Chapter 14 - The Laplace Transform

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

Transcript of Chapter 14 - The Laplace Transform

ProblemsSection 14-2: Laplace Transform P14.2-1 L A f1 ( t ) = A F1 ( s ) f1 ( t ) = cos ( t ) P14.2-2

As s F ( s ) = s2 + 2 F1 ( s ) = 2 2 s +

L t n = P14.2-3

n! s n +1

F ( s ) = L t 1 =

1 1! = 2 1+1 s s

Linearity: L a1 f1 ( t ) + a2 f 2 ( t ) = a1 F1 ( s ) + a2 F2 ( s ) Here a1 = a2 = 1 L f1 ( t ) = L e 3t = L f 2 ( t ) = L[t ] = so F ( s ) = 1 1 + 2 s +3 s 1 = F1 ( s ) s+3

1 = F2 ( s ) s2

P14.2-4

f ( t ) = A (1 e bt ) u ( t ) = A f1 (t ) f1 ( t ) = (1 e bt ) u ( t ) = 1u ( t ) e bt u ( t ) = f 2 ( t ) + f 3 ( t ) F2 ( s ) = 1 1 , F3 ( s ) = s s +b

Ab 1 1 F ( s ) = AF1 ( s ) = A F2 ( s ) + F3 ( s ) = A = s( s +b ) s s +b

1

Section 14-3: Impulse Function and Time Shift Property P14.3-1

(1e A Ae sT F ( s ) = AL u ( t ) AL u ( t T ) = =A s s sP14.3-2

f ( t ) = A u ( t ) u ( t T )

sT

)

f ( t ) = u ( t ) u ( t T ) e at L u ( t ) u ( t T ) =

F ( s ) = L eat u ( t ) u ( t T ) 1 e( s a )T F (s) = ( s a )

1 e sT s at L e g ( t ) =G ( s a )

P14.3-3 (a)

F (s) =

2

( s +3)

3

(b) (c)

f ( t ) = ( t T ) F ( s ) = e sT L ( t ) = e sT

F (s) =

5

( s + 4 ) +( 5 )2

2

=

5 5 = 2 ( s + 8 s + 16 ) + 25 s + 8 s + 412

P14.3-4

g ( t ) = e t u ( t 0.5 ) = e ( t + (0.5 0.5) )u ( t 0.5 ) = e 0.5 e (t 0.5)u ( t 0.5 )

L e 0.5 e (t 0.5)u ( t 0.5 ) = e0.5 L e (t 0.5)u ( t 0.5 ) = e0.5 e0.5 s L e t u ( t ) =

e0.5 e0.5 s e0.5 0.5 s = s +1 s +1

P14.3-5 sT e sT t T t e L L t u( t ) = 2 u ( t T ) = e sT L u ( t ) = T Ts T T

1

P14.3-6

5 5 f ( t ) = t + 5 u ( t ) ( t 4.2 ) u ( t 4.2 ) 3 3 4.2 s 1) 5 5 4.2 s 5 15 s + 5 ( e F (s) = 2 + e 2= s 3 s2 3s 3s

P14.3-7

F (s) = 0

3 e st f (t ) e dt = 0 3 e dt = s st 2 st

2

=0

3(1e 2 s ) s

P14.3-8

5 2 t 0 0 since pi = 1 3 j . No final value of

f ( t ) exists.

14-1

Section 14-6: Solution of Differential Equations Describing a Circuit P14.6-1 KVL:4 di + v = 2 e 210 t for t 0 dt The capacitor current and voltage are related by

50 i + 0.001

i = ( 2.5 106 )

dv dt

v1 = 2 e 210 t V , i (0) = 1 A, v (0) = 8 V

4

Taking the Laplace transforms of these equations yields 50 I ( s ) + 0.001 [ s I ( s ) i (0) ] + V ( s ) = I ( s ) = ( 2.5 106 ) sV ( s ) v( 0 ) Solving for I(s) yields I (s) = where A = ( s + 10 ) I ( s )4 s = 10 4

2 s + 2104

s 2 + 1.4 104 s 1.6 108 ( s +104 ) ( s + 2 104 ) ( s + 4 104

)

=

A B C + + 4 4 s +10 s + 2 10 s + 4104 2108 2 = = 3 108 3 =s = 2 10 4

s 2 + 1.4 104 s 1.6 108 = ( s + 2 104 )( s + 4 104 )4 =

s = 10

4

B = ( s + 2104 ) I ( s ) C = ( s + 4104 ) I ( s ) Then I (s) =

s = 2 10

s 2 + 1.4 104s 1.6 108 ( s + 104 ) ( s+ 4104 ) s 2 + 1.4 104s 1.6 108 ( s +104 ) ( s + 2 104 )

.4 108 1 = 8 2 10 5 8.8 108 22 = 8 6 10 15

s = 4 10

4

=

=s = 4 10 4

23 15 22 15 + + 4 4 s +10 s + 210 s + 4 104

i (t ) =

4 4 4 1 10 e 10 t + 3e 2x10 t + 22 e 4x10 t u ( t ) A 15

14-1

P14.6-2

We are given v ( t ) = 160 cos 400 t . The capacitor is initially uncharged, so v C ( 0 ) = 0 V . Then i ( 0) = 160 cos ( 400 0 ) 0 = 160 A 1

KCL yields103 dvC dt + vC 100 =i

Apply Ohms law to the 1 resistor to get v v C i= vC = v i 1 Solving yields di + 1010 i = 1600 cos 400t ( 6.4 104 ) sin 400t dt Taking the Laplace transform yields

s I ( s ) i (0) + (1010 ) I ( s ) =so I ( s) = Next

1600s s 2 +( 400 )2

( 6.410 ) ( 400 ) 2

s 2 +( 400 )

2

160 1600s 2.5107 + s + 1010 ( s + 1010 ) s 2 + (400) 2

1600s 2.5107 A B B* = + + ( s + 1010 ) s 2 + (400)2 s + 1010 s + j 400 s j 400 where

A =

1600 s 2.5107 s 2 + ( 400 )2 s = 1010

= 23.1 ,

B =Then

1600s 2.5 x 107 ( s +1010 ) ( s j 400 )

=s = j 400

2.56 107 1.4 8.69 10 68.45

=11.5 j 27.2 and B* = 11.5 + j 27.2

I (s) = Finally

136.9 11.5 j 27.2 11.5 + j 27.2 + + s + 1010 s + j 400 s j 400

i ( t ) = 136.9e 1010t + 2 (11.5 ) cos 400t 2 ( 27.2 ) sin 400t for t > 0 = 136.9e 1010t + 23.0 cos 400t 54.4sin 400t for t > 0

14-2

P14.6-3

vC (0) = 0

vc +15103 i = 10 cos 2t 1 3 d vc i = 10 30 dt Taking the Laplace Transform yields: sVC ( s ) vC ( 0 ) + 2VC ( s ) = 20 where A= Then 20 s s2 +4 =s = 2

d vc + 2 vc = 20 cos 2t dt

20s B B* s A VC ( s ) = = + + s2+ 4 ( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s j 2 =s = j2

40 20s = 5, B = 8 ( s + 2 )( s j 2 )

5 5 5 5 5 = + j and B* = j 1 j 2 2 2 2

5 5 5 5 +j j 5 2 2 2 2 VC ( s ) = + + s+2 s+ j2 s j2

vC ( t ) = 5e2t + 5 ( cos 2t + sin 2t ) V

P14.6-4vc + 12i L + 2 diL dt = 8 and i L = C d vc dt

Taking the Laplace transform yieldsVc ( s ) + 12 I L ( s ) + 2 sI L ( s ) iL ( 0 ) = I L ( s ) = C sVc ( s ) vc ( 0 )

8 s

vc (0) = 0, iL (0) = 0

14-3

Solving yieldsVc ( s ) = 4C C s s 2 + 6s + 2 72 s ( s + 3)2

(a) C =

1 F 18

Vc ( s ) =

=

c a b + + s s + 3 ( s + 3)2 24 8 8 + + s s + 3 ( s + 3)2

a = 8, b = 8, and c = 24 Vc ( s ) =

vc ( t ) = 8 8 e3t 24 t e3t V, t 0(b) C =1 F 10

Vc ( s ) =

c 40 a b = + + s ( s +1) ( s +5 ) s s +1 s + 52 8 10 + + s s +1 s + 5

a = 8, b = 10, and c = 2 VC ( s ) =

vc ( t ) = 8 10 et + 2 e5t V, t 0

P14.6-5

vc (0 ) = 10 V, i L ( 0 ) = 0 A d vc di and 400 i + 1 + vc = 0 dt dt

i = ( 5 106 )

Taking Laplace transforms yields 1 400 I ( s ) = ( 5 106 ) ( sVc ( s ) 10 ) 10 40 = I (s) = 2 2 5 s + 400 s + 2 10 400 I ( s ) + ( s I ( s ) 0 ) + Vc ( s ) = 0 ( s + 200 ) + 4002

soi (t ) = 1 200t e sin ( 400t ) u ( t ) A 40

14-4

P14.6-6 After the switch opens, apply KCL and KVL to get d R1 i ( t ) + C v ( t ) + v ( t ) = Vs dt

Apply KVL to getv (t ) = L d i (t ) + R 2 i (t ) dt

Substituting v ( t ) into the first equation gives d d d R1 i ( t ) + C L i ( t ) + R 2 i ( t ) + L i ( t ) + R 2 i ( t ) = Vs dt dt dt d2 d R 1 C L 2 i ( t ) + R1 C R 2 + L i ( t ) + R1 + R 2 i ( t ) = Vs then dt dt Dividing by R1 C L :

(

)

(

)

R1 C R 2 + L d R1 + R 2 Vs i (t ) + i (t ) + i (t ) = R1 C L dt R1 C L R1 C L dt 2 d2 With the given values:

d2 dt2

i ( t ) + 25

d i ( t ) + 156.25 i ( t ) = 125 dt

Taking the Laplace transform: 2 125 d s I ( s ) dt i ( 0 + ) + s i ( 0 + ) + 25 s I ( s ) i ( 0 + ) + 156.25 I ( s ) = 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 divisionv (0 ) = 9 20 = 14.754 V 9 + (16 || 4 )

Then, using current division

4 v (0 ) i (0 ) = = 0.328 A 16 + 4 9

The capacitor voltage and inductor current are continuous so v ( 0 + ) = v ( 0 ) and i ( 0 + ) = i ( 0 ) . After the switch opens v (t ) = L d i (t ) + R2 i (t ) dt v ( 0 + ) 9 i ( 0 + ) 14.754 9 ( 0.328 ) d i (0 +) = + = + = 29.508 dt 0.4 0.4 0.4 0.4

Substituting these initial conditions into the Laplace transformed differential equation gives s 2 I ( s ) ( 29.508 + 0.328 s ) + 25 s I ( s ) 0.328 + 156.25 I ( s ) = 125 s

( s 2 + 25 s + 156.25) I ( s ) = 125 + ( 29.508 + 0.328 s ) + 25 ( 0.328) sso I (s) = 0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125 s s 2 + 25 s + 156.25 s ( s + 12.5 )

(

)

=

0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 1252

=

0.471 23.6 0.8 + + s + 12.5 ( s + 12.5 )2 s

Taking the inverse Laplace transform

i ( t ) = 0.8 + e12.5 t ( 23.6 t 0.471) A for t 0so0.328 A for t 0 i (t ) = 12.5 t ( 23.6 t 0.471) A for t 0 0.8 + e

(checked using LNAP 10/11/04)

P14.6-7

KCL:

v1 5

+ i = 7 e 6 t

KVL: 4

di di + 3 i v1 = 0 v1 = 4 + 3 i dt dt di 4 + 3i 35 di Then dt + i = 7 e6 t + 2 i = e 6 t 5 4 dt

Taking the Laplace transform of the differential equation: s I ( s ) i(0) + 2 I ( s) = 35 1 35 1 I ( s) = 4 s +6 4 ( s + 2)( s + 6)

Where we have used i (0) = 0 . Next, we perform partial fraction expansion. 1 A B 1 = + where A = ( s + 2) ( s + 6) s + 2 s + 6 s+6 ThenI (s) = 35 35 35 1 35 1 i (t ) = e 2t e 6 t A, t 0 16 s + 2 16 s + 6 16 16

=s =2

1 1 1 = and B = 4 s + 2 s = 6 4

P14.6-8

Apply KCL at node a to get 1 d v1 v 2 v1 = 48 dt 24 2 v1 + d v1 dt = 2 v2

Apply KCL at node b to get v 2 50 cos 2 t 20 + v 2 v1 24 + v2 30 + d v2 1 d v2 = 0 v1 + 3 v 2 + = 60 cos 2t 24 dt dt

Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get

( 2+ s ) V1 (s) 2V2 (s)

= 10 and V1 ( s) + ( 3+ s ) V2 ( s ) =

25s 2 + 60s +100 s2 + 4

Solve these equations using Cramers rule to get

V2 ( s ) =

( 2+ s )

25s 2 + 60s +100 +10 ( 2+ s ) ( 25s 2 + 60 s +100 ) +10 ( s 2 + 4 ) s2 +4 = ( 2+ s ) (3+ s) 2 ( s 2 + 4) ( s +1)( s + 4) = 25s 3 +120s 2 + 220 s + 240 ( s 2 + 4 ) ( s +1)( s + 4 )

Next, partial fraction expansion gives

A A* B C + + + V2 ( s ) = s + j 2 s j 2 s +1 s