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  • 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

    AdAA

    =

    = = + + =

    =

    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