Chapter 14 - The Laplace Transform

62
Problems Section 14-2: Laplace Transform P14.2-1 () ( ) () ( ) () () 1 1 2 2 1 1 2 2 cos Af t AF s As Fs s s f t t Fs s ω ω ω = = + = = + L P14.2-2 () 1 1 1 2 1 ! 1 F n n n t s t s s s + + = = = L L 1 ! = P14.2-3 ( ) ( ) ( ) ( ) () () () [] () () 1 1 2 2 1 1 2 2 1 2 3 1 1 2 2 2 2 Linearity: a a Here 1 1 3 1 1 1 so 3 t f t af t F s aF s a a f t e Fs s f t t F s s Fs 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 ut Af t f t e ut ut e ut f t f t F s F s s sb Ab Fs AF s AF s F s A s sb ssb = = = = = + = = + 3 = = + = = + + 1

description

Resolução da 7ª edição do livro do Dorf e Svoboda, Introdução Aos Circuitos Elétricos Cap. 14

Transcript of Chapter 14 - The Laplace Transform

Page 1: Chapter 14 - The Laplace Transform

Problems

Section 14-2: Laplace Transform P14.2-1

( ) ( )

( ) ( ) ( )( )

1 1

2 21 1 2 2cos

A f t A F sAsF ss sf t t F s

sωω

ω

⎫=⎡ ⎤⎣ ⎦ ⎪ ⇒ =⎬ += ⇒ = ⎪+ ⎭

L

P14.2-2

( ) 11 1 2

1! 1Fnnnt s t

s s s+ +⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦L L 1

!=

P14.2-3

( ) ( ) ( ) ( )

( ) ( )

( ) [ ] ( )

( )

1 1 2 2 1 1 2 2

1 2

31 1

2 22

2

Linearity: a a

Here 11

31

1 1so3

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

+ = +⎡ ⎤⎣ ⎦= =

⎡ ⎤= = =⎡ ⎤⎣ ⎦ ⎣ ⎦ +

= = =⎡ ⎤⎣ ⎦

= ++

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 ss s b

AbF s AF s A F s F s As s b s s b

− −

= − =

= − = − = +

−= =

+⎡ ⎤

3

∴ = = + = − =⎡ ⎤⎣ ⎦ ⎢ ⎥+ +⎣ ⎦

1

Page 2: Chapter 14 - The Laplace Transform

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 As 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

( ) ( )

( ) ( )( )

( )

( )

11

sTs a T

at

eu t u t T es F ss ae g t G s a

−−

⎫−− − =⎡ ⎤ −⎪⎣ ⎦ ⇒ =⎬ −⎪⎡ ⎤ = −⎣ ⎦ ⎭

L

L

P14.3-3 (a) ( )

( )323

F ss

=+

(b) ( ) ( ) ( ) ( )sT sTf t t T F s e t eδ δ− −= − ⇒ = =⎡ ⎤⎣ ⎦L

(c) ( )

( ) ( ) ( )2 2 22

5 558 48 16 254 5

F ss 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 0.5

0.5 ( 0.5) 0.5 ( 0.5) 0.5 0.50.5 0.51 1

sst 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 tT T T T

− −−−⎡ ⎤ ⎡ ⎤− − = − = − = −⎡ ⎤⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

L L LT

s

1

Page 3: Chapter 14 - The Laplace Transform

P14.3-6

( ) ( ) ( ) (5 55 4.23 3

f t t u t t u t⎛ ⎞ ⎛ ⎞= − + − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

)4.2

( ) ( )4.24.2

2 2

15 5 15 5 53 3 3

ss

s eF s e

s s s s

−−

+ −⎛ ⎞ ⎛ ⎞= − + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

P14.3-7

2 22

0 00

3 3(1 )( ) ( ) 3 st s

st st e eF s f t e dt e dtss

− −∞ − − −= = = =

−∫ ∫

P14.3-8

( )5 2 0 20 otherwis

t tf 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 22 2

s ss 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 ses s s s

− −− −

= − − = − − = − − − − −⎡ ⎤ ⎡⎣ ⎦ ⎣

⎡ ⎤⎡ ⎤

⎤⎦

∴ = = − − = − −⎡ ⎤ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

L

2

Page 4: Chapter 14 - The Laplace Transform

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 ss s s s s ss s

+ += = = +

+ + + + + +⎡ ⎤+ + +⎣ ⎦

Bs C+

where

( )21

3 231 3 s

sAs =−

+= =+ +

Then

( )( ) ( ) ( ) 2

22

23 2 43 3 ( )

1 2 4 3 3 31 2 4s 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 3133 3 3 3 3

1 11 3 1 3 1 3

s sF s

s ss s s

− + − += + = + +

+ + 2+ + + + + +

Taking the inverse Laplace transform yields

( ) 2 2 1cos 3 sin 3 , 03 3 3

t t tf t e e t e t t− − −= − + ≥

1

Page 5: Chapter 14 - The Laplace Transform

P14.4-2

( ) ( ) ( )( )2 2 *

3 22 1 2 1

3 4 2 1 1 1 1 1s 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 41 1 1

2 1 3 4 3 21 1 2 21

3 22

s sbs s

s s ja js s j s j

a j

− += =

+ + =−

− + −= =

+ + + −=− +

= − −

=− +

Then

( )3 32 2 42 2

1 1

j jF s

s j s j s

− + − −= + +

1+ − + + +

Next

( ) ( )22 15 23 2 2 and tan 126.9322

m θ −

⎛ ⎞⎜ ⎟

= − + = = =⎜ ⎟⎜ ⎟−⎝ ⎠

°

From Equation 14.5-8 ( ) ( ) ( )5 cos 127 4t tf t e t e u− −⎡ ⎤= + ° +⎣ ⎦ t

P14.4-3

( ) ( ) ( )2 25 1( )

1 21 2 1s A B CF s

s ss s s−

= = + ++ −+ − +

where ( )2

1 2

5 1 5 12 and 12 1s s

s sB Cs s=− =

− −= = =

− +=

Then ( ) ( )( )

221

1

91 12s

s

dA s F sds s=−

=−

−⎡ ⎤= + = =⎣ ⎦ −−

Finally ( )

( ) ( )22

1 2 1( ) 2 + 1 21

t t tF s f t e t e e u ts ss

− −− ⎡ ⎤= + + ⇒ = − +⎣ ⎦+ −+

2

Page 6: Chapter 14 - The Laplace Transform

P14.4-4

( )( ) ( ) ( ) ( ) ( )222

1 111 2 2 1 11 1 1

A Bs CY sss s s ss s

+= = = +

+⎡ ⎤+ + + + ++ + +⎣ ⎦

where 21

1 12 2 s

As s =−

= =+ +

Next

( ) ( ) ( )

( ) ( )

222

2

1 1 1 2 2 ( 1)1 2 21 2 2

1 1 2

B s C s s Bs C ss s ss s s

B s B C s C 2

+= + ⇒ = + + + + +

+ + ++ + +

⇒ = + + + + + +

Equating coefficients: 2s : 0 1 1

: 0 2 1B B

s B C C= + ⇒ = −= + + ⇒ =−

Finally ( )( )

( ) ( )21 cos

1 1 1t tsY s y t e e t u t

s s− −1 + ⎡ ⎤= − ⇒ = −⎣ ⎦+ + +

P14.4-5

( ) ( )( ) ( )

( )( ) ( )2 22

2 3 11 211 2 5 1 4 1 4

s sF 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+3s s+1 2 1 2

A B CF ss s s s

= = + ++ + +

where

( ) ( )( ) ( ) ( ) ( ) ( )

( )0 1 10

2 3 2 33, 1 4

1 2 2s s ss

s sA sF s B s F s

s s s s= =− =−=

+ += = = = + = =

+ + +−

and

( ) ( ) ( )( )2 2

2 32 1

1s s

ss F s C

s s=− =−

++ = =

+=

Finally

( ) ( ) ( ) ( )23 4 1 3 41 2

t tF s f t e e u ts s s

− −−= + + ⇒ = − +

+ +

3

Page 7: Chapter 14 - The Laplace Transform

P14.4-7

( ) ( )( )

( ) [ ] ( ) ( ) ( ) ( ) (

2 2

2 2 1

2 2

12

1 , tan2

cos sin cos cos

j j

at at at

cs ca d c jd c jdF ss 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

ωω ωω

θω ω

ω ω ω θ ω θ

−−

− − −

+ − ⎡ ⎤+ −= = +⎢ ⎥+ − + ++ + ⎣ ⎦

⎡ ⎤= + = + =⎢ ⎥+ − + +⎣ ⎦

⎡ ⎤∴ = − = + + = +⎣ ⎦

P14.4-8 (a)

( ) ( )( )

( )( )

( ) ( )

( ) ( ) ( )

22

2 21

2

2 8 38 3 14 13 2 2 9

3 8 2 2, 8, 3 & 3 7.33

36.33 tan 38.4 , 8 6.33 10.85

810.85 cos 3 42.5t

ssF ss 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 900

So ( ) (3 4) sin 4 . Next, 1 . Fin

s

t s

eF s F ss 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 25

11 1s A B CF s

s ss s s−

= = + +++ +

where

( ) ( ) ( )20 1

5 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

Page 8: Chapter 14 - The Laplace Transform

Then

( )( )2

5 6 41 1

F ss s s−

= + ++ +

Finally ( ) ( )5 6 4 , 0t tf t e t e− − t= − + + ≥

(b)

( )( ) ( ) ( ) ( )

2

3 24

33 3s A B CF s

ss s= = + +

++ + 33s+

where

( ) ( ) ( ) ( )2

3 32 3 3

1 3 4, 32 s s

d dA s F s B s F sds ds=− =−

⎡ ⎤ ⎡ ⎤= + = = + =⎣ ⎦ ⎣ ⎦ 24−

6

and ( ) ( )3

33 3

sC s F s

=−= + =

Then

( ) ( ) ( ) ( )2 34 24 36

3 3 3F s

s s s−

= + ++ + +

Finally ( ) ( )2 34 24 18 , 0tf t t t e t−= − + ≥

5

Page 9: Chapter 14 - The Laplace Transform

Section 14-5: Initial and Final Value Theorems P14.5-1

(a) ( ) ( )

2 2

2 22 3 4 20 lim lim

3 2s s sf sF ss s ss s− +

= = =+ +→∞ →∞

2=

(b) ( ) ( ) 4lim 2

20f sF s

s∞ = = =

P14.5-2

Initial value: ( ) ( ) ( ) 2

2 2

16 160 lim lim lim 14 12 4 12s s s

s s s sv sV ss s s s→∞ →∞ →∞

+ += = =+ + + +

=

Final value:

( )2

2 20 0

16 16lim lim 04 12 4 12s s

s sv ss 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

3 210(0) lim ( ) lim 0

3 2s s

s sv sV ss s s→∞ →∞

+= =+ +

=

Final value:

( ) ( ) ( )( )20 0

10lim lim 10

3 2 1s s

s sv 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: ( ) ( )

2

22 140 lim lim 2

2 10ss

s sf s F ss 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

Page 10: Chapter 14 - The Laplace Transform

Section 14-6: Solution of Differential Equations Describing a Circuit P14.6-1 KVL:

42 1050 0.001 2 tdii vdt

− ×+ + = e for t ≥ 0

The capacitor current and voltage are related by

( )62.5 10 dvidt

−= ×

42 101 = 2e Vtv − × , (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 ss

I s sV s v−

+ − + =+ ×

= × −⎡ ⎤⎣ ⎦

Solving for I(s) yields

( ) ( ) ( ) ( )2 4 8

4 44 4 4

1.4 10 1.6 1010 2 10 4 1010 2 10 4 10

s s A BI ss s ss s s

+ × − ×= = +

+ + × + ×+ + × + × 4C

+

where

( ) ( ) ( )( )

( ) ( ) ( ) ( )

( ) ( )

2 4 8 84

4 84 4= 104

= 10

2 4 8 84

4 84 4= 2 104

= 2 10

24

4= 4 10

1.4 10 1.6 10 2 10 2 103 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

ss

ss

s

s sA s I ss s

B s I s

C s I s

−−

− ×− ×

− ×

+ × − × − × −= + = = =

×+ × + ×

+ × − × ×= + × = = =

×+ + ×

+ ×= + × = ( ) ( )

4 8 8

84 44

= 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

Page 11: Chapter 14 - The Laplace Transform

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

1i

× −= =

KCL yields C C3 10

100d v v

idt

− + =

Apply Ohm’s law to the 1 Ω resistor to get C

C 1

v vi 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 ss

×− + = −

+ +

so

( )7

2 2

160 1600s 2.5 10( )1010 1010 (400)

I ss s s

− ×= +

+ ⎡ ⎤+ +⎣ ⎦

Next

( )7 *

2 2

1600 2.5 10 1010 400 4001010 (400)

s A Bs s j s js s

− ×= + +

+ + −⎡ ⎤+ +⎣ ⎦

B

where

( )

7

22

= 1010

1600 2.5 10 23.1400

s

sAs

− ×= =+

− ,

( ) ( )7 7

*

5 = 400

1600 2.5 10 2.56 10 1.4 11.5 27.2 and 11.5 27.21010 400 8.69 10 68.4s j

s xB js s j

°

°

− × ∠= = = −

+ − × ∠B j= +

Then

( ) 136.9 11.5 27.2 11.5 27.2 1010 400 400

j jI ss 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

Page 12: Chapter 14 - The Laplace Transform

P14.6-3

C (0) 0v =

3c

cc3 c

15 10 10cos 22 20cos 21 10

30

v i td v v td v dti

dt−

⎫+ × =⎪ ⇒ + =⎬⎛ ⎞= ×⎜ ⎟ ⎪⎝ ⎠ ⎭

Taking the Laplace Transform yields:

( ) ( ) ( ) ( )( )( )

*

C C C C2 2

200 2 204 22 4

s s AsV s v V s V ss ss s

− + = ⇒ = = + +2 2

B Bs 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 js s s j j− −

−= = = − = = = + = −

+ + − − 2B j

Then

( ) ( ) ( )2C C

5 5 5 55 2 2 2 2 5 5 cos 2 sin 22 2 2

tj j

V s v t e t ts s j s j

−+ −−= + + ⇒ = − + +

+ + − V

P14.6-4

L cc L L12 2 8 and

d i d vv i i Cdt dt

− + + = − = −

Taking the Laplace transform yields

( ) ( ) ( ) ( )L Lc L812 2 0V s I s sI s is

− + + − = −⎡ ⎤⎣ ⎦

( ) ( ) ( )c cL 0I s C sV s v= − −⎡ ⎤⎣ ⎦

c L(0) 0, (0) 0v i= =

14-3

Page 13: Chapter 14 - The Laplace Transform

Solving yields

( )c2

4 6

2

CV sCs s s

=⎛ ⎞+ +⎜ ⎟⎝ ⎠

(a) 1 F18

C = ( )( ) ( )2 2

72 33 3

cca bV s

s ss s s= = + +

++ +

( )( )2

248 88, 8, and 243 3

ca b c V ss s s

−−= = − = − ⇒ = + +

+ +

( ) 3 3 8 8 24 V, 0t tcv t e t e t− −= − − ≥

(b) 1 F10

C = ( ) ( ) ( )40 1 5 1 5c

ca bV ss s s s s s

= = ++ + + +

+

( )C28 108, 10, and 2

1 5a b c V s

s s s−

= = − = ⇒ = + ++ +

( ) 5 8 10 2 V, 0t t

cv t e e t− −= − + ≥ P14.6-5

( )c L(0 ) 10 V, 0 0 Av i− −= =

( )6 cc5 10 and 400 1 0d v d ii i

dt dt−= × + + =v

Taking Laplace transforms yields

( ) ( ) ( )( )( ) ( )( ) ( )

( )( )

6c

22 5 2c

1 4005 10 10 10 40400 2 10400 0 0 200 400

I s sV sI s

s sI s s I s V s s

−⎛ ⎞−⎜ ⎟⎫= × − −⎪ ⎝ ⎠⇒ = =⎬ + + ×+ − + = + +⎪⎭

so

( ) ( ) ( )2001 sin 400 A40

ti t e t u t−= −

14-4

Page 14: Chapter 14 - The Laplace Transform

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 ( ) ( ) ( ) ( ) ( )2

1 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 t1R C L dt R C L R C Ldt

⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

With the given values: ( ) ( ) ( )2

2 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 sdt 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 || 4v − = =

+

Then, using current division

Page 15: Chapter 14 - The Laplace Transform

( ) ( )040 0.328 A16 4 9

vi

−⎛ ⎞− = =⎜ ⎟+⎝ ⎠

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 idt 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 ss

⎡ ⎤− + + − + =⎡ ⎤⎣ ⎦⎣ ⎦

( ) ( ) ( )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

sI s

s s s

ss 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 00.8 23.6 0.471 A for 0t

ti t

e t t−

≤⎧= ⎨ + − ≥⎩

(checked using LNAP 10/11/04)

Page 16: Chapter 14 - The Laplace Transform

P14.6-7

KCL: 1 675

tvi e−+ =

KVL: 1 14 3 0 4di dii v v idt dt

3+ − = ⇒ = +

Then 6 64 3 357 2

5 4t t

di i didt i e i edt

− −+

+ = ⇒ + =

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 ss 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 Bs s s s s s=− = −

= + = = = = −+ + + + + + 6

1 1

Then 2 635 3535 1 35 1( ) ( ) A, 0

16 2 16 6 16 16t tI s i t e e

s s− −= − ⇒ = −

+ +t ≥

P14.6-8 Apply KCL at node a to get

1 2 1 11 2

1 2 248 24

d v v v d vv v

dt dt−

= ⇒ + =

Apply KCL at node b to get

2 2 1 2 2 21 2

50cos 2 1 0 3 6020 24 30 24

v t v v v d v d vv 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

( ) ( )2

1 2 1 2 225 60 1002 ( ) 2 ( ) 10 and ( ) 3 ( )

4s ss V s V s V s s V s

s+ +

+ − = − + + =+

Solve these equations using Cramer’s rule to get

Page 17: Chapter 14 - The Laplace Transform

( )( )

( )( )( ) (

( )( )( ))

( )( )( )

2

2 22

2 2

3 2

2

25 60 1002 10 2 25 60 100 10 442 (3 ) 2 4 1 4

25 120 220 2404 1 4

s ss s s s ssV s

s s s s s

s s ss s s

⎛ ⎞+ ++ +⎜ ⎟ + + + + ++⎝ ⎠= =

+ + − + + +

+ + +=

+ + +

Next, partial fraction expansion gives

( )*

2V2 2 1 4

A A B Css j s j s s

= + + ++ − + +

where

( ) ( ) ( )

( ) ( )

( ) ( )

3 2

2

*

3 2

21

3 2

24

25 120 220 240 240 240 6 61 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 js s s j

A j

s s sBs s

s s sCs s

=−

=−

=−

+ + + − −= =+ + − −

= −

+ + += = =

+ +

+ + + −= = =−+ +

= +

Then

( )26 6 6 6 23 3 16 3

2 2 1j jV s

s j s j s s 4+ −

= + + ++ − + +

Finally 4

223 16( ) 12cos 2 12sin 2 V 03 3

t tv t t t e e t− −= + + + ≥

Page 18: Chapter 14 - The Laplace Transform

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 ss s s s s

ti t

e t−

− −= = = −+ +

− <⎧= ⎨ − >⎩

+

P14.7-2

( )LLL

L

L

8003

L

10 8( ) ( ) .015( ) 30 ( 8002000 4000 53

8( ) 0.15 0.003 0.005 0.00215( ) 8008005

33

( ) 5 2 mA, 0

L

t

V s V sV ss V ss s

V sI ss s s ss s

i t e t−

− − −= + + ⇒ =

+

+= = + = −

⎛ ⎞ ++⎜ ⎟⎝ ⎠

= − >

)

14-1

Page 19: Chapter 14 - The Laplace Transform

P14.7-3

6

1000

8( )( )0.006 0102000.5

6000 8500 ( ) 0.5 ( ) 0

8 12000 12 4( )( 1000) 1000

( ) 12 4 V, 0

cc

c c

c

tc

V sV s ss

s

V s s V ss s

sV ss s s s

V t e t−

−− + + =

⎛ ⎞− + + − =⎜ ⎟⎝ ⎠

+= = −+ +

= − >

P14.7-4

( )

6

1500

6( ) ( ) 0.5 8( ) 02000 4000 10

6 8500 ( ) 250 ( ) 0.5 ( ) 0

6000 8 4 4( )1500 1500

( ) 4 4 V, 0

cc

c

c c c

c

tc

V s V s ss V ss

V s V s s V ss s

sV ss s s s

v t e t−

− ⎛ ⎞ ⎛ ⎞+ + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+= = +

+ +

= + >

14-2

Page 20: Chapter 14 - The Laplace Transform

P14.7-5

Node equations:

( ) ( ) ( ) ( ) ( )a C aa C

1 66 6

V s V s V sV s V s

s s s−

+ = ⇒ = +6

6s+ +

( ) ( ) ( )( )

C CC

C

6 661 36 63 0

4 4

V s V sV s ss ss V ss s

⎛ ⎞− +− ⎜ ⎟+ +⎝ ⎠+ + + +2

− =

)

After quite a bite of algebra:

( ) ( )( )(2

C6 56 132

2 3s sV s

s s s+ +

=5+ + +

Partial fraction expansion:

( )( )( )2

44 16 56 132 93 3( )c 3 2 5 2 3

s sV ss 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

Page 21: Chapter 14 - The Laplace Transform

P14.7–6

Write a node equation in the frequency domain:

( ) ( ) ( )

2 2

1 1o oo

1 2

22

1010 5 10 5 105 01 11

1

R Rs

R C RV s V ss C V sR R s ss sCs R CR C

R⎡ ⎤

+ −⎢ ⎥⎢ ⎥+ − + = ⇒ = − = − +⎢ ⎥⎛ ⎞ +⎢ ⎥+⎜ ⎟⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠

Inverse Laplace transform:

( ) 22 2 1000o

1 110 5 10 10 5 V for 0t R C tR R

v t e e tR R

− −⎡ ⎤⎛ ⎞

⎡ ⎤= − + − = − − >⎢ ⎥⎜ ⎟ ⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

14-4

Page 22: Chapter 14 - The Laplace Transform

P14.7-7 Here are the equations describing the coupled coils:

( ) ( )

( ) ( )

1 21 1 1 1 2 1 2

2 12 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 sdt dtdi div t L M V s s I s sI s sI s sI sdt 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 Is sV 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):

( ) ( )( )2 215 8 3 1.6 0.64 2.36 = +

0.26 1.54 + 0.26 + 1.545 9 2s sI s

s s s ss s+ +

= =+ ++ +

Taking the inverse Laplace transform:

0.26 1.542( ) 0.64 2.36 A for 0t 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 s

s s+ − + + = ⇒ = −

14-5

Page 23: Chapter 14 - The Laplace Transform

( ) ( ) ( )( ) ( ) ( )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

12 2 2 6 26 6 13 3

2

I s I s I ss s s s

⎛ ⎞⎛ ⎞+ − − = ⇒ =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ +

Calculate for Vo(s):

( ) ( )o 2

11 6 1 6 22

1 12 22 2

V s I ss ss s

⎛ ⎞⎜ ⎟ −

= − = − =⎜ ⎟⎜ ⎟+ +⎝ ⎠

4s

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:

( ) ( ) ( )

2612 3 354 5 30 0 44

55

ss 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

Page 24: Chapter 14 - The Laplace Transform

P14.7-10 Steady-state for t<0:

From the equation for vo(t):

( ) ( )o

26 12 6 Vv e− ∞∞ = + =

From the circuit:

( ) ( )o3 18

3v

R∞ =

+

Therefore:

( )36 18 63

RR

= ⇒ =+

Ω

Steady-state for t>0:

( ) ( )1 18 6 62 0 12

I s I sC s s s s

C

⎛ ⎞ −+ + − = ⇒ =⎜ ⎟

⎝ ⎠ +

( ) ( )o1 18 1 6 18 12 12 18 12

1 12 2

V s I sCs s Cs s s s ss s s

C C

⎛ ⎞⎜ ⎟− −

= + = + = + + =⎜ ⎟⎜ ⎟+ +⎜ ⎟⎝ ⎠

61

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 F2

CC

= ⇒ = .

(Checked using LNAP, 12/29/02)

14-7

Page 25: Chapter 14 - The Laplace Transform

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

( )o10 5

100V s

s s= +

+

Next, we determine from the 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 00 0

vv

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 ss s

Rs

− −=

Solving gives

( )

6

16 6

1 1

103 22 1

10 10

sR

V ss

s s sR R

+

= = +⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Apply KCL at the inverting input of the op amp to get

Page 26: Chapter 14 - The Laplace Transform

( ) ( ) ( ) ( ) ( )o 2o 62

1

2 11 11000 1000 1000 10

V s V s R RV sV s V s

R ss

R

⎛ ⎞⎜ ⎟

− ⎛ ⎞ ⎛ ⎞⎜ ⎟= ⇒ = + = + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎛ ⎞⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

2

The expressions for Vo(s) must be equal, so

26

1

10 5 2 11100 1000 10

Rs s s

sR

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟+ = + +⎜ ⎟⎜ ⎟+ ⎛ ⎞⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Equating coefficients gives

221 5 4

1000R

R+ = ⇒ = kΩ and 6

11

10 100 10 kRR

= ⇒ = 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

C

V s V sA B B C sV ss R s L s+

= + + +

so

Page 27: Chapter 14 - The Laplace Transform

( )2

CA L s R R LC s V ss R L s

+ +=

Then

( )C 22 1 1

AA R L CV s

R LC s L s R s sRC LC

= =+ + + +

and

( ) ( )CL

2 1 1

AV s B BLCI s

L s s ss s s

RC LC

= + = +⎛ ⎞

+ +⎜ ⎟⎝ ⎠

a.) When 12 , 4.5 H, F, 5 mA and 2 mA9

R L C A B= Ω = = = = − , then

( ) ( )L 2

5 4010 2 3 7 7

144.5 22

I ss s ss s s s

−= + = +

++ +−

+

Taking the inverse Laplace transform gives

( ) 4 0.5L

5 53 mA for 07 7

t 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 1510 25 5 5

I ss 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 0.110 125 5 10 5 10 5 10

s sI ss 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−= − − + ≥

Page 28: Chapter 14 - The Laplace Transform

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.

( )2Ai tR

= and ( )C 2Av t =

Consequently,

( )02AiR

− = and ( )C 02Av − =

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 s

R s− + − − =

( ) ( ) ( )( )C L C1 0

2AI s R I s I s

C s s− + − =

Or, in matrix form

( )( )

L

C

21

2

A L AL s R R

I s R sR R I s A

C ss

⎛ ⎞++ −⎛ ⎞ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟− + ⎜ ⎟⎜ ⎟⎝ ⎠ −⎝ ⎠ ⎜ ⎟⎝ ⎠

( )( )

( )C

22

2 2 21 11

A A L A AL s R Rs R s LI s

s sL s R R R RC LCC s

⎛ ⎞ ⎛ ⎞+ − + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= =⎛ ⎞ + ++ + −⎜ ⎟⎝ ⎠

Page 29: Chapter 14 - The Laplace Transform

a.) When 13 , 2 H, F and 12 V24

R L C A= Ω = = = ,

( ) ( )( )C 2

3 33 2 4 4

8 12 2 6 2 8I s

s s s s s s= = = −

+ + + + + +.

Taking the inverse Laplace transform gives

( ) ( )2 6C

3 3 A4 4

t ti t e e u t− −⎛ ⎞= −⎜ ⎟⎝ ⎠

b.) 12 , 2 H, F and 12 V8

R L C A= Ω = = = ,

( )( )C 22

3 34 4 2

I ss s s

= =+ + +

Taking the inverse Laplace transform gives

( ) ( )2C 3 ti t t e u t−= A

c.) 110 , 2 H, F and 12 V40

R L C A= Ω = = =

( )( ) ( )C 2 22

3 3 3 44 20 42 16 2 1

I ss 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)

Page 30: Chapter 14 - The Laplace Transform

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 08 4 4v 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 0.758 4 40

V s V s ss V s V ss

− ⎛ ⎞+ + + =⎜ ⎟⎝ ⎠

Solving gives

( ) ( ) ( ) ( )10 10 4 2 2 4 1 15 4 2

5 5 5 5s 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.

Page 31: Chapter 14 - The Laplace Transform

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 3

2 00 0.010 10 40 10

i −− = =

× + ×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

CC C63 3

12 1.612 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 ss s

s

− − ⎛ ⎞ ⎛ ⎞+ + = ⇒ − + − +⎜ ⎟ ⎜ ⎟×× × ⎝ ⎠ ⎝ ⎠C =

( )( ) ( ) ( ) ( )C C48 80 48 1.6 600 9.6 80.08 5 0.128

0.08 5 62.5 62.5s 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 C

50 50 9.6 8 12 10 V for 040 40

t tv t v t e e t− −= − = − − = − + ≥

so

( )o 62.5

2 V for 012 10 V for 0t

tv t

e t−

− ≤⎧= ⎨− + ≥⎩

(checked using LNAP 10/11/04)

Page 32: Chapter 14 - The Laplace Transform

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 5 032 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

( )oc32 20 30 20 5 4 1 V

32 96 120 30V s

s s s−⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ s

=

Page 33: Chapter 14 - The Laplace Transform

( ) ( )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

0.7580 48 8048

s sI ss

s

−= =

++

( ) ( )80 0.25 80 0.75 0.2548 80

V s I ss 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 1.67V 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

tv t

e t−

≤⎧= ⎨ − ≥⎩

(checked using LNAP 7/1/04)

Page 34: Chapter 14 - The Laplace Transform

P14.7-17

Mesh Equations:

( ) ( ) ( )( ) ( ) ( )4 1 4 16 0 62 2C C C 6I s I s I s I s I

s s s s⎛ ⎞− − − + = ⇒ − = + +⎜ ⎟⎝ ⎠

s

( ) ( )( ) ( ) ( ) ( ) ( )106 3 4 09C C CI s I s I s I s I s I s− + + = ⇒ = −

Solving for I C(s):

( ) ( )4 2 1 633 24

C CI s I ss s s

⎛ ⎞− = − + ⇒ =⎜ ⎟⎝ ⎠ −

So Vo(s) is ( ) ( )o244 3

4

CV s I ss

= =−

Back in the time domain: ( ) 0.75

o 24 ( )tv t e u t= V for t ≥ 0 P14.7-18

KVL:

( )8 204 8 4 Ls I ss s

⎛ ⎞+ = + +⎜ ⎟⎝ ⎠

so

( ) ( )( )22

1 122 5 1 4

L

ssI ss s s

+ ++= =

+ + + +

Taking the inverse Laplace transform:

( ) ( )L1cos 2 sin 2 A2

t ti t e t e t u t− −⎛ ⎞= +⎜ ⎟⎝ ⎠

Page 35: Chapter 14 - The Laplace Transform

Section 14-8: Transfer Function and Impedance P14.8-1

1

2 111 2

1 2 1 1 2 21

1

( ) where and1 1 1

RZ RC sH s Z Z 2R

Z Z R C sRC s

= = =R C s

=+ + ++

( ) ( )( ) ( )

( ) ( )( ) ( )

2 11 1 1 2 2 2

1 2 1 2

2 22 2

1 2 1 2

1Let and then

1 1

1When constant, as required.

1

R sR C R C H s

R s s R

R s RH sR R s R R

ττ τ

τ τ

ττ τ τ

τ

+= = =

+ + +

+= = ⇒ = = =

+ + +

1 1 2 2we require R C R C∴ = P14.8-2

1 21Let and then the input impedance isZ R Z R Ls

Cs= + = +

( )( ) 2

1 22

1 2

1 1 1 2 1

LR R Ls LCs RC sZ Z Cs RZ s R

Z Z LCs RCsR R LsCs

⎛ ⎞⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟= = =+ +⎜ ⎟+ + + ⎜ ⎟

⎝ ⎠

+

+

2Now require : 2 then LRC RC L R C ZR

+ = ⇒ = = R

P14.8-3 The transfer function is

( )

2

2 1

2 1 2

21

2 1

11

1

RR C s R C

H s R R RR s

R C s R R C

+= = +

+ ++

Using 1 22 , 8 and 5 FR R C= Ω = Ω =gives

( ) 0.10.125

H ss

=+

The impulse response is ( ) ( ) ( )0.1250.1 Vth t H s e u t-1L −⎡ ⎤= =⎣ ⎦ . The step response is

1

Page 36: Chapter 14 - The Laplace Transform

( )

( ) ( ) ( )0.1250.1 0.8 0.8 0.8 1 V0.125 0.125

tH se 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:

( ) ( )( )

42 2

12 12128 164

tH s t e u ts ss

−⎡ ⎤= = =⎣ ⎦ + ++L

The Laplace transform of the step response is:

( )( ) ( )2 2

312 34

44 4H 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 4 124 4

s s ks s k s k s k⎛ ⎞⎟⎜= + − + + = + + + + ⇒ =−⎟⎜ ⎟⎟⎜⎝ ⎠

34

The step response is ( ) ( )1 43 33 V

4 4tH s

e t u ts

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 c110000

RV s V s⎛ ⎞⎟⎜= + ⎟⎜ ⎟⎟⎜⎝ ⎠

Now, writing node equations,

2

Page 37: Chapter 14 - The Laplace Transform

( ) ( ) ( ) ( ) ( ) ( )c i o a o

c 0 and 01000 5000

V s V s V s V s V sCsV s

Ls− −

+ = + =

Solving these node equations gives

( )

1 500011000 10000

1 50001000

RC LH s

s sC L

⎛ ⎞⎟⎜ + ⎟⎜ ⎟⎜⎝ ⎠= ⎛ ⎞⎛⎟ ⎟⎜ ⎜+ +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝⎞⎠

Comparing these two equations for the transfer function gives

( ) ( )1 12000 or 50001000 1000

s s s sC C

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

( ) ( )5000 50002000 or 5000s s s sL L

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

1

10001

100005000

15 106

CR

L+

⎛⎝⎜

⎞⎠⎟ = ×

The solution isn’t unique, but there are only two possibilities. One of these possibilities is

( )1 2000 0.5 F1000

s s CC

μ⎛ ⎞⎟⎜ + = + ⇒ =⎟⎜ ⎟⎟⎜⎝ ⎠

( )sL

s L+⎛⎝⎜

⎞⎠⎟ = + ⇒ =

50005000 1 H

( )6

61 50001 15 10

10000 11000 0.5 10R R

⎛ ⎞⎟⎜ + = × ⇒ = Ω⎟⎜ ⎟⎟⎜⎝ ⎠×5 k

(Checked using LNAP, 12/29/02)

3

Page 38: Chapter 14 - The Laplace Transform

P14.8-6 The transfer function of the circuit is

( )

2

2 1

1

2

11

1

RR C s R C

H sR 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 1250 250 250

tH se u t H s

s s s s s− − −⎛ ⎞= − − = − − = ⇒ =⎜ ⎟+ +⎝ ⎠

Ls +

Comparing these results gives

( )2 62

1 1 1250 40 k250 250 0.1 10

RR C C −

= ⇒ = = =×

Ω

( )1 61

1 1 11000 10 k1000 1000 0.1 10

RR C C −

= ⇒ = = =×

Ω

(Checked using LNAP, 12/29/02)

P14.8-7

( ) ( ) ( )a i4 2

4 2 2V s V s V s

s s⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

i

( ) ( ) ( ) ( ) ( )o b b a

122 2 2 25 512 2 2 2 26

sV s V s V s V s V ss s s s

s

⎛ ⎞⎜ ⎟ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎝ ⎠

i

The transfer function is:

( ) ( )( ) ( )

o2

i

202

V sH s

V s s= =

+

4

Page 39: Chapter 14 - The Laplace Transform

The Laplace transform of the step response is:

( )( ) ( )o 2 2

20 5 5 1022 2

V ss 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 44 6

1 44 66

CCs LH s k kLs 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

( ) ( )( )123 2

H ss s s

=+ +

Comparing the two representations of the transfer functions let 1 13 F6 1

CC= ⇒ =

8,

4 2 2LL= ⇒ = H and . 2 3 12 2 V/Vk k× × = ⇒ =

(Checked using LNAP, 12/29/02)

P 14.9.9 From the circuit:

5

Page 40: Chapter 14 - The Laplace Transform

( ) ( )( )

o

i1212

RsV s R L s LH s RV s R L s sL

++= = =

++ + +

From the given step response:

( ) ( ) ( ) ( ) ( )4 0.5 0.5 2 20.5 14 4

tH 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

RLL L R

R LL

⎫= ⎪ +⎪ ⇒ = ⇒ = =⎬+ ⎪=⎪⎭

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 sCs Cs Cs

R R I s I sCs Cs

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

2

Solving for I2(s): ( )( )

2

1 2

1

2 12( )

V sCsI s

R RCs Cs Cs

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1

Then gives ( ) ( )o 2V s R I s=

( ) ( )( ) [ ] [ ]

( )

0

12 1

1 22 21 1

2 2 1 1 4 122 2

V s RCs sH sV s R Cs RCs RC R CR C s s

RR C RR C

= = =+ + − ⎡ ⎤+⎢ ⎥+ +

⎢ ⎥⎣ ⎦

6

Page 41: Chapter 14 - The Laplace Transform

P14.8-11 Let

2

1

1

1 1

x x

RRCsZ

RCsRCs

Z R L s

⎛ ⎞⎜ ⎟⎝ ⎠= =

++

= +

Then

( )

( )

2 22

1 1 2 x x x xx x

2 x

x x21 x

x x

1

11

RV Z RRCs

RV Z Z L RCs L R RC s R RR L sRCs

V L CL R RCV R Rs s

L RC L RC

+= = =+ + ++ +

+

=+ +

+ +

+ +

P14.8-12

Node equations:

( ) ( )1 out1 in 1 11 1

1

out11 2 2 out

2

2

0 1

0 1

V VV V sC R C s V R C sV VR

VV V R C sVR

sC

−− + = ⇒ + = +

− − = ⇒ = −

1 1 in out

Solving gives:

( ) 1 1 2 2out2

2in 1 2 1 2 2 2

1 1 1 2 1 2

1

1 1 1

sR C s R CVH s

V R R C C s R C s s sR C R R C C

−−

= = =+ + + +

7

Page 42: Chapter 14 - The Laplace Transform

P14.8-13

Node equations in the frequency domain:

1 11

1 2 3

0iV V V VVR R R

0− −+ + =

01

1 2 3 3

1 1 1 iV VV1R R R R R

⎛ ⎞⇒ + + − =⎜ ⎟

⎝ ⎠

1

2 0 1 2 2 02

0V sC V V sC R VR−

− = ⇒ = −

After a little algebra:

( ) 0 3

2 2 3 2 1 3 2 1 2 1i

V RH sV sC R R sC R R sC R R R

−= =

+ + +

P14.8-14

( )( ) 2

1 1

( ) 1 1o

i

V s Cs LCH s RV s Ls R s sCs L LC

= = =+ + + +

L, H C, F R, Ω H(s)

2

0.025

18 ( )( )2

20 209 20 4 5s s s s

=+ + + +

2

0.025

8 ( )22 2

20 204 20 2 4s s s

=+ + + +

1

0.391

4 ( )( )2

2.56 2.564 2.56 0.8 3.2s s s s

=+ + + +

2

0.125

8 ( )22

4 44 4 2s s s

=+ + +

8

Page 43: Chapter 14 - The Laplace Transform

a) ( ) ( )( )20

4 5H 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 ts s

H ss s s s s s s

e e u t

− −

− −

= − ⇒ = −+ +

−= = = + +

+ + + +

+ −

L

L ⇒

b) ( )( )2 2

202 4

H ss

=+ +

( ) ( ) 22 2

5(4) (s) 5 sin 4 ( )( 2) 4

th t H h t e t u ts

−= = ⇒ =+ +

L

( ) ( ) ( )

( )( )

( )( )

1 22 2

2 21 2 1 2

1 2

2 22

2

( ) 20 1 step response( 4 20) 4 20

20 4 20 1 4 201, 4

1 421 2 step response2 42 4

1step response 1 cos 4 sin 4 ( )2

t

H s K s Ks s s s s s s

s s s K s K s K s KK K

ss ss

e t t u t−

+= = = +

+ + +

= + + + + = + + + +

⇒ = − = −

−− += + +

+ ++ +

⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

L

L

+

)

c)

( ) ( )(2.56

0.8 3.2H 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 3.21 4step response 1 ( )3 3

t t

h t H s h ts s

H ss s s s s s s

e e u t

− −

− −

= = − ⇒ = −+ +

= = = + ++ + + +

⎛ ⎞= + −⎜ ⎟⎝ ⎠

L

L

d) ( )( )2

42

H ss

=+

( )( )( )

2

2

4 ( )

step response 1 1 2 ( )

t

t

h t te u t

t e u t

=

= − +

9

Page 44: Chapter 14 - The Laplace Transform

P14.8-15 For an impulse response, take ( )1 1 V s = . Then

( ) ( )( ) ( )

*

0

3 23 2 3 2 3 2 3 2

s A B BV ss 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

( )00.462 0.47 119.7 0.47 119.7

3 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

( ) ( )o10 1.5 3.75 V4

v 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 o3 3

4 30 04 10 10 10 10 10j

ω ω∠ °+ +

× − × ×V V

3 =

or ( ) ( )o10 30 1 0j ω∠ °+ + =V

( )o10 30 7.07 165

1 jω ∠ °

= − = ∠ °+

V

Finally,

10

Page 45: Chapter 14 - The Laplace Transform

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 o3 3

6

1

014 10 10 1010

V s V ss

s

+ +× ××

=

( ) ( )3

o10 100 04

s V ss+ + =

( ) ( )o250 2.5 2.5

100 100V s

s s s s−

= = ++ +

Finally, ( ) ( ) ( )1002.5 1 Vt

ov 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,

( )( )

( )

22

2 22

2

11 || 1 1

R L s R L sC sR L sC s C L s C R sR L s

C s

× + ++ = =

+ ++ +

Next, using voltage division,

11

Page 46: Chapter 14 - The Laplace Transform

( ) ( )( ) ( )

22

2 2o2

2i 2 1 212

2

2

1 12

1 2 1 22

1 1

11

1

2 44 29

R L sC L s C R s R L sV s

H s R L sV s R L s R C L s C R sRC L s C R s

RsR C R LC s

L R R C R R s ss sR LC R LC

++ + +

= = =+ + + + ++

+ +

++

= =+ + + ++ +

Using ( )i1V ss

= gives

( ) ( )( )

( )

( ) ( )

o 22

2 2

2 22 2

2 4 0.1379 0.1379 1.44834 294 29

0.1379 0.1379 1.44832 5

0.1379 2 50.1379 0.34492 5 2 5

H s s sV ss s s ss s s

ss s

ss s s

+ − += = = +

+ ++ +

− += +

+ +

+= − +

+ + + +

Taking the inverse Laplace transform

( ) ( ) ( )( )( )

2o

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 1.66750 20 50 20

100050 20

H s s s s s s sV s

s s s s s s s

s s s

+ + + + − += = + − =

+ + + +

=+ +

Consequently,

( ) ( )( ) ( )( )

o

i

100050 20

V sH s

V s s s= =

+ +

12

Page 47: Chapter 14 - The Laplace Transform

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 sV 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 sL s R L s R K L s

= ⇒ =+ + + i

K

Then, the transfer function of the circuit is

( ) ( )( ) ( )( )

1 2o

i 2 1

2 1

R KL LV s R KH s

V s L s R L s K R Ks sL L

= = =⎛ ⎞⎛+ +

+ +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝

⎞⎟⎟⎠

Comparing the two transfer functions gives

( )( ) ( ) 1 2

2 1

100050 20

R KL L

H ss s R Ks s

L L

= =+ + ⎛ ⎞⎛

+ +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝

⎞⎟⎟⎠

We require 1 2

1000 R KL L

= and either

2

50 RL

= and 1

20 KL

= or 2

20 RL

= and 1

50 KL

= . 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

Page 48: Chapter 14 - The Laplace Transform

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,

( )222

2 22

2

111||

11

R L s R C L sC sR L s

C s C L s C R sR L s

C s

⎛ ⎞× +⎜ ⎟ +⎛ ⎞ ⎝ ⎠+ = =⎜ ⎟ + +⎛ ⎞⎝ ⎠ + +⎜ ⎟⎝ ⎠

Next, using voltage division twice,

( ) ( )( )

( )

( ) ( )

( )

( )

22

22 2o

2 2i 2 1 2 1 2 1 2

122

2

1 22

1 22

1 2

1 11

111

81 10

R C L sC 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

RR R LC

R R s ss sLCR R L

+

+ += = × =

+ + + + ++++ +

+= =

+ ++ +

+16

Using ( )i1V ss

= gives

( ) ( )( ) ( )( )o 2

2 118 8 3 62

2 8 210 16H s

V ss s s s s ss s s

−= = = = + +

8s+ + ++ + +

Taking the inverse Laplace transform

( ) ( )2 8o

1 2 1 V2 3 6

t tv t e e u t− −⎛ ⎞= − +⎜ ⎟⎝ ⎠

(checked using LNAP 10/15/04)

14

Page 49: Chapter 14 - The Laplace Transform

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 16 805 5 5

H s s s s sI s

s s s s s s

⎛ ⎞ + − + += = − + = =⎜ ⎟

⎜ ⎟+ + +⎝ ⎠25s s +

Consequently,

( ) ( )( ) ( )

o2

i

805

I sH 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

11

11

1

11|| 1 1

R RC sRC s R C sR

C s

×= =

++

Next, using voltage division,

( ) ( ) ( )

1

1 1a i

1 1 2 1 22

1

1

1

RR C s R

V s V s V sR R R R R C sRR C s

+= =

+ +++

i

( ) ( ) ( ) ( )( ) ( ) ( )1 2ao o i

3 3 1 2 1 2 3 1 2

1 2

KK 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 sL R R C

= ⇒ = =+ ⎛ ⎞+ + + +⎛ ⎞

+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

V s

Then, the transfer function of the circuit is

15

Page 50: Chapter 14 - The Laplace Transform

( ) ( )( )

2o

i 3 1

1 2

KR C LI s

H sV s 2R R R

s sL R R C

= =⎛ ⎞+⎛ ⎞

+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Comparing the two transfer functions gives

( )( ) 2

23 1

1 2

805

KR C L

H s2R R Rs

s sL R R

= =⎛ ⎞++ ⎛ ⎞

+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠C

We require

( )1 2

1 2

40 105 240 10

R RC

R R C C+ +

= = ⇒ =×

5 mF ,

3 205 4 HR

LL L

= = ⇒ =

and

( )2

80 80 V/V10 0.025 4

K K KR 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 2.56 133 2 3 2 3 2

s sH ss ss s s

+ += − = =

13s ++ ++ + + + + +

The Laplace transform of the step response is

( )

( ) ( ) ( ) ( )2 222 2 2

6 13 1 1 1 3 3 26 13 26 13 3 2 3 2 3 2

H s s s s ss s s s s ss s s s s s

+ += = − = − = − + ×

+ ++ + 2 2+ + + + + +

16

Page 51: Chapter 14 - The Laplace Transform

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 6 933 3

H s ss s s ss s

⎡ ⎤ += − + = − =⎢ ⎥

++ +⎢ ⎥⎣ ⎦23s s +

The transfer function is

( )( )2

93

H ss

=+

Taking the inverse Laplace transform gives the impulse response:

( ) ( )3

o 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

22

22

2

11|| 1 1

R RC sRC s R C sR

C s

×= =

++

Next, using voltage division,

( ) ( ) ( ) ( )( )( ) ( )3 2 33 3 2 3

o a a2 2 3 2 3 1 2 3 2 3

321

K R R R C sR R R R C sV 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

Page 52: Chapter 14 - The Laplace Transform

Then, the transfer function of the circuit is

( ) ( )( )

( )( )( )

2o

i 1 2 3

2 3

15 0.5

5 2.

K sL R CV s s

H sV s s sR R R

s sL R R C

⎛ ⎞+⎜ ⎟⎜ ⎟ +⎝ ⎠= = =

+ +⎛ ⎞+⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

5

Using ( )i1V ss

= gives

( ) ( ) ( )( )( )o

5 0.5 0.2 1.8 1.65 2.5 5 2.5

H s sV 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

Page 53: Chapter 14 - The Laplace Transform

Section 14-9: Convolution Theorem P14.9-1

( ) ( ) ( ) ( ) ( ) ( ) 1 11 1s se ef t u t u t F s u t u t

s s s

− −−= − − ⇒ = − − = − =⎡ ⎤⎣ ⎦L

( ) ( ) ( )

( ) ( ) ( ) ( ) (

2 21 2 1 1

2

1 1 2*

2 1 1 2 2

s s se e ef t f t F ss 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 ss s

= − − ⇒ = −⎡ ⎤⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( ) (2 4

1 12 2 2

4 8 4 4 8 2 2 4 4s 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

12 1 1

2 1 2 2

/2

1

1 1

1 1

1 11 1

1 1

1 1 , 0t RC

v t t u t V ss

V s Cs RCH sV s R s

Cs RC

v t h t v t V s H s

1RC RC RV s V s H s

s s ss sC

RC R

v t t e tRC

= ⇒ =

= = =+ +

⎡ ⎤= ∗ = ⎣ ⎦⎛ ⎞ − −⎜ ⎟⎛ ⎞= = = + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎜ ⎟+ +⎝ ⎠

= − − ≥

L

C

1

Page 54: Chapter 14 - The Laplace Transform

P14.9-4

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

12

22

2 2

( )

2 2

1 1 where and

1 1So

Solving the partial fractions yields: 1 , 1 , 11So , 0

at

h t f t H s F s H s F ss s

A B CH s F ss a s s s as

A a B a C at eh t f t t

a a a

∗ = = =⎡ ⎤⎣ ⎦ +

⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

=− = =

−∗ = + + ≥

La

2

Page 55: Chapter 14 - The Laplace Transform

Section 14-10: Stability P14.10-1 a. From the given step response:

( ) ( ) ( ) ( )

1003 714 100

tH se u t

s s−⎡ ⎤= − =⎢ ⎥ +⎣ ⎦

L5

s

From the circuit:

( ) ( )55

RH sR LH s

RR Ls s s sL

= ⇒ =++ + ⎛ ⎞+⎜ ⎟

⎝ ⎠

Comparing gives

75 15 5 0.2 H100

RRL

R LL

⎫= ⎪ = Ω⎪ ⇒⎬+ =⎪=⎪⎭

b. The impulse response is

( ) ( )1 10075 75100

th t e u ts

− −⎡ ⎤= =⎢ ⎥+⎣ ⎦L

c.

( )100

75 3 45100 100 4 2jω

ω=

= = ∠−+

H °

( ) ( )o3 1545 5 0 45 V

4 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

( ) ( )( ) ( )( ) ( )

( )( )

22 2

5 5 10 20 205 5 1 22 2 2

tH se 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

20 202 2s j

s j

H ss jω

ω

ωω=

=

= = =+ +

H

The phasor of the output is

14-1

Page 56: Chapter 14 - The Laplace Transform

( )( )

( )( )

( )o 2 220 205 45 5 45 12.5 45 V

2 2 2 2 45jω = ∠ ° = ∠ ° = ∠−

+ ∠ °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 52

3030 ( )5

tH s t e u ts

− −⎡ ⎤= =⎣ ⎦ +L . The circuit is stable

so we can determine the network function as

( ) ( )( ) ( )2 2

30 305 5s j

s j

H ss jω

ω

ωω=

=

= = =+ +

H

The phasor of the output is

( )( )

( )( )

( )o 2 230 3010 0 10 0 8.82 62 V

5 3 5.83 31jω = ∠ ° = ∠ ° = ∠−

+ ∠ °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

( ) ( ) (1024008 32

H ss s

=+ + 0

The poles of the transfer function are 1 8 rad/ss = − and 2 320 rad/ss = − , so circuit is stable. Consequently,

( ) ( ) ( ) ( )102400 408 320 1 1

8 320s j

H sj j j j

ωω

ω ωω ω== = =

+ + ⎛ ⎞ ⎛+ +⎜ ⎟ ⎜⎝ ⎠ ⎝

H⎞⎟⎠

14-2

Page 57: Chapter 14 - The Laplace Transform

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

( ) ( )( )2406 2

sH ss 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 12 6

s j

j jH sj j j j

ω

ω ωωω 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

Page 58: Chapter 14 - The Laplace Transform

P14.10-6

( ) ( ) ( ) ( )( )( )

90 36 104 32 36 3604 3290 90 90

tH s sse u ts s s s s

− ++⎡ ⎤= + = + = =⎣ ⎦ + + +L

s s

so

( ) ( )( )

1036

90s

H ss+

=+

The pole of the transfer function , so circuit is stable. Consequently, 1 90 rad/ss = −

( ) ( ) ( )( )

110 1036 490 1

90s j

jjH s

j jω

ωω

ωωω=

⎛ ⎞+⎜ ⎟+ ⎝ ⎠= = =+ ⎛ ⎞+⎜ ⎟

⎝ ⎠

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

Page 59: Chapter 14 - The Laplace Transform

( ) ( ) ( ) ( ) ( )5 205 1.67 1.67 25

3 5 20t tH s

e e u ts s s

− −⎡ ⎤= − = − =⎢ ⎥ + + + +⎣ ⎦L

5 20s s

)

so

( ) ( )(25

5 2sH 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 0.255 20 1 1

5 2s j

j jH sj j j j

ω

0

ω ωωω 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 V30 30 1 6 1 1.51 15 20

j jj 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

( ) ( ) ( )( )

( )( ) ( )

52 2

10 5 5010 50 1010 505 5 5 5

t s sH s e t u ts 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 0.45 1

5

s j

j jH sj j

ω

ω ωωω ω=

= = =+ ⎛ ⎞+⎜ ⎟

⎝ ⎠

H

Using ( ) ( ) ( )o sω ω ω=V H V at ω = 10 rad/s gives

14-5

Page 60: Chapter 14 - The Laplace Transform

( ) ( ) ( )( )o 2 2

0.4 10 4812 0 9.6 37 V1 2101

5

j jjj

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

Page 61: Chapter 14 - The Laplace Transform

Section 14.12 How Can We Check…? P14.12-1

( ) ( ) 2.1 15.93 6 2t tL L

dv t i t e edt

− −= = − −

( ) ( ) 2.1 15.91 0.092 0.57575

t tC C

di t v t e edt

− −= = − −

( ) ( ) 2.1 15.9

1 12 12 6 2t tR Lv t v t e e− −= − = + +

( )( ) ( )( ) 2.1 15.9

2

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, ( ) ( ) ( ) ( ) ( )1 212 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 ss s

= =− −

1

Page 62: Chapter 14 - The Laplace Transform

KVL for left mesh: 12 1 18 18 206 03 3 324 4 4

s s s s s

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

+ + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠

= (ok)

KVL for right mesh: 18 20 20 186 3 43 3 3 34 4 4 4

s s s s

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

− − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0= (ok)

The analysis is correct.

P14.12-3

Initial value of IL (s): 2

lim 2 15

sss s s

+=

→∞ + + (ok)

Final value of IL (s): 2

lim 2 00 5

sss s s

+=

→ + + (ok)

Initial value of VC (s): ( )( )2

lim 20 20

5s

ss s s s

− +=

→∞ + + (not ok)

Final value of VC (s): ( )( )2

lim 20 28

0 5s

ss 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 ( )20LI s

s− instead of ( )20 8

LI ss s

− + . After correcting this error

( ) 2

20 2 85C

sV ss s s s

+⎛ ⎞= − +⎜ ⎟+ +⎝ ⎠.

Initial value of VC (s): ( )( )2

lim 20 2 8 85

ss

s ss s s

⎛ ⎞− +⎜ ⎟+ =⎜ ⎟→ ∞ + +⎝ ⎠

(ok)

Final value of VC (s): ( )( )2

lim 20 2 8 00 5

ss

s ss s s

⎛ ⎞− +⎜ ⎟+ =⎜ ⎟→ + +⎝ ⎠

(ok)

2