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### Transcript of Chapter 12 Problem Solutions - Pennsylvania State php. 310/Ch12s.pdf · PDF fileChapter...

• Chapter 12 Problem Solutions 12.1 (a)

5

5

5 5

15 10120

1 (5 10 )120 (120)(5 10 ) 5 10

0.008331

fAAA

=+

=+

+ = =

(b) 3

3

3 3

5 101201 (5 10 )

120 (120)(5 10 ) 5 100.008133

=+

+ = =

12.2 (a) 0.151f

AAA

= =+

T A= (i) T = (ii) 4 380 dB 10 1.5 10A A T= = = (iii) 15T =

(i) 1 6.667fA = =

(ii) 6.662fA = (iii) 6.25fA = (b) (i) T = (ii) 32.5 10T = (iii) 25T =

(i) 1 4.00fA = =

(ii) 3.9984fA = (iii) 3.846fA = 12.3 (a)

1 1251f

AAA

= =+

0.0080 = (b)

• [ ][ ]

(125)(0.9975) 124.6875

124.68751 (0.008)

124.6875 1 (0.008) A124.6875 1 0.9975

49,875

fA

AA

AA

A

= =

=+

+ =

= =

12.4

(a) 4

4

3

101001 (10 )

9.9 10

=+

=

(b)

4

100 ( 0.10) 0.001 0.10%10

f f

f

f

f

dA A dAA A A

dAA

=

= = =

12.5

4

4

4

3

0.001 ( 0.10)5 10

500

5 105001 (5 10 )

1.98 10

f f

f

f

f

dA A dAA A A

A

A

=

=

=

=+

=

12.6 (a) For Fig. P12.6(a)

1

1 1 1

200 101 200 1 10

o o

i o

v vv v

= =+ +

1 1

200 10 401 200 1 10f

A

= = + +

21 1 1 150 (1 200 )(1 10 ) 1 210 2000 = + + = + +

21 12000 210 49 0 + =

1

1

210 44100 4(2000)(49)2(2000)

0.1126

+=

=

For Fig P12.6(b)

2

2

(200)(10)401 (200)(10)0.0245

=+

=

Fig. P12.6(a) (b)

• 1 180180 10 (8.4634)(4.704)

1 (180)(0.1126) 1 (10)(0.1126)39.81

39.81 40 0.475%40

f

f

f

f

A

A

=

= = + + =

=

Fig. P12.6(b) (180)(10) 39.91

1 (180)(10)(0.0245)39.91 40 0.225%

40

f

f

f

A

dAA

= =+

=

(c) Fig. P12.6(b) is a better feedback circuit. 12.7 (a) ( 10)( 15)( 20) 3000O

O S

V V VV V V

= = = +

So 3000( )O O SV V V= + We find

30001 3000

Ovf

S

VAV

= =+

For 3000120 0.0081 3000vf

A

= = =+

(b) Now ( 9)( 13.5)( 18) 2187OV V V = = Then

2187 2187 118.241 2187 1 2187(0.008)vf

A

= = = + +

120 118.24% change 100 1.47% change120

=

12.8

5(10 )(4) (50) 8B Bf f kHz= = 12.9 (a) 53 3(50) (10 )(4) 8dB dBf f kHz = =

(b) 53 3(10) (10 )(4) 40dB dBf f kHz = = 12.10

30(50)(20 10 ) 5A = so

50 2 10A =

12.11

Low freq. 001

fAAA

=+

5000100 0.00981 (5000)

= =+

Freq. response

• 1 2

1 2

1 2

1 2 1 2

1 2 1 2

5000

1 1

(5000)(0.0098)1 11 1

5000

1 1 49

5000

1 49

5000

50

f

f fj jf fAA

Af fj jf f

f fj jf f

f f jf jfj jf f f f

f f jf jfj jf f f f

+ + = =

+ + + +

= + + +

= + + + +

= + + +

Also 0 100

1 1 1

ff

A B A B A B

AA

f f f f f fj j j j j jf f f f f f

= = + + + + +

So

1 2 1 2

100 10011 1

50 50 50A B A Bf f f f f f jf jfj j j j j jf f f f f f f f

= + + + + + +

Then

1 2

1 1 1 150 50A Bf f f f

+ = +

and 1 2

1 150A Bf f f f

=

1 10f = and 2 2000f = 1 1 1 1 0.002 0.000010 0.002010

50(10) 50(2000)A Bf f+ = + = + =

and

6

1 1 1(50)(10)(2000) 10

B

A B A

ff f f

= =

Then 61 0.002010

10B

B

ff

+ =

6 2 3

6 2 3

10 1 2.01 1010 2.01 10 1 0

B B

B B

f ff f

+ = +

+ =

3 6 6

6

3 4

6

2.01 10 4.0401 10 4(10 )(1)2(10 )

2.01 10 2.0025 102(10 )

B

B

f

f

=

=

+ sign 31.105 10 HzBf = + sign 29.05 10 HzAf =

• 12.12 (a) Fig. P12.6(a)

1

1

1

1 1

3

1

200

110

200 1 (10)(0.1126)1 (0.1126)1

200 (4.704)1 22.52

940.73 940.73 123.5223.52 1

(23.52)40 (23.

1(23.52)

f

dB

fjf

A

fjf

fjf

f fj jf f

fjf

f

+ = + + + = + +

= = + +

= =+

52)(100) 2.352 kHz

Fig P12.6(b)

1

1

1

3

1

(200)(10)

12000

(0.0245)(200)(10)1 1 491

2000 1 (50)(100) 5 KHz50 1

(50)

f

dB

fjfA

fjf fjf

ffj

f

+= =

+ + ++

= = +

(b) Overall feedback wider bandwidth. 12.13

0 1 2 1

00

0

1000(100) (1) (100)(10) (1)(1) 10001

i n

i n

v A A v A vSv v vN

= +

= + = + = =

12.14 (a)

• (b)

Circuit (b) less distortion 12.15 (a) Low input R Shunt input Low output R Shunt output Or a Shunt-Shunt circuit (b) High input R Series input High output R Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a) 4 5(max) (1 ) 10(1 10 ) (max) 10i i iR R T R k= + = +

• 34

10(min) 101 1 10

ii

RR k

T= =

+ +

Or (min) 1iR =

(b) 4 4(max) (1 ) 1(1 10 ) (max) 10o o oR R T R k= + = +

44

1(min) 101 1 10

oo

RR k

T= =

+ +

Or (min) 0.1oR = 12.17

Overall Transconductance Amplifier, ogi

iAv

= Series output = current signal and Shunt input = current

signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.

12.18

1

1 2

||||

ix

i

R RV V

R R R

= +

12.19

3

50 48 2mV

5 2.5 10 V/V0.002

0.048 0.0096 V/V5

5 100 V/V0.05

b

b

i f

ov

f

o

orf

i

V V V

VA

VVVVAV

= = =

= = =

= = =

= = =

12.20

• 2 2

1 1

1 20 19vfR RAR R

+ = =

d S iv i R=

0

1 2

( )S d s ds

v v v v viR R

= + (1)

0 0 0

0 2

( ) 0L d s dv A v v v vR R

+ = (2)

00

0 2 0 2

( )1 1 L d S dA v v vvR R R R

+ = +

0

0 20

0 2

( )

1 1

L d S dA v v vR R

v

R R

+=

+

From (1):

0

2 0 2

1 2

0 2

0

0 22

2 21 2 1 2

0 0

( )1

1 1

111 1 1 1

1 1

L d S d

S d S dS

L

S S d

d S i

A v v vR R Rv v v v

iR R

R R

AR RRi v v

R RR R R RR R

v i R

+ = +

+ = + + + + +

=

02 2

1 2 0 0 2 1 2 0 2

2 2

0 0

02 2 2

0 1 1 0 0 0 1 1 0 0

0 2

1 1 1 1 1 11 11

1 1

1 1 1 1 1 11

Li

S S

LS i S

S

AR RRR R R R R R R R R

i vR RR R

AR R Ri R vR R R R R R R R R R

i R R

+ + + + + + = + + + + + + + = + +

+ 0 02 201 1 1 1

1 1 (1)i L SR RR RR A vR R R R

+ + + + = + +

Let 2 190 kR ,= 1 10 kR =

5

7

0.1 0.10.1 190 100 20 10 2010 10

(1.000219 10 ) (20.01)

S S

S S

i v

i v

+ + + + = + =

55 10 k 500 MSif ifS

vR Ri

=

Output Resistance

• 0

0 2 1

1

1 2

0 1

0 0 0 1 2 2 1

1

||||

||||1 1 1

( || ) ||

|| 10 ||100 9.09

X L d XX

i

id X

i

L iX

X f i i

i

V A v VIR R R R

R Rv V

R R RA R RI

V R R R R R R R R R

R R

= +

+

= +

= = + +

+ +

= =

5

0

4

50 0

1 1 10 9.09 10.1 0.1 9.09 190 190 9.09

10 4.566 10 0.005022.19 10 k 0.0219

f

f f

R

R R

= + + + +

= + += =

12.21 a.

0

1 2

( )S d S dv v v v vR R

= and 0dvvA

=

• 0

1 2 2 1 2

0 0

2 1 2

0 2

1 2 2 1

2

10

2

1

1 1

1 1

1 1 11 1

1

11 1

S Sd

S

S

v v v vR R R R R

v vR A R R

v RvR R R A R

RRv

v RA R

+ = + +

= + +

+ = + +

+ =

+ +

which can be written as 0

2

1

1 / 1vf

S

v AAv RA

R

= =

+ +

b. 2

1

1

1 RR

=+

c. 5

5

10201 (10 )

=+

So

5

5

10 120 0.0499910

= =

Then 2 21 1

1 11 1 19.0040.04999

R RR R

= = =

d. 49 10A 4

4

9 10 19.999561 (9 10 )(0.04999)f

A = =+

434.444 10 2.222 10 % 0.005%

20f f

f f

A AA A

= = =

12.22

[ ]

1000 90.9 A/A1 1 (1000)(0.01)

1 90.91 1 (0.01)(1000)

(1 ) 10 1 (0.01)(1000) 110 k

iif

i i

iif if

i i

of o i i of

AAA

RR R

AR R A R

= = =+ +

= = = + +

= + = + =

12.23

• 50 47.5 2.5 A

5 2000 A/A0.0025

0.0475 0.0095 A/A5

5 100 A/A0.05