CHAPTER SIXTEEN SOLUTIONS - mhhe. · PDF fileCHAPTER SIXTEEN SOLUTIONS Engineering Circuit...

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

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: 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 → ∞.

10-3∠∠∠∠ 0o A jωωωωL -j/ ωωωωC



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



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



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).



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 1 3080C400 2 15

1 229 1 1C 15.775 F; L 43.88mH; R 396.732,000 901 C 2o

jj

Cµ

ω α−

− +=− +

∴ = = = = = = Ω



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 j

j j jj j

ω ωω ω ω

ω ω ω ω ωω ω ω ω

ω ω ω ω ω ω



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)

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

6 6 5 3 3

2 2

26

1 I 1000Y 10 10 , V 10 /10 10Y 1000

10 10V , V1000 10000.001 101000 1000

j j

j

ωω ω

ω ωω ω

− − − − −

− −

−

= + − = = + −

∴ = = + − + −



8. (a)

(b) 2

2 2 6

102000Z ( ) 2 2.294400 10

oin o

o o

ωωω ω

= + + = Ω+ +

2 2 6

42 6 2 2

2 2 6

5(100 / ) 0.1Z 25 (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 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



9. (a)

(b) 4 6 1 1Y 10 10 , 1000 Z 9997 1.43210.9975 Y

j ω ωω

− − = + − = ∴ = = ∠ °Ω

1 1 2 2 2

6 6

2

50 , 1000 1,002,500 1001.249

1 10 1 10L 0.9975 H; R 10C 1,002,500 2 C 100

d o d o

o

s s

k

α ω ω α ω ω

ω α

− −

+

= = ∴ = + = ∴ =

= = = = = = Ω



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−

−

∴ × × × × = ∴ =

×∴ = = = =×



11. (a) (b)

4

4 5 4 83

8 2 44 8

8 2 4

Apply 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

−

− − −−

− −− −

− −

± ∴ = −

∴ = = + + − −×

× + × +∴ = + + × =

− × + ×∴ =

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



12. 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)

23

B / 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

= = =

∴ =

− − ∴ = + = × +



13.

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 ωωω

ωω

++

−+

= =

= Ω

= = ×

∴ = = − = = = =

∴ = ∴ = = × = =

= = + − = + 12,154 C 18.456C

1050.1 1 1C 5.418mF, R C 12.304 , L 15.185 mH

18.456 15 12,154C−

− =

∴ = = = = Ω = =



14. 30 krad/s, Q 10, R 600 ,= = = Ωo oω

(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) (e)

1

1

Q1 1 10Z ( 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

600Z 351.906 54.090328 301 1030 28

−

−

= + − = = ×

× = = = ∴ = + × − ×

= = ∠ °Ω + −

oin

o

ino o

in

j j j

j

j

ω

ω

approx-true 360 351.906magnitude: 100% 100% 2.300%true 351.906

53.1301 54.0903angle: 100% 1.7752%54.0903

−= =

°− ° = −°



15. 3400Hz, Q 8, R 500 , I 2 10 A B 50Hzo o Sf −= = = Ω = × ∴ = (a) (b)

3 2 2 400V 2 10 500 / 1 N 0.5 1 N 4, N 350 / 2

400 25 3 443.3 and 356.7 Hz

f

f

− −= × × + = ∴ + = = ± =

∴ = ± =

3 2 2

2

1 1I 0.5 10 1 N 4, N 15, N 15R 5001 N400 25 15 496.8 and 303.2 Hz

R

v

f

−= = × = × ∴ + = = = ±+

∴ = ± =



16. 6 310 , Q 10, R 5 10 , . .= = = ×o o p rω

(a) 3

6

R 5 10Q L 0.5mHL 10 10o

oω×= ∴ = =×

(b) (c)

62

6

2

2 2 2 22 2

4 2 2

10Approx: 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

− −∠ = ° ∴ = ° = = =

= + − ∴ ° = = −

+ +∴ − = − − = = =

j

ω ω

ω ωω ω

ω ω ω ωω



17.

(a) 6

4 8

1C 3 7 10nF 10 rad/s10 10

oω− −

= + = ∴ = =

(b)

(c) 3

313

15 10 15 9015 10 N 1.5 V 8.321 33.69 V10 10 1 1.5o j

ω ω × ∠ °− = × ∴ = = ∴ = = ∠ °× +

6 8 3

9 6 9

3

3 31,0

Q CR 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

−

− −

−

= = × == =

∠ ° = × = ×

∴ = × × × = ∠ °



18. (a)

(b) 25 10,000Z ( ) (5 100) (5 100) 1002.510in oj j jω += + − = = Ω

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

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



19. , 1000Hz, Q 40, Z ( ) 2 B 25Hzo o in of j kω= = = Ω ∴ = (a) Zin(jω) = Zin = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω (b) 0.9 1.1 900 1100 Hzo of f f f< < ∴ < <

2000 1000, N , 1010, N 0.81 N 12.5

−= = ∴ =+

f fj



20. 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



21. (a) (b)

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

= = = = =

− ∴ = +

9 rad/s800(Approx: V ( 9) 194.03V

1740 600Exact: V

3 (6 600 / )24,000V ( 9) 204.86 13.325 V

9[3 (54 66.67)]

out

out

out

j

j j

jj

ω

ω ω ω

−

=

= =

= ×+ −

∴ = = ∠ −+ −



22. 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= × =



23. Apply 1 A, in at top. V 10 VR∴ = (a)

(b) 3 3L 346.4 10Q 34.64

R 10o

oω −×= = =

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 , 346.4 krad/s

in in

in o o

o o

s ss s

j jω ω ω ω ωω ω

− −

− −

×= = + + × + = + +

= + − × ∴ = ×

∴ = × =



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

11 1 1

1

6

max

VSC : 1.5 V 10 0.105V V 50 V125

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 1600 V

−

= + − ∴ =

∴↓ = =

= ∴ = ∴ = = 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



25. ,0Series, 500Hz, Q 10, X 500o o Lf = = = Ω (a) (b)

2 2

,0

1 2500 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) 10,000V 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



26.

1 4

8 2 8

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

120 R 10,000L10,00

−= ∞ = − ± − = − + Ω ∴

= = ∴ = + = = = ×

×= = ∴ = × = = = + +

∴− = − −

in

d o o

s s j s j

α ω ω ω

α σ σσ

1 170,000R R R R 1.23080C 4 10,000

1.2308 1L 30.77 H, C 4.779 F40,000 170,000 1.2308

= − − ∴ = Ω

∴ = = = =×

µ µ



27. = 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 = ∴ = = −

Bo L PL

c PC

o o

in o

i

k

j

8.333( 99,000)1 1.667

=−n j

j



28. 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.



29.

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



30. (a) (b)

3

2

2

2

3

1/ 2 0.2 10 50 rad/s

Q 50 2.5 / 2 62.5, 2 62.5 7812.550 10Q 50, 10 50 25

101000Q 100, 100 1 10 , R 7.8125 25 10 3731

50 0.2 150 1Q 50 3731 0.2 10 37.31; B 1.3400, B 0.6700

37.31 2V 10

−

−

ω × × =

= × = × = Ω

×= = × = Ω

= = × = Ω = = Ω× ×

= × × × = = = =

∴ =

Bo

leftL

rightL

c p

o

o

k

k

3 3731 3.731V− × =

3

3

V 10 [(2 125) (10 500) (1 100)]

10 3.7321 0.3950 V1 1 12 125 10 500 1 100

−

−+

= + + −

= = ∠ − °+ +

+ + −

j j j

j j j

3.731 V

2.638 V

50

↔↔↔↔ 1.34 rad/s

|V| (volts)

ω (rad/s)



31. (a) (b)

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

×∴ = = ∴ = × × =

Bo 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−× =



32. (a) (b) same ordinate; divide numbers on abscissa by 50

3

6

50 20 10K 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 nF 257 nF0.5 0.02

×= = = =

∴ µ → × × = µ µ → × = µ

→ =×

m f



33. (a)

(b) 2( / 5 10) 0.1( 50)K 2, K 5 Z ( )20( / 5 5) 25

+ += = ∴ → =+ +m f in

s sss s

(c) 1 10.1 0.2 , 0.2 0.4 , 0.5F 0.05F, 0.5I 0.5IΩ → Ω Ω → Ω → →

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) 10I 10 10 Z ( )

0.2 2 / 0.2 2 0.2 2 10 20( 5)

∴ = ∴ = ↓ Ω Ω− − + + +∴ →= + = + = = ∴ =+ + + + +in in

s s s sss s s s s



34. (a) (b)

(c) 6

6 1010 rad/s, Q stays the same, B 21.40 krad/s46.73

ω = ∴ = =o o

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 10mH 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

− −

−

−

−

−

ω + =

= × = ∴ = = Ω

× ×+ = ∴ = =

∴ = × = Ω = = = × = Ω×

∴ = = Ω ∴ = × ×

Bo

L L S

L

L P C C P

P o

k k

k 3.673 10 46.73× =

6 4K 10 /10 100, K 1 R s stay the same; 2 mH 20 H, 8mH 80 H,1 F 10nF′= = = ∴ → µ → µ µ →f m



35. (a) (b)

3

0.1K 250, K 400 0.1F 1 F250 400

2 2505 1250 , 2H 1.25H, 4 I 10 I400

= = ∴ → = µ×

×Ω → Ω → = →

m f

x x

3 61250

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 0

I 0.2

−

−−

− − −

−− −

ω = ∴ = ↓ =

−∴ = ∴→ =

∴ = + + − = + =

×∴ = + = × ∴ = = = − 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



36. (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



37. (a) H /( ) 0.2 H 20 log 0.2 13.979dB= ∴ = = −dBs (b) H( ) 50 H 20 log 50 33.98dB= ∴ = =dBs (c) (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

12 26 6 13 292 380H( 10) H 20 log 20 log 6.451dB2 10 20 10 1 5 10 5 60 220

+= + ∴ = + = =+ + + + − +dB

jjj j j j j



38. (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

(b) 2 2

2000( 1) 0.2 (1 )H( ) , 0.2 14dB( 100) (1 /100)

+ += = → −+ +

s s s sss s

(c)

2200 45 200 ( 5)( 40) 200(1 / 5)(1 / 40)H( ) 45 , 200 46dB+ + + + + += + + = = = →s s s s s ss ss s s s

1 10 100



39.

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 s ssI s s s

s ss s s s

sss s



40.

(a) 8

3 3

5 10 ( 100) 2.5 (1 /100)H( ) , 2.5 8dB( 20)( 1000) (1 / 20)(1 /1000)

× + += = →+ + + +

s s s sss 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

Corners: 20, 31.13dB

1 ( /100)100, 36.69dB H 20 log 2.5

[1 ( / 20) ][1 ( /1000) ]1000, 44.99dB

ω =

+ ωω = = ω

+ ω + ωω =

dB



41.

(a) 8

3 3

5 10 ( 100) 2.5 (1 /100)H( ) ,( 20)( 1000) (1 / 20)(1 /1000)

× + += =+ + + +

s s s sss 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

− −

− − −

− = − °

= ∠ = °+ − − = − °



42. (a) (b) (c)

2

2 2

2

2

20 400 20 400H( ) 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



43. (a)

(c) 0.520, H( 20) H 15.68 dB H( 20) 80.541 4 0.5 dB

jj jj

ω = = ∴ = − ∠ = − °− +

225

V 25 25 0.025H( )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)



44. 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

−

−

− × × −= = − = =+ × × +

− − ∴ = − − = + + − →



45. (a) 20log10(0.1) = -20 dB (b) (c)

51 1 1

15 51

5 6

5 6

1st 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

−

−

= = ∞ = ∴ = − = −

−= = = ∴ =

+

×∴ = = −+ × +

−∴ = − 2

0 10 0.110 10 (1 /10)

ss s s

− = − + + +



46. 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.



47. 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 LC

LCCRRCHc 2

4 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.



48. 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:



49. 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.



50. 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.



51. 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).



52. 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.



53. 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

CHAPTER SIXTEEN SOLUTIONS - mhhe. · PDF fileCHAPTER SIXTEEN SOLUTIONS Engineering Circuit...

Documents

Transcript of CHAPTER SIXTEEN SOLUTIONS - mhhe. · PDF fileCHAPTER SIXTEEN SOLUTIONS Engineering Circuit...