chapter 16 solutions to exercises - circuit analysis and design
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
1. We have a parallel RLC with R = 1 kΩ, C = 47 μF and L = 11 mH. (a) Qo = R(C/L)½ = 65.37
(b) fo = ωo/ 2π = (LC)-½ / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source:
10-3∠0o A jωL -j/ ωC
The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω2 + jω/RC).
Since V = 10-3 Z, we note that |V| > 0 as ω → 0 and also as ω → ∞.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
2. (a) R = 1000 Ω and C = 1 μF. Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 = 25 μH (b) L = 12 fH and C = 2.4 nF R = Qo (L/ C)½ = 447.2 mΩ (c) R = 121.7 kΩ and L = 100 pH C = (Qo / R)2 L = 270 aF
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
3. We take the approximate expression for Q of a varactor to be
Q ≈ ωCjRp/ (1 + ω2 Cj2 Rp Rs)
(a) Cj = 3.77 pF, Rp = 1.5 MΩ, Rs = 2.8 Ω (b) dQ/dω = [(1 + ω2 Cj
2 Rp Rs)(Cj Rp) - ωCjRp(2ωCj2 Rp Rs)]/ (1 + ω2 Cj
2 RpRs) Setting this equal to zero, we may subsequently write
CjRp (1 + ω2 Cj2 Rp Rs) - ωCjRp(2ωCj
2 Rp Rs) = 0 Or 1 – ω2 Cj
2 Rp Rs = 0. Thus, ωo = (Cj2 RpRs)–½ = 129.4 Mrad/s = 21.00 MHz
Qo = Q(ω = ωo) = 366.0
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
4. Determine Q for (dropping onto a smooth concrete floor): (a) A ping pong ball: Dropped twice from 121.1 cm (arbitrarily chosen). Both times, it bounced to a height of 61.65 cm.
Q = 2πh1/ (h1 – h2) = 12.82 (b) A quarter (25 ¢). Dropped three times from 121.1 cm. Trial 1: bounced to 13.18 cm Trial 2: bounced to 32.70 cm Trial 3: bounced to 16.03 cm. Quite a bit of variation, depending on how it struck. Average bounce height = 20.64 cm, so
Qavg = 2πh1/ (h1 – h2) = 7.574 (c) Textbook. Dropped once from 121.1 cm. Didn’t bounce much at all- only 2.223 cm. Since the book bounced differently depending on angle of incidence, only one trial was performed.
Q = 2πh1/ (h1 – h2) = 6.4
All three items were dropped from the same height for comparison purposes. An interesting experiment would be to repeat the above, but from several different heights, preferrably ranging several orders of magnitude (e.g. 1 cm, 10 cm, 100 cm, 1000 cm).
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
5.
2 2
80Np/s, 1200 rad/s, Z( 2 ) 400
1200 80 1202.66 rad/s Q 7.5172
( )( ) ( )( 2 )Now, Y( ) C Y( 2 ) C2
80( 80 2400)Y( 160 1200) C Y( 160 1200)160 1200
d d
oo o
d d dd
d
j
s j s j js js j
jj jj
α ω α ω
ωωα
α ω α ω α α ωα ωα ω
= = − + = Ω
= + = ∴ = =
+ − + + − − += ∴ − + =
− +
− − +∴ − + = ∴ − + =
− +
2
1 180C400 2 15
1 229
30
C32,000 901
j
1 115.775 F; L 43.88mH; R 396.7C 2o
j
Cμ
ω α−
− +=
− +
∴ = = Ω= = = =
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
6.
2
2
2 6 2 2 2 6
3 5 3 2
1 1 2 0.1Y 0.2 0.22 0.1 1 1000 / 4 0.01 1000 10
2 0.1 1000 0.1 10000.2 04 0.1 10 4 0.01 100.1 10 4000 10 9.9 96,000 98.47 rad/s
−= + + = + +
+ + +
− + −= + + ∴ + =
+ + + +
+
∴ + = + ∴ = ∴
inj jω
j j jj j
=
ωω ω ω
ω ω ω ω ωω ω ω ω
ω ω ω ω ω ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
7. 6 6Parallel: R 10 , L 1, C 10 , I 10 0 As μ−= = = = ∠ °
(a) 3 6 61 1000 rad/s; Q RC 10 1000LCo o oω ω + −= = = = =
(b) 6 6 5 3 3
2 2
26
1 I 100010 10 , V 10 /10 10Y 1000
10 10V , V1000 10000.001 101000 1000
j j
j
ωω ω
ω ωω ω
− − − − −
− −
−
⎡ ⎤⎛ ⎞ ⎛ ⎞− = = + −⎜ ⎟ ⎜ ⎟= +
ω V
995 996 997 998 999 1000 1001 1002 1003 1004 1005 999.5 1000.5
0.993 1.238 1.642 2.423 4.47 10.0 4.47 2.428 1.646 1.243 0.997 7.070 7.072
Y ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
∴ = =⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
8. (a)
2 2
42 6 2 2
2 2 6
5(100 / ) 0.1
6
5 (100 / ) 10 0.01500 10 100 10 100(20 ) 10 (1000 )2 2 2
100 5 1000 20 1000 400 10100 10 0 10 100 40,000, 99 960,000
400 10960,000 / 99
= + ++ +
Z 2
(b) 2
2 2 6
102000Z ( ) 2 2.294400 10
oin o
o o
ωωω ω
= + + =+ +
Ω
9
− −= + + = + + = + +
+ + + + + +
−∴ + = ∴ + = + =
+ +∴ = =
in
o
j jj j
j j j j jj j j j
ω ωω ω
ω ω ω ωω ω ω ω ω ω
ω ω ω ω ωω ωω 8.47 rad/s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
9. (a)
1 1 2 2 2
2
50 , 1000 1,002,500 1001.249
1 10LC 1,002,500
d o d o
o
s s ω α ω ω
ω α
− −
+
= = ∴ = + = ∴ =
= = = = = = Ω6 61 100.9975 H; R 10
2 C 100k
(b) 4 6 1 1Y 10 10 , 1000 Z 9997 1.43210.9975 Y
j ω ωω
− −⎛ ⎞= + − = ∴ = = ∠ °Ω ⎜ ⎟⎝ ⎠
α ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
10.
min max
o
3 3
max min
214
max min max 3
3 3
535kHz, 1605kHz, Q 45 at one end andQ 45 for 535 1605 kHz
1 11/ 2 LC 535 10 ,1605 102 L C 2 L C
1L / L 3; L C 8.8498 102 535 10
RC 45,535 10 1605 10 . Use2
o
o
oo
f
f ff
ππ π
πωωπ
−
= = =≤ ≤ ≤
= ∴ × = × =
⎛ ⎞∴ = = = ×⎜ ⎟× ×⎝ ⎠
≤ × ≤ ≤ × max
3 3
14max
max min12
2 1605 10 20 10 C 45 C 223.1pFL8.8498 10L 397.6 H, L 44.08 H
223.1 10 9
oω
π
μ μ−
−
∴ × × × × = ∴ =
×∴ = = = =
×
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
11. (a) A 4
4 5 4 83
8 2 44 8
8 2 4
pply 1V. I 10 A1Y I 10 (1 [10 ( 10 )])10
4.4 101000 48.4 10 4.4 10 1000Y 10 11 104.4 4.4
1000 48.4 10 4.4 10Y ( )4.4
−
− − −−
− −− −
− −
± ∴ = − (b)
∴ = = + + − −×
× + × +∴ = + + × =
− × + ×=
R
in in
in
in
ss
s sss s
jjj
ω ωωω
∴
8 2
14
At , 1000 48.4 10 , 45.45 krad/s
4.4 10Z ( ) 104.4
− −
−−
= = × =
⎛ ⎞×= = Ω⎜ ⎟
⎝ ⎠
o o o
oin o
o
jj kj
ω ω ω ω
ωωω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
12. 01 24 4.9 rad/sLC
ω = = = or f0 = 0
2ωπ
= 780 mHz
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
13. ( )
06
1 1 200 rad/s1 25 10
1.01LC
ω−
= = =×
or f0 = 0
2ωπ
= 31.99 Hz
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
14. (a) 31 1 10 5
2 2 200R
RC Cα
α= ∴ = = = Ω
0 6
1 1 1000 rad/s10LC
ω−
= = = or f0 = 0
2ωπ
= 159.2 Hz
Zin(ω0) = R = 5 Ω (b) We see from the simulation result that the ratio of the test source voltage to its current is 5 Ω at the resonant frequency; the small error is due to the series resistance PSpice required.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
15. (a) -11 = 50 s 2RC
α = and 2 20 5000 rad/sdω ω α= − =
Zin(ω0) = R so find R.
( )2 2 20
1 1 40 Fd
CL L
μω ω α
= = =+
.( )2 2
1 250 2 2(50)
dLR
C
ω α
α
+= = = Ω
(b) The resonant frequency is 01 5000 rad/s or 795.8 HzfLC
= = .
We see from the simulation result that the ratio of the test source voltage to its current is 250 Ω at the resonant frequency; the small error is due to the series resistance PSpice required.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
16. 1000 rad/s, Q 80, C 0.2 Fo oω μ= = =
(a) 6
2 6 3 6
1 10 80L 5H, Q RC R 400C 0.2 10 10 0.2 10o o
o
kωω −= = = = ∴ = = Ω
×× ×
(b) B = =
23
/ Q 1000 / 80 12.51 B 6.25 rad/s2
Z R / 1 400 10 / 1B / 2 6.25
o o
o oj
ω
ω ω ω ω
=
∴ =
− −⎛ ⎞∴ = + = × + ⎜ ⎟⎝ ⎠
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
17.
1 2
21 2
o
32
1
103rad/s, 118,
Z( 105) 10
103 118
110.245110.245 , B 118 103 15 rad/s, Q 7.350B 15
7.350 1 17.350 RC RC 66.67 10 ,LC110.245 12,154
1 1Y( 105) 0.1 105C 15C 105CR 105L
o
oo
oo
j
j j j
ω ω
ω ω ω
ωω
ωω
++
−+
= =
= Ω
= = ×
∴ = = − = = = =
∴ = ∴ = = × = =
⎛ ⎞= = + − = +⎜ ⎟⎝ ⎠
12,154 C 18.456C105
0.1 1 1C 5.418mF, R C 12.304 , L 15.185 mH18.456 15 12,154C
−
⎛ ⎞− =⎜ ⎟⎝ ⎠
∴ = = = = Ω = =
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
18. 30 krad/s, Q 10, R 600 ,= = =o o Ωω
(a) B 3 krad/sQ
o
o
ω = =
(b) 28 30N 1.3333B / 2 1.5
oω ω− −= = = −
(c) Zin(j28 000) = 600 / (1 – j1.333) = 360 ∠ 53.13o Ω (d)
1
1
Q1 1 10 (e)
Z ( 28,000) 28,000C ,C600 28,000L R 30,000 600
R 600 1 30,000 10 1 28 10 30 10L , ZQ 30,000 10 L 600 600 30 600 28 600
600Z28 301 1030 28
−⎡ ⎤= + − = =⎢ ⎥ ×⎣ ⎦
351.906 54.0903
−× ⎡ ⎤⎛ ⎞= = = ∴ = + × −⎜ ⎟⎢ ⎥× ⎝ ⎠⎣ ⎦
= = °Ω⎛ ⎞+ −⎜ ⎟⎝ ⎠
oin
o
ino o
in
j
j
j
ω
ω
∠
j j
approx-true 360 351.906magnitude: 100% 100% 2.300%true 351.906
53.1301 54.0903angle: 100% 1.7752%54.0903
−= =
° − °= −
°
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
19. 3400Hz, Q 8, R 500 , I 2 10 A B 50Hzo o Sf −= = = Ω = × ∴ = (a) 3 2 2 4002 10 500 / 1 N 0.5 1 N 4, N 3
50 / 2400 25 3 443.3 and 356.7 Hz
f
f
− −= ×V × + = ∴ + = = ± =
(b)
∴ = ± =
3 2 2
2
1 1I 0.5 10 1 N 4, N 15, N 15R 5001 N400 25 15
R
v
f
−= = × = × ∴ + = = = ±+
496.8 and 303.2 Hz= ± =∴
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
20. 6 310 , Q 10, R 5 10 , . .= = = ×o o p rω
(a) 3
6
R 5 10Q L 0.L 10 1
5mHooω
×= ∴ =
×
0=
(b)
62
6
2
2 2 2 22 2
4 2 2
10 (c)
Approx: 2 5 / 1 N N 2.291 1.1146 Mrad/S10 / 20
1 1Exact: Y 1 Q 0.5 0.2 1 100 ( in Mrad/S)R
1 16.25 1 100( 2 1/ ), 2 0.0525, 2.0525
12.0525 1 0, 22
−= + ∴ = = ∴ =
⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ∴ = + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦
∴ = + − + − + = + =
− + = =
oo
o
j
ω ω
ωω ω ωω ω ω
ω ω ω ωω ω
ω ω ω ( )2.0525 2.0525 4 1.2569, 1.1211 Mrad/s+ − = =ω
1
22
1Approx: Y 30 tan N 30 , N 0.5774 , 1.0289 Mrad/s1/ 20
1 1 1Exact: Y 1 10 (in Mrad/s) tan 30 0.5774 105000
1 0.05774 0.05774 40.05774, 0.05774 1 0, 1.0293 Mrad/s2
− −∠ = ° ∴ = ° = = =
⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ∴ ° = = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
+ +
ω
− = − − = = =
j
∴
ω
ω ωω ω
ω ω ω ωω
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
21.
(a) 6
4 8
1C 3 7 10nF 10 rad/s10 10
oω− −
= + = ∴ = =
(b) 6 8 3
9 6
33 3
1,0
9
R 10 10 5 5 10 50B / Q 20 krad/s
1 0Parallel current source is 3 10 At , I 10 3Z
V 3 10 5 10 15 90 V
o o
o o
o sj j
j
ωω
ω ω
−
− −
−
= = × == =
∠ °
Q C = × = ×
(c) 3
313
15 10 15 9015 10 N 1.5 V 8.321 33.69 V10 10 1 1.5o j
ω ω × ∠ °− = × ∴ = = ∴ = = ∠ °
× +
∴ = × × × = ∠ °
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
22. (a)
6 6
6 2 6
2 4 6
2 6
6 2
6 2
6 2 6
(5 0.01 )(5 10 / ) (5 0.01 )(5 10 )Z ( )10 0.01 10 / 0.01 10 10
0.05 25 10 5 10Z ( )0.01 10 10
5 10 0.05 10,025Z ( )10 0.01 10
10,025 10At ,5 10 0.05 10 0.01
+ + + += =
+ + + ++ + + ×
=+ +
× − +
∴ =− +
= =× − −
in
in
in
o oo
o
s s s sss s s s
s s sss s
jjj
ω ωωω ω
ω ωω ωω
(b) 25 10,000Z ( ) (5 100) (5 100) 1002.510in oj j jω +
= + − = = Ω
9 2 7 22
2 9
, 10.025 10 100.25 5 10 0.5
99.75 9.975 10 , 10,000 rad/s
× − = × −
= × =
o oo
o o
ω ωω
ω ω∴
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
23. , 1000Hz, Q 40, Z ( ) 2 B 25Hzo o in of j kω= = = Ω ∴ = 2000 1000, N , 1010, N 0.8
1 N 12.5−
= = ∴+
f fj
(a) Zin(jω) = Zin = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω (b) 0.9 1.1 900 1100 Hzo of f f f< < ∴ < <
=
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
24. Taking 2–½ = 0.7, we read from Fig. 16.48a: 1.7 kHz – 0.6 kHz = 1.1 kHz Fig. 16.48b: 2×107 Hz – 900 Hz = 20 MHz
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
25. Bandwidth = 2 2 60 2 110f π ω ω= = − , where ω1 = ( ) 32 5.5 10π . π
(a) 2 1 Bω ω= + , therefore f2 = 5.5 + 103 kHz = 1.0055 MHz (b) ( )( )0 1 2 5.5 1005.5f f 74.37 kHz f= = =
(c) 3
00 6
74.37 1010
fQ = = 0.074 B
×=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
26. Bandwidth = 9210 Hz 1f f= − , where f1 = Hz. 675.3 10×
(a) 2 1f f B= + , therefore f2 = 1.0753 GHz
(b) ( )( )6 90 1 2 75.3 10 1.0753 10f= = × × =f f 284.6 MHz
(c) 6
00 9
284.6 1010
fQB
×= = = 0.2846
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
27. (a) To complete the sketch, we need to first find ω0, which we obtain in part (b). (b) 0 1 2 02000 rad/s or 318.3 Hzfω ω ω= = = (c) B = 2 1 3000 rad/s or 477.5 Hzω ω− =
(d) 0
2 1
2000 0.6673000
Q ωω ω
= = =−
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
28. (a) We begin by labelling the series string with the capacitor as string 1, and the other as string 2. We next find the parallel equivalent of each, and determine the frequency where Xp1 + Xp2 = 0.
Then 1
2 21 1
1p
R XXX+
= , and similarly 2
2 22 2
2p
R XXX+
= .
For 1 2
0 we have p pX X+ =2 2 2 2
1 1 2 2
1 2
0R X R XX X+ +
+ = [1]
At ω0, 10
1 XCω
= − ∴( )
242
222 201 1
121
0
105330
10330
R XX
ω
ω
++
=−
.
At ω0, 2 0 X Lω= ∴2 42 2
02 22
2 0
5 1010
R XX
2ωω
−
−
++= .
Enforcing Eq. [1], then, leads to ( )22 12
0 8 2
10 25 (330)10550.5 krad/s
(330)10 25(33)ω
−= =
−
or f0 = 87.61 kHz. (b) We see the simulation result agrees reasonably, with a resonant frequency of 87.6 kHz
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
29. (a) We design for a bandwidth of 5.5 kHz, a low-frequency cut-off of 500 Hz, and a resonant impedance of 1 kΩ (no value was specified). Thus, we need to specify values for R, L, and C.
( )2 1
0 1 2
30
0 3
6 kHz
0.5 (6) 3 kHz
3 105.5 10
f f B
f f f
fQB
= + =
= = =
×= =
×
( )( )
00 0 3 3
0
1 so 28.9 nF5.5 10 2 10
QQ RC CR
ωω π
= = = =×
L = ( )
( )3 3
2 60
5.5 10 101 292 mH2 3 10Cω π
×= =
× and, of course, R = 1 kΩ
(b) From the simulation, we observe a bandwidth of 5.5 kHz, a lower frequency cutoff of approximately 500 Hz, and a peak impedance of 1000 Ω, as desired.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
30. (a) ( )( )0 6 6
1 1 1 1 4.38 kHz2 2 400 10 3.3 10
fLCπ π − −
= = =× ×
(b) 00
1 1 400 1.1010 3.3
L LQR RLC
ω= = = =
(c) Z at resonance = R = 10 Ω (d) Z at 0.438 kHz =
( )( ) ( )( )6
6
110 2 438 400 10 10 109.01 2 438 3.3 10
j jππ
−−
⎡ ⎤⎢ ⎥+ × − = −
×⎢ ⎥⎣ ⎦Ω
(e) Z at 43.8 kHz =
( )( ) ( )( )4
4
110 2 438 400 10 10 108.98 j j= −2 438 3.3 10
ππ
−−
⎡ ⎤⎢ ⎥+ × − Ω
×⎢ ⎥⎣ ⎦
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
31. Bandwidth = 3 MHz, f1 = 17 kHz. (a) 2 1 3.017 MHzf f B= + = (b) 0 1 2 226.5 kHzf f f= =
(c) 00 0.0755fQ
B= =
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
32. (a) Z0 = 1 Ω by definition
(b) 3
01 10 707 rad/s = 112.5 Hz
2LCω = = =
(c) PSpice simulation verifies an impedance of 1 Ω at f = 112.6 Hz.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
33. (a) Z0 = 1 kΩ by definition
(b) 6
01 10 707 krad/s = 112.5 kHz
2LCω = = =
(c) PSpice simulation verifies an impedance of 1 kΩ at f = 112.8 kHz.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
34. (a)
2
20A 6 , 3 6 2, 40V in series with 2 1 3
L1 6010 rad/s, Q 20R 3LC
10 1B 0.5, B 0.25, V ( ) 40Q 800V20 2
10V ( ) 800 / 10.25
oo o
out o o
out
j
j
ωω
ω
ωω
Ω = + = Ω
= = = = = Ω
= = = = =
−⎛ ⎞
∴ = + ⎜ ⎟
⎝ ⎠ (b)
9 rad/s800(Approx: V ( 9) 194.03V
1740 600Exact: V
3 (6 600 / )24,000V ( 9)
9[3 (54 66.67)]
out
out
out
j
j j
jj
204.86 13.325 V
ω
ω ω ω
−
= =
= ×+ −
=
∴ = = ∠ −+ −
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
35. 7Series: R 50 , L 4mH, C 10−= Ω = = (a) 3 71/ 4 10 50 krad/soω − −= × = (b) 350 10 / 2 7.958kHzof π= × =
(c) 3 3L 50 10 4 10Q 4
R 50o
oω −× × ×
= = =
(d) 3B / Q 50 10 / 4 12.5 krad/so oω= = × =
(e) 21 1 (1/ 2 ) 1/ 2 50 1 1/ 64 1/ 8 44.14 krad/so o oQ Qω ω ⎡ ⎤ ⎡ ⎤= + − = + − =⎣ ⎦⎣ ⎦
(f) 2 50 65 / 64 1/ 8 56.64 krad/sω ⎡ ⎤= + =⎣ ⎦
(g) 7 3Z ( 45,000) 50 (180 10 / 45) 50 42.22 65.44 40.18in j j j−= + − = − = ∠ − °Ω (h) 7
45,000Z / Z 10 / 45,000 50 4.444c R j= × =
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
36. Apply 1 A, in at top. V 10VR∴ = (a)
8 83 3
3 8 3 8
2 11
10 1.2 10V Z 10 10 (0.5 10 1) 10 105
Z ( ) 10 (10 1.2 10 / ) 10 1.2 10 /
1.2 10 ,
in in
in o o
o o
s ss s
j j
(b) 3 3L 346.4 10Q 34.64
R 10o
oω −×
= = =
346.4 krad/s
ω ω ω ω ω
ω ω
− −
− −
×+ + × + = + +
= + − × ∴ = ×
= =
= × =∴
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
37. Find the Thévenin equivalent seen by the inductor-capacitor combination:
11 1 1
1
6
max
VSC : 1.5 V 10 0.105V V 50V125
50I 0.4A125
1.5OC :V 0 V 1.5V R 3.750.4
1000 41/ 4 0.25 10 1000,Q 1066.73.75
1000 1B / Q 0.9375, B 0.4688 rad/s1066.7 2
V Q V 1066.7 1.5 1600V
−
⎛ ⎞= + − ∴ =⎜ ⎟⎝ ⎠
∴↓ = =
= ∴ = ∴ = = Ω
×∴ = × × = = =
= = = =
= = × =
SC
OC th
o o
o o
C o th
ω
ω
Therefore, keep your hands off!
To generate a plot of |VC| vs. frequency, note that VC(jω) =
CjLj
Cj
ωω
ω
−+
−
75.35.1
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
38. ,0Series, 500Hz, Q 10, X 500o o Lf = = = Ω
2 2
,0
1 2(a) (b)
500 = =L 2 (500)L L 0.15915 H, C 0.6366 FL (2 500)
X 500Q 10 R 50R R
+∴ = = = =×
= = = ∴ = Ω
oo
Lo
πω π μω π
6
6
1 10 0.5 250,0001 = +I 50 2 I 502 2
10 0.5I 1/ 50 ( 250,000 / ), V I2
250,000 /V V (2 450) 4.757 V50 ( 250,000 / )
V (2 500) 10,000 V V (2 550) 4.218V
⎛ ⎞ ⎛ ⎞×× − = + −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠×
∴ = + − =
−= ∴ × =
+ −
× = × =
c
C c
c c
j f j j f jf f
j f fj f
j fj f f
πππ π
ππ
π
π π
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
39.
1 4
8 2
8
1
X : 0, ,0 : 20,000 80,000 , Z ( 10 ) 20 0 SERIES120,000, 80,000 (64 4)10 82,462 rad/s, 68 10
LCR R 1 L 68 10 120,000 40,000, 170,000; Z( ) R L2L L LC R 40,000 C
8
20 R 10,
−= ∞ = − ± − = − + Ω ∴
= = ∴ = + = = = ×
×= = ∴ = × = = = + +
1000L10,00
∴− = −
in
d o o
s s j s j
α ω ω ω
α σ σ
−
σ
1 170,000R R R R 1.23080C 4 10,000
130.77 H, C 4.779 F170,000 1.2308
= − − ∴ = Ω
1.2308L40,000
∴ = = = =×
μ μ
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
40. = 4.287 ∠ 59.04o kΩ
5 33 7 5
25 7
5 7
3
101/ 10 10 rad/s, Q 100, R 10,0001
1Q 500, R 500 0.2 50,00010 0.2
50 10 8.333 Q CR 10 8333 83.33100,000B 1200 rad/s, Z ( ) 8333
83.33(99 100)1099,000 N 1.6667, Z
600
−− −
−
−
ω = = = = Ω
= = = × = Ω×
= Ω ∴ = ω = × =
= = ω = Ω
−ω = ∴ = = −
o L PL
c PC
o o
in o
i
k
j
8.333( 99,000)1 1.667
=−n j
j
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
41. Req = Qo/ ωo C = 50 / 105-7 = 5000 Ω. Thus, we may write 1/5000 = 1/8333 + 1/Rx so that Rx = 12.5 kΩ.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
42.
3 5
3 42
3 4
4 3
2
6
4
13mH 1.5mH 1mH, 2 F 8 F 10 F, 10 krad/s10
3 10 10Q 100, R 100 0.3 30.3
1.5 10 10Q 60, R 60 0.25 9000.25
692.3900 3000 692.3 Q 69.2310
692.3R 0.1444469.2310Q 125, R
10 0.1 8
− −
−
−
−
= μ + μ = μ ∴ω = =
× ×= = = × = Ω
× ×= = = × = Ω
= Ω ∴ = =
∴ = = Ω
= =× ×
o
p
p
L
LS
pc
k
2
4 52
, min
125 0.1 1562.5 10 F
1562.5Q 10 10 15625 156.25 R 0.064(156.25)
R 0.14444 0.064 0.2084 Z , 10 krad/s
−
= × = Ω μ
∴ = × × = ∴ = = Ω
∴ = + = Ω = ω =
c SC
S tot in o
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
43. (a) 3
2
2
2
3
1/ 2 0.2 10 50 rad/s
Q 50 2.5 / 2 62.5, 2 62.5 7812.5
50 10Q 50, 10 50 2510
1000Q 100, 100 1 10 , R 7.8125 25 10 373150 0.2 1
50 1Q 50 3731 0.2 10 37.31; B 1.3400, B 0.670037.31 2
V 10
−
−
ω × × =
= × = × = Ω
×= = × = Ω
= = × = Ω = = Ω× ×
= × × × = = = =
(b)
∴ =
o
leftL
rightL
c p
o
o
k
k
3 3731− × = 3.731V
3
3
V = +10 [(2 125) (10 500) (1 100)]
101 1 1
2 125 10 500 1 100
−
−
+ −
= = ∠ − °+ +
+ + −
j j j
j j j
3.7321 0.3950 V+
3.731 V
2.638 V
50
↔ 1.34 rad/s
|V| (volts)
ω (rad/s)
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
44. (a) 6 3
42
,
,
1000 2000 rad/s, Q 2000 2 10 25 10 1000.25
R 20 10R 25,000 /100 2.5 ; Q 40L 2000 0.25
20,000R 12.5 R 12.5 2.5 151600
2000 0.25 1Q 33.33 V 1 33.33 16.667 V15 2
−ω = = × × × × =
×
∴ = = Ω = = =ω ×
∴ = = Ω ∴ = + = Ω
× (b)
∴ = = ∴ = ×
o c
C S Lo
L S tot
o x × =
20,000 50020,000 500 12,4922 499.68820,000 500
25,000( 250)25,000 250 2.4998 249.97525,000 250
Z 12.4922 2.4998 499.688 250 249.975 14.9920 0.2870
I 1/ 14.9920 0.2870 66.6902mA V 250
×= = + Ω
+−
− = = −−
∴ = + + − − = − Ω
= − = ∴ = ×in
x
jj jj
jj jjj j j j
j 366.6902 10−× = 16.6726V∴
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
45. 2
2 2, , and 1 1
P PS S
R Q XQ CR R XQ Q
ω= = =+ +
2
2
1 1 1, S P S PS P
QX X C CC Cω ω
+= − = − ∴ =
Q
(a) ω = 103 rad/s, Q = 5 Therefore, RS = 5/26 = 192 Ω, CS = 26/25 μF = 1.06 μF (b) ω = 104 rad/s, Q = 50 Therefore, RS = 5/2501 = 2 Ω, CS = 2501/2500 μF = 1.0004 μF (c) ω = 105 rad/s, Q = 500 Therefore, RS = 5000/250001 = 20 mΩ, CS = 250001/250000 μF = 1.0 μF
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
46. ( )2
22
1 1 , and P S P SQR R Q X X
Q+
= + =
2
21P SQC C
Q=
+
(a) ω = 103 rad/s, Q = 0.2 Therefore, RP = 5(1 + 0.04) = 5.2 kΩ, CP = 38.5 nF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 5(1 + 0.0004) = 5.002 kΩ, CP = 400 pF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 5(1 + 4×10–6) = 5 kΩ, CP = 4 pF
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
47. 2 2
2 2, , and . 1 1
P PS S S
R Q XR QQ R X L LL Q Qω
= = = =+ + 21P Q+
(a) ω = 103 rad/s, Q = 142.4×103
Therefore, RS = 470/(1 + Q2) = 23.2 nΩ, LS = 3.3 μH (b) ω = 104 rad/s, Q = 14.24×103
Therefore, RS = 470/(1 + Q2) = 23.2 μΩ, LS = 3.3 μH (c) ω = 105 rad/s, Q = 1.424×103
Therefore, RS = 470/(1 + Q2) = 232 μΩ, LS = 3.3 μH
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
48. ( )2
22
1 1 , and P S P SQR R Q X X
Q⎛ ⎞+
= + = ⎜ ⎟⎝ ⎠
2
2
1P S
QL LQ
⎛ ⎞+= ⎜ ⎟
⎝ ⎠
(a) ω = 103 rad/s, Q = 7.02×10–6
Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 67 mF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 670 μF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 6.70 μF
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
49. (a) For the left parallel circuit, 7 6
470 4710 10
RQLω −= ≈ = . Since Q > 5, the series
equivalent is a 10/47 Ω resistor in series with 1 μH. For the right parallel circuit, ( )7 810 10 200 20Q CRω −= ≈ = . Again, Q > 5, so the series equivalent is a 10/20 Ω = 500 mΩ resistor in series with 10 nF. We may therefore approximate the network as a 700 mΩ resistor in series with a 10 nF capacitor, in series with a 1 μH inductor, in series with the 10 μH inductor of interest. At the resonant frequency the network connected in series with the inductor has an impedance of 700 mΩ. The inductor present an impedance of 100 Ω. Thus, |Vx| = 1 V.
(b) ZL = ( ) 7 6470 ( 10 10 )
0.213 9.995 470 10
jj
j
−
= + Ω2
2
22
1
0.499 9.975 1L
Rj C j
Rj C
ω
ω+
. = = − Ω+
Z
Z3 = j100 Ω.
Thus, ( )3
1 3
1001 0 0.99745 0.0071 V0.714 0.02x
L
j jj
= ∠V = = ++ + +
ZZ Z Z
So that |Vx| = 0.99977 V . Our approximation was pretty accurate, at least at this frequency.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
50. 3
6
50 20 10(a) K = = 0.5 K 0.02100 10
1 0.59.82 H 0.5 9.82 24.55 H, 31.8 H 31.8 795 H0.02 0.02
2.572.57 nF0.5 0.02
×= =
∴ μ → × × = (b) same ordinate; divide numbers on abscissa by 50
257 nF→ =×
m f
μ μ → × = μ
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
51. (a) 1 1Apply 1 V I 10A 0.5 I 5A ; 5A 0.2 can be replaced by 1 V in series with 0.2
1 ( 1) 2 4 20 20( 5)I 10 100.2 2 / 0.2 2 0.2 2 10
∴ = ∴ = ↓ Ω Ω
− − + + 10Z ( )20( 5)
+ ∴
(b) 2( / 5 10) 0.1( 50)K 2, K 5 Z ( )20( / 5 5) 25
+ += = ∴ → =
+ +m f ins sss s
(c) 1 10.1 0.2 , 0.2 0.4 , 0.5F 0.05F, 0.5I 0.5IΩ → Ω Ω → Ω → →
→= + = + = =+ + + +in
s s ss s s s
∴ =+ in
sss
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
52. (a) 3 6 4
44 3 4
,8 , 2
4 3
2 2, ,4 6
4 6
1/ (2 8)10 10 10 rad/s
10Q 10 / 8 10 10 125 R 0.64125
10 10 102 8 10 mH Q 156.250.64
1R 0.64 156.25 15.625 ; Q 100, R 100 1 1010 10
R 20 15.625 10 4.673 Q 10 10 4
− −
−
−
−
−
(b)
(c) 6
6 1010 rad/s, Q stays the same, B 21.40 krad/s46.73
ω = ∴ = =o o
ω + =
= × = ∴ = = Ω
× ×+ = ∴ = =
∴ = × = Ω = = = × = Ω×
∴ = = Ω ∴ = × ×
o
L L S
L
L P C C P
P o
k k
k 3.673 10 46.73× =
6 4K = =f m10 /10 100, K 1 R s stay the same; 2 mH 20 H, 8mH 80 H,1 F 10nF′= ∴ → μ → μ μ →
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
53.
3
0.1(a) K = =250, K 400 0.1F 1 F
250 4002 2505 1250 , 2H 1.25H, 4 I 10 I
400
∴ → = μ×
×Ω → Ω → = →
m f
x x (b) 3 6
1250
33
6 3 6 3
33 3
110 . Apply 1 V I 10 , I1250
1 101000 I 10 I1.25
1 0.8 0.8I 10 (1 10 ) 10 ; 1012500.8 10 1 1000I 10 0.2 10 Z 5 V
I 0.2
−
−−
− − −
−− −
ω =
0
∴ = ↓ =
−
∴ = ∴→ =
∴ = + + − = + =
×= + = × ∴ = = = − Ω =
x
x L
in
in th ocin
s
sss
s s s s js s
j j j kj j
1 μF
1.25 H
1250 Ω 103
∴
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
54. (a) I 2 0 A, 50 V 60 25 V= ∠ ° ω = ∴ = ∠ °s out (b) I 2 40 A, 50 V 60 65 V= ∠ ° ω = ∴ = ∠ °s out (c) I 2 40 A, 200, OTSK= ∠ ° ω = ∴s (d) K 30, I 2 40 A, 50 V 1800 65 V= = ∠ ° ω = ∴ = ∠ °m S out (e) K 30, K 4, I 2 40 A, 200 V 1800 65 V= = = ∠ ° ω = ∴ = ∠m f s out °
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
55. (a) H /( ) 0.2 H 20 log 0.2 13.979dB= ∴ = = −dBs (b) H( ) 50 H 20 log 50 33.98dB= ∴ = =dBs (c) 12 26 6 13 292 380 (d) 37.6/ 20H 37.6dB H( ) 10 75.86= ∴ = =dB s (e) 8/ 20H 8dB H( ) 10 0.3981−= − ∴ = =dB s (f) 0.01/ 20H 0.01dB H( ) 10 1.0012= ∴ = =dB s
H j( 10) H 20 log 20 log 2 10 20 10 1 5 10 5 60 220
+= + ∴ = + =
+ + + + − +dBj
j j j j j6.451dB=
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
56. (d) MATLAB verification- shown adjacent to Bode plots below.
(a) 20( 1) 0.2(1 )H( ) , 0.2 14dB100 1 /100+ +
= = → −+ +s ss
s s
1 10 100
(b) 2 2
2000( 1) 0.2 (1 )H( ) , 0.2 14dB( 100) (1 /100)
+ += = → −
+ +s s s ss
s s
(c) 2200 45 200 ( 5)( 40) 200(1 / 5)(1 / 40)( ) 45 , 200 46dB+ + + + + +
= + + = = = →s s s s s ss s
s s s sH
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
57.
2
V (20 2 )(182 200 / ) 200 /H( )202 2 200 / 182 200 /
400( 10) 200(10 )2( 101 100) (1 )(100 )
20(1 /10)H( ) , 20 26dB(1 )(1 /100)
+ += = ×
+ + ++ +
= =+ + + +
+= →
+ +
C
R
s ss sI s s
s ss s s s
sss s
s
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
58.
(a) 8
3 3
5 10 ( 100) 2.5 (1 /100)H( ) , 2.5 8dB( 20)( 1000) (1 / 20)(1 /1000)
× + += =
+ + + +s s s ss
s s s s→
(b) (c)
2 9
3 3
Corners: 20, 34dB;
100, 34dB;
1000, 54dB
Intercepts: 0dB, 2.5 1, 0.42.5 ( /100) 2.5 (20)101, 8dB; 0dB, 1 22,360 rad/s
( / 20)( /1000) 100
ω =
ω =
ω =
ω = ω =
ω ω ωω = = = ∴ω =
ω ω ωω
2
2 2 3
orners: 20, 31.13dB
1 ( /100)100, 36.69dB H 20 log 2.5
[1 ( / 20) ][1 ( /1000) ]1000, 44.99dB
ω =
+ ωω = = ω
+ ω + ω
ω =
dB
C
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
59.
(a) 8
3 3
5 10 ( 100) 2.5 (1 /100)H( ) ,( 20)( 1000) (1 / 20)(1 /1000)
× + += =
+ + + +s s s ss
s s s s
(b) (c)
2 : 901010 : 90 45 45 log 58.520100 100100 : 90 45 45 log 45 45 log 58.520 100
200 200200 : 90 90 45 45 log 3 45 45 log 17.9100 100
1000 : 90 9
ω
ω
ω
ω
ω
= °
⎛ ⎞= ∠ = ° − ° + ° = °⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= ∠ = ° − ° + ° + ° + ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= ∠ = ° − ° + ° + ° − ° + ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= ∠ = ° −
= ∠
10000 90 3 45 45 log 451000
10,000 : 90 90 90 3 90 180ω
⎛ ⎞° + ° − ° + ° = − °⎜ ⎟⎝ ⎠
= ∠ = ° − ° + ° − × ° = − °
1 1 1
1 1 1
1 1 1
1 1 1
1
2 : 90 tan 0.02 tan 0.1 3tan 0.002 85.09
10 : 90 tan 0.1 tan 0.5 3tan 0.01 67.43
100 : 90 tan 1 tan 5 3tan 0.1 39.18
200 : 90 tan 2 tan 10 3tan 0.2 35.22
1000 : 90 tan 10 t
ω = ∠
ω
ω
ω
ω
− − −
− − −
− − −
− − −
−
= ° + − − = °
= ∠ = ° + − − = °
= ∠ = ° + − − = °
= ∠ = ° + − − = °
= ∠ = ° + − 1 1
1 1 1
an 50 3tan 1 49.56
10,000 : 90 tan 100 tan 500 3tan 10 163.33ω
− −
− − −
− = − °
= ∠ = ° + − − = − °
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
60. 2
2 2
2
2
20 400 20 400(a) (b) (c)
H( ) 1
1 2 0.5( / 20) ( / 20)400
20, 0.520 log 400 52dBCorrection at is 20 log 2 0 dB
o
o
s sss s s
s ss
ω ζ
ω ζ
+ += + + =
+ × +=
∴ = ==
=
5 : H 52 2 20 log 5 24.0dB(plot)
H 20 log 1 16 4 23.8dB (exact)
100 : H 0dB (plot)
H 20 log 1 0.04 0.2 0.170 dB (exact)
dB
dB
dB
dB
j
j
ω
ω
− × =
= − + =
= =
= − + = −
= =
Hdb
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
61. (a)
225
V 25 25 0.025
(c) 0.520, H( 20) H 15.68 dB H( 20) 80.541 4 0.5 dB
jj jj
ω = = ∴ = − ∠ = −− +
°
H( )V 10 25 1000 / 10 25 1000 11 2
8 10 10110, 1/ 8 correction 20 log 2 12 dB8
0.025 32 dB
= = = =+ + + + ⎛ ⎞⎛ ⎞ ⎛ ⎞+ +⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞∴ = = ∴ = − × =⎜ ⎟⎝ ⎠
→ −
R
o
s sss s s s s s
ω ζ(b)
HdB ang(H)
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
62. 3 6
1 2 3 3 6
1/(50 10 10 ) 201st two stages, H ( ) H ( ) 10; H ( )1/(200 10 10 ) 5
20 400H( ) ( 10)( 10)5 1 / 5
400 52 dB
s s ss s
ss s
−
−
− × × −= = − = =
+ × × +
− −⎛ ⎞∴ = − − =⎜ ⎟+ +⎝ ⎠− →
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
63. (a) 20log10(0.1) = -20 dB (b) (c)
51 1 1
15 51
5 6
5 6
t stage: C 1 F, R , R 10 H (S) R C 0.11/ R C
2nd stage: R 10 , R 10 , C 1 F H ( )1/ R C
1/(10 10 ) 10H ( )1/(10 10 ) 10
3rd stage: same as 2nd1H( ) ( 0.1 )
A A fA A fA A
B fBB fB fB B
fB fB
B
s s
ss
ss s
s s
μ
μ
−
−
= = ∞ = ∴ = − = −
−= = = ∴ =
+
×
1s
∴ = = −+ × +
−∴ = − 2
0 10 0.110 10 (1 /10)
ss s s
−⎛ ⎞⎛ ⎞ = −⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
64. An amplifier that rejects high-frequency signals is required. There is some ambiguity in the requirements, as social conversations may include frequencies up to 50 kHz, and echolocation sounds, which we are asked to filter out, may begin below this value. Without further information, we decide to set the filter cutoff frequency at 50 kHz to ensure we do not lose information. However, we note that this decision is not necessarily the only correct one.
Our input source is a microphone modeled as a sinusoidal voltage source having a peak amplitude of 15 mV in series with a 1-Ω resistor. Our output device is an earphone modeled as a 1-kΩ resistor. A voltage of 15 mV from the microphone should correspond to about 1 V at the earphone according to the specifications, requiring a gain of 1000/15 = 66.7.
If we select a non-inverting op amp topology, we then need 65.7 1- 66.7 1
==RR f
Arbitrarily choosing R1 = 1 kΩ, we then need Rf = 65.7 kΩ. This completes the amplification part. Next, we need to filter out frequencies greater than 50 kHz.
Placing a capacitor across the microphone terminals will “short out” high frequencies.
We design for ωc = 2πfc = 2π(50×103) = filtermicCR
1 . Since Rmic = 1 Ω, we require
Cfilter = 3.183 μF.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
65. We choose a simple series RLC circuit. It was shown in the text that the “gain” of the
circuit with the output taken across the resistor is ( )[ ] 2
122222 -1
RC CRLC
AV
ωω
ω
+= .
This results in a bandpass filter with corner frequencies at
LCLCCR-RC
Lc 24
22 ++=ω and
LCLCCRRC
Hc 24
22 ++=ω
If we take our output across the inductor-capacitor combination instead, we obtain the opposite curve- i.e. a bandstop filter with the same cutoff frequencies. Thus, we want
2π(20) = LC
LCCR-RC2
4 22 ++ and 2π(20×103) =
LCLCCRRC
24
22 ++
Noting that
Hcω – Lcω = R/L = 125.5 krad/s, we arbitrarily select R = 1 kΩ, so that L =
7.966 mH. Returning to either cutoff frequency expression, we then find C = 7.950 μF
PSpice verification. The circuit performs as required, with a lower corner frequency of about 20 Hz and an upper corner frequency of about 20 kHz.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
66. We choose a simple RC filter topology:
Where RC1
1 VV
in
out
ωj+= and hence
( )2in
out
RC1
1 VV
ω+= . We desire a cutoff
frequency of 1 kHz, and note that this circuit does indeed act as a low-pass filter (higher frequency signals lead to the capacitor appearing more and more as a short circuit). Thus,
( ) 21
RC1
1 2
c
=+
=ω
where ωc = 2πfc = 2000π rad/s.
A small amount of algebra yields 1 + [2π(1000)RC]2 = 2 or 2000πRC = 1. Arbitrarily setting R = 1 kΩ, we then find that C = 159.2 nF. The operation of the filter is verified in the PSpice simulation below:
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
67. We are not provided with the actual spectral shape of the noise signal, although the reduction to 1% of its peak value (a drop of 40 dB) by 1 kHz is useful to know. If we place a simple high-pass RC filter at the input of an op amp stage, designing for a pole at 2.5 kHz should ensure an essentially flat response above 25 kHz, and a 3 dB reduction at 2.5 kHz. If greater tolerance is required, the 40 dB reduction at 1 kHz allows the pole to be moved to a frequency even closer to 1 kHz. The PSpice simulation below shows a
filter with R = 1 kΩ (arbitrarily chosen) and C = nF 63.66 )1000)(105.2(2
13 =
×π.
At a frequency of 25 kHz, the filter shows minimal gain reduction, but at 1 kHz any signal is reduced by more than 8 dB.
We therefore design a simple non-inverting op amp circuit such as the one below, which
with Rf = 100 kΩ and R1 = 1 kΩ, has a gain of 100 V/V. In simulating the circuit, a gain of approximately 40 dB at 25 kHz was noted, although the gain dropped at higher frequencies, reaching 37 dB around 80 kHz. Thus, to completely assess the suitability of design, more information regarding the frequency spectrum of the “failure” signals would be required.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
68. We select a simple series RLC circuit with the output taken across the resistor to serve as a bandpass filter with 500 Hz and 5000 Hz cutoff frequencies. From Example 16.12, we know that
(500)2 4LC CR2LC
12LR- 22 πω =++=
Lc
and
(5000)2 4LC CR2LC
12LR 22
Hπω =++=c
With -
LH cc ωω = 2p(5000 – 500) = R/L, we (arbitrarily) select R = 1 kΩ, so that L = 35.37 mH. Substituting these two values into the equation for the high-frequency cutoff, we find that C = 286.3 nF. We complete the design by selecting R1 = 1 kΩ and Rf = 1 kΩ for a gain of 2 (no value of gain was specified). As seen in the PSpice simulation results shown below, the circuit performs as specified at maximum gain (6 dB or 2 V/V), with cutoff frequencies of approximately 500 and 5000 KHz and a peak gain of 6 dB.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
69. For this circuit, we simply need to connect a low-pass filter to the input of a non- inverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the cutoff frequency is
(3000)2 RC1 πω ==c
Selecting (arbitrarily) R = 1 kΩ, we find C = 53.05 nF. The PSpice simulation below shows that our design does indeed have a bandwidth of 3 kHz and a peak gain of 10 V/V (20 dB).
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
70. We require four filter stages, and choose to implement the circuit using op amps to isolate each filter sub- circuit. Selecting a bandwidth of 1 rad/s (no specification was given) and a simple RLC filter as suggested in the problem statement, a resistance value of 1 Ω leads to an inductor value of 1 H (bandwidth for this type of filter = ωH – ωL = R/L). The capacitance is found by designing each filter’s respective resonant frequency ( LC1 ) at the desired “notch” frequency. Thus, we require CF1 = 10.13 μF, CF2 = 2.533 μF, CF3 = 1.126 μF and CF4 = 633.3 nF. The Student Version of PSpice® will not permit more than 64 nodes, so that the total solution must be simulated in two parts. The half with the filters for notching out 50 and 100 Hz components is shown below; an additional two op amp stages are required to complete the design.
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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006
71. Using the series RLC circuit suggested, we decide to design for a bandwidth of 1 rad/ s (as no specification was provided). With ωH – ωL = R/ L, we arbitrarily select R = 1 Ω so that L = 1 H. The capacitance required is obtained by setting the resonant frequency of the circuit ( LC1 ) equal to 60 Hz (120π rad/s). This yields C = 7.04 μF.
1 H
1 Ω
7.04 μF
vin vout
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