Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition

1.

Chapter Sixteen Solutions

10 March 2006

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
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2.


(a) R = 1000 Ω and C = 1 μF.
Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 =
(b) L = 12 fH and C = 2.4 nF
R = Qo (L/ C)½

=

(c) R = 121.7 kΩ and L = 100 pH
C = (Qo / R)2 L =

10 March 2006

25 μH

447.2 mΩ

270 aF



3.


10 March 2006

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 Cj2 Rp Rs)(Cj Rp) - ωCjRp(2ωCj2 Rp Rs)]/ (1 + ω2 Cj2 RpRs)
Setting this equal to zero, we may subsequently write
CjRp (1 + ω2 Cj2 Rp Rs) - ωCjRp(2ωCj2 Rp Rs) = 0
Or

1 – ω2 Cj2 Rp Rs = 0. Thus, ωo = (Cj2 RpRs)–½

= 129.4 Mrad/s = 21.00 MHz

Qo = Q(ω = ωo) = 366.0



4.


10 March 2006

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).




10 March 2006

5.

α = 80Np/s, ω d = 1200 rad/s, Z(−2α + jω d ) = 400 Ω
ω o = 12002 + 802 = 1202.66 rad/s ∴ Qo =
Now, Y( s ) = C

( s + α − jω d )( s + α + jω d )
(−α )(−α + j 2ω d )
∴ Y(−2α + jω d ) = C
s
−2α + jω d

∴ Y(−160 + j1200) = C
∴C =

ωo
= 7.517
2α

−80(−80 + j 2400)
1
−1 + j 30
∴ Y(−160 + j1200) =
= 80C
−160 + j1200
400
−2 + j15

1
229
1
1
= 15.775− μ F; L = 2 = 43.88 mH; R =
= 396.7 Ω
ωo C
32, 000 901
2α C




10 March 2006

6.
Yin =

1
1
2 − j 0.1ω
jω
+ 0.2 +
=
+ 0.2 +
2
2 + j 0.1ω
1 + 1000 / jω 4 + 0.01ω
1000 + j10

2 − j 0.1ω
ω 2 + j1000ω
1000ω
−0.1ω
+ 0.2 +
∴
+ 2
=0
2
6
2
2
4 + 0.1ω
10 + ω
4 + 0.01ω ω + 106
∴ 0.1ω 3 + 105ω = 4000ω + 10ω 3 ∴ 9.9ω 2 = 96, 000 ∴ω = 98.47 rad/s
=




7.

Parallel: R = 106 , L = 1, C = 10−6 , Is = 10∠0° μ A

(a)

ωo =

(b)

10 March 2006

1⎞
I
⎡
1000 ⎞ ⎤
⎛
⎛ ω
−
Y = 10−6 + j ⎜ 10−6 − ⎟ , V = = 10−5 /10−3 ⎢10−3 + j ⎜
ω⎠
ω ⎟⎥
Y
⎝
⎝ 1000
⎠⎦
⎣

1
= 1000 rad/s; Qo = ω o RC = 103+ 6−6 = 1000
LC

∴V =

10−2
10−2
, V =
2
1000 ⎞
⎛ ω
1000 ⎞
⎛ ω
0.001 + j ⎜
−
10−6 + ⎜
−
ω ⎟
⎝ 1000
⎠
ω ⎟
⎝ 1000
⎠

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




10 March 2006

8.
(a)

5(100 / jω )
j 0.1ω
+2+
5 + (100 / jω )
10 + j 0.01ω
500
100
100(20 − jω )
j10ω
j10ω
j10ω (1000 − j )
=
+2+
=
+2+
=
+2+
2
ω + 400
ω 2 + 106
100 + j 5ω
1000 + jω 20 + jω
1000 + jω
Zin =

104 ω
−100ω
+ 2
= 0 ∴ω 2 + 106 = 100ω 2 + 40, 000, 99ω 2 = 960, 000
2
6
ω + 400 ω + 10
∴ω o = 960, 000 / 99 = 98.47 rad/s
∴

(b)

Zin (ω o ) =

10ω 2
2000
+ 2 + 2 o 6 = 2.294 Ω
2
ω o + 400
ω o + 10




10 March 2006

9.
(a)

2
2
α = 50 s −1 , ω d = 1000 s −1 ∴ω o = α 2 + ω d = 1, 002,500 ∴ω o = 1001.249

1
106
1
106
+
L= 2 =
= 0.9975 H; R =
=
= 10 k Ω
2α C 100
ω o C 1, 002,500
(b)

1
1
⎛
⎞
Y = 10−4 + j ⎜ 10−6 ω −
⎟ , ω = 1000 ∴ Z = = 9997∠1.4321° Ω
0.9975ω ⎠
Y
⎝




10 March 2006

10.
f min = 535 kHz, f max = 1605 kHz, Qo = 45 at one end and
Qo ≤ 45 for 535 ≤ f ≤ 1605 kHz
f o = 1/ 2π LC ∴ 535 × 103 =

1
1
,1605 × 103 =
2π L max C
2π L min C
2

1
⎛
⎞
∴ L max / L min = 3; L max C = ⎜
= 8.8498 × 10−14
3 ⎟
⎝ 2π × 535 ×10 ⎠

ω o RC ≤ 45,535 × 103 ≤

ωo
≤ 1605 × 103. Use ω o max
2π

∴ 2π × 1605 ×103 × 20 × 103 C = 45 ∴ C = 223.1pF
∴ L max =

8.8498 ×10−14
L
= 397.6 μ H, L min = max = 44.08 μ H
−12
223.1× 10
9




10 March 2006

11.
(a)

Apply ± 1V. ∴ IR = −10−4 A
1
+ 10−4 + (1 − [105 (−10−4 )])10−8 s
4.4 × 10−3 s
1000
48.4 × 10−8 s 2 + 4.4 × 10−4 s + 1000
−4
−8
∴ Yin =
+ 10 + 11× 10 s =
4.4 s
4.4 s
−8 2
−4
1000 − 48.4 × 10 ω + j 4.4 ×10 ω
∴ Yin ( jω ) =
j 4.4ω
∴ Yin = Iin =

(b)

2
At ω = ω o , 1000 = 48.4 × 10−8 ω o , ω o = 45.45− krad/s
−1

⎛ j 4.4 ×10−4 ω o ⎞
Zin ( jω o ) = ⎜
⎟ = 10 k Ω
j 4.4ω o ⎠
⎝



12.

ω0 =


10 March 2006

ω
1
= 24 = 4.9 rad/s or f0 = 0 = 780 mHz
2π
LC



13.

ω0 =

1
=
LC

1

(

1
25 × 10−6
1.01

)


= 200 rad/s or f0 =

10 March 2006

ω0
= 31.99 Hz
2π




1
103
=
=5Ω
2α C 200

(a) α =

1
2 RC

ω0 =

14.

10 March 2006

ω
1
1
=
= 1000 rad/s or f0 = 0 = 159.2 Hz
2π
LC
10−6

∴ R=

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.



15.

(a) α =


10 March 2006

1
2
= 50 s -1 and ωd = ω0 − α 2 = 5000 rad/s
2RC

Zin(ω0) = R so find R.

(

)

2
L ωd + α 2
1
1
=
= 250 Ω
C= 2 = 2
= 40 μ F . R =
2α C
2(50)
ω0 L ωd + α 2 L

1

(

)

1
= 5000 rad/s or f 0 = 795.8 Hz .
LC
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.
(b) The resonant frequency is




16.

ω o = 1000 rad/s, Qo = 80, C = 0.2 μ F

(a)

1
106
80
= 5 H, Qo = ω o RC ∴ R = 3
= 400 k Ω
L= 2 =
6
ω o C 0.2 ×10
10 × 0.2 ×10−6

(b)

10 March 2006

B = ω o / Q o = 1000 / 80 = 12.5
1
∴ B = 6.25 rad/s
2

ω − ωo

⎛ ω − ωo ⎞
∴ Z = R / 1+ j
= 400 × 10 / 1 + ⎜
⎟
B/2
⎝ 6.25 ⎠

2

3




10 March 2006

17.

ω1 = 103rad/s, ω 2 = 118,
Z( j105) = 10 Ω

ω o2 = ω1ω 2 = 103 ×118
ωo

110.245+
∴ω o = 110.245 , B = 118 − 103 = 15 rad/s, Qo =
=
= 7.350
B
15
7.350
1
1
∴ 7.350 = ω o RC ∴ RC =
= 66.67 × 10−3 , LC = 2 =
+
110.2451
ω o 12,154
+

1
1 ⎞
12,154 ⎞
⎛
⎛
+ j ⎜ 105C −
C ⎟ = 18.456 C
⎟ = 15C + j ⎜105C −
R
105L ⎠
105
⎝
⎝
⎠
0.1
1
1
∴C =
= 5.418 mF, R = C = 12.304 Ω, L =
= 15.185− mH
18.456
15
12,154C
Y( j105) = 0.1 =




10 March 2006

18.

ω o = 30 krad/s, Qo = 10, R = 600 Ω,

(a)

B=

ωo

(b)

N=

ω − ωo

(c)

Zin(j28 000) = 600 / (1 – j1.333) = 360 ∠ 53.13o Ω

(d)

Qo
1
10
⎡ 1
⎤
Zin ( j 28, 000) = ⎢
+ j 28, 000C − j
⎥ , C = ω R = 30, 000 × 600
28, 000L ⎦
⎣ 600
o

Qo

= 3 krad/s

B/ 2

=

28 − 30
= −1.3333
1.5
−1

R
600
1 30, 000 ×10
⎡ 1
⎛ 28 10 30 10 ⎞ ⎤
L=
, =
=
∴ Zin = ⎢
+ j⎜ ×
−
⎟⎥
600
ω o Qo 30, 000 ×10 L
⎝ 30 600 28 600 ⎠ ⎦
⎣ 600
600
= 351.906∠54.0903°Ω
Zin =
⎛ 28 30 ⎞
1 + j10 ⎜ − ⎟
⎝ 30 28 ⎠
(e)

−1

approx-true
360 − 351.906
= 100%
= 2.300%
true
351.906
53.1301° − 54.0903°
= −1.7752%
angle: 100%
54.0903°
magnitude: 100%




19.

f o = 400 Hz, Qo = 8, R = 500 Ω, IS = 2 ×10−3 A ∴ B = 50 Hz

(a)

V = 2 × 10−3 × 500 / 1 + N 2 = 0.5 ∴1 + N 2 = 4, N = ± 3 =

10 March 2006

f − 400
50 / 2

∴ f = 400 ± 25 3 = 443.3 and 356.7 Hz

(b)

IR =

v
R

=

1
1+ N

2

×

1
= 0.5 × 10−3 ∴ 1 + N 2 = 4, N 2 = 15, N = ± 15
500

∴ f = 400 ± 25 15 = 496.8 and 303.2 Hz



20.

Qo =

(b)

Approx: 2 = 5 / 1 + N

10 March 2006

ω o = 106 , Qo = 10, R = 5 × 103 , p.r.

(a)


R
5 ×103
= 0.5 mH
∴L =
ωo L
10 × 106
2

∴ N = 2.291 =

ω − 106
106 / 20

∴ω = 1.1146 Mrad/S

2
⎛ ω ωo ⎞⎤
1⎡
1⎞
⎛
Exact: Y = ⎢1 + jQo ⎜
− ⎟ ⎥ ∴ 0.5 = 0.2 1 + 100 ⎜ ω − ⎟ (ω in Mrad/S)
R⎣
ω⎠
⎝
⎝ ωo ω ⎠⎦
1
1
∴ 6.25 = 1 + 100(ω 2 − 2 + 1/ ω 2 ), ω 2 − 2 + 2 = 0.0525, ω 2 + 2 = 2.0525

ω 4 − 2.0525ω 2 + 1 = 0, ω 2 =
(c)

(

ω

)

1
2.0525 + 2.05252 − 4 = 1.2569, ω = 1.1211 Mrad/s
2

Approx: ∠Y = 30° ∴ tan −1 N = 30°, N = 0.5774 =
Exact: Y =
∴ω −

1

ω

ω

ω −1
1/ 20

, ω = 1.0289 Mrad/s

1 ⎡
1 ⎞⎤
1⎞
⎛
⎛
⎢1 + j10 ⎜ ω − ω ⎟ ⎥ (in Mrad/s) ∴ tan 30° = 0.5774 = 10 ⎜ ω − ω ⎟
5000 ⎣
⎝
⎠⎦
⎝
⎠

= 0.05774, ω 2 − 0.05774ω − 1 = 0, ω =

0.05774 + 0.057742 + 4
= 1.0293 Mrad/s
2




10 March 2006

21.
(a)
(b)

C = 3 + 7 = 10 nF ∴ω o =

1
−4

10 10

−8

= 106 rad/s

Q o = ω o CR = 10610−85 5 × 103 = 50
B = ω o / Q o = 20 krad/s
Parallel current source is

1∠0°
= jω 3 × 10−9 At ω o , I s = j106 −9 × 3
Z3

∴ V1,0 = j 3 × 10−3 × 5 × 103 = 15∠90° V

(c)

ω − ω o = 15 × 103 ∴ N =

15 ×103
15∠90°
= 1.5 ∴ V1 =
= 8.321∠33.69° V
3
10 × 10
1 + j1.5




10 March 2006

22.
(a)

(5 + 0.01s )(5 + 106 / s ) (5 + 0.01s )(5s + 106 )
=
10 + 0.01s + 106 / s
0.01s 2 + 10 s + 106
0.05s 2 + 25s + 104 s + 5 × 106
Zin ( s ) =
0.01s 2 + 10 s + 106
5 ×106 − 0.05ω 2 + j10, 025ω
∴ Zin ( jω ) =
106 − 0.01ω 2 + j10ω
10, 025ω o
10ω o
2
2
At ω = ω o ,
= 6
, 10.025 × 109 − 100.25 ω o = 5 × 107 − 0.5 ω o
6
2
2
5 × 10 − 0.05ω o 10 − 0.01ω o
Zin ( s ) =

2
∴ 99.75ω o = 9.975 × 109 , ω o = 10, 000 rad/s

(b)

Zin ( jω o ) = (5 + j100) (5 − j100) =

25 + 10, 000
= 1002.5 Ω
10




23.

, f o = 1000 Hz, Qo = 40, Zin ( jω o ) = 2k Ω ∴ B = 25 Hz

(a)

Zin(jω) =

10 March 2006

2000
f − 1000
, N=
, f = 1010, ∴ N = 0.8
1 + jN
12.5

Zin = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω

(b)

0.9 f o < f < 1.1 f o ∴ 900 < f < 1100 Hz



24.


10 March 2006

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



25.


10 March 2006

Bandwidth = 2π f 0 = 2π 106 = ω2 − ω1 , where ω1 = 2π ( 5.5 )103 .
(a) ω2 = ω1 + B , therefore f2 = 5.5 + 103 kHz = 1.0055 MHz
(b) f 0 =
(c) Q0 =

f1 f 2 =

( 5.5 )(1005.5) =

74.37 kHz

f 0 74.37 ×103
=
= 0.074
106
B



26.


10 March 2006

Bandwidth = 109 Hz = f 2 − f1 , where f1 = 75.3 ×106 Hz.
(a) f 2 = f1 + B , therefore f2 = 1.0753 GHz
(b) f 0 =
(c) Q0 =

f1 f 2 =

( 75.3 ×10 )(1.0753 ×10 ) =
6

9

284.6 MHz

f 0 284.6 × 106
=
= 0.2846
109
B



27.


10 March 2006

(a) To complete the sketch, we need to first find ω0, which we obtain in part (b).

(b) ω0 = ω1ω2 = 2000 rad/s or f 0 = 318.3 Hz
(c) B = ω2 − ω1 = 3000 rad/s or
(d) Q =

ω0

ω2 − ω1

=

477.5 Hz

2000
= 0.667
3000



28.


10 March 2006

(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.
2
2
R12 + X 12
R2 + X 2
, and similarly X p2 =
.
Then X p1 =
X1
X2

For X p1 + X p2 = 0 we have

1
At ω0, X 1 = −
ω0C

At ω0, X 2 = ω0 L

∴

∴

2
2
R12 + X 12 R2 + X 2
+
=0
X1
X2

R12 + X 12
=
X1

52 +

[1]

1024

ω02 ( 330 )
−1012
330ω0

2

.

2
2
2
R2 + X 2 52 + 10−4 ω0
=
.
10−2 ω0
X2

Enforcing Eq. [1], then, leads to ω0 =

1022 − ( 25 ) (330)1012
(330)108 − 25(33) 2

= 550.5 krad/s

or f0 = 87.61 kHz.
(b) We see the simulation result agrees reasonably, with a resonant frequency of 87.6 kHz



29.


10 March 2006

(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.
f 2 = f1 + B = 6 kHz
f0 =

f1 f 2 =

( 0.5 ) (6) =

3 kHz

f0
3 × 103
Q0 =
=
B 5.5 × 103
Q0 = ω0 RC so C =

L=

(

Q0
1
=
= 28.9 nF
3
ω0 R 5.5 × 10 ( 2π )103

(

)

)

5.5 ×103 103
1
=
= 292 mH
ω02C
2π 3 × 106

(

)

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.



30.

(a) f 0 =

(b) Q0 =

1
2π

1
1
=
LC 2π

ω0 L

=

R


1

(

)(

400 × 10−6 3.3 × 10−6

)

10 March 2006

= 4.38 kHz

1 L 1 400
=
= 1.10
LC R 10 3.3

(c) Z at resonance = R = 10 Ω
(d) Z at 0.438 kHz =
⎡
1
10 + j ⎢ 2π ( 438 ) 400 × 10−6 −
2π ( 438 ) 3.3 × 10−6
⎢
⎣

(

)

(

(e) Z at 43.8 kHz =
⎡
1
10 + j ⎢ 2π ( 438 ) 400 × 10−4 −
2π ( 438 ) 3.3 × 10−4
⎢
⎣

(

)

(

)

⎤
⎥ = 10 − j109.01 Ω
⎥
⎦

)

⎤
⎥ = 10 − j108.98 Ω
⎥
⎦



31.


10 March 2006

Bandwidth = 3 MHz, f1 = 17 kHz.
(a) f 2 = f1 + B =
(b) f 0 =
(c) Q0 =

f1 f 2 =

3.017 MHz
226.5 kHz

f0
= 0.0755
B



32.


10 March 2006

(a) Z0 = 1 Ω by definition

(b) ω0 =

1
103
=
= 707 rad/s
LC
2

= 112.5 Hz

(c) PSpice simulation verifies an impedance of 1 Ω at f = 112.6 Hz.



33.


10 March 2006

(a) Z0 = 1 kΩ by definition

(b) ω0 =

1
106
=
= 707 krad/s
LC
2

= 112.5 kHz

(c) PSpice simulation verifies an impedance of 1 kΩ at f = 112.8 kHz.




10 March 2006

34.
(a)

20A 6Ω, 3 6 = 2, 40 V in series with 2 + 1 = 3 Ω
1
ω L 60
= 10 rad/s, Qo = o =
= 20 Ω
R
3
LC
10
1
B=
= 0.5, B = 0.25, Vout ( jω o ) = 40Qo = 800 V
20
2

ωo =

⎛ ω − 10 ⎞
∴ Vout ( jω ) = 800 / 1 + ⎜
⎟
⎝ 0.25 ⎠

(b)

2

ω = 9 rad/s
800
= 194.03V
17
40
600
Exact: Vout =
×
3 + j (6ω − 600 / ω ) jω
24, 000
∴ Vout ( j 9) =
= 204.86∠ − 13.325− V
9[3 + j (54 − 66.67)]
(Approx: Vout ( j 9) =




10 March 2006

35.

Series: R = 50 Ω, L = 4 mH, C = 10−7

(a)

ω o = 1/ 4 × 10−3−7 = 50 krad/s

(b)

f o = 50 ×103 / 2π = 7.958 kHz

(c)

Qo =

(d)

B = ω o / Qo = 50 × 103 / 4 = 12.5 krad/s

(e)

ω1 = ω o ⎡ 1 + (1/ 2Qo ) 2 − 1/ 2Qo ⎤ = 50 ⎡ 1 + 1/ 64 − 1/ 8⎤ = 44.14 krad/s
⎣
⎦

(f)

ω 2 = 50 ⎡ 65 / 64 + 1/ 8⎤ = 56.64 krad/s
⎣
⎦

(g)

Zin ( j 45, 000) = 50 + j (180 − 107 −3 / 45) = 50 − j 42.22 = 65.44∠ − 40.18°Ω

(h)

Zc / Z R

ωo L
R

=

⎣

45,000

50 ×103 × 4 × 10−3
=4
50

⎦

= 107 / j 45, 000 × 50 = 4.444



36.

10 March 2006

Apply 1 A, in at top. ∴ VR = 10 V

(a)


Vin = Zin = 10−3 s + 10 +

108
1.2 ×108
(0.5 ×10 + 1) = 10−3 s + 10 +
s
5s
8
8
−3
−3
Zin ( jω ) = 10 + j (10 ω − 1.2 × 10 / ω ) ∴10 ω o = 1.2 × 10 / ω o

2
∴ω o = 1.2 × 1011 , ω o = 346.4 krad/s

(b)

Qo =

ωo L
R

=

346.4 ×103−3
= 34.64
10



37.


10 March 2006

Find the Thévenin equivalent seen by the inductor-capacitor combination:

⎛ V
⎞
SC : 1.5 = V1 + 10 ⎜ 1 − 0.105 V1 ⎟ ∴ V1 = 50 V
⎝ 125
⎠
50
∴↓ ISC =
= 0.4 A
125
1.5
= 3.75 Ω
OC :V1 = 0 ∴ VOC = 1.5 V ∴ R th =
0.4
1000 × 4
∴ω o = 1/ 4 × 0.25 × 10−6 = 1000, Qo =
= 1066.7
3.75
1000
1
= 0.9375, B = 0.4688 rad/s
B = ω o / Qo =
1066.7
2
VC max = Qo Vth = 1066.7 ×1.5 = 1600 V
Therefore, keep your hands off!

To generate a plot of |VC| vs. frequency, note that VC(jω) = 1.5

−

j
ωC

3.75 + jωL −

j
ωC




38.

Series, f o = 500 Hz, Qo = 10, X L ,0 = 500 Ω

(a)

500 = ω o L = 2π (500)L ∴ L = 0.15915+ H, C =

10 March 2006

Qo = 10 =

(b)

X L ,0
R

=

1
2π
=
= 0.6366 μ F
2
ω o L (2π × 500) 2

500
∴ R = 50 Ω
R

⎛
⎛
1
106 × 0.5π ⎞
250, 000 ⎞
1 = I ⎜ 50 + j 2π f ×
−j
⎟ = I ⎜ 50 + j f − j
⎟
f
2π
2π f ⎠
⎝
⎠
⎝
6
10 × 0.5π
∴ I = 1/ 50 + j ( f − 250, 000 / f ), Vc =
I
j 2π f
− j 250, 000 / f
∴ Vc (2π × 450) = 4.757 V
VC =
50 + j ( f − 250, 000 / f )
Vc (2π × 500) = 10, 000 V Vc (2π × 550) = 4.218 V




10 March 2006

39.
X : s = 0, ∞, 0 : s = −20, 000 ± j80, 000 s −1 , Zin (−104 ) = −20 + j 0 Ω ∴ SERIES

α = 20, 000, ω d = 80, 000 ∴ω o = (64 + 4)108 = 82, 462 rad/s,

1
2
= ω o = 68 × 108
LC

R
R
1 L 68 ×108
1
= α = 20, 000 ∴ = 40, 000,
× =
= 170, 000; Z(σ ) = R + σ L +
2L
L1
LC R 40, 000
σC
1
170, 000
1
=R− R−
R ∴ R = 1.2308 Ω
4
10, 000
10, 000C
1
1.2308
∴L =
= 30.77 μ H, C =
= 4.779 μ F
170, 000 ×1.2308
40, 000
∴−20 = R − 10, 000L −



40.
ωo

1/ 10

−3− 7


10 March 2006

105−3
= 10 rad/s, Q L =
= 100, R PL = 10, 000 Ω
1
5

1
= 500, R PC = 5002 × 0.2 = 50, 000 Ω
10 × 0.2
50 10 = 8.333 k Ω ∴ Q o = ωo CR = 105−7 × 8333 = 83.33
Qc =

5− 7

100, 000
= 1200 rad/s, Zin ( jωo ) = 8333 Ω
83.33
(99 − 100)103
8.333
ω = 99, 000 ∴ N =
= −1.6667, Zin ( j 99, 000) =
600
1 − j1.667
= 4.287 ∠ 59.04o kΩ
B=



41.


10 March 2006

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.




10 March 2006

42.
3mH 1.5 mH = 1mH, 2 μF + 8 μF = 10 μF, ∴ωo =

1
10−3−5

= 10 krad/s

3 × 10−3 ×104
= 100, R p = 1002 × 0.3 = 3 k Ω
0.3
1.5 × 10−3 × 104
Q=
= 60, R p = 60 × 0.25 = 900 Ω
0.25
692.3
900 3000 = 692.3 Ω ∴ Q L = 4−3 = 69.23
10
692.3
∴ R LS =
= 0.14444 Ω
69.232
106
Q= 4
= 125, R pc = 1252 × 0.1 = 1562.5 Ω 10 μF
10 × 0.1× 8
1562.5
∴ Qc = 104 × 10−5 × 15625 = 156.25 ∴ R SC =
= 0.064 Ω
(156.25) 2
Q=

∴ R S ,tot = 0.14444 + 0.064 = 0.2084 Ω = Zin

min

, ωo = 10 krad/s




10 March 2006

43.
(a)

ωo

1/ 2 × 0.2 ×10−3 = 50 rad/s

QleftL = 50 × 2.5 / 2 = 62.5, 2 × 62.52 = 7812.5 Ω
50 ×10
= 50, 10 × 502 = 25 k Ω
10
1000
Qc =
= 100, 1002 ×1 = 10 k Ω, R p = 7.8125 25 10 = 3731Ω
50 × 0.2 ×1
50
1
Qo = 50 × 3731× 0.2 ×10 −3 = 37.31; B =
= 1.3400, B = 0.6700
37.31
2
−3
∴ V o = 10 × 3731 = 3.731V
Q rightL =

3.731 V

2.638 V

|V| (volts)

↔
1.34 rad/s

50

(b)

ω (rad/s)

V = 10−3 [(2 + j125) (10 + j500) (1 − j100)]
=

10−3
= 3.7321∠ − 0.3950+ ° V
1
1
1
+
+
2 + j125 10 + j 500 1 − j100




10 March 2006

44.
(a)

1000
= 2000 rad/s, Qc = 2000 × 2 ×10−6 × 25 ×103 = 100
0.25
R
20 ×104
∴ R C , S = 25, 000 /1002 = 2.5 Ω; Q L =
=
= 40
ωo L 2000 × 0.25
ωo

20, 000
= 12.5 Ω ∴ R tot = 12.5 + 2.5 = 15 Ω
1600
2000 × 0.25
1
∴ Qo =
= 33.33 ∴ Vx = 1× 33.33 × = 16.667 V
15
2
∴ R L,S =

(b)

20, 000 × j 500
= 12, 4922 + j 499.688 Ω
20, 000 + j 500
25, 000(− j 250)
= 2.4998 − j 249.975
25, 000 − j 250 =
25, 000 − j 250
20, 000 j 500 =

∴ Zin = 12.4922 + 2.4998 + j 499.688 − j 250 − j 249.975 = 14.9920 − j 0.2870 Ω
∴ I = 1/ 14.9920 − j 0.2870 = 66.6902 mA ∴ Vx = 250 × 66.6902 ×10−3 = 16.6726 V



45.

Q = ωCR, RS =

XS = −


10 March 2006

RP
Q2 X P
, and X S =
1 + Q2
1 + Q2

1
1
1 + Q2
, XP = −
∴ CS = C P
Q2
ωCS
ωCP

(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



46.

(

)

RP = RS 1 + Q 2 , and X P = X S
C P = CS


10 March 2006

1 + Q2
Q2

Q2
1 + Q2

(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



47.

Q=


10 March 2006

RP
Q2 X P
R
Q2
, RS =
, and X S =
. LS = LP
1 + Q2
ωL
1 + Q2
1 + Q2

(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



48.


10 March 2006

⎛ 1 + Q2 ⎞
RP = RS 1 + Q 2 , and X P = X S ⎜
⎟
2
⎝ Q ⎠
⎛ 1 + Q2 ⎞
LP = LS ⎜
⎟
2
⎝ Q ⎠

(

)

(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



49.


10 March 2006

R
470
≈ 7 −6 = 47 . Since Q > 5, the series
ω L 10 10
equivalent is a 10/47 Ω resistor in series with 1 μH.
(a) For the left parallel circuit, Q =

For the right parallel circuit, Q = ωCR ≈ 10710−8 ( 200 ) = 20 . 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 =

( 470 ) ( j10710−6 )
470 + j10

1
jωC2
= 0.213 + j9.995 Ω . Z L =
= 0.499 − j9.975 Ω
1
R2 +
jωC2
R2

Z3 = j100 Ω.

Thus, Vx =

Z3
j100
= 0.99745 + j 0.0071 V
(1∠0 ) =
0.714 + j 0.02
Z1 + Z L + Z3

So that |Vx| = 0.99977 V . Our approximation was pretty accurate, at least at this
frequency.




10 March 2006

50.
(a)

(b)

50
20 × 103
= 0.5 K f =
= 0.02
100
106
1
0.5
∴ 9.82 μH → 0.5 × 9.82 ×
= 24.55 μH, 31.8 μH →
× 31.8 = 795 μH
0.02
0.02
2.57
= 257 nF
2.57 nF →
0.5 × 0.02
Km =

same ordinate; divide numbers on abscissa by 50




10 March 2006

51.
(a)

Apply 1 V ∴ I1 = 10A ∴ 0.5 I1 = 5A ↓; 5A 0.2 Ω can be replaced by 1 V in series with 0.2 Ω
∴ Iin → = 10 +

s + 10
1 − (−1)
2s
4s + 20 20( s + 5)
= 10 +
=
=
∴ Zin ( s) =
20( s + 5)
s + 10
0.2 + 2 / s
0.2s + 2 0.2s + 2
2( s / 5 + 10) 0.1( s + 50)
=
20( s / 5 + 5)
s + 25

(b)

K m = 2, K f = 5 ∴ Zin ( s ) →

(c)

0.1Ω → 0.2 Ω, 0.2 Ω → 0.4 Ω, 0.5F → 0.05 F, 0.5 I1 → 0.5 I1




10 March 2006

52.
(a)

ωo

1/ (2 + 8)10−310−6 = 104 rad/s
−3

Q L ,8 = 10 / 8 ×10 10 = 125 ∴ R L , S
4

4

2 + 8 = 10 mH ∴ Q L =

104
=
= 0.64 Ω
1252

104 ×10 × 10−3
= 156.25
0.64

1
= 100, R C , P = 1002 × 1 = 10 k Ω
10 ×10−6
∴ R P = 20 15.625 10 = 4.673 k Ω ∴ Qo = 104 × 10−6 × 4.673 × 103 = 46.73
∴ R L , P = 0.64 × 156.252 = 15.625 k Ω; QC =

4

(b)

K f = 106 /104 = 100, K m = 1 ∴ R ′s stay the same; 2 mH → 20 μH, 8mH → 80 μH,1μF → 10 nF

(c)

ωo = 106 rad/s, Qo stays the same, ∴ B =

106
= 21.40 krad/s
46.73




10 March 2006

53.
(a)

K m = 250, K f = 400 ∴ 0.1F →
5Ω → 1250 Ω, 2H →

0.1
= 1μF
250 × 400

2 × 250
= 1.25 H, 4 Ix → 103 Ix
400

1.25 H

1 μF

(b)

ω = 103. Apply 1 V ∴ I x = 10−6 s, ↓ I1250 =

1250 Ω

103

1
1250

1 − 10−3 s
∴1000 I x = 10 s ∴→ I L =
1.25s
1
0.8
0.8
(1 − 10−3 s ) = 10−6 s +
; s = j103
∴ Iin = 10−6 s +
+
1250 s
s
−3
0.8 ×10
1 1000
∴ Iin = j10−3 +
= j 0.2 × 10−3 ∴ Zth =
=
= − j 5 k Ω Voc = 0
Iin
j
j 0.2
−3




10 March 2006

54.
(a)

Is = 2∠0° A, ω = 50 ∴ Vout = 60∠25° V

(b)

Is = 2∠40° A, ω = 50 ∴ Vout = 60∠65° V

(c)

Is = 2∠40° A, ω = 200, ∴ OTSK

(d)

K m = 30, IS = 2∠40° A, ω = 50 ∴ Vout = 1800∠65° V

(e)

K m = 30, K f = 4, Is = 2∠40° A, ω = 200 ∴ Vout = 1800∠65° V




10 March 2006

55.
(a)

H /( s) = 0.2 ∴ H dB = 20 log 0.2 = −13.979 dB

(b)

H( s ) = 50 ∴ H dB = 20 log 50 = 33.98dB

(c)

H( j10) =

(d)

H dB = 37.6 dB ∴ H( s) = 1037.6 / 20 = 75.86

(e)

H dB = −8dB ∴ H( s ) = 10−8/ 20 = 0.3981

(f)

H dB = 0.01dB ∴ H( s ) = 100.01/ 20 = 1.0012

12
26
6
13
292 + j 380
+
∴ H dB = 20 log
+
= 20 log
= 6.451dB
2 + j10 20 + j10
1 + j 5 10 + j 5
−60 + j 220



56.
(a)


10 March 2006

(d)

MATLAB verification- shown adjacent to Bode plots below.
20( s + 1) 0.2(1 + s )
H( s ) =
=
, 0.2 → −14 dB
s + 100 1 + s /100

1

10

100

2000( s + 1) s
0.2 s(1 + s)
=
, 0.2 → −14 dB
2
( s + 100)
(1 + s /100) 2

(b)

H( s ) =

(c)

200 s 2 + 45s + 200 ( s + 5)( s + 40) 200(1 + s / 5)(1 + s / 40)
H( s ) = s + 45 +
=
=
=
, 200 → 46 dB
s
s
s
s




10 March 2006

57.
H( s ) =

VC (20 + 2 s)(182 + 200 / s)
200 / s
=
×
IR
202 + 2 s + 200 / s
182 + 200 / s

400( s + 10)
200(10 + s )
=
2
2( s + 101s + 100) (1 + s)(100 + s )
20(1 + s /10)
H( s ) =
, 20 → 26 dB
(1 + s )(1 + s /100)
=




10 March 2006

58.
5 ×108 s ( s + 100)
2.5s (1 + s /100)
=
, 2.5 → 8dB
3
( s + 20)( s + 1000)
(1 + s / 20)(1 + s /1000)3

(a)

H( s ) =

(b)

Corners: ω = 20, 34 dB;
ω = 100, 34 dB;
ω = 1000, 54 dB
Intercepts: 0 dB, 2.5ω = 1, ω = 0.4
2.5ω (ω /100)
2.5ω2 (20)109
ω = 1, 8dB; 0 dB,
=
= 1 ∴ω = 22,360 rad/s
(ω / 20)(ω /1000)3
100ωω3

(c)

Corners: ω = 20, 31.13dB
ω = 100, 36.69 dB H dB = 20 log 2.5ω

1 + (ω /100) 2
[1 + (ω / 20) 2 ][1 + (ω /1000) 2 ]3

ω = 1000, 44.99 dB




10 March 2006

59.
(a)

(b)

H( s ) =

5 × 108 s ( s + 100)
2.5s (1 + s /100)
=
,
3
( s + 20)( s + 1000)
(1 + s / 20)(1 + s /1000)3

ω = 2 : ∠ = 90°
⎛
⎝

ω = 10 : ∠ = 90° − ⎜ 45° + 45° log

10 ⎞
⎟ = 58.5°
20 ⎠

⎛
⎝

100 ⎞ ⎛
100 ⎞
⎟ + ⎜ 45° + 45° log
⎟ = 58.5°
20 ⎠ ⎝
100 ⎠
200 ⎞ ⎛
200 ⎞
⎛
ω = 200 : ∠ = 90° − 90° + ⎜ 45° + 45° log
⎟ − 3 ⎜ 45° + 45° log
⎟ = 17.9°
100 ⎠ ⎝
100 ⎠
⎝

ω = 100 : ∠ = 90° − ⎜ 45° + 45° log

1000 ⎞
⎛
⎟ = −45°
1000 ⎠
⎝
ω = 10, 000 : ∠ = 90° − 90° + 90° − 3 × 90° = −180°

ω = 1000 : ∠ = 90° − 90° + 90° − 3 ⎜ 45° + 45° log

(c)

ω = 2 : ∠ = 90° + tan −1 0.02 − tan −1 0.1 − 3 tan −1 0.002 = 85.09°
ω = 10 : ∠ = 90° + tan −1 0.1 − tan −1 0.5 − 3 tan −1 0.01 = 67.43°
ω = 100 : ∠ = 90° + tan −1 1 − tan −1 5 − 3 tan −1 0.1 = 39.18°
ω = 200 : ∠ = 90° + tan −1 2 − tan −1 10 − 3 tan −1 0.2 = 35.22°
ω = 1000 : ∠ = 90° + tan −1 10 − tan −1 50 − 3 tan −1 1 = −49.56°
ω = 10, 000 : ∠ = 90° + tan −1 100 − tan −1 500 − 3 tan −1 10 = −163.33°




10 March 2006

60.
(a)

20 400 s 2 + 20 s + 400
+ 2 =
s
s
s2
1 + 2 × 0.5( s / 20) + ( s / 20) 2
= 400
s2
∴ω o = 20, ζ = 0.5
H( s ) = 1 +

Hdb

20 log 400 = 52dB
Correction at ω o is 20 log 2 ζ = 0 dB

(b)

ω = 5 : H dB = 52 − 2 × 20 log 5 = 24.0 dB (plot)
H dB = 20 log 1 − 16 + j 4 = 23.8dB (exact)

ω = 100 : H dB = 0 dB (plot)
H dB = 20 log 1 − 0.04 + j 0.2 = −0.170 dB (exact)

(c)




10 March 2006

61.
(a)

(b)

H( s ) =

VR
25
25s
=
=
=
2
V5 10 s + 25 + 1000 / s 10 s + 25s + 1000

⎛ 1 ⎞⎛ s ⎞ ⎛ s ⎞
1 + 2 ⎜ ⎟⎜ ⎟ + ⎜ ⎟
⎝ 8 ⎠⎝ 10 ⎠ ⎝ 10 ⎠

2

⎛ 1⎞
∴ω o = 10, ζ = 1/ 8 ∴ correction = −20 log ⎜ 2 × ⎟ = 12 dB
⎝ 8⎠
0.025 → −32 dB
HdB

(c)

0.025s

ω = 20, H( j 20) =

ang(H)

j 0.5
∴ H dB = −15.68 dB ∠H( j 20) = −80.54°
1 − 4 + j 0.5




10 March 2006

62.
1st two stages, H1 ( s ) = H 2 ( s ) = −10; H 3 ( s ) =

−1/(50 ×103 ×10−6 )
−20
=
3
−6
s + 1/(200 × 10 × 10 ) s + 5

⎛ −20 ⎞ −400
∴ H( s ) = (−10)(−10) ⎜
⎟=
⎝ s + 5 ⎠ 1+ s / 5
−400 → 52 dB



63.
(a)


10 March 2006

1st stage: C1 A = 1 μ F, R1 A = ∞, R fA = 105 ∴ H A (S) = − R fA C1 A s = −0.1 s
2nd stage: R 1B = 105 , R fB = 105 , C fB = 1 μ F ∴ H B ( s ) =

−1/ R1B C fB
s + 1/ R fB C fB

1/(105 × 10−6 )
10
=−
5
−6
s + 1/(10 × 10 )
s + 10
3rd stage: same as 2nd

∴ H B (s) =

0.1s
⎛ −10 ⎞ ⎛ −10 ⎞
∴ H( s ) = (−0.1s ) ⎜
⎟⎜
⎟=−
(1 + s /10) 2
⎝ s + 10 ⎠ ⎝ s + 10 ⎠

20log10(0.1) = -20 dB
(b)

(c)



64.


10 March 2006

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.
Rf
= 66.7 - 1 = 65.7
If we select a non-inverting op amp topology, we then need
R1
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.
1
We design for ωc = 2πfc = 2π(50×103) =
. Since Rmic = 1 Ω, we require
Rmic C filter
Cfilter = 3.183 μF.



65.


10 March 2006

We choose a simple series RLC circuit. It was shown in the text that the “gain” of the
ωRC
.
circuit with the output taken across the resistor is AV =
1
2
2
2 2 2 2
1 - ω LC + ω R C
This results in a bandpass filter with corner frequencies at

[(

ωc =
L

-RC + R 2 C 2 + 4 LC
2 LC

and

ωc =
H

)

]

RC + R 2 C 2 + 4 LC
2 LC

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
-RC + R 2 C 2 + 4 LC
2π(20) =
and 2π(20×103) =
2 LC

RC + R 2 C 2 + 4 LC
2 LC

Noting that ω cH – ω cL = 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.



66.


10 March 2006

We choose a simple RC filter topology:

Vout
1
=
Vin 1 + jωRC

Vout
1
. We desire a cutoff
=
2
Vin
1 + (ωRC )
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,
1
1
=
=
where ωc = 2πfc = 2000π rad/s.
2
2
1 + (ω c RC )

Where

and

hence

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:



67.


10 March 2006

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
1
filter with R = 1 kΩ (arbitrarily chosen) and C =
= 63.66 nF .
2π (2.5 × 10 3 )(1000)
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.



68.


10 March 2006

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
R
1
ωc L = +
R 2C 2 + 4LC = 2π (500)
2L 2LC
and
R
1
ωcH =
+
R 2C 2 + 4LC = 2π (5000)
2L 2LC

With ω cH - ω cL = 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.



69.


10 March 2006

For this circuit, we simply need to connect a low-pass filter to the input of a noninverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the
cutoff frequency is
1
ωc =
= 2π (3000)
RC
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).



70.


10 March 2006

We require four filter stages, and choose to implement the circuit using op amps to isolate each filter subcircuit. 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 ( 1 LC ) 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.



71.


10 March 2006

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 ( 1 LC ) equal to 60 Hz (120π rad/s). This yields C = 7.04 μF.

vin

vout

1Ω
1H

7.04 μF


Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

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