Solution manual fundamentals of electric circuits 3rd edition chapter17

Using MATLAB, synthesize the periodic waveform for which the Fourier trigonometric Fourier series is 1 cos... Determine the Fourier coefficients an and bn of the first three harmonic t

Chapter 17, Problem 1

Evaluate each of the following functions and see if it is periodic If periodic, find its period

(a) f ( t ) = cos π t + 2 cos 3 π t + 3 cos 5 π t

(b) y(t) = sin t + 4 cos 2 π t

(a) This is periodic with ω = π which leads to T = 2π/ω = 2

(b) y(t) is not periodic although sin t and 4 cos 2πt are independently

periodic

(c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],

g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t

which is harmonic or periodic with the fundamental frequency

ω = 1 or T = 2π/ω = 2π

(d) h(t) = cos 2 t = 0.5(1 + cos 2t) Since the sum of a periodic function and

a constant is also periodic, h(t) is periodic ω = 2 or T = 2π/ω = π

(e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic

ω = 0.2π or T = 2π/ω = 10

(f) p(t) = 10 is not periodic

(g) g(t) is not periodic

Chapter 17, Problem 2

Using MATLAB, synthesize the periodic waveform for which the Fourier

trigonometric Fourier series is

1 cos

Chapter 17, Problem 3

Give the Fourier coefficients a0, an, and bn of the waveform in Fig 17.47 Plot the

amplitude and phase spectra

2

n cos(

n sin n

n sin n

2

n sin(

n cos n

n cos n

10 x

π ] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)]

1 n

π π

+ ∑∞

=

5 0 ] x 2 x [ 2

1 dt ) t ( z

π

= π

− π

= π

− π

= ω

sin n

1 ntdt cos 2

1 ntdt cos 1

1 dt n cos ) t ( z

= π

+ π

−

= π

− π

= ω

even n

0,

odd n , n

6 nt

cos n

2 nt

cos n

1 ntdt sin 2

1 ntdt sin 1

1 dt n cos ) t ( z

6 5

0 ,

π π

3 3

t n sin n

3 1 3

t n sin n

π π

3 1 3

n cos

3

2

n

1

+0

t n 2 sin 3

n 4 cos 1 n

3 3

t n 2 cos 3

n sin n

3 1

We can now use MATLAB to check our answer,

1 52

2 5

2 0

Chapter 17, Problem 9

Determine the Fourier coefficients an and bn of the first three harmonic terms of the

rectified cosine wave in Fig 17.52

2 ,

T

183 3

10 4

/ sin )

4 ( 4

10 0 4 / cos 10 8

2 )

(

0 2

0 0

dt n t

40 cos

/ sin

2 5 ] 1 2 / [cos

π π

For n>1,

2

) 1 ( sin ) 1 (

20 2

) 1 ( sin ) 1 (

20 4

) 1 ( sin ) 1 (

20 4

) 1 ( sin )

+ +

=

−

− +

+ +

n

n n

n n

t n n

π

π π

π π

π π

0 sin

10 2 sin 4

20 a , 244 4 2 / sin

20 5

=

= π π + π π

=

Thus,

321

32

1

−

T

0

01

10

2/jn2

/jnt

ojn

4

1 dt e

) t ( y

−

− π

− π

−

22

2/jn

jn

2 e

jn

2 ) 1 2 / jn ( 4 / n

− π

+

− π π

+ π

− π

jn

2 e

jn

2 e

jn

2 ) 1 2 / jn ( e n

4 jn

2 n

4 4

222

2

But

2 / n sin j 2 / n sin j 2 / n cos e

, 2 / n sin j 2 / n sin j 2 / n cos

e 2 / n sin n 2 / n sin ) 1 2 / jn ( j 1 n

1 )

t (

−∞

=

π π + π

− π + π

Chapter 17, Problem 12

* A voltage source has a periodic waveform defined over its period as

v(t) = t(2 π − t) V, 0 < t < 2 π

Find the Fourier series for this voltage

* An asterisk indicates a challenging problem

0

2 T

4 )

3 / t t ( 2

2 0 3

=

− π

π

=

− π π

0 2

T 0

n

t 2 ) nt cos(

n

2 1 dt ) nt cos(

) t t 2 ( T

2

+

− π

0 2

4 ) n 2 cos(

n n

1 ) 1 1 ( n

= π π

2 ( T

0

2

π π

− +

π

−

− π

0 2

2 3

0

n

1 )) nt cos(

nt ) nt (sin(

n

1 n

2

Hence, f(t) = ∑∞

=

− π

1

n 2

2

) nt cos(

n

4 3

2

0 n

4 n

4 π + π =

−

=

π 2 ),

sin(

20

0 ,

t t

0

an = (2/T) ∫T ω

0h ( t ) cos( n ot ) dt = [2/(2π)] ⎢⎣ ⎡ ∫0π + ∫ππ − π ⎥⎦ ⎤

2

dt ) nt cos(

) t sin(

20 dt

) nt cos(

t sin 10

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)]

sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t]

sin(t – π) = sin t cos π – cost sin π = –sin t

sin(t – π)cos(nt) = –sin(t)cos(nt)

2

dt )] t ] n 1 sin([

) t ] n 1 [sin([

20 dt )]

t ] n 1 sin([

) t ] n 1 [sin([

+

+ +

2 n

1

) t ] n 1 cos([

2 n

1

) t ] n 1 cos([

n 1

) t ] n 1 cos([

π +

−

−

+ +

) ] n 1 cos([

3 n

1

) ] n 1 cos([

3 n 1

3 n 1 3 5

= [2/(2π)][ ∫0π10 sin t sin nt dt + ∫π2π20 ( − sin t ) sin nt dt

But, sin A sin B = 0.5[cos(A–B) – cos(A+B)]

sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)]

bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ ( 1 + n )]0π + [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ + π2π

)]

n 1

n 1

) ] n 1 sin([

60 30

10 2

)

(

n

n nt n

) 4 / n sin(

1 n

10 ) nt 2 cos(

) 4 / n cos(

1 n

10 2

+

=

10 sin

1 10 cos 1

4 10

)

(

n

nt n

nt n

t

f

(a) in a cosine and angle form

(b) in a sine and angle form

Chapter 17, Solution 15

(a) Dcos ωt + Esin ωt = A cos(ωt - θ)

where A = D2 + E2 , θ = tan-1(E/D)

n

1 ) 1 n (

2

1 n tan nt 10 cos n

1 ) 1 n (

16 10

(b) Dcos ωt + Esin ωt = A sin(ωt + θ)

where A = D2 + E2 , θ = tan-1(D/E)

+ +

+

1

3 1 6

2

n 4 tan nt 10 sin n

1 ) 1 n ( 16 10

Chapter 17, Problem 16

The waveform in Fig 17.55(a) has the following Fourier series:

V 5

cos 25

1 3 cos 9

1 cos

If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e v2*(t) = v2(t) + 1

Comparing v2*(t) with v1(t) shows that

v2*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1

Hence v2*(t) = 2v1((t + 1)/2)

But v2*(t) = v2(t) + 1

v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2) = -1 + 1 − π 82 ⎢ ⎣ ⎡ cos π ⎜ ⎝ ⎛ + t 2 1 ⎟ ⎠ ⎞ + 9 1 cos 3 π ⎜ ⎝ ⎛ + t 2 1 ⎟ ⎠ ⎞ + 25 1 cos 5 π ⎜ ⎝ ⎛ + t 2 1 ⎟ ⎠ ⎞ + L ⎥ ⎦ ⎤

2

5 2

t 5 cos 25

1 2

3 2

t 3 cos 9

1 2 2

t cos

2

t 5 sin 25

1 2

t 3 sin 9

1 2

t sin

82

Chapter 17, Problem 17

*

Determine if these functions are even, odd, or neither

(a) 1 + t (b) t2 − 1 (c) cos n π t sin n π t

(d) sin2π t (e) t

e−

Chapter 17, Solution 17

We replace t by –t in each case and see if the function remains unchanged

(a) 1 – t, neither odd nor even

(b) t2 – 1, even

(c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt, odd

(d) sin2 n(-t) = (-sin πt)2 = sin2 πt, even

(e) et, neither odd nor even

Chapter 17, Problem 18

Determine the fundamental frequency and specify the type of symmetry present in the functions in Fig 17.56

Figure 17.56

For Probs 17.18 and 17.63

Chapter 17, Solution 18

(a) T = 2 leads to ωo = 2π/T = π

f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric

Obtain the Fourier series for the periodic waveform in Fig 17.57

π π

Chapter 17, Problem 20

Find the Fourier series for the signal in Fig 17.58 Evaluate f(t) at t = 2 using the

first three nonzero harmonics

Figure 17.58

For Probs 17.20 and 17.67

Chapter 17, Solution 20

This is an even function

bn = 0, T = 6, ω = 2π/6 = π/3

ao = ∫ = ⎢⎣ ⎡ ∫ − ∫3 ⎥⎦ ⎤

2

2 1

2 / T

6

2 dt ) t ( T

2

= 1 3 ⎢⎣ ⎡ ( 2 t − 4 t ) + 4 ( 3 − 2 ) ⎥⎦ ⎤

2 1

an = ∫T/4 π

0 ( t ) cos( n / 3 ) dt T

4

= (4/6)[ ∫2 − π

1( 4 t 4 ) cos( n t / 3 ) dt + ∫3 π

2 4 cos( n t / 3 ) dt ] =

3

2

2

1 2

t n sin n

3 6

16 3

t n sin n

3 3

t n sin n

t 3 3

t n cos n

9 6

n cos 3

n 2 cos n

1 24 2

At t = 2,

f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)

+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -]

= 2 + 2.432(0.5 + 0 + 0.2222 + -) f(2) = 3.756

Chapter 17, Problem 21

Determine the trigonometric Fourier series of the signal in Fig 17.59

2

t n cos ) t 1 ( 2 4

4 dt ) t n cos(

) t ( T

+

1

t n cos 2

n cos 1 n

8 2

1

Chapter 17, Problem 22

Calculate the Fourier coefficients for the function in Fig 17.60

Figure 17.60

For Prob 17.22

Chapter 17, Solution 22

Calculate the Fourier coefficients for the function in Fig 16.54

Figure 16.54 For Prob 16.15 This is an even function, therefore bn = 0 In addition, T=4 and ωo = π/2

0 2 1

0

2 T

4

2 dt ) t ( T

4

4 dt ) nt cos(

) t ( T 4

1

0 2

n

t 2 ) 2 / t n cos(

n

4 4

=

n

8 ) 1 ) 2 / n (cos(

n

162

π +

− π π

Chapter 17, Problem 23

Find the Fourier series of the function shown in Fig 17.61

2

4 dt ) t n sin(

) t ( T

4

0 2

2 sin( n t ) n t cos( n t ) n

2

π π

− π π

) n sin(

n

) 1 ( 2

Chapter 17, Problem 24

In the periodic function of Fig 17.62,

(a) find the trigonometric Fourier series coefficients a2 and b2,

(b) calculate the magnitude and phase of the component of f(t) that has

−

=

11

1 9

1 7

1 5

4

f(t) = 1 + t/π, 0 < t < π

bn = ∫π + π

π 0 ( 1 t / ) sin( nt ) dt 2

n

1 ) nt cos(

n

1 2

= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1] a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183 (b) ωn = nωo = 10 or n = 10

a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π)

Thus the magnitude is A10 = 2

10

a10 + = 1/(5π) = 0.06366 and the phase is φ10 = tan−1(bn/an) = −90°

=

π

− π

1 n

) nt sin(

)]

n cos(

2 1 [ n

f(π/2) = ∑∞

=

π π

− π

1 n

) 2 / n sin(

)]

n cos(

2 1 [ n

For n = 1, f1 = (2/π)(1 + 2) = 6/π For n = 2, f2 = 0

For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π) For n = 4, f4 = 0

For n = 5, f5 = 6/(5π), Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) -

= (6/π)[1 − 1/3 + 1/5 − 1/7 + -]

f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5

Determine the Fourier series representation of the function in Fig 17.63

Figure 17.63

For Prob 17.25

Chapter 17, Solution 25

This is a half-wave (odd) function since f(t−T/2) = −f(t)

ao = 0, an = bn = 0 for n = even, T = 3, ωo = 2π/3

For n = odd,

5.1

3

4 tdt n cos ) t ( 3

4

=

102

nt 2 sin n 2

t 3 3

nt 2 cos n 4

9 3

2 1 3

n 2 cos n

322

0

5.1

3

4 dt ) t n sin(

) t ( 3 4

=

1

0 2

nt 2 cos n 2

t 3 3

nt 2 sin n 4

9 3

2 3

n 2 sin n

322

odd

n 1

n

22

22

3

nt 2 sin 3

n 2 cos n

2 3

n 2 sin n 3

3

nt 2 cos 3

n 2 sin n

2 1 3

n 2 cos n 3

Chapter 17, Problem 26

Find the Fourier series representation of the signal shown in Fig 17.64

Figure 17.64

For Prob 17.26

Chapter 17, Solution 26

T = 4, ωo = 2π/T = π/2

ao = ∫ = ⎢⎣ ⎡ ∫ + ∫ + ∫34 ⎥⎦ ⎤

3 1

1 0

T

4

1 dt ) t ( T

1

= 1

an = ( t ) cos( n t ) dt T

π π +

π π

2 2

t n sin n

4 2

t n sin n

2 2

n sin n 4

bn = ∫T ω

0 ( t ) sin( n ot ) dt T

2

2

t n sin 1 dt 2

t n sin 2 dt 2

t n sin 1 4 2

−

π π

2 2

t n cos n

4 2

t n cos n

2 2

= [ cos( n ) 1 ]

n

4

− π π

π π

− π π

+

1

n

) 2 / t n sin(

) 1 ) n (cos(

) 2 / t n cos(

)) 2 / n sin(

) 2 / n 3 (sin(

(a) specify the type of symmetry it has,

2 1

t n cos n

t 2 2

t n sin n

4 dt

2

t n sin t 4

= π

∫

2

n cos n

2 2

n sin n

= 4(−1)(n−1)/2/(n2π2), n = odd

−2(−1)n/2/(nπ), n = even a3 = 0, b3 = 4(−1)/(9π2) = –0.04503

(c) b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = 4/(25π2)

Frms = + ∑ ( + 2)

n

2 n

2

2

1 a

Frms2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(81π2)) + (1/4) + (16/(625π2))] = (1/19.729)(2.6211 + 0.27 + 0.00259)

4 dt ) t n cos(

) t ( T

4

=

1

0 2

n

t ) t n cos(

n

1 ) t n sin(

=

1

0 2

n

t ) t n sin(

n

1 ) t n cos(

n

1 4

− π π

− = 4/(nπ), n = odd

1

k 2 2

) n sin(

n

4 ) n cos(

+ π

− 2 cos( nt ) nt sin( nt ) 0n

2

= 4/(n2π)

− π

) nt sin(

) t (

n

1 ) nt cos(

ot

ojn

T

1 dt e

) t ( T

o

T

2 c

(b) The first term on the right hand side of (1) vanishes if f(t) is odd Hence,

Chapter 17, Problem 31

Let an and bn be the Fourier series coefficients of f(t) and let ωo be its fundamental

frequency Suppose f(t) is time-scaled to give h(t) = f(α t) Express the '

2 ' T

2 ' /

T ' T ),

⎯→

⎯ α

= α

o

' T

2 tdt ' n cos ) t ( h ' T

2 ' a

Let α t = λ , d t = d λ / α , α T ' = T

nT

Similarly, bn' = bn

Chapter 17, Problem 32

Find i(t) in the circuit of Fig 17.68 given that

1 1

I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n)

n 4 1 n

1 )

3 / n ( tan n 1 3

0 n 1

2 2

1 2

2

−

∠ +

=

∠ +

+

1 n

1 2

2 cos( 3 n tan ( 2 n ))

n 4 1 n 3

1 3

1

Chapter 17, Problem 33

In the circuit shown in Fig 17.69, the Fourier series expansion of vs(t) is

1 4

For the DC case, the inductor acts like a short, Vo = 0

For the AC case, we obtain the following:

so

oo

so

V V n

5 n 2 j 1

0 4

V jn n j

V 10

V V

=

π + π +

−

) 5 n

2 ( j n

4 n

5 n 2 j 1

1 n

4 A

n

5 n 2 j 1

V V

22n

n

so

− π +

π

= Θ

−

=

Θ

− π +

π

n

5 n

2 tan

; ) 5 n

2 ( n

4

2222

2n

V ) t n sin(

A )

t ( v

=

Chapter 17, Problem 34

Obtain vo(t) in the network of Fig 17.70 if

tan ) 2 / ( ) 4 / n [(

4 n n 20

) n / ) 2 n ((

tan ) 2 n ( n n

) 2 / ) 4 / n ((

20

2 1 2

2

2 1 2

2 2 2

−

− π

− π

∠ +

=

−

∠

− +

π

− π

2 1 2

2 n tan 2 4

n nt cos 4 n n 20

If vs in the circuit of Fig 17.71 is the same as function f2(t) in Fig 17.56(b), determine the dc component and the first three nonzero harmonics of vo(t)

Figure 17.71

For Prob 17.35

Chapter 17, Solution 35

PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part

of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior

If vs in the circuit of Fig 17.72 is the same as function f2(t) in Fig 17.57(b), determine the dc component and the first three nonzero harmonics of vo(t)

Figure 16.64 For Prob 16.25

Figure 16.50(b) For Prob 16.25

The signal is even, hence, bn = 0 In addition, T = 3, ωo = 2π/3

1 3

4

n

2 )

3 / n 2 sin(

n

6 )

3 / n 2 sin(

n

3 3

π vs(t) = ∑∞

=

π π

π

−

1 n

) 3 / t n cos(

) 3 / n 2 sin(

n

1 2 3 4

Now consider this circuit,

+

−

+

vo

Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3)

Therefore, vo = Zvs/(Z + 1 + j2nπ/3) Simplifying, we get

vo =

) 18 n

4 ( j n 12

v 9 j

2 2

s

− π +

π

−

For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V

We can now solve for vo(t)

3

t n 2 cos A 8

31 n

π

π π

n 2

3 3

n tan 90

and 6

3

n 4 n

16

) 3 / n 2 sin(

n

6

n 2

2 2 2

2 n

where we can further simplify An to this,

81 n

4 n

) 3 / n 2 sin(

9 A

4 4 n

+ π π

π

=

Chapter 17, Problem 36

* Find the response io for the circuit in Fig 17.72(a), where v(t) is shown in Fig

1 2 0

−

100 5

For dc component, ω0 = 0 which leads to I0 = 0

For the nth harmonic,

nn2

2

) 50 n

( j n

10 n

100 j n j 5

0 n

10

− π +

π

= π

− π +

°

∠ π

=

where

2 2 1

n

π φ

If the periodic current waveform in Fig 17.73(a) is applied to the circuit in Fig 17.73(b),

) n cos 1 ( 2 90 ) n cos 1 ( n

2 jn 1

jn

22

π +

− π π +

π

1

2 2 1

1

o n

If the square wave shown in Fig 17.74(a) is applied to the circuit in Fig 17.74(b), find

the Fourier series for vo(t)

π +

=

1k

n

1 2 2

1 ) t ( v

π

= ω ω

+

ω

j 1

j

n

no

1o

12

2

oo

n 1

n tan 2

90 n

2 n tan n

1

90 n V

π +

• π

∠ π +

∠ π

−

1 k n ), n tan t n cos(

n 1

2 )

1 k

), t n sin(

n

1 10

T = 2, ωo = 2π//T = π, ωn = nωo = nπ For the DC component, io = 5/(20 + 40) = 1/12

For the kth harmonic, Vs = (10/(nπ))∠0°

100 mH becomes jωnL = jnπx0.1 = j0.1nπ

50 mF becomes 1/(jωnC) = −j20/(nπ)

Let Z = −j20/(nπ)||(40 + j0.1nπ) =

π +

+ π

−

π +

π

−

n 0 j 40 n

20 j

) n 0 j 40 ( n

20 j

j0.1nπ

+

−

−j20/(nπ)

=

) 20 n

0 ( j n 40

800 j n n

0 j n 40 20 j

n 0 j 40 ( 20 j

2 2 2

− π

= π +

π +

−

π +

−

Zin = 20 + Z =

) 20 n

0 ( j n 40

) 1200 n

( j n 802

2 2

2 2

− π +

π

− π +

π

I =

)]

1200 n

2 ( j n 802 [ n

) 200 n

( j n 400 Z

V

2 2

2 2

in

s

− π +

π π

− π + π

=

Io =

) 20 n

0 ( j n 40

I 20 j )

n 0 j 40 ( n

20 j

I n

20 j

2

2π − +

π

−

= π +

+ π

2 ( j n 802 [ n

200 j

2

2π − +

π π

−

=

2 2

2 2

2 2 1

) 1200 n

( ) 802 ( n

)}

n 802 /(

) 1200 n

{(

tan 90

200

− π + π

π

− π

+

1 k

n

n sin( n t ) I

200 20

°

=

n 802

1200 n

2 tan 90

2 2 1 n

In =

) 1200 n

( ) n 804 ( n

1

2 2

1 dt ) t ( v T

0

2 1

an = ∫ π = ∫1 − π

0

T

0v ( t ) cos( n t ) dt 2 ( 1 t ) cos( n t ) dt T

2 =

1

0 2

n

t ) t n cos(

n

1 ) t n sin(

n

1 2

− π π

=

2 2

2 2 2

2

) 1 n (

4 odd

n , n 4

even n

, 0 ) n cos 1 ( n

2

− π

=

= π

=

= π

− π

0

T

0v ( t ) sin( n t ) dt 2 ( 1 t ) sin( n t ) dt T

− π π

−

n

2 )

t n cos(

n

t ) t n sin(

n

1 ) t n cos(

n

1 2

1

0 2

2 1

n

) 1 n (

16 n

4 A

, n 2

) 1 n ( tan

− π

+ π

=

− π