacing the part of the circuit to the left of 
terminals a-b by its Thevenin equivalent circuit as shown 
 
( )
1 0.25
0.75
80 48 8048
s sI s
s
s
−
= = ++
 
 
( ) ( )80 0.25 80 0.75 0.25
48 80
V s I s
s s s s
⎛ ⎞= + = +⎜ ⎟ +⎝ ⎠ s 
 
 
( ) ( ) ( )
60 0.25 1.25 0.25 0.75 0.75 0.25 1 0.75
48 80 1.67 1.67 1.67
V s
s s s s s s s s s s s
− −= + = + = + + = ++ + + + 
 
Taking the inverse Laplace transform gives 
 
( ) 1.671 0.75 V for 0tv t e t−= − ≥ 
Then 
( ) 1.670.25 V for 01 0.75 V for 0t
t
v t
e t−
≤⎧= ⎨ − ≥⎩ 
 
(checked using LNAP 7/1/04) 
 
 
 
 
P14.7-17 
 
 
Mesh Equations: 
( ) ( ) ( )( ) ( ) ( )4 1 4 16 0 62 2C C C 6I s I s I s I s Is s s s⎛ ⎞− − − + = ⇒ − = + +⎜ ⎟⎝ ⎠ s 
( ) ( )( ) ( ) ( ) ( ) ( )106 3 4 0 9C C CI s I s I s I s I s I s− + + = ⇒ = − 
Solving for I C(s): 
 
 
( ) ( )4 2 1 6 33 2
4
C CI s I ss s s
⎛ ⎞− = − + ⇒ =⎜ ⎟⎝ ⎠ −
So Vo(s) is ( ) ( )o 244 3
4
CV s I s
s
= =
−
 
Back in the time domain: 
( ) 0.75o 24 ( )tv t e u t= V for t ≥ 0 
 
 
P14.7-18 
 
 
 
KVL: 
( )8 204 8 4 Ls I ss s
⎛ ⎞+ = + +⎜ ⎟⎝ ⎠ 
so 
( ) ( )( )22
1 12
2 5 1 4
L
ssI s
s s s
+ ++= =+ + + + 
Taking the inverse Laplace transform: 
 
( ) ( )L 1cos 2 sin 2 A2t ti t e t e t u t− −
⎛ ⎞= +⎜ ⎟⎝ ⎠ 
 
 
 
 
Section 14-8: Transfer Function and Impedance 
 
P14.8-1 
1
2 11
1 2
1 2 1 1 2 2
1
1
 ( ) where and1 1 1
R
Z RC sH s Z Z 2R
Z Z RC sR
C s
= = =
R C s
=+ + ++
 
( ) ( )( ) ( )
( ) ( )( ) ( )
2 1
1 1 1 2 2 2
1 2 1 2
2 2
2 2
1 2 1 2
1
Let and then 
1 1
1
When constant, as required.
1
R s
RC R C H s
R s s R
R s RH s
R R s R R
ττ τ τ τ
ττ τ τ τ
+= = = + + +
+= = ⇒ = = =+ + +
 
1 1 2 2we require RC R C∴ = 
 
 
P14.8-2 
1 2
1Let and then the input impedance isZ R Z R Ls
Cs
= + = + 
( )
( ) 2
1 2
2
1 2
1 1
 1 2 1
LR R Ls LCs RC s
Z Z Cs RZ s R
Z Z LCs RCsR R Ls
Cs
⎛ ⎞⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟= = =+ +⎜ ⎟+ + + ⎜ ⎟⎝ ⎠
+
+ 
2Now require : 2 then LRC RC L R C Z
R
+ = ⇒ = = R 
 
 
P14.8-3 
The transfer function is 
( )
2
2 1
2 1 2
2
1
2 1
1
1
1
R
R C s R C
H s R R R
R s
R C s R R C
+
= = +
+ +
+
 
Using 1 22 , 8 and 5 FR R C= Ω = Ω =
gives 
( ) 0.1
0.125
H s
s
=
+
 
 
 
The impulse response is ( ) ( ) ( )0.1250.1 Vth t H s e u t-1L −⎡ ⎤= =⎣ ⎦ . 
 
The step response is 
1 
 
( )
( ) ( ) ( )
0.1250.1 0.8 0.8 0.8 1 V
0.125 0.125
tH s e u t
s s s s s
-1 -1 -1L = L = L −
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥+ +⎣ ⎦⎢ ⎥⎣ ⎦ ⎣ ⎦
 
 
(Checked using LNAP, 12/29/02) 
P14.8-4 
The transfer function is: 
 
( ) ( ) ( )
4
2 2
12 1212
8 164
tH s t e u t
s ss
−⎡ ⎤= = =⎣ ⎦ + ++L 
The Laplace transform of the step response is: 
( )
( ) ( )2 2
3
12 34
44 4
H s k
s ss s s
−= = + +
++ + s
 
The constant k is evaluated by multiplying both sides of the last equation by . ( )24s s+
( ) ( ) ( )2 23 312 4 3 4 3 4 12
4 4
s s ks s k s k s k
⎛ ⎞⎟⎜= + − + + = + + + + ⇒ =−⎟⎜ ⎟⎟⎜⎝ ⎠
3
4
 
The step response is 
( ) ( )1 43 33 V
4 4
tH s e t u t
s
L − −
⎡ ⎤ ⎛ ⎞⎛ ⎞⎟⎜ ⎟⎢ ⎥ ⎜= − + ⎟⎟⎜ ⎜ ⎟⎟⎟⎜⎢ ⎥ ⎜ ⎟⎜ ⎝ ⎠⎝ ⎠⎣ ⎦
 
 
P14.8-5 
The transfer function can also be calculated form the circuit itself. The circuit can be represented 
in the frequency domain as 
 
 
We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R 
and the op amp comprise a non-inverting amplifier. Thus 
 
( ) ( )a c1 10000
RV s V s
⎛ ⎞⎟⎜= + ⎟⎜ ⎟⎟⎜⎝ ⎠ 
Now, writing node equations, 
2 
 
( ) ( ) ( ) ( ) ( ) ( )c i o a oc 0 and 01000 5000
V s V s V s V s V s
CsV s
Ls
− −
+ = + = 
 
Solving these node equations gives 
( )
1 50001
1000 10000
1 5000
1000
R
C LH s
s s
C L
⎛ ⎞⎟⎜ + ⎟⎜ ⎟⎜⎝ ⎠= ⎛ ⎞⎛⎟ ⎟⎜ ⎜+ +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝
⎞
⎠
 
 
Comparing these two equations for the transfer function gives 
 
( ) ( )1 12000 or 5000
1000 1000
s s s s
C C
⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠ 
 
( ) ( )5000 50002000 or 5000s s s s
L L
⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠ 
 
1
1000
1
10000
5000
15 106
C
R
L
+⎛⎝⎜
⎞
⎠⎟ = × 
 
The solution isn’t unique, but there are only two possibilities. One of these possibilities is 
 
( )1 2000 0.5 F
1000
s s C
C
μ⎛ ⎞⎟⎜ + = + ⇒ =⎟⎜ ⎟⎟⎜⎝ ⎠ 
 
( )s
L
s L+⎛⎝⎜
⎞
⎠⎟ = + ⇒ =
5000
5000 1 H 
 
( )
6
6
1 50001 15 10
10000 11000 0.5 10
R R
⎛ ⎞⎟⎜ + = × ⇒ = Ω⎟⎜ ⎟⎟⎜⎝ ⎠×
5 k 
 (Checked using LNAP, 12/29/02) 
 
3 
P14.8-6 
The transfer function of the circuit is 
( )
2
2 1
1
2
1
1
1
R
R C s R C
H s
R s
R C
+= − = −
+
 
The give step response is ( ) ( ) ( )2504 1 Vtov t e u t−= − − . The correspond transfer function is 
calculated as 
 
( ) ( ) ( ){ } ( ) ( )250 4 4 1000 10004 1 250 250 250tH s e u t H ss s s s s− − −⎛ ⎞= − − = − − = ⇒ =⎜ ⎟+ +⎝ ⎠L s + 
 
Comparing these results gives 
 
( )2 62
1 1 1250 40 k
250 250 0.1 10
R
R C C −
= ⇒ = = =× Ω 
( )1 61
1 1 11000 10 k
1000 1000 0.1 10
R
R C C −
= ⇒ = = =× Ω 
 
(Checked using LNAP, 12/29/02) 
 
P14.8-7 
 
 
( ) ( ) ( )a i4 24 2 2V s V s V ss s
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠ i 
( ) ( ) ( ) ( ) ( )o b b a
12
2 2 2 25 512 2 2 2 26
sV s V s V s V s V s
s s s s
s
⎛ ⎞⎜ ⎟ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎝ ⎠
i
 
 
The transfer function is: 
( ) ( )( ) ( )
o
2
i
20
2
V s
H s
V s s
= = + 
4 
The Laplace transform of the step response is: 
( ) ( ) ( )o 2 2
20 5 5 10
22 2
V s
s ss s s
− −= = + +++ + 
Taking the inverse Laplace transform: 
( ) ( ) ( )2o 5 5 1 2 Vtv t e t u t−⎡ ⎤= − +⎣ ⎦ 
(checked using LNAP 8/15/02) 
 
 
P 14.9-8 
 
 
From the circuit: 
( ) ( ) ( )
11 4
4 6
1 44 6
6
CCs LH s k k
Ls s s
Cs L C
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
1
 
From the given step response: 
( ) ( ) ( ) ( )( )3 2 2 4 6 122 4 6 3 2 3t tH s e e u ts s s s− −⎡ ⎤= + − = + − =⎣ ⎦ 2s s s+ + + +L 
so 
( ) ( )( )
12
3 2
H s
s s s
= + + 
 
Comparing the two representations of the transfer functions let 1 13 F
6 1
C
C
= ⇒ =
8
, 
4 2 2L
L
= ⇒ = H and . 2 3 12 2 V/Vk k× × = ⇒ =
 
(Checked using LNAP, 12/29/02) 
 
 
P 14.9.9 
From the circuit: 
5 
( ) ( )( )
o
i
1212
RsV s R L s LH s RV s R L s s
L
++= = = ++ + +
 
From the given step response: 
 
( ) ( ) ( ) ( ) ( )4 0.5 0.5 2 20.5 1 4 4tH s s se u t H ss s s s s− 4s+ +⎡ ⎤= + = + = ⇒ =⎣ ⎦ + + +L 
 
Comparing these two forms of the transfer function gives: 
 
2 12 2 4 6 H, 12 
12 4
R
LL L R
R L
L
⎫= ⎪ +⎪ ⇒ = ⇒ = =⎬+ ⎪= ⎪⎭
Ω 
(Checked using LNAP, 12/29/02) 
 
 
P14.8-10 
 
Mesh equations: 
 
( ) ( ) ( )
( ) ( )
1 1
2 1
1 1 1 
1 10 
V s R I s I s
Cs Cs Cs
R R I s I s
Cs Cs
⎛ ⎞= + + −⎜ ⎟⎝ ⎠
⎛ ⎞= + + −⎜ ⎟⎝ ⎠
2
 
Solving forI2(s): ( )
( )
2
1 2
1
2 12
( )
V s
CsI s
R R
Cs Cs Cs
⎛ ⎞⎜ ⎟⎝ ⎠= ⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1
 
Then gives ( ) ( )o 2V s R I s=
 
( ) ( )( ) [ ] [ ]
( )
0
1
2 1
1 22 2
1 1
2 2 1 1 4 12
2 2
V s RCs sH s
V s RCs RCs RC RCRC s s
RRC RRC
= = =+ + − ⎡ ⎤+⎢ ⎥+ +⎢ ⎥⎣ ⎦
 
 
 
6 
P14.8-11 
Let 
2
1
1
1 1
x x
R
RCsZ
RCsR
Cs
Z R L s
⎛ ⎞⎜ ⎟⎝ ⎠= = ++
= +
 
Then 
( )
( )
2 2
2
1 1 2 x x x x
x x
2 x
x x21 x
x x
1 
1
1
 
R
V Z RRCs
RV Z Z L RCs L R RC s R RR L s
RCs
V L C
L R RCV R Rs s
L RC L RC
+= = =+ + ++ + +
= + ++ +
+ +
 
 
P14.8-12 
 
Node equations: 
( ) ( )1 out1 in 1 11 1
1
out1
1 2 2 out
2
2
 0 1
0 1
V VV V sC R C s V R C sV V
R
VV V R C sV
R
sC
−− + = ⇒ + = +
− − = ⇒ = −
1 1 in out
 
Solving gives: 
( ) 1 1 2 2out 2
2in 1 2 1 2 2 2
1 1 1 2 1 2
1
 1 1 1 
s
R C s R CVH s
V R R C C s R C s s s
R C R R C C
−−= = =+ + + +
 
 
7 
 
P14.8-13 
 
Node equations in the frequency domain: 
 
1 11
1 2 3
0iV V V VV
R R R
0− −+ + = 
 
0
1
1 2 3 3
1 1 1 iV VV
1R R R R R
⎛ ⎞⇒ + + − =⎜ ⎟⎝ ⎠
 
 
1
2 0 1 2 2 0
2
0V sC V V sC R V
R
− − = ⇒ = − 
 
After a little algebra: 
( ) 0 3
2 2 3 2 1 3 2 1 2 1i
V RH s
V sC R R sC R R sC R R R
−= = + + + 
 
 
P14.8-14 
 
( )
( ) 2
1 1
( ) 1 1
o
i
V s Cs LCH s RV s Ls R s s
Cs L LC
= = =
+ + + +
 
 
 
L, H C, F R, Ω H(s) 
 
2 
 
0.025 
 
18 ( )( )2
20 20
9 20 4 5s s s s
=+ + + + 
 
2 
 
0.025 
 
8 ( )22 2
20 20
4 20 2 4s s s
=+ + + + 
 
1 
 
0.391 
 
4 ( )( )2
2.56 2.56
4 2.56 0.8 3.2s s s s
=+ + + +
 
2 
 
0.125 
 
8 ( )22
4 4
4 4 2s s s
=+ + + 
 
8 
a) ( ) ( )( )
20
4 5
H s
s s
= + + { } ( ) ( )
{ }
( )
4 5
5 4
20 20 ( ) ( ) = 20 20 ( )
4 5
( ) 20 1 5 4 step response 
( 4) ( 5) 4 5
step response = 1 4 5 ( )
t t
t t
h t H s h t e e u t
s s
H s
s s s s s s s
e e u t
− −
− −
= − ⇒ = −+ +
−= = = + ++ + + +
+ −
L
L ⇒ 
 
b) ( ) ( )2 2
20
2 4
H s
s
= + + ( ){ } ( ) 22 2
5(4) (s) 5 sin 4 ( )
( 2) 4
th t H h t e t u t
s
−= = ⇒ =+ +L 
{ }
( ) ( ) ( )
{ } ( )( )
( )
( )
1 2
2 2
2 2
1 2 1 2
1 2
2 22
2
( ) 20 1 step response
( 4 20) 4 20
20 4 20 1 4 20
1, 4
1 421 2 step response
2 42 4
1step response 1 cos 4 sin 4 ( )
2
t
H s K s K
s s s s s s s
s s s K s K s K s K
K K
s
s ss
e t t u t−
+= = = ++ + +
= + + + + = + + + +
⇒ = − = −
−− += + + + ++ +
⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
L
L
+
)
 
 
c) 
( ) ( )(
2.56
0.8 3.2
H s
s s
= + + 
( ){ } ( ) ( )
{ }
.8t 3.2t
3.2 .8
1.07 1.07( ) 1.07 e e u(t)
.8 3.2
4 1
( ) 2.56 1 3 3step response
( .8) ( 3.2) .8 3.2
1 4step response 1 ( )
3 3
t t
h t H s h t
s s
H s
s s s s s s s
e e u t
− −
− −
= = − ⇒ = −+ +
−
= = = + ++ + + +
⎛ ⎞= + −⎜ ⎟⎝ ⎠
L
L
 
d) ( ) ( )2
4
2
H s
s
= + 
( )
( )( )
2
2
4 ( )
step response 1 1 2 ( )
t
t
h t te u t
t e u t
−
−
=
= − + 
 
 
 
 
 
9 
P14.8-15 
For an impulse response, take ( )1 1 V s = . Then 
 
( ) ( )( ) ( )
*
0
3 2
3 2 3 2 3 2 3 2
s A B BV s
s s j s j s s j s j
+= = ++ − + + + − + ++ 
Where 
 
( ) ( ) *0 0 3 20 0.462, ( 3 2) 0.47 119.7 and 0.47 119.7s jsA sV s B s j V s B=− +== = = + − = ∠− ° = ∠ ° 
 
Then 
( )0 0.462 0.47 119.7 0.47 119.73 2 3 2V s s s j s j
∠− ° ∠= + ++ − + +
° 
The impulse response is 
 
( ) ( )30 ( ) 0.462 2(0.47) cos 2 119.7 Vt ov t e t u t−⎡ ⎤= + −⎣ ⎦ 
 
 
P14.8-16 
a. 
 
A capacitor in a circuit that is at steady state 
and has only constant inputs acts like an open 
circuit. Then 
 
( ) ( )o 10 1.5 3.75 V4v t = − = − 
 
 
b. Here’s the circuit represented in the frequency 
domain, using phasors and impedances. Writing a 
node equation at the inverting input node of the op 
amp gives 
 
( ) ( )o o
3 3
4 30 0
4 10 10 10 10 10j
ω ω∠ ° + +× − × ×
V V
3 = 
or 
( ) ( )o10 30 1 0j ω∠ °+ + =V 
( )o 10 30 7.07 1651 jω
∠ °= − = ∠ °+V 
Finally, 
10 
vo(t) = 7.07 cos(100t +165°) V. 
 
c. Here’s the circuit represented in the frequency 
domain, using The Laplace transform (assuming 
zero initial conditions). Writing a node equation at 
the inverting input node of the op amp gives 
( ) ( )o o
3 3
6
1
014 10 10 1010
V s V ss
s
+ +× ××
= 
( ) ( )3 o10 100 04 s V ss + + = 
( ) ( )o
250 2.5 2.5
100 100
V s
s s s s
−= = ++ + 
Finally, 
( ) ( ) ( )1002.5 1 Vtov t e u t−= − 
 
 
P14.8-17 
Represent the circuit in the frequency domain using 
the Laplace transform as shown. (Set the initial 
conditions to zero to calculate the step response.) 
 
First, 
( ) ( )( )
2
2
2 2
2
2
1
1 || 1 1
R L s R L sC sR L s
C s C L s C R sR L s
C s
× + ++ = = + ++ +
 
 
Next, using voltage division, 
 
11 
( ) ( )( ) ( )
2
2
2 2o
2
2i 2 1 2
12
2
2
1 1
2
1 2 1 22
1 1
1
1
1
2 4
4 29
R L s
C L s C R s R L sV s
H s R L sV s R L s R C L s C R sR
C L s C R s
Rs
R C R LC s
L R R C R R s ss s
R LC R LC
+
+ + += = =+ + + + +++ +
+ += =+ + + ++ +
 
Using ( )i 1V s s= gives 
( ) ( ) ( )
( )
( ) ( )
o 22
2 2
2 22 2
2 4 0.1379 0.1379 1.4483
4 294 29
0.1379 0.1379 1.4483
2 5
0.1379 2 50.1379 0.3449
2 5 2 5
H s s sV s
s s s ss s s
s
s s
s
s s s
+ − += = = + + ++ +
− += + + +
+= − ++ + + +
 
 
Taking the inverse Laplace transform 
 
( ) ( ) ( )( )
( )
2
o
2
0.1379 0.1379cos 5 0.3448sin 5
0.1379 0.3713 cos 5 111.8 V
t
t
v t e t t
e t
−
−
= + − +
= + − ° 
 
(checked using LNAP 10/15/04) 
 
P14.8-18 
First, we determine the transfer function corresponding to the step response. Taking the Laplace 
transform of the given step response 
 
( ) ( ) ( )( ) ( ) (( )( )
)
( )( )
o
50 20 0.667 20 1.667 501 0.667 1.667
50 20 50 20
1000
50 20
H s s s s s s s
V s
s s s s s s s
s s s
+ + + + − += = + − =+ + + +
= + +
 
Consequently, 
( ) ( )( ) ( )( )oi
1000
50 20
V s
H s
V s s s
= = + + 
 
12 
Next, we determine the transfer function of the 
circuit. Represent the circuit in the frequency 
domain using the Laplace transform as shown. (Set 
the initial conditions to zero to calculate the transfer 
function.) 
 
Apply KVL to the left mesh to get 
 
( ) ( ) ( ) ( ) ( )ii 1 a a a
1
V s
V s L s I s K I s I s
K L s
= + ⇒ = + 
 
Next, using voltage division, 
 
( ) ( ) ( ) ( )( ) ( )o a o2 2 1
R RV s K I s V s V s
L s R L s R K L s
= ⇒ =+ + + i
K
 
 
Then, the transfer function of the circuit is 
( ) ( )( ) ( )( ) 1 2oi 2 1
2 1
RK
L LV s R KH s
V s L s R L s K R Ks s
L L
= = = ⎛ ⎞⎛+ + + +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝
⎞⎟⎟⎠
 
Comparing the two transfer functions gives 
( )( ) ( )
1 2
2 1
1000
50 20
RK
L L
H s
s s R Ks s
L L
= =+ + ⎛ ⎞⎛+ +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝
⎞⎟⎟⎠
 
 
We require 
1 2
1000 RK
L L
=
 
and either 
2
50 R
L
= and 
1
20 K
L
= or 
2
20 R
L
= and 
1
50 K
L
= . These 
equations do not have a unique solution. One solution is 
 
L1 = 0.1 H, L2 = 0.1 H, R = 5 Ω and K = 2 V/A 
 
(checked using LNAP 10/15/04) 
 
 
13 
P14.8-19 
Represent the circuit in the frequency domain using the 
Laplace transform as shown. (Set the initial conditions 
to zero to calculate the step response.) 
 
First, 
( )22 2
22
2
2
1
11||
11
R L s R C L sC s
R L s
C s C L s C R s
R L s
C s
⎛ ⎞× +⎜ ⎟ +⎛ ⎞ ⎝ ⎠+ = =⎜ ⎟ + +⎛ ⎞⎝ ⎠ + +⎜ ⎟⎝ ⎠
 
Next, using voltage division twice, 
 
( ) ( )( )
( )
( ) ( )
( )
( )
2
2
2
2 2o
2 2
i 2 1 2 1 2 1 2
12
2
2
1 2
2
1 22
1 2
1 1
1
11
1
8
1 10
R C L s
C L s C R s RV s C sH s
V s R C L s R R C L s R R C s R RL sR C sC L s C R s
R
R R LC
R R s ss s
LCR R L
+
+ += = × =+ + + + ++++ +
+= = + ++ ++
16
 
Using ( )i 1V s s= gives 
( ) ( ) ( ) ( )( )o 2
2 11
8 8 3 62
2 8 210 16
H s
V s
s s s s s ss s s
−
= = = = + +
8s+ + ++ + + 
 
Taking the inverse Laplace transform 
 
( ) ( )2 8o 1 2 1 V2 3 6t tv t e e u t− −
⎛ ⎞= − +⎜ ⎟⎝ ⎠ 
 
(checked using LNAP 10/15/04) 
 
14 
P14.8-20 
First, we determine the transfer function corresponding to the step response. Taking the Laplace 
transform of the given step response 
 
( ) ( ) ( )
( ) ( )
( ) ( )
2
o 2 2
3.2 5 3.2 5 163.2 3.2 16 80
5 5 5
H s s s s s
I s
s s s s s s
⎛ ⎞ + − + += = − + = =⎜ ⎟⎜ ⎟+ + +⎝ ⎠ 25s s +
 
Consequently, 
( ) ( )( ) ( )
o
2
i
80
5
I s
H s
V s s
= = + 
 
Next, we determine the transfer function of the 
circuit. Represent the circuit in the frequency 
domain using the Laplace transform as shown. 
(Set the initial conditions to zero to calculate the 
transfer function.) 
 
First 
1
1
1
1
1
1
1|| 1 1
R RC sR
C s R C sR
C s
×
= = ++
 
 
 
Next, using voltage division, 
( ) ( ) ( )
1
1 1
a i
1 1 2 1 2
2
1
1
1
R
R C s R
V s V s V sR R R R R C sR
R C s
+= = + +++
i 
 
( ) ( ) ( ) ( )( ) ( ) ( )1 2ao o i3 3 1 2 1 2 3 1 2
1 2
K
K R R C LKV s
iI s I s V sL s R L s R R R R R C s R R R
s s
L R R C
= ⇒ = =+ ⎛ ⎞+ + + +⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
V s
 
 
Then, the transfer function of the circuit is 
15 
( ) ( )( )
2o
i 3 1
1 2
K
R C LI s
H s
V s 2R R Rs s
L R R C
= = ⎛ ⎞+⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
Comparing the two transfer functions gives 
( ) ( )
2
2
3 1
1 2
80
5
K
R C L
H s
2R R Rs s s
L R R
= = ⎛ ⎞++ ⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠C
 
 
We require 
( )
1 2
1 2
40 105 2
40 10
R R
C
R R C C
+ += = ⇒ =× 5 mF , 
 
 3 205 4 H
R
L
L L
= = ⇒ = 
and 
( )280 80 V/V10 0.025 4
K K K
R C L
= = ⇒ = . 
 
(checked using LNAP 10/15/04) 
 
 
P14.8-21 
First, 
 
( ) ( ) ( ) ( ) ( ) ( ) ( )6.5cos 2 22.6 6.5 cos 22.6 cos 2 6.5 sin 22.6 sin 2 6cos 2 2.5sin 2t t t+ ° = ° − ° = −t t 
 
Consequently, the impulse response can be written as 
 
( ) ( ) ( )( ) ( )2o 6cos 2 2.5sin 2 Vtv t e t t u t−= − 
The transfer function is 
 
( ) ( ) ( ) ( )2 2 2 22 2 2
3 2 6 13 66 2.5
6 133 2 3 2 3 2
s sH s
s ss s s
+ += − = = 13s ++ ++ + + + + + 
 
The Laplace transform of the step response is 
 
( )
( ) ( ) ( ) ( )2 222 2 2
6 13 1 1 1 3 3 2
6 13 26 13 3 2 3 2 3 2
H s s s s s
s s s s s ss s s s s s
+ += = − = − = − + ×+ ++ + 2 2+ + + + + + 
16 
Taking the inverse Laplace transform gives the step response: 
 
( ) ( ) ( )( )( ) ( ) ( )( )2 2o 1 1.5sin 2 cos 2 1 1.803 cos 2 123.7 Vt tv t e t t u t e t− −= + − = + − ° 
 
 
 
P14.8-22 
Taking the Laplace transform of the step response, 
 
( )
( ) ( ) ( )2 2
1 3 1 1 6 9
33 3
H s s
s s s ss s
⎡ ⎤ += − + = − =⎢ ⎥++ +⎢ ⎥⎣ ⎦ 23s s +
 
 
The transfer function is 
( ) ( )2
9
3
H s
s
= + 
 
Taking the inverse Laplace transform gives the impulse response: 
 
( ) ( )3o 9 Vtv t t e u t−= 
 
(checked using LNAP 10/15/04) 
 
P14.8-23 
Represent the circuit in the frequency domain 
using the Laplace transform as shown. (Set the 
initial conditions to zero to calculate the transfer 
function.) First, 
( ) ( )a
1
iV sI s
L s R
= + 
The equivalent impedance of the parallel capacitor 
and inductor is 
2
2
2
2
2
1
1|| 1 1
R RC sR
C s R C sR
C s
×
= = ++
 
 
Next, using voltage division, 
( ) ( ) ( ) ( )( )( ) ( )3 2 33 3 2 3o a a2 2 3 2 3 1 2 3 2 3
3
21
K R R R C sR R R R C s
V s K I s K I s V sR R R R R C s L s R R R R R C sR
R C s
++= = =+ + + + +++
i
 
17 
 
Then, the transfer function of the circuit is 
( ) ( )( )
( )
( )( )
2o
i 1 2 3
2 3
1
5 0.5
5 2.
K s
L R CV s s
H s
V s s sR R R
s s
L R R C
⎛ ⎞+⎜ ⎟⎜ ⎟ +⎝ ⎠= = = + +⎛ ⎞+⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5 
Using ( )i 1V s s= gives 
( ) ( ) ( )( )( )o
5 0.5 0.2 1.8 1.6
5 2.5 5 2.5
H s s
V s
s s s s s s s
+ −= = = + ++ + + + 
 
Taking the inverse Laplace transform 
 
( ) ( ) ( )5 2.5o 0.2 1.8 1.6 Vt tv t e e u t− −= − + 
 
(checked using LNAP 10/15/04) 
 
 
 
 
 
 
 
18 
Section 14-9: Convolution Theorem 
 
 
P14.9-1 
( ) ( ) ( ) ( ) ( ) ( ) 1 11 1 s se ef t u t u t F s u t u t
s s s
− −−= − − ⇒ = − − = − =⎡ ⎤⎣ ⎦L 
( ) ( ) ( )
( ) ( ) ( ) ( ) (
2 2
1 2 1 1
2
1 1 2*
2 1 1 2 2
s s se e ef t f t F s
s s
t u t t u t t u t
− − −
− − −⎡ ⎤⎛ ⎞ ⎡ ⎤− − +⎡ ⎤= = =⎢ ⎥⎜ ⎟ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦
)= − − − + − −
L L L
 
 
P14.9-2 
( ) ( ) ( ) ( ) 22 22 2 sef t u t u t F s
s s
−
= − − ⇒ = −⎡ ⎤⎣ ⎦ 
( ) ( ) ( ) ( ) ( ) ( ) (2 41 1 2 2 24 8 4 4 8 2 2 4 4
s se e )4f f F s F s t u t t u t t u t
s s s
− −
− − ⎡ ⎤∗ = = − + = − − − + − −⎡ ⎤ ⎢ ⎥⎣ ⎦ ⎣ ⎦
L L 
 
 
P14.9-3 
( ) ( ) ( )
( ) ( )( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )
1 1 2
2
1
1
2 1 1
2 1 2 2
/
2
1
1 1
1 1
1 1
1 1
1 1
1 1 , 0t RC
v t t u t V s
s
V s Cs RCH s
V s R s
Cs RC
v t h t v t V s H s
1
RC RC RV s V s H s
s s ss s
C
RC R
v t t e t
RC
−
−
= ⇒ =
= = =
+ +
⎡ ⎤= ∗ = ⎣ ⎦
⎛ ⎞ − −⎜ ⎟⎛ ⎞= = = + +⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟+ +⎝ ⎠
= − − ≥
L
C
 
 
 
 
1 
P14.9-4 
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
1
2
22
2 2
( )
2 2
1 1 where and
1 1So
Solving the partial fractions yields: 1 , 1 , 1
1So , 0
at
h t f t H s F s H s F s
s s
A B CH s F s
s a s s s as
A a B a C a
t eh t f t t
a a a
−
−
∗ = = =⎡ ⎤⎣ ⎦ +
⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
=− = =
−∗ = + + ≥
L
a
 
 
 
 
 
2 
Section 14-10: Stability 
 
P14.10-1 
 a. From the given step response: 
 
( ) ( ) ( ) ( )1003 714 100tH s e u ts s−⎡ ⎤= − =⎢ ⎥ +⎣ ⎦L 5s 
From the circuit: 
( ) ( ) 55
R
H sR LH s
RR Ls s s s
L
= ⇒ = ++ + ⎛ ⎞+⎜ ⎟⎝ ⎠
 
Comparing gives 
75 15 
5 0.2 H100
R
RL
R L
L
⎫= ⎪ = Ω⎪ ⇒⎬+ =⎪= ⎪⎭
 
b. The impulse response is 
( ) ( )1 10075 75
100
th t e u t
s
− −⎡ ⎤= =⎢ ⎥+⎣ ⎦L 
c. 
( )
100
75 3 45
100 100 4 2jω
ω = = = ∠−+H ° 
( ) ( )o 3 1545 5 0 45 V4 2 4 2ω
⎛ ⎞= ∠ ° ∠ ° = ∠− °⎜ ⎟⎝ ⎠V 
( ) ( )2.652 cos 100 45 Vov t t= − ° 
 
(Checked using LNAP, 12/29/02) 
 
P14.10-2 
The transfer function of this circuit is given by 
 
( ) ( )( ) ( ) ( ) ( ) ( ) ( )2 2 2
5 5 10 20 205 5 1 2
2 2 2
tH s e t u t H s
s s s s s s
− − −⎡ ⎤= − + = + + = ⇒ =⎣ ⎦ + + + +L 22 
 
This transfer function is stable so we can determine the network function as 
 
( ) ( ) ( ) ( )2 2
20 20
2 2s j
s j
H s
s jω ω
ω ω= =
= = =+ +H
 
The phasor of the output is 
14-1 
 
( ) ( ) ( ) ( ) ( )o 2 2
20 205 45 5 45 12.5 45 V
2 2 2 2 45j
ω = ∠ ° = ∠ ° = ∠−+ ∠ °
V ° 
 
The steady-state response is 
( ) ( )o 12.5cos 2 45 Vv t t= − ° 
 
(Checked using LNAP, 12/29/02) 
 
 
P 14.11-3 
 The transfer function of the circuit is ( ) ( )
1 5
2
3030 ( )
5
tH s t e u t
s
− −⎡ ⎤= =⎣ ⎦ +L . The circuitis stable 
so we can determine the network function as 
 
( ) ( ) ( ) ( )2 2
30 30
5 5s j
s j
H s
s jω ω
ω ω= =
= = =+ +H
 
The phasor of the output is 
 
( ) ( ) ( ) ( ) ( )o 2 2
30 3010 0 10 0 8.82 62 V
5 3 5.83 31j
ω = ∠ ° = ∠ ° = ∠−+ ∠ °V ° 
 
The steady-state response is 
( ) ( )o 8.82cos 3 62 Vv t t= − ° 
 
 
P14.10-4 
 
( ) ( ) ( ) ( )( )8 320 40 1.03 41 10240040 1.03 41 8 320 8 320t tH s e e u ts s s s− −⎡ ⎤= + − = + − =⎣ ⎦ + + + +L s s s
)
 
so 
( ) ( ) (
102400
8 32
H s
s s
= + + 0 
 
The poles of the transfer function are 1 8 rad/ss = − and 2 320 rad/ss = − , so circuit is stable. 
Consequently, 
( ) ( ) ( ) ( )
102400 40
8 320 1 1
8 320
s j
H s
j j j j
ωω ω ωω ω== = =+ + ⎛ ⎞ ⎛+ +⎜ ⎟ ⎜⎝ ⎠ ⎝
H ⎞⎟⎠
 
14-2 
 
The network function has poles at 8 and 320 rad/s and has a low frequency gain equal to 32 dB = 
40. Consequently, the asymptotic magnitude Bode plot is 
 
 
 
 
P14.10-5 
 
( ) ( ) ( ) ( )( )2 6 60 60 24060 2 6 2t tH s e e u ts s s− −⎡ ⎤= − = − =⎣ ⎦ 6s s+ + + +L 
so 
( ) ( )( )
240
6 2
sH s
s s
= + + 
 
The poles of the transfer function are 1 2 rad/ss = − and 2 6 rad/ss = − , so circuit is stable. 
Consequently, 
( ) ( ) ( ) ( )
240 20
2 6 1 1
2 6
s j
j jH s
j j j j
ω
ω ωω ω ωω ω== = =+ + ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
H 
 
The network function has poles at 2 and 6 rad/s. The asymptotic magnitude Bode plot has a gain 
equal to 40 = 32 dB between 2 and 6 rad/s. Consequently, the asymptotic magnitude Bode plot is 
 
 
14-3 
 
P14.10-6 
 
( ) ( ) ( ) ( ) ( )( )90 36 104 32 36 3604 32 90 90 90tH s sse u ts s s s s− ++⎡ ⎤= + = + = =⎣ ⎦ + + +L s s 
so 
( ) ( )( )
10
36
90
s
H s
s
+= + 
 
The pole of the transfer function , so circuit is stable. Consequently, 1 90 rad/ss = −
 
( ) ( ) ( )( )
110 1036 4
90 1
90
s j
jj
H s
j j
ω
ω
ωω ωω=
⎛ ⎞+⎜ ⎟+ ⎝ ⎠= = =+ ⎛ ⎞+⎜ ⎟⎝ ⎠
H 
 
The network function has a zero at 10 rad/s and a pole at 90 rad/s. The low frequency gain is 
equal to 4 = 12 dB. Consequently, the asymptotic magnitude Bode plot is 
 
 
 
 
 
 
P14.10-7 
 
14-4 
( ) ( ) ( ) ( ) ( )5 205 1.67 1.67 253 5 20t tH s e e u ts s s− −⎡ ⎤= − = − =⎢ ⎥ + + + +⎣ ⎦L 5 20s s
)
 
so 
( ) ( )(
25
5 2
sH s
s s
= + + 0 
 
The poles of the transfer function are 1 5 rad/ss = − and 2 20 rad/ss = − , so the circuit is stable. 
Consequently the network function of the circuit is, 
 
( ) ( ) ( ) ( )
25 0.25
5 20 1 1
5 2
s j
j jH s
j j j j
ω
0
ω ωω ω ωω ω== = =+ + ⎛ ⎞⎛+ +⎜ ⎟⎜⎝ ⎠⎝
H ⎞⎟⎠
s
 
Using ( ) ( ) ( )o ω ω ω=V H V at ω = 30 rad/s gives 
 
( ) ( ) ( ) ( )( )o
0.25 30 9012 0 8.2 47 V
30 30 1 6 1 1.51 1
5 20
j j
j jj j
ω = ∠ = =+ +⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
V ∠− °
 
 
Back in the time domain, the steady state response is 
 
( ) ( )o 8.2 cos 30 47 Vv t t= − ° 
 
(checked using LNAP 10/12/04) 
 
 
 
P14.10-8 
 
( ) ( ) ( ) ( )
( )
( ) ( )
5
2 2
10 5 5010 50 1010 50
5 5 5 5
t s sH s e t u t
s s s s
− + −⎡ ⎤= − = − = =⎣ ⎦ + + +L 2+ 
 
The poles of the transfer function are 1 5 rad/ss = − and 2 5 rad/ss = − , so the circuit is stable. 
Consequently the network function of the circuit is, 
 
( ) ( ) ( )2 2
10 0.4
5 1
5
s j
j jH s
j j
ω
ω ωω ω ω== = =+ ⎛ ⎞+⎜ ⎟⎝ ⎠
H 
Using ( ) ( ) ( )o sω ω ω=V H V at ω = 10 rad/s gives 
 
14-5 
( ) ( ) ( ) ( )o 2 2
0.4 10 4812 0 9.6 37 V
1 2101
5
j j
jj
ω = ∠ = = ∠−+⎛ ⎞+⎜ ⎟⎝ ⎠
V °
 
 
Back in the time domain, the steady state response is 
 
( ) ( )o 9.6 cos 10 37 Vv t t= − ° 
 
(checked using LNAP 10/12/04) 
 
 
14-6 
Section 14.12 How Can We Check…? 
 
P14.12-1 
 
( ) ( ) 2.1 15.93 6 2t tL Ldv t i t e edt − −= = − − 
 
( ) ( ) 2.1 15.91 0.092 0.575
75
t t
C C
di t v t e e
dt
− −= = − − 
 
( ) ( ) 2.1 15.91 12 12 6 2t tR Lv t v t e e− −= − = + + 
 
( ) ( ) ( )( ) 2.1 15.92 12 1 0.456 0.1236L C t tR
v t v t
i t e e− −
− += = + − 
 
( ) ( ) 2.1 15.93 1 0.548 0.4526C t tR
v t
i t e e− −= = + + 
Thus, 
( ) ( ) ( ) ( ) ( )1 212 0 and +L R R C Rv t v t i t i t i t− + + = = 3 
 
as required. The analysis is correct. 
 
P14.12-2 
 
 
( ) ( )1 218 20 and 3 3
4 4
I s I s
s s
= =
− −
 
1 
KVL for left mesh: 12 1 18 18 206 03 3 32
4 4 4
s s s s s
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠
= (ok) 
KVL for right mesh: 18 20 20 186 3 43 3 3 3
4 4 4 4
s s s s
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0= (ok) 
The analysis is correct. 
 
 
P14.12-3 
Initial value of IL (s): 2
lim 2 1
5
ss
s s s
+ =→∞ + + (ok) 
Final value of IL (s): 2
lim 2 0
0 5
ss
s s s
+ =→ + + (ok) 
Initial value of VC (s): 
( )
( )2
lim 20 2
0
5
s
s
s s s s
− + =→∞ + + (not ok) 
Final value of VC (s): 
( )
( )2
lim 20 2
8
0 5
s
s
s s s s
− + = −→ + + (not ok) 
Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s) 
was calculated as ( )20 LI ss− instead of ( )
20 8
LI ss s
− + . After correcting this error 
( ) 220 2 85C
sV s
s s s s
+⎛ ⎞= − +⎜ ⎟+ +⎝ ⎠ . 
Initial value of VC (s): 
( )
( )2
lim 20 2 8 8
5
s
s
s ss s s
⎛ ⎞− +⎜ ⎟+ =⎜ ⎟→ ∞ + +⎝ ⎠
 (ok) 
Final value of VC (s): 
( )
( )2
lim 20 2 8 0
0 5
s
s
s ss s s
⎛ ⎞− +⎜ ⎟+ =⎜ ⎟→ + +⎝ ⎠
 (ok) 
 
2 
	Ch14sec2
	Problems 
	Section 14-2: Laplace Transform 
	P14.2-1 
	P14.2-2 
	P14.2-3 
	P14.2-4 
	Ch14sec3
	Section 14-3: Impulse Function and Time Shift Property 
	P14.3-1 
	P14.3-2 
	P14.3-3
	P14.3-4 
	P14.3-5 
	P14.3-6
	P14.3-7
	P14.3-8
	Ch14sec4
	Section 14-4: Inverse Laplace Transform 
	P14.4-1 
	P14.4-2 
	P14.4-3 
	P14.4-4 
	P14.4-5 
	P14.4-6 
	P14.4-7 
	P14.4-8
	P14.4-9
	Ch14sec5
	Section 14-5: Initial and Final Value Theorems 
	P14.5-1 
	P14.5-2
	P14.5-3
	P14.5-4
	Ch14sec6
	Section 14-6: Solution of Differential Equations Describing a Circuit 
	P14.6-1
	P14.6-2
	P14.6-3 
	P14.6-4
	P14.6-5
	P14.6-6
	P14.6-7
	P14.6-8
	Ch14sec7
	Section 14-7: Circuit Analysis Using Impedance and Initial Conditions 
	P14.7-1 
	P14.7-2 
	P14.7-3 
	P14.7-4 
	P14.7-5 
	P14.7–6 
	P14.7-7 
	P14.7-8 
	P14.7-9
	P14.7-10
	P14.7-11 
	P14.7-12
	P14.7-13
	P14.7-14
	P14.7-15
	P14.7-16
	P14.7-17 
	P14.7-18 
	Ch14sec8
	Section 14-8: Transfer Function and Impedance 
	P14.8-1 
	P14.8-2 
	P14.8-3
	P14.8-4 
	P14.8-5 
	P14.8-6 
	P14.8-7 
	P14.9-8 
	P14.9.9 
	P14.8-10
	P14.8-11
	P14.8-12 
	P14.8-13
	P14.8-14
	P14.8-15 
	P14.8-16
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	P14.8-18 
	P14.8-19
	P14.8-20 
	P14.8-21 
	P14.8-22 
	P14.8-23
	Ch14sec9
	Section 14-9: Convolution Theorem 
	P14.9-1 
	P14.9-2 
	P14.9-3 
	P14.9-4 
	Ch14sec10
	Section 14-10: Stability 
	P14.10-1 
	P14.10-2 
	P14.11-3 
	P14.10-4 
	P14.10-5 
	P14.10-6 
	P14.10-7 
	P14.10-8 
	Ch14sec12
	Section 14.12 How Can We Check…? 
	P14.12-1 
	P14.12-2 
	P14.12-3