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