Chapter 13 - Frequency Response

download Chapter 13 - Frequency Response

of 63

  • date post

    10-Apr-2015
  • Category

    Documents

  • view

    243
  • download

    1

Embed Size (px)

description

Introdução Aos Circuitos Elétricos - 7th ed - Dorf Svoboda - Resolução - Capitulo 13

Transcript of Chapter 13 - Frequency Response

ProblemsSection 13-2: Gain, Phase Shift, and the Network Function P13.2-1

R 2 ||

1 j C

=

R2 1 + j C R 2R2 1 + j C R 2 R2 R1 + 1 + j C R 2 R2 R1 + R 2 1 + j C R p

H ( ) =

Vo ( ) = Vi ( )

= When R1 = 40 , R2 = 10 and C = 0.5 F

where Rp = R1 || R2.

H ( ) =

0.2 1 + j 4(checked using ELab on 8/6/02)

P13.2-2

H ( ) =

Vo ( ) j C = Vi ( ) R + R + 1 1 2 j C = 1 + j C R 2 1 + j C R1 + R 2

R2 +

1

(

)

When R1 = 40 k, R2 = 160 k and C = 0.025 F

H ( ) =

1 + j ( 0.004 ) 1 + j ( 0.005) (checked using ELab on 8/6/02)

13-1

P13.2-3H ( ) =

R2 Vo ( ) = Vi ( ) R1 + R 2 + j L R2 = R1 + R 2 L 1 + j R1 + R 2

When R1 = 4 , R2 = 6 and L = 8 HH ( ) = 0.6 1 + j ( 0.8 )

(checked using ELab on 8/6/02)

P13.2-4H ( ) = R 2 + j L Vo ( ) = Vi ( ) R + R 2 + j L L 1 + j R R2 2 = R + R2 L 1 + j R + R2

Comparing the given and derived network functions, we require R2 = 0.6 R + R2 R 2 = 12 L R + R2 = 20 L

L 1 + j R R2 2 R + R2 L 1 + j R + R2

1+ = ( 0.6 ) 1+

j j

12

20

Since R2 = 60 , we have L =

60 = 5 H , then R = ( 20 )( 5 ) 60 = 40 . 12 (checked using ELab on 8/6/02)

13-2

P13.2-5

R 2 ||

1 j C

=

R2 1 + j C R 2R2 1 + j C R 2 R2 R+ 1 + j C R 2 R2 R + R2 1 + j C R p

H ( ) =

Vo ( ) = Vi ( )

= where Rp = R || R2. Comparing the given and derived network functions, we require

R2 0.2 = 1 + j C R p 1 + j 4 R + R2

R2 = 0.2 R + R2 CR =4 p

Since R2 = 2 , we have Finally, C =

2 ( 2 )( 8 ) = 1.6 . = 0.2 R = 8 . Then R p = R+2 2+8

4 = 2.5 F . 1.6(checked using ELab on 8/6/02)

P13.2-6

Vi ( ) R + j L 1 Vo ( ) = ( A I a ( ) ) j C I a ( ) =

A Vo ( ) CR = L Vi ( ) ( j ) 1 + j R

13-3

When R = 20 , L = 4 H, A = 3 A/A and C = 0.25 F H ( ) = 0.6 ( j ) (1 + j ( 0.2 ) ) (checked using LNAP on 12/29/02)

P13.2-7

In the frequency domain, use voltage division on the left side of the circuit to get: 1 1 j C VC ( ) = Vi ( ) = Vi ( ) 1 1 + j C R1 R1 + j C Next, use voltage division on the right side of the circuit to get: 2 A R3 2 3 V o ( ) = Vi ( ) A VC ( ) = A VC ( ) = 3 1 + j C R1 R 2 + R3 Compare the specified network function to the calculated network function: 2 2 A A 4 2 1 3 3 = = 4 = A and = 2000 C 1 + j C R1 1 + j C 2000 3 100 1+ j 100 Thus, C = 5 F and A = 6 V/V. (checked using ELab on 8/6/02)

13-4

P13.2-8 H ( ) = Vo ( ) = Vi ( ) R2 1 j C R1

R2 R1 = 1+ j C R2When R 1 = 10 k , R 2 = 50 k , and C = 2 F, then R2 R1 = 5 and R2 C = 1 5 so H ( ) = 10 1+ j 10

P13.2-9 H ( ) = Vo ( ) = Vi ( ) R2 R1 1 j C2 1 j C1

R2 1 + j C2 R2 = R1 1 + j C1R1 R 1 + j C1R1 H ( ) = 2 R1 1 + j C2 R2

When R1 = 10 k, R2 = 50 k , C1 = 4 F and C2 = 2 F, then so R2 R1 = 5 , C1R1 = 1 1 and C2 R 2 = 25 10

1 + j 25 H ( ) = 5 1+ j 10

13-5

gain = H ( ) = ( 5 )

1+ 1+

2625

2100

phase shift = ( ) = 180 + tan 1 tan 1 25 10

P13.2-10

R3

1 = jC

1 R3 jC = 1 1 + jC R 3 R3 + jC R3R3 1+ j C R3 R + R + j R2 R3C = 2 3 R1 R1 + j R1 R3C R2 + R3 R1 R2 R2 = 2 R1 = 20 k R1

H ( ) =

R2 +

5 = lim H ( ) = 0

2 = lim H ( ) =

then

R3 = 5R1 R2 = 30 k

P13.2-11 H ( ) = R2 + 1 j C = 1 + j C R 2 j C R1

R1

H ( ) = 180 + tan 1 ( C R 2 ) 90 H ( ) = 135 tan 1 (CR2 ) = 45 C R 2 = 1 R2 = 10 = lim H ( ) =

1 = 10 k 10 1073

R2 R R1 = 2 = 1 k R1 10

13-6

P13.2-12R2 1 j C2 1+ j C 2 R2 H ( ) = = j C 1 R1 + 1 1 R1 + j C1 j C1 R 2 || =

(1 + j C R )(1 + j C1 1

( C

1

R2 ) j

2

R2 )

C R = 0.1 1 2 ( C1 R 2 ) j 0.1) j ( 1 1 or = C1 R1 = p 125 (1 + j C1 R1 )(1 + j C 2 R 2 ) 1 + j 1 + j p 125 1 1 or C 2 R 2 = p 125 Since C1 = 5 F, R1 = 8 k and R 2 = 20 k

C1 R1 = ( 5 106 )( 8 103 ) =1 = C 2 R2 125P13.2-13

40 1 1 = 1000 25 125

p = 25 rad/s

C2 =

1 1 = = 0.4 106 = 0.4 F 3 125 R 2 125 ( 20 10 )

1 j C2 V ( ) = H ( ) = o 1 Vs ( ) R1 + j C1 R2 =

( C1R2 ) j

(1 + j R1C1 ) (1 + j R2C2 )

When R1 = 5 k, C1 = 1 F, R2 = 10 k and C2 = 0.1 F, thenH ( ) =

( 0.01) j 1 + j 200 1 + j 1000

13-7

so

0 500 2500 Then

H( ) 0 1.66 0.74

H ( ) 90 175 116

v ( t ) = (0) 50 + (1.66) ( 30 ) cos(500t + 115 + 175) (0.74) ( 20 ) cos(2500t + 30 + 116) o =49.8cos(500t 70) 14.8cos (2500t +146) mV When R1 =5 k, C1 =1 F, R 2 =10 k and C2 = 0.01 F, thenH ( )= 0.01j 1+ j 1+ j 200 10,000

So

0 500 2500 Then

H ( ) 0 1.855 1.934

H ( ) 90 161 170

v (t ) = (0) ( 50 ) + (1.855) ( 30 ) cos(500t + 115 161) (1.934) ( 20 ) cos(2500t + 30 + 170) o = 55.65 cos(500t 46) 38.68cos(2500t + 190) mV

13-8

P13.2-14 a)

2 V (8 div) div = 8 V Vs = 2 2 V (6.2 div) div = 6.2 V Vo = 2 V 6.2 gain = o = = 0.775 Vs 8 1 H ( ) = Vo ( ) 1 j C = = 1 1 + j C R Vs ( ) R+ j C 1 1 + 2C 2 R 2 1 then C = R 1 1 g2

b)

Let g = H ( ) =

In this case = 2 500 = 3142 rad s , H ( ) = 0.775 and R = 1000 so C = 0.26 F.

c)

tan( H ( )) RC Recalling that R = 1000 and C=0.26F, we calculate H ( )= tan 1 ( R C ) so =

2 (200) 2 (2000)

H ( ) 0.95 0.26

H ( ) 18 73

tan 45 = 3846 rad s H ( )= 45 requires = 6 1000 .2610

(

)

(

( )

)

( ) = 135 requires =

tan ( (135)) = 3846 rad s 6 ) (1000)(0.2610

A negative frequency is not acceptable. We conclude that this circuit cannot produce a phase shift equal to 135.

13-9

d)

tan (60) C = (2 500) (1000) = 0.55 F tan (H( )) C= R C = tan ((300)) = 0.55 F (2 500 ) (1000) A negative value of capacitance is not acceptable and indicates that this circuit cannot be designed to produce a phase shift at 300 at a frequency of 500 Hz.

e)

C =

tan( ( 120 )) = 0.55 F (2 500)(1000)

This circuit cannot be designed to produce a phase shift of 120 at 500 Hz.

P 13.3-15R2 1 j C2 1+ j C 2 R2 H ( ) = = j C 1 R1 + 1 1 R1 + j C1 j C1 R 2 || =

(1 + j C R )(1 + j C1 1

( C

1

R2 ) j

2

R2 )

C R = 0.1 1 2 C 1 R 2 ) j ( ( 0.1) j 1 1 or = C1 R1 = p 125 (1 + j C1 R1 )(1 + j C 2 R 2 ) 1 + j 1 + j p 125 1 1 or C 2 R 2 = 125 p Since C1 = 5 F, R1 = 8 k and R 2 = 20 k

C1 R1 = ( 5 106 )( 8 103 ) =1 = C 2 R2 125 C2 =

40 1 1 = 1000 25 125

p = 25 rad/s

1 1 = = 0.4 106 = 0.4 F 3 125 R 2 125 ( 20 10 )

13-10

P13.2-16 I 1 ( ) =

V s ( ) N1 R1 + N2 2

( R2 + j L))

N1 V o ( ) = R 2 + j L I 2 ( ) = R 2 + j L I 1 ( ) N2

(

)

(

N1 R + j L V s ( ) N2 2 =

(

)

N1 R1 + N2

2

( R2 + j L)1+ j L R2 L

H ( ) =

Vo ( ) = Vs ( )

N1 R + j L N2 2

(

)=

N2 R N1 2 2

N1 R1 + N2

2

( R2 + j L)

N2 1+ j 2 R1 + R 2 N2 N1 R + R2 N1 1

Comparing to the given network function:

k=

N2 R N1 2 N2 R + R2 N1 1 2

N2 R + R2 N1 1 R2 , z= and p = . L L

2

13-11

P13.2-17 Mesh equations:

V s ( ) = R1 + j L1 I 1 ( ) + j M I 2 ( )

( ) 0 = ( R 2 + j L 2 ) I 2 ( ) + j M I 1 ( )I 1 ( ) = R2 + j L2 I 2 ( ) j M R2 + j L2 + j M I 2 ( ) j M V s ( )

Solving the mesh equations

V s ( ) = R 1 + j L1

(

)

I 2 ( ) =

( R1 + j L1 )( R 2 + j L 2 ) + 2 M 2j M R2 j M R2

j M

V o ( ) = R 2 I 2 ( ) = =

( R1 + j L1 )( R 2 + j L 2 ) + 2 M 2

V s ( ) V s ( )

R1 R 2 + 2 M 2 L1 L 2 + j R1 L 2 + L1 R 2

(

)

(

)

H ( ) =

V o ( ) V s ( )

=

M R2 R1 R 2 + 2

(M

2

L1 L 2 1 + j

)

j R1 L 2 + L1 R 2 R1 R 2 + 2 M 2 L1 L 2

(

)

Comparing to the given network function:

k=

M R2 R1 R 2 + 2 M 2 L1 L 2

(

)

and p =

R1 R 2 + 2 M 2 L1 L 2 R1 L 2 + L1 R 2

(

)

13-12

P13.2-18 Using voltage division twice gives: A V2 ( ) = Vi ( ) and R2 A R2

A R2 R1 + R 2 1 + j C R 2 A= = R2 C R1 R 2 R1 + R 2 + j C R1 R 2 R1 + 1 + j R1 + R 2 1 + j C R 2

Vo ( ) = A V2 ( )

j L R 4 R 4 + j L j L R 4 L j = = j L R 4 R 3 R 4 + j L ( R 3 + R 4 ) R 3 L ( R3 + R 4 ) R3 + 1 + j R 4 + j L RR3 4

Combining these equations gives

ALR 2 Vo ( ) j = Vi ( ) R 3 ( R1 + R 2 ) L ( R3 + R 4 ) C R1 R 2 1 + j 1 + j R3 R 4 R1 + R 2 ALR 2 Comparing to the given network function gives k = and either R 3 R1 + R 2 H ( ) =

(

)

p1 =

L R3 + R 4

(

R 3R 4

)

and p 2 =

R1 + R 2 CR1R 2

or p 1 =

R1 + R 2 CR1R 2

and p 2 =

L R3 + R 4

(

R 3R 4

)

.

13-13

P13.2-19 Represent the circuit in the frequency domain.

Apply KCL at the top node of the left capacitor, C1, to getVa Vs R1 + j C 1 Va = 0 Va = 1 Vs 1 + j C 1 R1

The op amp, together with resistors R2 and