Chapter 14 - Alexander`s Fundamentals of Electric circuits Ch.14 Solution PDF

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Alexander`s Fundamentals of Electric circuits Ch.14 Solution...

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Chapter 14, Solution 1. Vo R jRC   Vi R  1 jC 1  j RC j  0 1 , where  0  H ()  1  j  0 RC H() 

H  H () 

 0 1  (  0 )

  H () 

2

   tan -1   2  0 

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that  0  1 RC . Thus, the sketches of H and  are shown below. H 1 0.7071

0

0 =





90 45 0

0 =



Chapter 14, Solution 2. Using Fig. 14.69, design a problem to help other students to better understand how to determine transfer functions. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Obtain the transfer function V o /V i of the circuit in Fig. 14.66. 10  + 2

+ _

Vo

1 F 8

Figure 14.66

Vo _

For Prob. 14.2.

Solution V H ( s)  o  Vi

2

1 s/8

10  20 

1 s/8



s 4 2  8/s 1  12  8 / s 6 s  0.6667

Chapter 14, Solution 3. 0.2F

 

0.1F

 

5 1  (0.2) s s 1 10  s(0.1) s 1

j C



The circuit becomes that shown below.

2

V1

5 s +

Vi

+ _

5

10 s

Vo _

10 5 10 1  s (5  ) 5( ) 10 5 s  s s  10( s 1) Let Z  //(5  )  s 15 5 s s s (s  3) 5 (3  s ) s s Z V1  V Z2 i s s Z 5 Vi Vo  V1  V1   s 1 s 1 Z  2 5 5/ s

10(s 1) s 10 s 5s s( s  3) H ( s)  Vo     2 Vi s  1 10( s  1) 2s (s  3)  10(s  1) s  8s  5 2 s( s  3) H(s) = 5s/(s2+8s+5)

Chapter 14, Solution 4. (a)

R ||

1 R    j C 1 jRC

V H()  o  Vi

(b)

R 1  j RC R  R R  jL (1  jRC) jL  1 j RC

H () 

R -  RLC  R  j L

H () 

R  j L jC (R  jL)  R  jL  1 jC 1  jC (R  jL)

H() 

-  2LC  jRC 1   2LC  j RC

2

Chapter 14, Solution 5. (a) Let Z  R // sL 

Vo 

sRL R  sL

Z Vs Z  Rs

sRL Vo Z s RL H (s)    R  sL  Vs Z  Rs R  sRL RRs  s (R  Rs )L s R  sL

1 Rx 1 R sC  (b) Let Z  R //  sC R 1 1  sRC  sC Z Vo  Vs Z  sL

R R Vo Z H(s)    1  sRC  R Vi Z  sL s2 LRC  sL  R sL  1  sRC

Chapter 14, Solution 6. The 2 H inductors become jω2 or 2s. Let Z = 2s||2 = [(2s)(2)/(2s+2)] = 2s/(s+1) We convert the current source to a voltage source as shown below. 2

Is  2

2S

+

+ _

Vo

Z

_

V o = [(Z)/(Z+2s+2)](2I s ) = H(s) = I o /I s = [2s/(s2+3s+1)].

or

Chapter 14, Solution 7. (a)

0.05  20 log10 H

2.5  10 -3  log 10 H H  102.510  1.005773 -3

(b)

- 6.2  20 log10 H - 0.31  log 10 H

H  10-0.31  0.4898 (c)

104.7  20 log 10 H 5.235  log10 H

H  105.235  1.718  10 5

Chapter 14, Solution 8. Design a problem to help other students to better calculate the magnitude in dB and phase in degrees of a variety of transfer functions at a single value of ω. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the magnitude (in dB) and the phase (in degrees) of H(  ) at  = 1 if H( ) equals (a) 0.05 (b) 125 10 j (c) 2  j 3 6  (d)  1 j 2  j Solution

(a)

(b)

(c)

H  0.05 H dB  20 log10 0.05  - 26.02 ,

φ = 0

H  125 H dB  20 log10 125  41.94 ,

φ = 0

H dB (d)

j10  4.472 63.43 2 j  20 log10 4.472  13.01 ,

H(1) 

φ = 63.43

3 6   3.9  j2.7  4.743 - 34.7 1 j 2  j H dB  20 log10 4.743  13.521, φ = –34.7˚ H(1) 

Chapter 14, Solution 9.

H ( ) 

10 10(1  j )(1  j 10)

H dB  20 log 10 1 - 20 log 10 1  j  20 log 10 1  j / 10

  - tan -1 ()  tan -1 ( / 10)

The magnitude and phase plots are shown below. 20 H dB 0.1

1

10



100 20 log 10

-20

1 1  j  / 10 20 log10

-40

1 1  j

 0.1 -45

1

10



100 arg

1 1  j / 10

-90 arg

-135 -180

1 1  j

Chapter 14, Solution 10. Design a problem to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Sketch the Bode magnitude and phase plots of: H j  

50 j  5  j

Solution

H( j ) 

50  j(5  j)

10 j   1j 1   5 

H dB 40 20 log1

20 10 0.1 -20

1

100

  1 20 log  j 1 5 

 1   20 log   j 

-40

 0.1 -45

     



1 10



100 arg

1 1  j / 5

-90 arg

-135 -180

1 j

Chapter 14, Solution 11.

H ( ) 

0.2 x10 (1  j 10) 2[ j  (1  j  2)]

H dB  20 log10 1  20 log10 1  j 10  20 log10 j  20 log10 1  j 2

  -90  tan -1  10  tan -1  2 The magnitude and phase plots are shown below. H dB 40 20

0.1 -20

1

10

100



1

10

100



-40  90 45 0.1 -45 -90

Chapter 14, Solution 12.

T ( ) 

10(1 j  ) j (1  j / 10)

To sketch this we need 20log 10 |T(ω)| = 20log 10 |10| + 20log 10 |1+jω| – 20log 10 |jω| – –1 –1 20log 10 |1+jω/10| and the phase is equal to tan (ω) – 90° – tan (ω/10). The plots are shown below.

|T|

(db)

20



0 0.1

1

10

100

1

10

100

-20

-40

arg T

90o



0 0.1 -90o

Chapter 14, Solution 13.

G ( ) 

0.1(1  j ) (1 100)(1  j )  2 ( j ) (10  j ) ( j  )2 (1  j  10)

GdB  40  20 log10 1  j  40 log10 j  20 log10 1  j 10

  -180  tan -1  tan -1  10

The magnitude and phase plots are shown below. G dB 40 20 0.1

1

10

100



1

10

100



-20 -40

 90 0.1 -90 -180

Chapter 14, Solution 14. 250 25

H( ) 

1  j 2  j10  j    j 1    25  5   

H dB  20 log10 10  20 log10 1  j   20 log10 j 

 20 log10 1  j2 5  ( j 5) 2  10 25     -90   tan -1   tan -1  1  2 5 

The magnitude and phase plots are shown below. H dB 40 20 0.1

1

10

100



1

10

100



-20 -40  90 0.1 -90 -180

Chapter 14, Solution 15.

H ( ) 

2 (1  j  ) 0.1(1  j )  (2  j )(10  j ) (1  j 2)(1  j 10)

H dB  20 log10 0.1  20 log10 1  j  20 log10 1  j 2  20 log10 1  j 10

  tan -1   tan -1  2  tan -1  10

The magnitude and phase plots are shown below. H dB 40 20 0.1 -20

1

10

100



1

10

100



-40  90 45 0.1 -45 -90

Chapter 14, Solution 16.

H(ω) =

H db = 20log 10 |0.1| – 20olg 10 |jω| – 20log 10 |1+jω+(jω/4)2| The magnitude and phase plots are shown below. H 20 20 log (j) 1

4

10

100

40



0.1

–20  j  20 log 1  j     4 

–40

–60 

 0.4 -90

1

4

10

40

90

100

-tan-1

 

1

-180

2

16

2

Chapter 14, Solution 17.

G( ) 

(1 4) j  (1 j )(1 j 2) 2

G dB  -20log10 4  20 log10 j  20 log10 1  j  40 log10 1  j 2   -90 - tan -1  2 tan -1  2

The magnitude and phase plots are shown below. G dB 20 0.1

1

10

100 

-12 -20 -40

 90 0.1 -90 -180

1

10

100



Chapter 14, Solution 18. The MATLAB code is shown below. >> w=logspace(-1,1,200); >> s=i*w; >> h=(7*s.^2+s+4)./(s.^3+8*s.^2+14*s+5); >> Phase=unwrap(angle(h))*57.23; >> semilogx(w,Phase) >> grid on

60

40

H (j w ) P h a s e

20

0

-2 0

-4 0

-6 0 -1 10

10 w

0

Now for the magnitude, we need to add the following to the above, >> H=abs(h); >> HdB=20*log10(H); >> semilogx(w,HdB); >> grid on

10

1

0

-5

HdB

-1 0

-1 5

-2 0

-2 5 -1 10

10 w

0

10

1

Chapter 14, Solution 19. H(ω) = 80jω/[(10+jω)(20+jω)(40+jω)] = [80/(10x20x40)](jω)/[(1+jω/10)(1+jω/20)(1+jω/40)] H db = 20log 10 |0.01| + 20log 10 |jω| – 20log 10 |1+jω/10| – 20log 10 |1+jω/20| – 20log 10 |1+jω/40| The magnitude and phase plots are shown below .

20 log j

20 db

0 db 0.1

1

100

10

ω –20 log |1+jω/40|

–20 db –20 log 1  20 log |1/80|

j 10

–40 db

–20 log |1+jω/20|

jω 90˚

0˚ 0.1

1

100

10

ω (1+jω/40)

–90˚ (1+jω/10) –180˚

(1+jω/20)

Chapter 14, Solution 20. Design a more complex problem than given in Prob. 14.10, to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Include at least a second order repeated root. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Sketch the magnitude phase Bode plot for the transfer function

Solution

20log(1/100) = -40

For the plots, see the next page.

The magnitude and phase plots are shown below. 20 log j 

40

1

20 log 1

20

j 10



0.1

5

1

10

50

100 20 log

-20 20 log

1 100

1 1  j

-40 20 log -60

180˚

1 j   1  5   

2

jω (1+jω/10)

90˚ (1+jω) 0.1

 1

5

10

50

100

–90˚ (1+jω/5)2 –180˚

–270˚

Chapter 14, Solution 21. H(ω) = 10(jω)(20+jω)/[(1+jω)(400+60jω–ω2)] = [10x20/400](jω)(1+jω/20)/[(1+jω)(1+(3jω/20)+(jω/20)2)] H dB  20 log(0.5)  20 log j  20 log 1 

j j 3  j    20 log 1  j  20 log 1    20 20  20 

2

The magnitude plot is as sketched below. 20log 10 |0.5| = –6 db

db 40 20log|jω| 20 log |1+jω/20|

20 1

20 log 0.5

10

20 

0.1

100

–20

–40

–20 log 1  j

–60 –20 log –80

Chapter 14, Solution 22. 20  20 log 10 k   k  10

A zero of slope  20 dB / dec at   2   1  j 2 1 A pole of slope - 20 dB / dec a t   20   1  j  20 1 A pole of slope - 20 dB / dec at   100   1  j 100

Hence, H() 

10 (1  j  2) (1  j  20)(1  j  100)

H () 

10 4 ( 2  j) ( 20  j )(100  j )

Chapter 14, Solution 23. A zero of slope  20 dB / dec at the origin

 

j

1 1  j 1 1   (1  j 10) 2

A pole of slope - 20 dB / dec at   1  

A pole of slope - 40 dB / dec a t   10 Hence, H () 

j (1  j)(1  j 10) 2

H () 

100 j (1  j )(10  j )2

(It should be noted that this function could also have a minus sign out in front and still be correct. The magnitude plot does not contain this information. It can only be obtained from the phase plot.)

Chapter 14, Solution 24. 40  20log10 K

  K  100

There is a pole at =50 giving 1/(1+j/50) There is a zero at =500 giving (1 + j/500). There is another pole at =2122 giving 1/(1 + j/2122). Thus, H(jω) = 100(1+jω)/[(1+jω/50)(1+jω/2122)] = [100(50x2122)/500](jω+500)/[(jω+50)(jω+2122)] or H(s) = 21220(s+500)/[(s+50)(s+2122)].

Chapter 14, Solution 25. 0 

1 LC



1 ( 40  10 -3 )(1  10 -6 )

 5 krad / s

Z(0 )  R  2 k  4   Z ( 0 4)  R  j  0 L  0 C  4

  5  103 4  Z( 0 4)  2000  j   40  10 -3  3 -6 (5  10 )(1 10 )   4 Z ( 0 4)  2000  j (50  4000 5) Z(0 4)  2  j0. 75 k   2   Z ( 0 2)  R  j  0 L  0 C  2

 (5  103 )  2  Z( 0 2)  2000  j  (40  10 -3 )  3 -6 (5 10 )(1 10 )  2  Z(ω 0 /2) = 200+j(100-2000/5) Z(0 2)  2  j0.3 k  1   Z(20 )  R  j 20 L  20 C  

  1  Z(2 0 )  2000  j  (2)(5 10 3 )(40  10 -3 )  (2)(5 10 3)(1 10 -6)   Z(20 )  2  j0.3 k  1   Z(40 )  R  j 40 L  40 C  

  1  Z(4 0 )  2000  j  (4)(5 10 3 )(40  10 -3 )  3 -6 (4)(5 10 )(1 10 )   Z(40 )  2  j0. 75 k

Chapter 14, Solution 26. Design a problem to help other students to better understand ω o , Q, and B at resonance in series RLC circuits. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A coil with resistance 3  and inductance 100 mH is connected in series with a capacitor of 50 pF, a resistor of 6 , and a signal generator that gives 110V-rms at all frequencies. Calculate  o , Q, and B at resonance of the resultant series RLC circuit. Solution Consider the circuit as shown below. This is a series RLC resonant circuit.

6

50 pF

3

+ _

100 mH

R=6+3=9

1 1   447.21 krad/s 3 LC 100 x10 x50 x10 12

o 

Q

o L

B

o

R

Q

3





3

447.21x10 x100x10  4969 9

447.21x 103  90 rad/s 4969

Chapter 14, Solution 27.

o 

1  40 LC

  LC 

1 40 2

R  10   R  10L L If we select R =1 , then L = R/10 = 100 mH and B

C 

1 1  2  6.25 mF 2 40 L 40 x 0.1

Chapter 14, Solution 28.

R  10  . L

R 10   0.5 H B 20

C

1 1   2 F 2  0 L (1000) 2 (0.5)

Q

 0 1000   50 B 20

Therefore, if R  10  then L  500 mH , C  2  F ,

Q  50

Chapter 14, Solution 29. We convert the voltage source to a current source as shown below.

12 k

is

is 

45 k

1 F

20 cos t , R = 12//45= 12x45/57 = 9.4737 k 12 1 1   4.082 krad/s = 4.082 krad/s o  LC 60 x10 3 x1x10 6

B

1 1   105.55 rad/s = 105.55 rad/s RC 9.4737 x10 3x10 6 4082  Q o   38.674 = 38.67 B 105.55

4.082 krads/s, 105.55 rad/s, 38.67

60 mH

Chapter 14, Solution 30. (a) f o = 15,000 Hz leads to ω o = 2πf o = 94.25 krad/s = 1/(LC)0.5 or LC = 1/8.883x109 or C = 1/(8.883x109x10–2) = 11.257x10–9 F = 11.257 pF. (b) since the capacitive reactance cancels out the inductive reactance at resonance, the current through the series circuit is given by I = 120/20 = 6 A. (c) Q = ω o L/R = 94.25x103(0.01)/20 = 47.12.

Chapter 14, Solution 31.

R  10  . R 10   0.05 H  50 mH  0Q (10)( 20) 1 1 C 2   0.2 F  0 L (100)(0.05) 1 1   0.5 rad/s B RC (10)(0.2) L

Chapter 14, Solution 32. Design a problem to help other students to better understand the quality factor, the resonant frequency, and bandwidth of a parallel RLC circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem A parallel RLC circuit has the following values: R = 60 , L = 1 mH, and C = 50 F Find the quality factor, the resonant frequency, and the bandwidth of the RLC circuit. Solution

1 1   4.472 krad/s 3 LC 10 x 50x 10 6 1 1 B   333.33 rad/s RC 60 x50 x10 6  4472 Q o   13.42 B 333.33

o 

Chapter 14, Solution 33. B = ω o /Q = 6x106/120 = 50 krad/s. ω 1 = ω o – B = 5.95x106 rad/s and ω 2 = ω o + B = 6.05x106 rad/s.

Chapter 14, Solution 34.

Q  o RC

Q

R o L

 

 

C

L

80 Q   56.84 pF 2fo R 2x5.6 x10 6 x40 x10 3

R 40x103  = 14.21 µH 2fo Q 2 x 5.6x10 6x80

Chapter 14, Solution 35.

1

o 

(b)

B

(c)

Q   oRC  1.443x10 3 x5x10 3 x60x10 6  432.9

LC



1

(a)

3

8x10 x60x10 6

 1.443 krad/s

1 1   3. 33 rad/s 3 RC 5x10 x60x10  6

Chapter 14, Solution 36. At resonance,

1 1   40  Y 25  10-3 Q 80 Q   0 RC   C    10 F  0 R ( 200  10 3 )(40) 1 1 1 0    L  2   2.5 H 10 0 C (4  10 )(10  10-6 ) LC Y

1 R

 

R

0 200  10 3   2.5 krad / s Q 80 B 1  0   200  1.25  198.75 krad/s 2 B 2  0   200  1.25 201.25 krad/s 2 B

Chapter 14, Solution 37. 0 

1 LC

 5000 rad / s

Y(0 ) 

1   Z(0 )  R  2 k R

 1 4    0.5  j18.75 mS  j  0 C  R  0L   4 1  (1.4212  j53.3)  Z( 0 4)  0.0005  j0.01875

Y( 0 4) 

 1 2    0.5  j7.5 mS  j  0 C  R 0 L   2 1  (8.85  j132. 74)  Z( 0 2)  0.0005  j0.0075

Y( 0 2) 

 1    0.5  j7.5 mS j  20 L  20 C   Z(20 )  (8.85  j132.74) 

Y (2 0 ) 
...