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