chapter 14: the laplace transform exercisesnayda/courses/dorffifthedition/ch14.pdf · chapter 14:...
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417
Chapter 14: The Laplace Transform Exercises Ex. 14.3-1 Ex. 14.3-2 Ex. 14.4-1 Ex. 14.4-2 Ex. 14.4-3 Ex. 14.4-4 Ex. 14.4-5
2 2 2 2
11cos ( ) and 2
1 11 [cos ] ( )2
j t j t att e e es a
s st F s
s j s j s s
ω ωω
ωω ω ω ω
+ − = + = − ∴ = + = ⇒ = − + + +
2 22
2
2
1 1 [ sin ] [sin ]
2 1
3( )
( 2)( 1)
t te t e ts s
s sF s
s s
− − + = + = + + +
+ +=+ +
4 4 2 3( ) [2 ( ) 3 ( )] 2 [ ( )] 3 [ ( )]
4t t
P s u t e u t u t e u ts s
− −= + = + = ++
2 221
( ) [sin( 2) ( 2)] [sin ] 1
s sF s t u t e t e
s
− − = − − = =
+
2
2
( ) [ ]
1Now, since [ ] , use a frequency shift 1
1( )
( 1)
tF s te
t s ss
F ss
−=
= → +
∴ =+
4.2 4.2
1 22 2
4.2 4.2
1 2 2
5 5 5 2( ) ( )
33
5 6 15 5 ( ) ( ) ( )
3
s s
s s
e eF s F s
s ss s
e se sso F s F s F s
s
− −
− −
∴ = − + = −
− + −= + =
Using eqn - =. ( ) ( )( )
14 1 3 03 3 1
0 0
2
0
2 2
f s f t e dt ee
s
es
st stst s
= = +−
= −−∞ −− −
f t f t f t
f u t
f t u t
( ) ( ) ( )
( )
( . ) ( . )
= +
= − +
− −
−
1 2
1
2
5
35
5
34 2 2 4 2
t
=
418
Ex. 14.4-6 Ex. 14.5-1 Ex. 14.5-2 Ex. 14.5-3
( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )2 2
2 22 2 2
5 2 0 2From Fig. 1
0 otherwise
5 5 5 5 2 2 2 2 2 ( 2)
2 2 2 2
5 1 2 5 1 1 2
2 2
s ss s
t tf t
So 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
ss s s
− −− −
< <=
= − − = − − = − − − − −
∴ = = − − = − −
( ) ( )( )
( )( )
( ) ( )
( ) ( )
2 2
2 21
2
2 8 38 3 124 13 2 9
3 8 2 2, 8, 3 & 3 6.33
3
6.33 tan 38.4 , 8 6.33 10.2
8
10.2 cos 3 38.40 0t
ssF s
s s s
a c ca d d
m
f t e t t
ω ω
θ − °
−
−−= = ×
+ + + +
−∴ = = = − =− ⇒ = =
− ∴ = = = + =
⇒ = + ≥
( )
( ) ( )( )( )
( )( )
( ) ( ) ( ) ( )
( ) ( )
2
1 2 2
1
1
1 1
( 1)
3
2 17
2 33 1First consider
22 17 1 16
1, 0, 4 3 3 4
| | 3 4, tan 3 4 0 90
( ) (3 4) sin 4
since 1
(3 4) sin 4 1 1
s
t
s
t
eF s
s s
F ss s s
So a c and d d
m d
So f t e t
Now F s e F s f t f t
f t e t t
ω ω
θ
−
− °
−
−
− −
=+ +
= = ×+ + + +
= = = − = ⇒ =−
∴ = = = − =−
=
= ⇒ = −
∴ = − ≥
F ss
s s
As
Bs
C
s
A sF s
C s F s
B multiply both sides by s s
s s Bs s s
equating coefficients B
s
s
= −+
= ++
++
= = − = −
= + = −−
=
⇒ +
− =− + + + +⇒ =
=
=−
2
2 2
0
21
2
2 2
5
1 1 1
51
5
11 5
14
1
5 5 1 1 4
6
|
|
∴ = − + + + +
⇒ =− + + ≥− −
F s s s s
f t e tet t
5 6 1 4 1
5 6 4 0
2( )
t
( )
( ) [ ] ( ) ( )
12 2
2 2
, tan
cos sin cos cos
j j
at at at
c jd c jd me me dF s where m c d cs a j s a j s a j s a j
f t e c t d t e c d t m e t
θ θθ
ω ω ω ω
ω ω ω θ ω θ
−
− − −
+ −= + = + = + =
+ − + + + − + +
∴ = − = + + = +
(a)
(b)
(a)
419
Ex. 14.6-1 Ex. 14.7-1
F ss
s
As
B
s
C
s
A d ds s F s
B d ds s F s
C s F s
F s s s s
t e te t e t
s
s
s
t t t
=+
=+
++
++
= + =
= + =−
= + =
∴ = + − + + +
= − + ≥
=−
=−
=−
− − −
4
3 3 3 3
1 2 3 4
3 24
3 36
4 3 24 3 36 3
4 24 18 0
2
3 2 3
2 2 2
3
3
3
3
3
2 3
3 3 2 3
(
:
)
Using Table 14 - 3 f
F ss
s s
fs
ss
s s
s s
fs
ss
s s
s s
= ++ +
=→∞
=→∞
++ +
=
∞ =→
=→
++ +
=
6 5
2 1
06 5
2 16
0 0
6 5
2 10
2
2
2
lim lim
lim lim
sF
sF
F ss s
fs
s s
fs
s sundefined no final value
s
s
=− +
=− +
=
∞ =− +
= ∴
→∞
→
6
2 1
06
2 10
6
2 1
2
2
0 2
lim
lim
(b)
(a)
(b)
KCL at v v i e
also : i di dt
t1
1
v
: 165 7 1
3 4 2
+ =
= +
−
2 2354
6 into 1 yields : didt
+ = −i e t
Taking the Laplace Transform of the D.E.
2 6
2
35 1s I(s) i(0) + 2 I(s) = where i(0) = 0
4 6
35 1 I(s) =
4 ( 2)( 6)
1 1 1 1 1Now where and
( 2) ( 6) 2 6 6 4 2 4
35 1 35 1 ( )
16 2 16 6
35 35( )
16
s s
t
s
s s
A BA B
s s s s s s
I ss s
i t e
=− = −
−
−+
⇒+ +
= + = = = = −+ + + + + +
∴ = −+ +
⇒ = − 6 16
te−
420
Ex. 14.7-2
v
v
1
2
0 10
0 25
( )
( )
=
=
KCL at v ( ) v v
v
KCL at v : v
Taking Laplace Transform of (1) (2)
2+s
1 1 1 2
2
:
( )
( cos ) ( ) ( )
cos ( )
&
( ) ( ) ( )
1 48 24 0
2 2 0 1
50 2 20 24 30 1 24 0
3 60 2 2
2 10 3
1 1 2
2 2 1 2 2
1 2 2
1 2
dv dt
dv dt v
v t v v dv dt
v v dv dt t
V s V s
+ − =
⇒ + − =
− + − + + =
⇒ − + + =
− =
− + + = + ++
V s s V ss s
s1 2
2
23
25 60 100
44( ) ( ) ( )
Using Cramers rule with (3) & (4)
( )( )
( )( )( ) ( )
( )( )( ) ( )( )( )
( )
2
2 22 3 2
2 2 2
*
2
25 60 1002 10
2 25 60 100 10 44 25 120 220 2402 (3 ) 2 4 1 4 4 1 4
V2 2 1 4
s ss
s s s ss s s sV s
s s s s s s s s
A A B CLet s
s j s j s s
+ ++ + + + + + ++ + + + = = =+ + − + + + + + +
= + + ++ − + +
A s s ss s s j
jj
A j
B s s s
s s
C s s s
s s
s j
s
s
= + + ++ + −
= − −−
= +
∴ = −
= + + ++ +
= =
= + + ++ +
= −−
=
=−
=−
=−
25 120 220 2401 4 2
240 24040
6 6
6 6
25 120 220 240
4 4
11515
233
25 120 220 240
4 1
32060
163
3 2
2
3 2
2 1
3 2
2 4
*
∴ = ++
+ −−
++
++
⇒ = + + + ≥− −
V sj
s jj
s j s s
v t t t e e tt t
2
24
6 62
6 62
23 3
1
16 3
4
12 2 12 2 23 3 16 3 0
( ) cos sin ( ) ( ) , V
421
Ex. 14.7-3 Ex. 14.8-1
KCL at top node:
( ) ( ) 22
3 2C
C
V s sV s
s+ = +
( ) 6 223
CV ss s
= −+
( ) ( )0.676 2 ( ) VtCv t e u t−= −
( ) ( )232
2 23
CC
V sI s
ss
= − =+
so ( ) 0.672( ) A
3t
Ci t e u t−=
Ex. 14.8-2
Taking Laplace Transform of the D.E.
s F s sf f sF s f F ss
2 0 0 5 0 610
3 = − + − + =
+'
Plugging in initial conditions
( ) )
, ,
)
( )
s s ss
ss s
s
F ss s
s s sA
s
Bs
Cs
A B C
F(ss s s
f t te e e for tt t t
22
2
2
2
3 3 2
5 610
32 10
2 16 403
2 16 403 2 3 3 3 2
10 14 16
10
3
143
162
10 14 16 0
+ + =+
+ + = + ++
⇒ = + ++ + +
=+
++
++
⇒ = − = − =
= −+
+ −+
++
∴ = − − + ≥− − −
F(
422
Mesh Equations:
( ) ( ) ( )( ) ( ) ( )4 1 4 16 0 6 6
2 2C C CI s I s I s I s I ss s s s
− − − − = ⇒ − = + +
( ) ( )( ) ( ) ( ) ( ) ( )106 3 4 0
9C C CI s I s I s I s I s I s− + + = ⇒ = −
Solving for Ic(s):
( ) ( )4 2 1 633 24
C CI s I ss s s
− = − + ⇒ = −
So Vo(s) is
( ) ( ) 244
34
o CV s I ss
= =−
Back in the time domain:
( ) 0.7524 ( )tov t e u t= V
Ex. 14.8-3
KVL:
( )8 204 8 4 Ls I s
s s + = + +
so
( ) ( )( )22
1 12
2 5 1 4L
ssI s
s s s
+ ++= =+ + + +
Taking the inverse Laplace transform:
( ) ( )1cos 2 sin 2 A
2t ti t e t e t u t− − = +
Ex 14.9-1
(a) ( )1 5 105 105 10 ( )
5 10t timpulse response e e u t
s s− − − = − = − + +
(b) ( )1 10 51 1( )
10 5t tstep response e e u t
s s− − − = − = − + +
Ex 14.9-2
( ) ( ) ( ) ( )( )
22 22
5 4 205 sin 4
4 202 4tH s e t u t
s ss− = = = + ++ +
( )-1 -1 22
1 4 1(1 (cos 4 sin 4 )) ( )
4 20 2tH s s
step response e t t u ts s s s
− += = − = − − + +
423
Ex. 14.10-1 Ex. 14.10-2 Ex. 14.11-1 Ex. 14.11-2 Ex. 14.12-1
( )( )
[ ]
c
1
1 1.25
Voltage divider yields
8 28 1 4V ( ) 8 2
H(s) = 8 2( ) 2 8 1 42
8 2
8 110 8 1.25
( ) ( ) t
sss s
sV s ss
s s
h t H s e− −
++= =+ ++
+
= =+ +
= =
(a)
(b)
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
2
* 1 1 1 2
1 2
1
1 1 2 1 21 2 1 2
2 2
t
t
h t e H s s
f t u t F s s
so h t f t H s F s u t es s s s
−
− − − −
= ⇒ = +
= ⇒ =
− = = = + = − + +
For the poles to be in the left half of the s-plane, the s-term needs to be positive.
∴ − > < ≤ <8 0 8 0 8k or k so k
To get oscillation, we want the s-term to go to zero
∴ =k will give oscillation8
V ss
s
s s
ss
s s
s s s s
s s
s s s s
0 2 2 2 2
2
2
2
2 2
012
25
5 10
10
5 10
012 2 20
10
012 10 125 2 20
10 125
01250
10 125
25
10 125
( ) .
.
.
.
= − ++ +
++ +
= − ++
=+ + − +
+ +
=+ +
=+ +
+ 125
These specifications are consistent.
424
Problems Section 14-3: Laplace Transform P14.3-1 P14.3-2 P14.3-3 P14.3-4 Section 14-4: Impulse Function and Time Shift Property P14.4-1 P14.4-2
( ) ( )
( ) ( ) ( ) ( )( )
1 1
2 21 1
2 2
cos
A f t A F s
f t t F s s s
AsF s
s
ω ω
ω
=
= ⇒ = +
∴ =+
From Table 14.4-1: f t t F sn
s
ss
nn
= =
∴ =
+,!
,1
2
1
but we have n = 1
F
( ) ( ) ( )
( ) ( ) ( ) ( )sTsT
f t A u t u t T
1 eAeF s A u t A u t T A s A
s s
−−
= − −
−= − − = − =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )( )
( )
at
at
sTat
s a T
Have f t 1 u t u t T e
F s e u t u t T
1 eNow u t u t T and e g t G s a
s
1 eF s
s a
−
−
= − −
= − −
− − − = = −
−⇒ =
−
( ) ( ) ( ) ( )
( ) ( )
( ) [ ] ( )
( )
1 1 2 2 1 1 2 2
1 2
31 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 ss
f t t F ss
F ss s
−
+ = +
= =
= = = +
= = =
= ++
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
1 1
2 3
2 3
1 ( )
f 1 1
1 1 ,
1 1
bt
bt bti
f t A e u t Af t AF s
t A e u t u t e u t f t f t
F s F ss s b
AbF s A
s s bs s b
−
− −
= − = =
= − = − = +
−= =+
∴ = − = ++
425
P14.4-3 P14.4-4 P14.4-5 Section 14-5: Inverse Laplace Transform P14.5-1
( )( )
( )( )
( )( )
( )
(2 3) 1 1 3 32 3 (2 3) 1 3 2 3So 2 2 21 11 3 1 3 1 3
(2 3) (2 3) cos 3 1 3 sin 3
ssF s
s ss s s
t t tf t e e t e
− +− += + = + +
+ ++ + + + + +
− − −∴ = − +
F ss s
=+ +
=+ +
5
4 5
5
4 252 2 2
using Table 14.4 -1
F s s = +2 33
using Table 14 - 3(a)
(b) ( ) ( ) ( ) ( ) ( )& using time shift, : +sT sTf t t T u t T get F s e t eδ δ− −= − − = =
(c)
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )0.5 0.5
0.5 0.5 0.5
0.5 , 0.5
;
1, here 1
1 11 1 1
t
s
t t
at
ss s
g t e u t
f t u t e F s
f t e e f t e e
e as a
eF s e e e e
s s s
τ
τ τ τ
τ
τ
τ τ
τ
−
−
− − −
−
−− −
= − =
− − =
− = =
= = +
= = = + + +
( ) ( ) ( )
( ) ( )
( ) ( )
1
1 1
1 2 2
; ;
1 1 1
s
sT
f t u t e F s
t Tf t T f t t
T
F s and F s es Ts
ττ τ
τ τ τ
τ
−
−
− − = − −− = = =−
− ∴ = =
F ss
s s s
s
s s
As
Bs C
s s
As
s s
s
s s s sBs C
s s
s B s B C s C
( )
( )
= ++ + +
= +
+ + +=
++ +
+ +
= ++ + =−
=
∴+
+ + +=
++ +
+ +
⇒ + = + + + + + +
3
3 6 4
3
1 1 3 1 2 4
3
1 3 12 3
3
1 2 4
2 3
1 2 4
3 2 3 4 3 8 3
3 2 2 2
2
2 2
2
Equating coefficients s
s C
2 :
:
0 2 3 2 3
1 4 3 2 3 1 3
= + ⇒ == − + ⇒ =
B B
C
426
P14.5-2 P14.5-3 P14.5-4
As s s
Bs s s s
Bs C
s s
s s Bs C s
B s B C s C
=+ + =−
=
⇒+ + +
=+
+ ++ +
⇒ = + + + + +
= + + + + + +
1
2 2 11
1
1 2 2
11 2 2
1 2 2 1
1 1 2 2
2
2 2
2
2
( )
Equating coefficients : s2 :
:
cos
0 1 1
0 2 1
11
1 12
= + ⇒ = −= + + ⇒ =−
∴ = 1+
− ++ +
⇒ = −− −
B B
s B C C
Y ss
s
sy t e e tt t
( ) ( ) ( )( )2 2 *
3 22 1 2 1
1 1 1 1 1 13 4 2
s s s s a a bF s
s s j s j s j s j ss s s
− + − += = = + +
+ + − + + + − + + ++ + +
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
2
2
2
*
2 14
1 1 1
2 1 3 43 2 2
1 1 21
3 2 2
3 2 2 3 2 2 4
1 1 1
2 2for first two terms : m= -3 2 2 5 2
-1= tan 2 3 2 126.9
5 cos 233 4
s sb
s s
s s ja j
s s j s j
a j
j jF s
s j s j s
t tf t e t e
θ °
°
− += =
+ + =−
− + −= = =− ++ + + −=− +
= − −
− + − −⇒ = + +
+ − + + +
+ =
− =
− −∴ = + +
( ) ( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
5 1( ) 2 21 21 2 1
5 12
2 1
5 11
2 21
2 21 9 2 111
21 1 2 1 1 2
22 +
s A B CF s
s ss s s
sB
s s
sC
ss
dA s F s s ssds
F s s s s
t t tf t e t e e
−= = + +
+ −+ − +
−= =− =−
−= ==+
= + =− − = −=−=−
∴ = − + + + + −
− −⇒ = − +
Y ss s s s s
As
Bs C
s
=
+ + +=
+ + +=
++ +
+ +1
1 2 2
1
1 1 1 1 1 12 2 2
427
P14.5-5 P14.5-6 Section 14-6: Initial and Final Value Theorems P14.6-1 P14.6-2 P14.6-3
lim lim
Re .
s s
i i
ss
s s
s s
s sv
p p j
→ →
++ +
= ++ +
= ∴ ∞ =
< = − ±
0 2 0
2
2
16
4 12
16
4 120 0
0 2 2 828
since
lim lim lims s s
sV ss s
s s
s s
s sv
→∞ →∞ →∞=
++ +
= ++ +
= ∴ = 16
4 12
16
4 121 0 1
2
2
2
fs
sF s∞ =→
= = lim0
42
2
Use partias
As
Bs
Cs
sF ss
s sA s F s
s
s sB
and s F ss
s sC
So F ss s s
yields f t e e u t
s s
s
t t
l fractions : 2 s+3
s s+1
+= +
++
+
=+
+ += = + =
++
= = −
+ =++
= =
= + −+
++
= − +
= =−
=−
− −
2 1 2
2 3
1 23 1
2 3
24
22 3
11
3 41
12
3 4
0 1
2
2
;
( ) ( )( ) ( ) ( )
( ) ( )( ) ( )
( ) ( ) ( )
2 2
2 3 using partial fractions yields
1 1 2 1 2
11 21 1 4 1 4
o f t cos 2 sin 2t t t
sF s
s s j s j
sF s
s s s
S e e t e t− − −
+=
+ + + + −
− += + +
+ + + + +
= − +
fs
sF ss
s s
s s
s
s0
2 3 4
3 2
22
2
2
2
2 =→∞
=→∞
− ++ +
= =lim lim(a)
(b)
initial value
final value
( ) ( )( ) ( )
2
3 2
2
20 0
10initial value : lim ( ) lim 0 (0) 0
3 2 1
final value
poles : (3 2 1) yields 0.333 0.471 Re 0
10So lim lim 10 10
3 2 1
s s
s s
s ssV s v
s s
s s p i pi i
s ssV s v
s s s
→∞ →∞
→ →
+= = ∴ =
+ +
+ + = − ± ∴ <
+= = ∴ ∞ =
+ +
428
P14.6-4 Section 14-7: Solution of Differential Equations Describing a Circuit P14.7-1
( ) ( )
2
2
21
2 14initial value: lim lim 2 0 2
2 10
final value: 2 10 yields 1 3 Re 0 no final value exists
s s
i
s ssF s f
s s
s s p i p
→∞ →∞
− −= = − ∴ = −
− +
− + = ± > ∴
I s s s
A
s 10
B
s 2 10
C
s 4 10
A s 1.4 10 s 1.6 10
s 2 10 s 4 10
2 10
3 10
23
B s 1.4 10 s 1.6 10
s 10 s 4 10
.4 10
2
2
4 4 4
2 4 8
4 4 s =
8
8
2 4 8
4 4 s =
8
= + × − ×+ + × + ×
=+
++ ×
++ ×
= + × − ×+ × + ×
= − ××
= −
= + × − ×+ + ×
= ×
−
− ×
14 10 16 10
10 2 10 4 10
4 8
4 4 4
104
2 104
. .
s s s
×=
= + × − ×+ + ×
= ××
=
∴ = −+
++ ×
++ ×
⇒ − + +
− ×
− − −
10
15
C s 1.4 10 s 1.6 10
s 10 s 2 10
8.8 10
6 10
2215
I s 2 3
s 10
1 5
s 2 10
22 15
s 4 10
i t = 1
1510e 3e 22e A
8
2 4 8
4 4 s =
8
8
4 4
104 t 2x104 4x104
4 104
t t
Take the transform of (1) & (2)
I(s) + 0.001 s I(s) i(0) + V(s) = 2
s + 2 10(3)
I(s) = 2.5 10 sV(s) - v 0 (4)
4
6
50 −×
× −
v = 2e Vs
t− ×2 104
i
also :
t
(
( )
( )
0) = 1A, v(0) = 8V
KVL : 50I+0.001 didt
+ v=2e
= 2.5 10 dvdt
− ×
−×
2 104
6
1
2
Solving for V(s) in (4) and plugging into (3) yields
429
P14.7-2 P14.7-3
( )( )
( ) ( )( )
( )
( ) ( )
2
2 22 2
7
2 2
6.4 10 4001600ssI(s) i(0) + 1010 I(s) =
s 400 400
160 1600s 2.5 x 10( )
1010 1010 (400)
I s = I1 2
x
s
I ss s s
I s
− −+ +
−⇒ = +
+ + +
+
v = 160 cos400t
At t = 0 : V = 0+c
⇒ = 160+i( )0 Α
KCL at top : 10
also : i = v v
(2) into (1) yields :
3
c
− + =− ⇒ = −
+ = −
dv dt v i
v v i
i t x t
c c
c
100 1
1 2
1010 1600 400 6 4 10 4004
( )
( )
didt
cos . sin
Taking the Laplace Transform yields:
Now let I sA
sB
s jB
s j
A s x
s
Bs x
s s j
x
xj
B j
I s I s I ss
js j
s
s j
2
=
=
=
++
++
−
= −
+= −
= −+ −
= ∠
∠= −
= +
∴ = + =+
+ −+
+
−
−
°
°
1010 400 400
1600 2 5 10
400 2231
1600 2 5 10
1010 400
2 56 10 14
8 69 10 68 4115 27 2
115 27 2
136 91010
115 27 2400
115
7
21010
7
400
7
5
1 2
*
*
..
. . .
. .. .
. .
. . . . +−
⇒ = + −
= + −
−
−
j
s j
i t e t t
i t e t t
t
t
27 2
400
136 9 2 115 400 2 27 2 400
136 9 23 0 400 54 4 400
1010
1010
.
. . cos . sin
. . cos . sin
1KVL : 15 10 cos 2 but 2 20 cos2
30c c
c cdv dv
v i t i v tdt dt
+ = = ⇒ + =
where v t u t
vs
c
==
10 2
0
cos
&
(0)
430
P14.7-4
Taking the Laplace Transform yields:
sV s v V s s s
V ss
s s
As
Bs j
Bs j
A s
s
Bs
s s jj j j
c c c
c
s
s j
− + = +
⇒ =+ +
=+
++
+−
=+
= − = −
=+ −
= + = +
−
−
0 2 20 4
20
2 4 2 2 2
20
4
408
5
202 2
5 1 5 2 5 2
2
2
2 2
2
( )
*
=
=
B j
V s s j s j j s j
v t e t t
c
ct
*
( )
cos sin
= −∴ = − + + − + + − −
⇒ = − + +−
V
5 2 5 2
5 2 5 2 5 2 2 5 2 5 2 2
5 5 2 22
From Prob. 9.9 -11 : v = 0, i (0) = 0
after source transformation
c L(0)
KVL v i di dt
also i Cdv dt
c L L
L c
:
:
+ + = −12 2 8 1
2
Laplace Tr
V s I s sI s i s
I s sV s v
c L L L
L c c
ansform of (1) & (2) yields:
C
+ + − = −
= −
12 2 0 8 3
0 4
48 2
6 1 22
into (3) yields : V c s
C
s s s C= −
+ +( )
C V s s s a s b s c s
a c so Vc s s s s
v t e te
C V s s s s a s b s c s
soV s
c
ct t
c
c
= = − + = + + + +
⇒ = − = = = − + + + +
∴ = − + +
= = − + + = + + + +
⇒ = − = = − = −
− −
b
a b c
1 18 72 3 3 3
8 8 24 8 8 3 24 3
8 8 24
1 10 40 1 5 1 5
8 10 2
2 2
2
3 3
:
, , ,
:
, , ,
8 5 1 2 5
8 10 2 5
+ + − +
∴ = − + −− −
b s s
v t e ect t
(a)
(b)
431
P14.7-5
Section 14-8: Circuit Analysis Using Impedance and Initial Conditions P14.8-1
By inspect iL Aion v 10V
t > 0
c ( ) ,
@
0 0 0− −= =
Ld i
dt
didt
i
Cdidt
2
2400 0
10
+ + =
=
c
L s I s sidi
dtR sI s i
I s
C
I ss
s s s s
poles are i
i t e t A
i
t
2
2 5 2 5
200
00
0 0
1 0 1 10 400 0
1 400 2 10
10
400 2 10
200 400
140
400
− −
+ − + =
=− + +
+ + ×= −
+ + ×= − ±
∴ = − −
Solving for I(s) and plugging in values yields
The p
sin
I s ss
ss
s s s s
i tmA
e mA t
L
L t
( ). .
( ). .
( )
=−
+=
−
+= =
+
=<
− > −
60 010
5 2000
60 002
400003 005
400
2 0
3 5 0400
t
t < 0
t > 0
432
P14.8-2
0V (s)
10s
2000
V (s)
4000
V (s) .015
5s
V (s)
83
s400015
I (s)V (s) 0.15
5s
815
s s400015
0.015
I (s)0.005
50.002
s400015
i (t) 5 3e mA, t 0
LL L
L
LL
L
L
4000
15t
=−
+ +− −
⇒ =+
= + =+
+
= −+
= − >−
t < 0
t > 0
433
P14.9-3
− + +−
=
− + + −
=
= ++
= −+
= − >−
0.006s
V (s)
2000
V (s)8s
10.5s
0
6000s
500 V (s) 0.5s V (s)8s
0
V (s)8s 12000
s(s 1000)12s
4s 1000
V (t) 12 4e V, t 0
cc
6
c c
c
c1000t
t < 0
t > 0
434
P14.8-4
V (s)6s
2000
V (s)
40000.5s
10V (s)
8s
0
500 V (s)6s
250 V (s) 0.5s V (s)8s
0
V (s)6000 8s
s s 15004s
4s 1500
v (t) 4 4e V, t 0
cc
6 c
c c c
c
c1500t
−+ +
−
=
−
+ + −
=
= ++
= ++
= + >−
t < 0
t > 0
435
P14.8-5
P14.8–6
Steady - state
i (0)10V10
1A, V(0) 106
6 46VL = = =
+
=
Ω
t 0= −
t 0>KVL rt. mesh: 6i di 0 (1)
KCL top node: dV dt 4i 6e (2)
Taking Laplace Transforms of (1) & (2) :
L L dt
L3t
v
v
− − =
+ + = −
V(s) (6 + s) I (s) 1 (3)
1 s V(s) 4I (s) 6 6 s 3 6s 24 s 3 (4)
L
L
− = −
+ + = + + = + +
Using Cramer's rule on (3) & (4) yields
V(s) =4 6+s
6s 24
s 34 1 s 6 s
6s 56s 132
s 3 s 2 s 5A
s 2B
s 3C
s 5
A s 2 V(s) 44 3, B s 3 V(s) 9, C s 5 V(s) 1 3
V(s) 44 3 s 2 9 s 3 1 3 s 5
v(t) 44 3e 9e (1 3)e V
2
s 2 s 3 s 5
2t 3t 5t
− + ++
− + + += + +
+ + +=
++
++
+
= + = = + = − = + =
∴ = + − + + +
⇒ = − +
=− =− =−
− − −
v = 0
KCL node v
v
RC
ddt
vv
R0
v
RC
dv
dt
v
R0
2
2
2 1
12 0
2 0
2
1
1
0 0
2
− + − + − =
⇒ + + =
vv
v
Taking Laplace Transform
V (s)
100010 sV (s) 5
V (s)
10000
V (s)10s
V (s)10s
15s 1000
v (t) 15e 10 , t 0
1 60
0
1 0
01000t
+ − + =
= ⇒ = − ++
∴ = − ≥
−
−
436
P14.8-7
from KCL I I I
from KVLs
I V
V V I
Plugging s I s Is
Plugging sI s I
:
:
&
3 1 2
3 1
1 2 2
1 2
1 2
3
52 4
1 5
3 2 2 95
2 1 1
= +
= +
= +
⇒ + + + = +
⇒ − + =
(1) and (3) into (4)
(1) and (2) into (5)
So I ss
s
s
s s
s
s s s s
As
s
s
s
s s
i t e e u t A
s
s
t t
2 2
0 26
154
20 26 1 54
1 3 25
9
2 1
15 8
5 9 2
3 16
0 26 154
3 16
1540 64
3 16
0 262 36
0 64 2 36
= + + = ++ +
= ++ +
= ++
=
= ++
=
= +
= −
= −
− −
∆ =
A+ 0.26
+ B
+ 1.54
B
I = 0.64+ 0.26
+ 2.36+ 1.54
2
.
. .
.
..
.
..
( ) . . ( )
.
.
. .
∴ =+ +
− += − + +∆
3 2 2
2 15 9 22s s
s ss s
v t Ldi
dtM
di
dtV s sI sI
v t Ldi
dtM
di
dtV s sI sI
1 11 2
1 1 2
2 22 1
2 1 2
3 9 1
2 8 2
( ) ( )
( ) ( )
= + ⇒ = + −
= + ⇒ = + −
437
Section 14-9: Transfer Function and Impedance P14.9-1 P14.9-2 P14.9-3
1 1 1 2 2 2let and R C R Cτ τ= =
∴ =+
+ + +
= = ⇒ =+
+ +=
+=
H s R s
R s s R
When H sR s
R R
R
R Rconstant
s
2 1
1 2 1 2
2 22 1
1 2
2
1 2
1
1 1
1
1
ττ τ
τ τ ττ
τ
Now requir L R RC
C
e : RC +
L = R then Z = R2
=
⇒
2
Z R Cs Z R Ls
Z sZ Z
Z Z
R sC R Ls
R sC R Ls
RLCs RC L R s
LCs RCs
L
L
L
1
1
1
2
2
1
1
1
1
2 1
=
+ = +
=+
=+ ++ + +
=+ + +
+ +
KCL at V V V sC V V R
VsCR V V
s CR
KCL atV V R V V s C V V sC
in in in out
in out
in out
:
:
− + − =
⇒ =+ −
+ − + − =
1 2
12
1 1 1 1 1
0
1
21
0 2
Plugging (1) into (2) yields :
V
V
R R Cs R R C s
R Cs R R C sout
in
=+ + +
+ +1 2
1 21 2 1 2
2 2
1 1 22 2
2 1 1 11
1 2 1 1 1 1
We want: ( )
1 2We have: ( ) where Z and 21 1 12 2
H s a
Z R C s R RH s Z
Z Z R C s R C s R C s
=
= = = =+ + + +
∴ =require R C R C1 1 2 2
438
P14.9-4
P14.9-5
From voltage division
V s V ss
s s
H sV s
V s
s
so
0 1
3
3 3
1
5
5
150 2 10
150 2 10 100 3 10
15 10 2
2 5 10 2
= + ×+ × + + ×
⇒ = = × +× +
−
− −
.
.
Now V s s
so V s ss
s
s
s s
As
B
S
A sV s B s V s
V s s s v t e V t
s s
t
1
0
5
5
5
5 4
0 04
0 5 10
04
05 10
100
10015 10 2
25 10 5
100 15 10 2
2 5 10 5 5 10
3 5 10 20
3 20 5 10 3 20 0
4
4
=
=× +
× +=
× +
× += +
+ ×
= = = + × = −
⇒ = + − + × ∴ = − >
= =− ×
− ×
. .
.
,
(a)
(b)
V s RCs Cs
I sCs
I
R RCs
ICs
I
= + +
−
= + +
−
1 1 2
2 1
1 1 11
01 1
2
( )
( )
(a)
Solving fo IV s
Cs
RCs
RCs Cs
V s
V sRCs
R Cs RCss
R C sRC R C
RR Cs
RR C
r I 2 ⇒ =
+
+
−
∴ =+ + −
=
+ + +
2
1 2
0
1
12 1
12
12 2
1
22
1 1
2 2 1 1
24
2
1
2
( )
(b)
439
P14.9-6 P14.9-7
V s
V s
Cs
R Ls Cs
LC
sRL
sLC
H s
e
s j
Hs s
Mag H
n
n
Nm
n n
n
n n
n
0
2
7
2 7
1
1
11
1136 10
625 1136 10
=+ +
=+ +
=
= ⋅
= ⋅
= ⋅+ ⋅ + ⋅
=
⋅
Mathcad spreadsheet:
N : 100 n := 0..N
min := 100 max := 100000 m := lnmaxmin
ω ω ωω
ω ω
ω
ω
: min
:
:.
.:
H s
So V(s) = 1s
where s1
=+
++
=+
+ + + +
= + ×+ × + + ×
= ++
+ ++
+ ++
= − +
∴ = − + −
−
− − −
− −
1
11
10
1
1 10 1
1 167 10
1 5 10 10 8 33 10
1 2 654644
0 826 604 0 826 604
3678 3509
1 2 65 165 3509 1209 3509
1
1 24
2
21
2 41
2
8 2
4 7 2 12 3
1 1
4644 3678
C s
C sLs
LCs
LCs
LCs LC s C s LCs
s
s s s
H ss s
j
s s
j
s s
j
v t e e t
v t
t t
.
.
. . . . .
. . cos . sin
*
04644 36781 2 65 2 05 3509 0 632= − + +− −. . cos .e e tt t
V s ss ( ) = 1
Use voltage divider
ZCs Ls
CsLs
Ls
LCsp =
+=
+
× −
( )11 1
10
2
8
LC = 1.67
440
P14.9-8
Now if = 2 f , then := 9.42.10
ng this into the transfer function: H:=1.136 10
yields a magnitude of H = 0.001
4
7
ω π ω
ω ω
rad
and pluggij j
sec
.
⋅⋅ + ⋅ ⋅ + ⋅ 2 7625 1136 10
V
V
Z
Z Z
RRCs
R L sR
RCs
R
LRCs L R RC s R R
V
V
R LRC
sL R RC
LRCs
R R
LRC
x xx x
x x
2
1
2
1 22
2
1 2
1
1
=+
= +
+ ++
=+ + + +
=+
++ +
ZR
Cs
RCs
RRCs
Z R L sx x
2
1
1
1 1=
+=
+
= +
a)
441
P14.9-9 P14.9-10
KCL at V V V sCV
Ror R C s V R C sV V
KCL at VV
RV sC V R C sV
Solving f
V
V
R C s
R R C C s C s
or T(sC
s
sC
s R C C
s s
inout
in out
out out
out
in
0 0 10
11 1 0 1 1
10
22 0 2 2
1 1
1 2 1 22
2
2
2
1 2 1 2
2 2
0 1
0 0
1
:
:
)
− + − = + = +
= + + = = −
= −+
−−
V
or
or V V from (1) & (2) get
+ R
=
1R
+ 1
R+
1R
= Q s
+ Q
+
out in
2
2
1 1
0
00
ωω ω
where R C Q
R C Q
o
o
1 1
2 2 1
=
=
ω
ω
Solving for V V yields
V
V
G Cs
G CsH for
( ) ( )1 3
1 0
2 1
2
1
→
=−+
= ≥H( ) =ω ω
node GV GV GV
node G sC V sCV
also V
3 2 0 1
4 0 2
3
2 3 1
4 1
3 4
: ( )
: ( )
: ( )
V
− + − =
+ − =
=
442
P14.9-11 P14.9-12
Vout =V1
KCL at V V V sCV V
Ror R C s V R C sV V
KCL at V V V sCV
RR C s V R C sV
So
V
V
R R C C s
R R C C s C s
T s
C R C Cs
inout
in out
out
out
in
0 0 10
11 1 0 1 1
1 1 0 21
22 2 2 2 0
1 2 1 22
1 2 1 22
1
2
2 2 1 2
2
0 1
0 1
1
:
: ( )
( )
− + − = + = +
− + = + =
=+
or
lving for V V from (1) and (2) yields
+ R
= s
s + 1
Rs +
1R
= s
s + Q
out in
1
2
2 1
2
0ω+
and R
0ωω ω
21 1 2 21where R C Q C Qo o= =
at node V
V V
R
V V
R
V V
Ri
1
1
1
1 2
2
1 0
3
0
:
− + − + − =
Rearranging @ node V
but
Plugging (2) into (1) and rearranging yields
V
V
1
0
i
yields
VR R R
V
R
V
R
node VV V
RV V sC V
V sC R V
R
sC R R sC R R sC R R R
i
:
@ : ,
( )
1
1 2 3
0
3 1
22 1
2
2 0 2 2
1 2 2 0
3
2 2 3 2 1 3 2 1 2 1
1 1 11
0 0
2
+ +
− =
− + − = =
∴ = −
= −
+ + +
443
P14.9-13 P14.9-14
H(s)2
s(s 2) (s 1)
A
s
B
s 2
C
s 1
D
(s 1)
A sH(s) 1
B (s 2) H(s) 2
D (s 1) H(s) 1
Cd
dss 1 H(s)
h(t) 1 2te e u(t)
h(0) 0 h(0) lim sH(s)
h( ) 1 h( ) limsH(s)
2 2
s 0
s 2
2s 1
2
s 1
t 2t
s
s 0
=+ +
= ++
++
++
⇒ = =
= + = −
= + = −
= + =
⇒ = − −
= ⇒ =
∞ = ⇒ ∞ =
=
=−
=
=−
− −
→∞
→
0
Node equations:
V V
R
V
RsC V V
V
R
V V
RsC V V
V
R
V
RV
R
RV or V
R
RV
V VR R
RV
R R
RV
Solving yields
V
V
R R sCR R R
R R sC R R R
a i aa b
b oa b
b ab a a b
a b a b
o
i
− + + − =
+ − − − =
− − = ⇒ = − = −
⇒ − = + = − +
=− + + +
+ + +
1 2
6
3 4
3 2
3
2
2
3
2 3
2
2 3
3
3 4 4 2 3
1 2 1 2 3
0 1
0 2
0 3
( )
( )
( )
444
P14.9-15 P14.9-16 P14.9-17
H(s) 400 / (s 400s 2 10 )
F(s)s a
f(t) e sin t Table 14.4 3
gives 400, 160000
H(s)400
(s 200) 160000h(t) e sin(400t) u(t)
2 5
2 2at
2
2200t
= + + ×
=+ +
⇒ = −
= =
∴ =+ +
∴ =
−
−
ωω
ω
ω ω
H(s)4 s 3
s(s 2)
As
Bs 2
C
s 2
Solving partial fractions yields : A 3, B 3, C 2
H(s)3s
3s 2
2
s 2h(t) 3 3e 2te
2 2
21t 2t
=+
+= +
++
+= = − = −
∴ = −+
−+
∴ = − −− −
T(s)V s
V s
1Cs
Ls R1
Cs
1LC
sRL
s1
LC
o
i 2= =
+ +=
+ +
L C R T(s) 2 .025 18 20 20
= s2 + 9s + 20 (s+4)(s+5)
2 .025 8 20 s2 + 4s + 20
1 .391 4 2.56 2.56 = s2 + 4s + 2.56 (s + .8)(s + 3.2)
2 .125 8 20 4 = s2 + 4s + 4 (s+2)
445
( )( )
( )
.8t 3.2t
3.2t .8t
.56T(s)
s .8 s 3.2
1.07 1.07impulse response T(s)
s .8 s 3.2
impulse response 1.07 e e u(t)
4 1T(s) 2.56 1 3 3step response
s s (s .8)(s 3.2) s s .8 s 3.2
1 4step response 1 e e u(t)
3 3
i
− −
− −
2=
+ +
= = −+ +
= −
−
= = = + ++ + + +
= + −
( )( )2t
2t
mpulse response 4te u(t)
step response 1 1 2t e u(t)
−
−
=
= − +
( ) ( ) ( )
( )( )
( )( )
1 22 2
2 21 2 1 2 1 2
2 22
2t
T(s) 20 1 K s K step response
s ss(s 4s 20) s 4s 20
20 s 4s 20 s K s K =s 1 K s 4 K 20 K 1, K 4
14s 21 2 step response
s s 2 4s 2 4
1step response 1 e cos4t sin4t u(t)
2−
+= = = +
+ + + +
= + + + + + + + + ⇒ = − = −
−− += + +
+ ++ +
= − +
a)
( )
( )
4t 5t
5t 4t
20T(s)
(s+4)(s+5)
20 20 impulse response = T(s) =
s+4 5
impulse response 20e 20e u(t)
T(s) 20 1 5 4 step response
s s(s 4)(s 5) s s 4 s 5
step response = 1+4e 5e u(t)
s− −
− −
=
−+
= −
−= = = + ++ + + +
−
2
2 2
2t
20T(s)
s 4s 20
5(4) impulse response T(s)
(s 2) 4
impulse response 5e sin 4t u(t)−
=+ +
= =+ +
=
b)
c)
d)
446
P14.9-18 P14.8-19 P14.9-20
h t e t
andb
bb
b
b
H ss s s s
t
n
nn n
n
n
n n
( ) sin( )
( )
=
= =
⇒
= = =
∴ = =
∴ = − ⇒ =
⇒ =+ +
=+ +
−2 2
1 2 2
2 1 2 1
21 2
1 1 2
2
1
2 1
2
2
2
2
2 2 2
Using Table 14.5-1 with
nbω ω
ω ω ω
ω
ξ ξ
ωξω ω
H jj j j
H j
( )( )
( )
ωω ω ω ω
ωω ω ω
=+ +
=− +
∴ =− +
=+
1
2 1
1
1 2
1
1 2
1
1
2 2
2 2 42
V V
(s+3
1 0
0 0
0 3 2
0
03
13 2
3 2 3 2 3 2 3 2
462
2 0 47 119 7
0 47 119 7
0 462 0 47 119 7
3 2
0 47 119 7
3 2
462 2 47 2
s ss
s s j s j
A
s
B
s j
B
s j
A sV s
B j V s
B
V ss s j s j
So v t e
s
s jo
o
o o
t
= =+
+ − + += +
+ −+
+ +
⇒ = =
= − = ∠−
= ∠
∴ = + ∠
+ −+ ∠
+ +
= +
=
=− +
−
,
.
) . .
. .
. . . . .
( ) . (. ) cos
*
*
t o− ≥119 7 0. t
H sV
V
R RCs
LS R RCs LCR Ls Ro( )
/( )= =
++ + + +1
1
1
=
R
s2
( )
( )
2
3
6 12 2
2 6
1have underdamped response s= 1 1 2 (1)
2RC
18 ms 125 Hz 250
8 10
Using (1): s
250 1 10 10 4 807.8
Critically damped when 1 1 2 0 R = 1 2 L = 1 2 10 = 500
j LC RC
T f or
j
R R
LC RC C
ω
α ω
π
−
−
⇒ − ± −
= ⇒ = = =×
= +
= − ⇒ = Ω
− = ⇒ Ω
(a)
(b)
(a)
(b)
(c)
447
Section 14-10: Convolution Theorem P14.10-1
P14.10-2 f * f P14.10-3
1f(t) * f(t) = [F(s) F(s)]
pulse:
−
∴ = − −F s
1 es
s
( )
( ) ( ) ( ) ( ) ( )
2s s 2s1 1
2
1 2
1 e 1 2e ef f =
s s
Now 1 s t u t
f f = t u t 2 t 1 u t 1 t 2 u t 2
− − −− −
−
− − +∴ ∗ =
=
∴ ∗ − − − + − −
( ) ( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2s
2s 4s1 1
2 2 2
f t 2 u t u t 2
2 2eF s
s s
4 8e 4ef f F s F s 4t u t 8 t 2 u t 2 4 t 4 u t 4
s s s
−
− −− −
= − −
= −
∗ = = − + = − − − + − −
( ) ( )( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
2
1
12 1 1
1 1 2
1 22
t/RC2
V s 1 Cs 1 RCH s
V s R 1 Cs s 1 RC
V t h t t V s H s
1Now if t tu t , V s
s
11 RCor V s H s V s
1s sRC
and v t t RC 1 e , t 0
v
v
−
−
= = =+ +
= ∗ =
= =
= = +
= − − ≥
f * f
448
P14.10-4 Section 14-11: Stability P14.11-1 P14.11-2 P14.11-3 P14.11-4
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
12
22
2 2
(at)
2 2
1 1h t f t H s F s where H s and F s
s as
1 1 A B CSo H s F s
s a s s ass
Solving the partial fractions yields: A 1 a , B 1 a, C 1 a
1 t eSo h t f t , t 0
aa a
−
−
∗ = = = +
= = + + + +
=− = =
−∗ = + + ≥
so 6 k < 0 or 0 k 6 will give this result <− ≤
∴ − > ≤ <3 k 0 or k < 3 so 0 k 3 for stability
H ss 2
s 1 i s 1 izeros : s 2
poles : s 1 i, 1 i
= +
+ + + −= −
= − + − −
This circuit is stable since the poles are in the left-hand s-plane.
To have stable operation, the roots of the denominator must be in the left-hand s-plane
We want the roots of the denominator to be in the left-hand side of the s-plane for stability,
449
P14.11-5
V s V s1
10
V s V s12s
10 V s V sV s
3s
T sV s
V s
s 5 1 s
s 5 1 s16
This circuit is stable for K > 1.
This circuit is critically damped when 5 1+K 416
0
i.e., when K = 1 + 275
but not for K = 1275
When K=2
T ss 15s
s 15s16
i 0 i 0i 0
0
0
i
2
2
2
2
20
−+
−+ − =
⇒ = =+ +
+ + +
−
−
=
− − −
= +
+ +⇒ =
K
K
K
ω 1
6, Q
1
15 6
When K = 1, this circuit will oscillate at =1
6rad
Since Q <12
, this circuit is overdamped.
=
− ω sec
K for oscillator K = 3
impulse response K=1,R=1k , C = 0.5 mF
⇒Ω
a)
b)
node 2: V V
R
V V
R(V V ) sC = 0
node 3: V V
R(V 0) sC=0 , also V KV
V
V
K/R C
s(3 )s
RC1
R C
a 1 aa 0
a0
0
1
2 2
22 2
− + − + −
− + − =
=+ − +K
V
V
10.25
s2s0.5
+ 1
0.25
= 4
s 4s +4=
4
(s+2)
v (t)=4te V
0
1 22 2
02t
=+ +
∴ −
450
PSpice Problems SP14-1 SP14-2
Circuit @ t > 0
Inut file: R1 1 0 0.2 L1 1 0 0.5 IC=-60 C1 1 0 3 IC=12 .tran 0.1 10 UIC .plot tran V(1) .probe .end
Circuit:
Probe output
Inut file: Vs 1 0 pulse(0 10 0 0 0 2 10) R1 1 2 1 L1 2 3 0.5 IC=0 C1 3 0 0.4 IC=0 .tran 0.1 10 UIC .plot tran V(3) .probe .end
Probe output
451
SP14-3 SP14-4
Inut file: Vs 1 0 pulse(0 2) L1 1 2 0.5 IC=1 R1 2 0 1
.tran 0.1 5 UIC
.probe
.end
Probe output
Input file: Vs 1 0 ac 1 R1 1 2 10 C1 2 3 100u C2 3 0 100u R2 3 4 50 R3 4 0 50 .ac dec 100 1 100k .probe .plot ac Vdb (4) .end
Probe output
452
SP14-5 SP14-6
Input file: Vs 1 0 pulse (0 1) L1 1 2 1 R1 2 0 807.8 C1 2 0 lu .tran 0.001 0.01 .probe .end
Probe output
* This value varies: a) 50 kΩ b) 62.5 kΩ c) 262.5 kΩ
Input file: Vs 1 0 pulse (0 1) R1 1 2 50k C1 2 3 2u R2 3 4 50k R3 4 5 262.5k * C2 2 5 2u XOA1 4 0 5 OA .subckt OA 1 2 3 *nodes listed in order − + o E 3 0 1 2 −1G .ends OA .tran 0.1 2 .probe .end
453
Verification Problems VP 14-1
( ) ( ) 2.1 15.93 6 2t tL L
dv t i t e e
dt− −= = − −
( ) ( ) 2.1 15.910.092 0.575
75t t
C C
di t v t e e
dt− −= = − −
( ) ( ) 2.1 15.91 12 12 6 2t t
R Lv t v t e e− −= − = + +
( ) ( ) ( )( ) 2.1 15.92
121 0.456 0.123
6L C t t
R
v t v ti t e e− −
− += = + −
( ) ( ) 2.1 15.93 1 0.548 0.452
6C t t
R
v ti t e e− −= = + +
Thus,
( ) ( ) ( ) ( ) ( )1 2 312 0 and +L R R C Rv t v t i t i t i t− + + = =
as required. The analysis is correct. VP 14-2
454
( ) ( )1 2
18 20 and
3 34 4
I s I ss s
= =− −
KVL for left mesh: 12 1 18 18 20
6 03 3 324 4 4
s s s s s
+ + − = − − −
KVL for right mesh: 18 20 20 18
6 3 4 03 3 3 34 4 4 4
s s s s
− − + − = − − − −
The analysis is correct. VP 14-3
Initial value of IL (s): 2
lim 21
5
ss
s s s
+ =→ ∞ + +
Final value of IL (s): 2
lim 20
0 5
ss
s s s
+ =→ + +
Initial value of VC (s): ( )
( )2
lim 20 20
5
ss
s s s s
− +=
→ ∞ + +
Final value of VC (s): ( )
( )2
lim 20 28
0 5
ss
s s s s
− += −
→ + +
Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s) was
calculated as ( )20LI s
s− instead of ( )20 8
LI ss s
− + . After correcting this error
( ) 2
20 2 8
5C
sV s
s s s s
+ = − + + + .
Initial value of VC (s): ( )
( )2
lim 20 2 88
5
ss
s ss s s
− + + = → ∞ + +
Final value of VC (s): ( )
( )2
lim 20 2 80
0 5
ss
s ss s s
− + + = → + +
455
Design Problems DP 14-1
Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response:
( )( )2
2
51 4
o
k R
LV sR ss sL LC
= =++ +
Equating the poles: 2
1,2
4
4 02
R R
L L LCs
− ± − = = − ±
Summarizing the results of these comparisons:
24, and 5
2
R k RR
L LLC= = =
Pick L = 1 H, then k = 0.625 V/V, R = 8 Ω and C = 0.0625 F. DP 14-2
Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response:
( )( )2 2
2
10 101 8 204 4
o
k R
LV sR s sss sL LC
= = =+ ++ ++ +
Equating coefficients: 1
8, 20, and 10R k R
L LC L= = =
Pick L = 1 H, then k = 1.25 V/V, R = 8 Ω and C = 0.05 F.
456
DP 14-3
Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response:
( ) ( ) ( ) 22
5 5 101 2 4 6 8o
k R
LV sR s s s ss sL LC
= = − =+ + + ++ +
Equating coefficients: 1
6, 8, and 10R k R
L LC L= = =
Pick L = 1 H, then k = 1.667 V/V, R = 6 Ω and C = 0.125 F. DP 14-4
Comparing the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response:
( ) ( ) ( ) 22
5 5 10 301 2 4 6 8o
k RsLV s
R s s s ss sL LC
+= ≠ + =+ + + ++ +
These two functions can not be made equal by any choice of k, R, C and L because the numerators have different forms.
457
DP14-5 DP14-6
v (t = 0 ) = C
C +C and v (t ) =
R
R Ro+ 1
1 2o
2
1 2
→∞+
C
C C
R
R R1
2 1
2
1 2+=
+→
a) Use voltage divider
V s
V
R
sC R
R
sC R
R
sC R
= C
C C
s+1/C R
s+R R
R R C C
o
1
2
2 2
1
1 1
2
2 2
1
1 2
1 1
1 2
1 2 1 2
( )
( )s= +
++
+
+ ++
1
1 1
To make natural response zero, eliminate the pole in V
V by causing it to cancel with the zero.
C R
R R
R R C C leads to
C
C
R
R
o
1
1 1
1 2
1 2 1 2
2
1
1
2
⇒ − = − ++
=1( )
b)
If v t u (t), V s s
V s C
C +C
s+ 1R C
s s+R R
R R C C
= K
s
K
s+R R
R R C C
where K R
R R and K =
C
C C
R
R R
1 1
o1
1 2
1 1
1 2
1 2 1 2
1 2
1 2
1 2 1 2
12
1 22
1
1 2
2
1 2
( ) ( )
( )
( ) ( )
= =
=+
+
+ ++
=+ +
−+
1c)
So v (t) = R
R R
C
C C
R
R R e t > 0
where = R R C C
R R
o2
1 2
1
1 2
2
1 2
t
1 2 1 2
1 2
++
+−
+
++
− τ
τ ( )
C
C C =
R
R R v (t) =
R
R R
C
C C1
1 2
2
1 2o
2
1 2
1
1 2+ +⇒
+=
+1) If
2) and 3)
For t < 0
458
v V
i(0) = 0c ( ) .0 0 4= −
For t > 0
mesh equations 2+8s I s I = 0.4
s
8s I + Ls +8
s I = 0
Solving for I I = 1.6
s (Ls +4Ls +8)
Thus s s+8L need char. eqn. with complex roots with significant damping
If L = 1H s s+8 = 0
So I(s) = 1.6
s (s +4s+8)
.2s
+ s
s s+8=
.2s
+ s+2
(s+2) s+2
So i(t) = .2 e e
1 2
1 2
2 2 2
2
2
2 2 2
t t
−
−
+ =
⇒ +
= − ++
−+
−+
− −−
8
4 0
4
2 4
4
2
4
8
4
2 2 4
2
2
:
. ( ) . ( ) .
( )
. cos . − ≥2 2t t (A), t 0sin