Chapter 17, Solution 1. 2 periodic ω πω · Hence f(t) = sin(n t) n 10 1 n 1 π π + ∑ ∞ = 5...

Chapter 17, Solution 1.

(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.


(a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1. ω = 1 = 2π/T or T = 2π.

(b) ω = 2 or T = 2π/ω = π. (c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)

ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600. (d) f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.


T = 4, ωo = 2π/T = π/2 g(t) = 5, 0 < t < 1 10, 1 < t < 2 0, 2 < t < 4

ao = (1/T) = 0.25[ + ] = ∫T

0dt)t(g ∫

1

0dt5 ∫

2

1dt10 3.75

an = (2/T) = (2/4)[ ∫ ωT

0 o dt)tncos()t(g ∫π1

0dt)t

2ncos(5 + ∫

π2

1dt)t

2ncos(10 ]

= 0.5[1

0

t2

nsinn25 ππ

+ 2

1

t2

nsinn2 ππ

10 ] = (–1/(nπ))5 sin(nπ/2)

an = (5/(nπ))(–1)(n+1)/2, n = odd 0, n = even

bn = (2/T) = (2/4)[ ∫ ωT

0 o dt)tnsin()t(g ∫π1

0dt)t

2nsin(5 + ∫

π2

1dt)t

2nsin(10 ]

= 0.5[1

0

t2

ncosn

5x2 ππ

− – 2

1

t2

ncosn

10x2 ππ

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


f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π

ao = (1/T) = (1/2) = ∫T

0dt)t(f ∫ −

2

0dt)t510(

2

0

2 )]2/t5(t10[5.0 − = 5

an = (2/T) = (2/2) ∫ ωT

0 o dt)tncos()t(f ∫ π−2

0dt)tncos()t510(

= – ∫ π2

0dt)tncos()10( ∫ π

2

0dt)tncos()t5(

= 2

022 tncos

n5

ππ

− + 2

0

tnsinn

t5π

π = [–5/(n2π2)](cos 2nπ – 1) = 0

bn = (2/2) ∫ π−2

0dt)tnsin()t510(

= – ∫ π2

0dt)tnsin()10( ∫ π

2

0dt)tnsin()t5(

= 2

022 tnsin

n5

ππ

− + 2

0

tncosn

t5π

π = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)

Hence f(t) = )tnsin(n110

1nπ

π+ ∑

∞

=

5


1T/2,2T =π=ωπ=

5.0]x2x1[21dt)t(z

T1a

T

0o −=π−π

π== ∫

∫∫∫π

π

ππ

ππ

=π

−π

=π

−π

=ω=2

20

0

T

0on 0ntsin

n2nt..sin

n1ntdtcos21ntdtcos11dtncos)t(z

T2a

∫∫∫π

π

ππ

ππ

=

=π=

π+

π−=

π−

π=ω=

22

00

T

0on

evenn 0,

oddn,n6

ntcosn2ntcos

n1ntdtsin21ntdtsin11dtncos)t(z

T2b

Thus,

ntsinn65.0)t(z

oddn1n∑∞

== π

+−=


.0a,functionoddanisthisSince

326)1x21x4(

21dt)t(y

21a

22,2T

n

20o

o

=

==+==

π=π

=ω=

∫

∑

∫∫∫

∞

==

=

=π

ππ

+=

=π−π

=π−π

−π−π

=

π−ππ

−−ππ

−=π

π−π

π−

=

π+π=ω=

oddn1n

evenn,0

oddn,n4

21

10

21

10

20 on

)tnsin(n143)t(y

))ncos(1(n2))ncos(1(

n2))ncos(1(

n4

))ncos()n2(cos(n2)1)n(cos(

n4)tncos(

n2)tncos(

n4

dt)tnsin(2dt)tnsin(4dt)tnsin()t(y22b


0a,6

T/2,12T 0 =π

=π=ω=

∫∫∫ π−+π=ω=−

10

4

4

2

T

0on ]dt6/tncos)10(dt6/tncos10[

61dtncos)t(f

T2a

[ ]3/n5sin3/nsin3/n2sin2n106/tnsin

n106/tnsin

n10 10

44

2 π−π+ππ

=ππ

−ππ

= −

∫∫∫ π−+π=ω=−

10

4

4

2

T

0on ]dt6/tnsin)10(dt6/tnsin10[

61dtnsin)t(f

T2b

[ ]3/n2sin23/ncos3/n5cosn106/ntncos

n106/tncos

n10 10

44

2 π−π+ππ

=ππ

+ππ

−= −

( )∑∞

=π+π=

1nnn 6/tnsinb6/tncosa)t(f

where an and bn are defined above.


π=π=ω=<<+= T/22,T1, t 1- ),t1(2)t(f o

2ttdt)1t(221dt)t(f

T1a

1

1

1

1

2T

0o =+=+==

−−∫∫

0tnsinn1tnsin

nttncos

n12tdtncos)1t(2

22dtncos)t(f

T2a

1

122

1

1

T

0on =

π

π+π

π+π

π=π+=ω=

−−∫∫

ππ

−=

π

π−π

π−π

π−=π+=ω=

−−∫∫ ncos

n4tncos

n1tncos

nttnsin

n12tdtnsin)1t(2

22dtnsin)t(f

T2b

1

122

1

1

T

0on

∑∞

=π

−π

−=1n

ntncos

n)1(42)t(f

Chapter 17, Solution 9. f(t) is an even function, bn=0.

4//2,8 ππω === TT

183.3104/sin)4(4

1004/cos1082)(1 2

0

2

00

===

+== ∫∫ π

ππ

π tdttdttfT

aT

o

[ ]dtntntdttntdtntfT

aT

on ∫∫∫ −++=+==2

0

2

0

2/

0

4/)1(cos4/)1(cos5]04/cos4/cos10[840cos)(4 ππππω

For n = 1,

102/sin25]12/[cos52

0

2

01 =

+=+= ∫ tdttdtta ππ

π

For n>1,

2)1(sin

)1(20

2)1(sin

)1(20

4)1(sin

)1(20

4)1(sin

)1(20 2

0

−−

++

+=

−−

++

+=

nn

nn

nn

tnn

anπ

ππ

ππ

ππ

π

0sin102sin420,3662.62/sin20sin10

32 =+==+= ππ

ππ

ππ

ππ

aa

Thus,

3213210 0,0,362.6,10,183.3 bbbaaaa ======= Chapter 17, Solution 10.

π=π=ω= T/2,2T o

π−−

π−=

−+==

π−π−π−π−ω−∫ ∫ ∫ 2

1

tjn10

tjnT

0

10

21

tjntjntojnn jn

e2jn

e421dte)2(dte4

21dte)t(h

T1c

[ ] ,even n ,0

oddn ,n

j6]6ncos6[

n2je2e24e4

n2jc jnn2jnj

n

=

=π

−=−ππ

=+−−π

= π−π−π−

Thus,

tjn

oddnn

en

6j)t(f π∞

=−∞=∑

π−

=


2/T/2,4T o π=π=ω=

∫ ∫ ∫

++==−

π−π−ω−T

0

01

10

2/tjn2/tjntojnn dte)1(dte)1t(

41dte)t(y

T1c

π−

π−−π−

π−= π−

−π−

π− 10

2/tjn01

2/tjn22

2/tjnn e

jn2e

jn2)12/tjn(

4/ne

41c

π

+π

−π

+−ππ

+π

−π

π−ππjn2e

jn2e

jn2)12/jn(e

n4

jn2

n4

41 2/jn2/jn2/jn

2222=

But

2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose 2/jn2/jn π−=π−π=π=π+π= π−π

[ ]2/nsinn2/nsin)12/jn(j1n

1c 22n ππ+π−π+π

=

[ ] 2/tjn22

ne2/nsinn2/nsin)12/jn(j1

n1)t(y π

∞

−∞=ππ+π−π+

π= ∑


A voltage source has a periodic waveform defined over its period as v(t) = t(2π - t) V, for all 0 < t < 2π Find the Fourier series for this voltage. v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1 ao =

(1/T) dt)tt2(21dt)t(f

2

0

2T

0 ∫∫π

−ππ

=3

2)3/21(24)3/tt(

21 23

2

032 π

=−ππ

=−ππ

= π

an = π

π

+π

π=−π∫

2

02

T

0

2 )ntsin(n

t2)ntcos(n21dt)ntcos()tt2(

T2

[ ] π+−

π−

2

0

223 )ntsin(tn)ntsin(2)ntcos(nt2

n1

232 n4)n2cos(n4

n1)11(

n2 −

=πππ

−−=

bn = ∫∫ −π

=− dt)ntsin()tnt2(1dt)ntsin()tnt2(T2 2T

0

2

ππ −+

π−−

π=

2

0

22302 ))ntcos(tn)ntcos(2)ntsin(nt2(

n1))ntcos(nt)nt(sin(

n1n2

0n

4n4

=π

+π−

=

Hence, f(t) = ∑∞

=

−π

1n2

2

)ntcos(n4

32

Chapter 17, Solution 13. T = 2π, ωo = 1

ao = (1/T) dttsin10[21dt)t(h

0

T

0 ∫∫π

π= + ]dt)tsin(20

2

∫π

ππ−

[ ]π

=π−−−π

= π

π

π 30)tcos(20tcos1021 2

0

an = (2/T) ∫ ωT

0 o dt)tncos()t(h

= [2/(2π)]

π−+∫ ∫

π π

π0

2dt)ntcos()tsin(20dt)ntcos(tsin10

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)

an =

−++−−++

π ∫ ∫π π

π0

2dt)]t]n1sin([)t]n1[sin([20dt)]t]n1sin([)t]n1[sin([10

21

=

−−

+++

+

−−

−++

−π

π

π

π 2

0 n1)t]n1cos([2

n1)t]n1cos([2

n1)t]n1cos([

n1)t]n1cos([5

an =

−π−

−+

π+−

−+

+π n1)]n1cos([3

n1)]n1cos([3

n13

n135

But, [1/(1+n)] + [1/(1-n)] = 1/(1–n2)

cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ

an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))]

= [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd

bn = (2/T) ∫ ωT

0 o dttnsin)t(h

= [2/(2π)][ + ∫π

0dtntsintsin10 ∫

π

π−

2dtntsin)tsin(20

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)/ π+

0)]n1(

+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ π

π+ 2)]n1(

=

+π+

+−

π−−

π n1)]n1sin([

n1)]n1sin([5 = 0

Thus, h(t) = ∑∞

= −π−

π 1k2 )1k4(

)kt2cos(6030

Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B.

f(t) = ∑∞

=

π

+−π

++

1n33 )nt2sin()4/nsin(

1n10)nt2cos()4/ncos(

1n102


(a) Dcos ωt + Esin ωt = A cos(ωt - θ) where A = 22 ED + , θ = tan-1(E/D)

A = 622 n1

)1n(+

+16 , θ = tan-1((n2+1)/(4n3))

f(t) = ∑+10∞

=

−

+−+

+1n3

21

622 n41ntannt10cos

n1

)1n(16

(b) Dcos ωt + Esin ωt = A sin(ωt + θ)

where A = 22 ED + , θ = tan-1(D/E)

f(t) = ∑+10∞

=

−

+

+++1n

2

31

622 1nn4tannt10sin

n1

)1n(16


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

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

1

v2*(t)

2

t

-2 -1 0 1 2 3 4 5

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+

+

π+

+

ππ

−2

1t5cos251

21t3cos

91

21tcos8

2

v2(t) =

+

π

+π

+

π

+π

+

π

+π

π−

25

2t5cos

251

23

2t3cos

91

22tcos8

2

v2(t) =

+

π

+

π

+

π

π 2t5sin

251

2t3sin

91

2tsin8

2−

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.


(a) T = 2 leads to ωo = 2π/T = π

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

(b) T = 3 leads to ωo = 2π/3 f2(t) = f2(-t), showing that f2(t) is even.

(c) T = 4 leads to ωo = π/2

f3(t) is even and half-wave symmetric. Chapter 17, Solution 19. This is a half-wave even symmetric function.

ao = 0 = bn, ωo = 2π/T π/2

an = ∫ ω

−

2/T

0 o dt)tncos(Tt41

T4

= [4/(nπ)2](1 − cos nπ) = 8/(n2π2), n = odd

= 0, n = even

f (t) = ∑∞

=

π

π oddn22 2

tncosn18

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

0dt4dt)4t4(

62dt)t(f

T2

=

−+− )23(4)t4t2(

31 2

1

2 = 2

an = ∫ π4/T

0dt)3/tncos()t(f

T4

= (4/6)[ + ] ∫ π−2

1dt)3/tncos()4t4( ∫ π

3

2dt)3/tncos(4

= 3

2

2

122 3

tnsinn3

616

3tnsin

n3

3tnsin

nt3

3tncos

n9

616

π

π+

π

π−

π

π+

π

π

= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]

Thus f(t) = ∑∞

=

π

π−

π

π+

1n22 3

tncos3ncos

3n2cos

n1242

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, Solution 21. This is an even function.

bn = 0, T = 4, ωo = 2π/T = π/2. f(t) = 2 − 2t, 0 < t < 1 = 0, 1 < t < 2

ao = ∫

−=−

1

0

1

0

2

2ttdt)t1(2

42 = 0.5

an = ∫∫

π

−=ω1

0

2/T

0 o dt2

tncos)t1(244dt)tncos()t(f

T4

= [8/(π2n2)][1 − cos(nπ/2)]

f(t) = ∑∞

=

π

π

−π

+1n

22 2tncos

2ncos1

n8

21


Calculate the Fourier coefficients for the function in Fig. 16.54. f(t)

4

t

-5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 17.61 For Prob. 17.22

This is an even function, therefore bn = 0. In addition, T=4 and ωo = π/2.

ao = === ∫∫1

0

21

0

2T

0ttdt4

42dt)t(f

T2 1

an = ∫∫ π=ω1

0

2T

0 o dt)2/tncos(t444dt)ntcos()t(f

T4

1

022 )2/tnsin(

nt2)2/tncos(

n44

π

π+π

π=

an = )2/nsin(n8)1)2/n(cos(

n16

22 ππ

+−ππ

Chapter 17, Solution 23. f(t) is an odd function.

f(t) = t, −1< t < 1 ao = 0 = an, T = 2, ωo = 2π/T = π

bn = ∫∫ π=ω1

0

2/T

0 o dt)tnsin(t24dt)tnsin()t(f

T4

= [ ]1022 )tncos(tn)tnsin(

n2

ππ−ππ

= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)

f(t) = ∑∞

=

+

π−

π 1n

1n

)tnsin(n)1(2


(a) This is an odd function.

ao = 0 = an, T = 2π, ωo = 2π/T = 1

bn = ∫ ω2/T

0 o dt)ntsin()t(fT4

f(t) = 1 + t/π, 0 < t < π

bn = ∫π

π+π 0

dt)ntsin()/t1(24

= π

π−

π+−

π 02 )ntcos(

nt)ntsin(

n1)ntcos(

n12

= [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

102 ba10+ = 1/(5π) = 0.06366

and the phase is φ10 = tan−1(bn/an) = −90°

(c) f(t) = ∑ π ∞

=

π−π1n

)ntsin()]ncos(21[n2

f(π/2) = ∑∞

=

ππ−π1n

)2/nsin()]ncos(21[n2

π

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.

(d) From part (c) f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -] (3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -] or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - - Chapter 17, Solution 25. This is an odd function since f(−t) = −f(t).

ao = 0 = an, T = 3, ωo = 2π/3.

b n = ∫∫ π=ω1

0

2/T

0 o dt)3/nt2sin(t34dt)tnsin()t(f

T4

= 1

022 3

nt2cosn2t3

3nt2sin

n49

34

π

π−

π

π

=

π

π−

π

π 3n2cos

n23

3n2sin

n49

34

22

f(t) = ∑∞

=

π

π

π−

π

π1n22 3

t2sin3n2cos

n2

3n2sin

n3

Chapter 17, Solution 26. T = 4, ωo = 2π/T = π/2

ao =

++= ∫∫∫∫

4

3

3

1

1

0

T

0dt1dt2dt1

41dt)t(f

T1 = 1

an = dt)tncos()t(fT

T

0 o∫ ω2

an =

π+π+π ∫∫∫

4

3

3

2

2

1dt)2/tncos(1dt)2/tncos(2dt)2/tncos(1

42

=

ππ

+π

π+

ππ

4

3

3

2

2

1 2tnsin

n2

2tnsin

n4

2tnsin

n22

=

π

−π

π 2nsin

2n3sin

n4

bn = ∫ ωT

0 o dt)tnsin()t(fT2

=

π

+π

+π

∫∫∫4

3

3

2

2

1dt

2tnsin1dt

2tnsin2dt

2tnsin1

42

=

ππ

−π

π−

ππ

−4

3

3

2

2

1 2tncos

n2

2tncos

n4

2tncos

n22

= [ ]1)ncos(n4

−ππ

Hence f(t) =

[ ]∑∞

=

π−π+ππ−ππ

+1n

)2/tnsin()1)n(cos()2/tncos())2/nsin()2/n3(sin(n41

Chapter 17, Solution 27. (a) odd symmetry. (b) ao = 0 = an, T = 4, ωo = 2π/T = π/2 f(t) = t, 0 < t < 1 = 0, 1 < t < 2

bn = 1

022

1

0 2tncos

nt2

2tnsin

n4dt

2tnsint

4

π

π−

ππ

=π

∫4

= 02

ncosn2

2nsin

n4

22 −π

π−

ππ

= 4(−1)(n−1)/2/(n2π2), n = odd

−2(−1)n/2/(nπ), n = even

a3 = 0, b3 = 4(−1)/(9π2) = 0.045 (c) b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = (25π2)

Frms = ( )∑ ++ 2n

2n

2o ba

21a

Frms

2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))]

= (1/19.729)(2.6211 + 0.27 + 0.00259) Frms = 14659.0 = 0.3829

Compare this with the exact value of Frms = 6/1dttT2 1

0

2 =∫ = 0.4082

Chapter 17, Solution 28. This is half-wave symmetric since f(t − T/2) = −f(t).

ao = 0, T = 2, ωo = 2π/2 = π

an = ∫∫ π−=ω1

0

2/T

0 o dt)tncos()t22(24dt)tncos()t(f

T4

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n14

π

π−π

π−π

π

= [4/(n2π2)][1 − cos(nπ)] = 8/(n2π2), n = odd

0, n = even

bn = 4 ∫ π−1

0dt)tnsin()t1(

= 1

022 )tncos(

nt)tnsin(

n1)tncos(

n14

π

π+π

π−π

π−

= 4/(nπ), n = odd

f(t) = ∑∞

=

π

π+π

π1k22 )tnsin(

n4)tncos(

n8 , n = 2k − 1

Chapter 17, Solution 29. This function is half-wave symmetric.

T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π

For odd n, an = ∫π−

0dt)ntcos()t(

T2 = [ ]π+

π−

02 )ntsin(nt)ntcos(n2 = 4/(n2π)

bn = [ ]ππ−

π−=−

π ∫ 020)ntcos(nt)ntsin(

n2dt)ntsin()t(2 = −2/n

Thus,

f(t) = ∑∞

=

−

π1k2 )ntsin(

n1)ntcos(

n22 , n = 2k − 1


∫ ∫ ∫−

− −ω−

ω−ω==

2/T

2/T

2/T2/T

2/T2/T oo

tojnn tdtnsin)t(fjtdtncos)t(f

T1dte)t(f

T1c (1)

(a) The second term on the right hand side vanishes if f(t) is even. Hence

∫ ω=2/T

0on tdtncos)t(f

T2c

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

∫ ω−=2/T

0on tdtnsin)t(f

T2jc


If oo /T2

'T2'/T'T),t(f)t(h αω=

απ

=π

=ω→α=α=

∫∫ ωα=ω='T

0o

'T

0on tdt'ncos)t(f

'T2tdt'ncos)t(h

'T2'a

Let T'T,/dtd,,t =ααλ=λ=α

n

T

0on a/dncos)(f

T2'a =αλλωλα

= ∫

Similarly, nn b'b =

Chapter 17, Solution 32. When is = 1 (DC component)

i = 1/(1 + 2) = 1/3

For n 1, ω≥ n = 3n, Is = 1/n2∠0°

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

= )n2tan(n41n3

1)3/n6(tann413

0n1

2212

2−∠

+=

∠+

°∠

−

Thus,

i(t) = ∑∞

=

−−+

+1n

1

22))n2(tann3cos(

n41n3

131


For the DC case, the inductor acts like a short, Vo = 0. For the AC case, we obtain the following:

so

ooso

VVn5n5.2j1

04Vjn

n2jV

10VV

=

π−π+

=π

+π

+−

)5n5.2(jn4

n5n5.2j1

1n4A

n5n5.2j1

VV

22nn

so

−π+π=

π−π+π

=Θ∠

π−π+

=

π−π

−=Θ−π+π

= −n

5n5.2tan;)5n5.2(n

4A22

1n22222n

V)tnsin(A)t(v1n

nno ∑∞

=Θ+π=

Chapter 17, Solution 34. For any n, V = [10/n2]∠(nπ/4), ω = n. 1 H becomes jωnL = jn and 0.5 F becomes 1/(jωnC) = −j2/n

2 Ω jn

−j2/n

+ −

+

Vo

−

V Vo = −j(2/n)/[2 + jn − j(2/n)]V = −j2/[2n + j(n2 − 2)][(10/n2)∠(nπ/4)]

)]n2/)2n((tan)2/()4/n[(

4nn

20)n2/)2n((tan)2n(n4n

)2/)4/n((20

21

22

212222

−−π−π∠+

=

−∠−+

π−π∠=

−

−

vo(t) = ∑∞

=

−

−−

π−

π+

+1n

21

22 n22ntan

24nntcos

4nn20


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).

1 Ω 1 H

1 F 1 Ω

+ −

+

vo

−

vS

Figure 17.72 For Prob. 17.35

1

f2(t)

2

t

-2 -1 0 1 2 3 4 5

Figure 17.57(b) For Prob. 17.35

The signal is even, hence, bn = 0. In addition, T = 3, ωo = 2π/3. vs(t) = 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5

ao = 34dt2dt1

32 5.1

1

1

0=

+ ∫∫

an =

π+π ∫∫

5.1

1

1

0dt)3/tn2cos(2dt)3/tn2cos(

34

= )3/n2sin(n2)3/tn2sin(

n26)3/tn2sin(

n23

34 5.1

1

1

0π

π−=

π

π+π

π

vs(t) = ∑∞

=

πππ

−1n

)3/tn2cos()3/n2sin(n12

34

Now consider this circuit,

j2nπ/3 1 Ω

-j3/(2nπ) 1 Ω

+ −

+

vo

−

vS

Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3) Therefore, vo = Zvs/(Z + 1 + j2nπ/3). Simplifying, we get

vo = )18n4(jn12

v9j22

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)

vo(t) = volts3

tn2cosA83

1nnn

Θ+

π+∑

∞

=

where

π−

π−=Θ

−

π+π

ππ= −

n23

3ntan90and

63

n4n16

)3/n2sin(n6

A 1on222

22

n

where we can further simplify An to this, 81n4n)3/n2sin(9

44n+ππ

A π=


vs(t) = ∑∞

==

θ−

oddn1n

nn )ntcos(A

where θn = tan−1[(3/(nπ))/(−1/(nπ))] = tan−1(−3) = 100.5°

An = 2

nsin9n1

2nsin

n1

n9 22

2222π

+π

=π

π+

π

ωn = n and 2 H becomes jωnL = j2n

Let Z = 1||j2n = j2n/(1 + j2n) If Vo is the voltage at the non-reference node or across the 2-H inductor. Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/1 + [j2n/(1 + j2n)] = j2nVs/(1 + j4n) But Vs = An∠−θn Vo = j2n An∠−θn/(1 + j4n)

Io = Vo/j = [2n An∠−θn]/ 2n161 + ∠tan−14n

= 2

2

n161

n22

nsin9n1

+

π+

π∠−100.5° − tan−14n

Since sin(nπ/2) = (−1)(n−1)/2 for n = odd, sin2(nπ/2) = 1

Io = 2

1

n161

n4tan5.100102

+

−°−∠π

−

io(t) = )n4tan5.100ntcos(n161

1102

oddn1n

1

2∑∞

==

−−°−+π

Chapter 17, Solution 37. From Example 15.1,

vs(t) = ∑∞

=

ππ

+1k

),tnsin(n1205 n = 2k − 1

For the DC component, the capacitor acts like an open circuit.

Vo = 5

For the nth harmonic, Vs = [20/(nπ)]∠0°

10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ)

vo = 22

1s

n25nn5tan90100

n20

100jn20100j

20n

100j

Vn

100j

π+ππ

+°−∠=

π−π−

=+

π−

π− −

vo(t) = ∑ π+°−π

π+π− )

n5tan90tnsin(

n25n

1100 1

22


∑∞

=+=π

π+=

1ks 1k2n,tnsin

n12

21)t(v

π=ωω+

ω= n,V

j1jV ns

n

no

For dc, 0V,5.0V,0 osn ===ω

For nth harmonic, os 90

n2V −∠π

=

22

1o

122

oo

n1

ntan290n2

ntann1

90nVπ+

π−∠=∠

π•

π∠π+

∠π=

−

−

1k2n),ntantncos(n1

2)t(v 1

1k 22o −=π−ππ+

= −∞

=∑


Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence

vs(t) = ∑∞

=

ππ

+1k

),tnsin(n1105 n = 2k − 1

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) 40 Ω

−j20/(nπ)

+ −

I Io

Z

20 Ω

VS j0.1nπ

Let Z = −j20/(nπ)||(40 + j0.1nπ) = π++

π−

π+π

−

n1.0j40n20j

)n1.0j40(n20j

= )20n1.0(jn40

800jn2n1.0jn4020j

n1.0j40(20j2222 −π+π

−π=

π+π+−π+−

Zin = 20 + Z = )20n1.0(jn40)1200n2(jn802

22

22

−π+π−π+π

I = )]1200n2(jn802[n

)200n(jn400ZV

22

22

in

s

−π+ππ−π+π

=

Io = )20n1.0(jn40

I20j

)n1.0j40(n20j

In20j

22 −π+π−

=π++

π−

π−

= )]1200n2(jn802[n

200j22 −π+ππ

−

= 2222

221

)1200n2()802(n)n802/()1200n2(tan90200

−π+π

π−π−°−∠ −

Thus

io(t) = ∑∞

=

θ−ππ

+1k

nn )tnsin(I200201 , n = 2k − 1

where π

−π+°=θ −

n8021200n2tan90

221

n

In = )1200n2()n804(n

1222 −π+π

Chapter 17, Solution 40. T = 2, ωo = 2π/T = π

ao = 2/12ttdt)t22(

21dt)t(v

T

1

0

21

0

T

0=

−=−= ∫∫

1

an = ∫∫ π−=π1

0

T

0dt)tncos()t1(2dt)tncos()t(v

T2

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n12

π

π−π

π−π

π

= 2222

22

)1n2(4oddn,

n4

evenn,0)ncos1(

n2

−π==

π

==π−

π

bn = ∫∫ π−=π1

0

T

0dt)tnsin()t1(2dt)tnsin()t(v

T2

= π

=

π

π+π

π−π

π−

n2)tncos(

nt)tnsin(

n1)tncos(

n12

1

022

vs(t) = ∑ ϕ−π+ )tncos(A21

nn

where 4422n

21

n )1n2(16

n4A,

n2)1n2(

−π+

π=

−π−tan=φ

For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts like an open circuit.

+

Vx

−

i

Vo2Vx

+

Vo

−

Vx

− +

+ − 3 Ω

1 Ω 0.5V

(a)

Vo

(1/4)F

2Vx

+

Vo

−

Vx

− +

+ − 3 Ω

1 Ω VS (b)

Applying KVL to the circuit in Figure (a) gives –0.5 – 2Vx + 4i = 0 (1) But –0.5 + i + Vx = 0 or –1 + 2Vx + 2i = 0 (2) Adding (1) and (2), –1.5 + 6i = 0 or i = 0.25 Vo = 3i = 0.75 For the nth harmonic, we consider the circuit in Figure (b).

ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ) At the supernode, (Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3 Vs = [1 + jnπ/4]Vx + Vo/3 (3) But –Vx – 2Vx + Vo = 0 or Vo = 3Vx Substituting this into (3), Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo

Vo = 12Vs/(8 + jnπ) = )8/n(tann64

A12122

n

π∠π+

φ−∠−

Vo = ))]n2/()1n2((tan)8/n([tan)1n2(

16n

4n64

12 11442222

−π−π∠−π

+ππ+

−−

Thus

vo(t) = ∑∞

=

θ+π+1n

nn )tncos(V43

where Vn = 442222 )1n2(

16n

4

n64

12−π

+ππ+

θn = tan–1(nπ/8) – tan–1(π(2n – 1)/(2n)) Chapter 17, Solution 41. For the full wave rectifier,

T = π, ωo = 2π/T = 2, ωn = nωo = 2n

Hence

vin(t) = ( )∑∞

= −π−

π 1n2 nt2cos

1n4142

For the DC component,

Vin = 2/π

The inductor acts like a short-circuit, while the capacitor acts like an open circuit.

Vo = Vin = 2/π

For the nth harmonic, Vin = [–4/(π(4n2 – 1))]∠0°

2 H becomes jωnL = j4n

0.1 F becomes 1/(jωnC) = –j5/n

Z = 10||(–j5/n) = –j10/(2n – j)

Vo = [Z/(Z + j4n)]Vin = –j10Vin/(4 + j(8n – 10))

=

−π°∠

−−+

−)1n4(

04)10n8(j4

10j2

= 22

1

)10n8(16)1n4()5.2n2(tan9040

−+−π

−−°∠ −

Hence vo(t) = ∑∞

=

θ++π 1n

nn )nt2cos(A2

where

An = 29n40n16)1n4(

2022 +−−π

θn = 90° – tan–1(2n – 2.5)


∑∞

==π

π+=

1ks 1-2kn ,tnsin

n1205v

π=ω=ωω

=→−ω=−

nn ,VRCjV)V0(Cj

R0V

onsn

oons

For n = 0 (dc component), Vo=0. For the nth harmonic,

22

5

9422o

oo

n210

10x40x10xn2090

n20

RCn901V

π=

π=−∠

ππ∠

=−

Hence,

∑∞

==π

π=

1k22

5o 1-2kn ,tncos

n1

210)t(v

Alternatively, we notice that this is an integrator so that

∑∫∞

==π

π=−=

1k22

5so 1-2kn ,tncos

n1

210dtv

RC1)t(v


(a) Vrms = )1020(2130)ba(

21a 222

1n

2n

2n

20 ++=++ ∑

∞

=

= 33.91 V

(b) Irms = )24(216 222 ++ = 6.782 A

(c) P = VdcIdc + ∑ Φ−Θ )cos(IV21

nnnn

= 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)]

= 180 + 32.76 – 9.659 = 203.1 W Chapter 17, Solution 44.

(a) [ ] W73.300535.319.27045cos1025cos6021vip oo =++=++==

(b) The power spectrum is shown below.

p 27.19 3.535 0 1 2 3 ω Chapter 17, Solution 45. ωn = 1000n jωnL = j1000nx2x10–3 = j2n 1/(jωnC) = –j/(1000nx40x10–6) = –j25/n

Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n

I = V/Z

For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23 I1 = 100/(10 – j23) = 3.987∠73.89°

For n = 2, V2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5 I2 = 50/(10 – j8.5) = 3.81∠40.36°

For n = 3, V3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333 I3 = 25/(10 – j2.333) = 2.435∠13.13° Irms = 0.5 222 435.281.3987.3 ++ = 3.014 A

p = VDCIDC + ∑=

φ−θ3

1nnnnn )cos(IV

21

= 0 + 0.5[100x3.987cos(73.89°) + 50x3.81cos(40.36°) + 25x2.435cos(13.13°)]

= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts Chapter 17, Solution 46. (a) This is an even function

Irms = ∫∫ =2/T

0

2T

0

2 dt)t(fT2dt)t(f

T1

f(t) = 2t1,01t0,t22

<<<<−

T = 4, ωo = 2π/T = π/2

Irms2 =

1

0

321

0

2 )3/ttt(2dt)t1(442

+−=−∫

= 2(1 – 1 + 1/3) = 2/3 or Irms = 0.8165 A (b) From Problem 16.14, an = [8/(n2π2)][1 – cos(nπ/2)], ao = 0.5

a1 = 8/π2, a2 = 4/π2, a3 = 8/(9π2), a4 = 0, a5 = 9/(25π2), a6 = 4/(9π2)

Irms = 66623.08116

62564

81641664

21

41A

21a 4

1n

2no =

++++

π+≅+ ∑

∞

=

Irms = 0.8162 A Chapter 17, Solution 47. Let I = IDC + I1 + I2

For the DC component

IDC = [5/(5 + 10)](3) = 1 A

Is

I

5 Ω

j8 10 Ω

For AC, ω = 100 jωL = j100x80x10–3 = j8

In = 5Is/(5 + 10 + j8)

For Is = 0.5∠–60° I1 = 10∠–60°/(15 + j8) or |I1| = 10/ 22 815 +

For Is = 0.5∠–120°

I2 = 2.5∠–120°/(15 + j8) or |I2| = 2.5/ 22 815 + p10 = (IDC

2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10

p10 = 11.838 watts

Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10–3 = 40

For the AC component, ωn = 10n, n = 1,2

1/(jωnC) = –j/(10nx100x10–6) = (–j/n) kΩ Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j)

V = ZI = [–j2/(2n – j)]I

For n = 1, V1 = [–j2/(2 – j)]16∠45° = 14.311∠–18.43° mV

For n = 2, V2 = [–j2/(4 – j)]12∠–60° = 5.821∠–135.96° mV

v(t) = 40 + 0.014311cos(10t – 18.43°) + 0.005821cos(20t – 135.96°) V

(b) p = VDCIDC + ∑∞

=

φ−θ1n

nnnn )cos(IV21

= 20x40 + 0.5x10x0.014311cos(45° + 18.43°) +0.5x12x0.005821cos(–60° + 135.96°)

= 800.1 mW Chapter 17, Solution 49.

(a) 5.2)5(21dt4dt1

21dt)t(z

T1Z

0

2T

0

2rms

2 =ππ

=

+

π== ∫ ∫∫

π π

π

581.1Zrms =

(b)

9193.2...251

161

91

411

181

41

n36

21

41)ba(

21aZ 2

1n 1n22n

2n

2o

2rms

2 =

+++++

π+=

π+=++= ∑ ∑

∞

=

∞

= 7086.1rms =Z

(c ) 071.8100x11.581

1.7086 %error =

−=


cn = ∫ π==ωω−T

0 ontj

1n2,dte)t(f

To

1

= ∫−π−1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt

cn = ∫−π−

−

π−

π+

π

1

1

tjn1

1

tjn dtejn21e

jn2t

−

= [ ]1

1

tjn222

tjnjn e)j(n2

1eenj

−

π−ππ−

−π++

π

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)

cn = π−

=ππ

+π−

n)1(j)nsin(

n2j2

n)1( n

22

nj

Thus

f(t) = ∑∞

−∞=

ω

n

tjnn

oec = ∑∞

−∞=

π

π−

n

tjnn enj)1(


π=π=ω= T/2,2T o

( ) 20

T

0

2222

03

tjntjn2tojn

n 2tjn2tn)jn(

e21dtet

21dte)t(f

T1c ∫ ∫ +π+π−

π−===

π−π−ω−

)jn1(n

2)n4jn4(n2j1c 22

2233n π+

π=π+π−

π=

∑∞

−∞=

ππ+π

=n

tjn22

e)jn1(n

2)t(f


cn = ∫ π==ωω−T

0 ontj

1n2,dte)t(f

To

1

= ∫−π−1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt

cn = ∫−π−

−

π−

π+

π

1

1

tjn1

1

tjn dtejn21e

jn2t

−

= [ ]1

1

tjn222

tjnjn e)j(n2

1eenj

−

π−ππ−

−π++

π

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)

cn = π−

=ππ

+π−

n)1(j)nsin(

n2j2

n)1( n

22

nj

Thus

f(t) = ∑∞

−∞=

ω

n

tjnn

oec = ∑∞

−∞=

π

π−

n

tjnn enj)1(

Chapter 17, Solution 53. ωo = 2π/T = 2π

cn = ∫ ∫ ω+−ω−− =1

0

t)jn1(T

0

tjnt dtedtee oo

= [ ]1en2j1

1en2j1

1 )n2j1(1

0

t)n2j1( −π+

−=

π+− π+−π+−

= [1/(j2nπ)][1 – e–1(cos(2πn) – jsin(2nπ))]

= (1 – e–1)/(1 + j2nπ) = 0.6321/(1 + j2nπ

f(t) = ∑∞

−∞=

π

π+n

tn2j

n2j1e6321.0

Chapter 17, Solution 54. T = 4, ωo = 2π/T = π/2

cn = ∫ ω−T

0

ntj dte)t(fT

o1

=

−+ ∫∫∫ π−π−π− 4

2

2/tjn2

1

2/tjn1

0

2/tjn dte1dte1dte241

= [ ]π−π−π−π−π− +−−+−π

jnn2j2/jnjn2/jn eeee2e2n2j

= [ ]π−π− +−π

jn2/jn e23e3n2j

f(t) = ∑∞

−∞=

ω

n

tjnn

oec

Chapter 17, Solution 55. T = 2π, ωo = 2π/T = 1

cn = ∫ ω−T

0

tjn dte)t(iT

o1

But i(t) = π<<π

π<<2t,0

t0),tsin(

cn = ∫∫π −−π π− −

π=

π 0

jntjtjt

0

tjn dte)ee(j2

121dte)tsin(

21

+

−+

−−

−=

+

+−π

=

+π−−π

π+−−

n11e

n11e

41

)n1(je

)n1(je

j41

)1n(j)n1(j

0

)n1(jt)n1(jt

[ ]nne1enne1e)1n(4

1 )n1(j)n1(j)n1(j)n1(j2 +−−+−+−−π

= +π−+π−−π−π

But ejπ = cos(π) + jsin(π) = –1 = e–jπ

cn = [ ])n1(2

e12neneee)1n(4

12

jnjnjnjnjn

2 −π+

=−+−−−−π

π−π−π−π−π−

Thus

i(t) = ∑∞

−∞=

ππ−

−π+

n

tjn2

jn

e)n1(2

e1

Chapter 17, Solution 56. co = ao = 10, ωo = π co = (an – jbn)/2 = (1 – jn)/[2(n2 + 1)]

f(t) = ∑∞

≠−∞=

π

+−

+

0nn

tjn2 e

)1n(2)jn1(10

Chapter 17, Solution 57. ao = (6/–2) = –3 = co cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2)

f(t) = ∑∞

≠−∞= −

+

0nn

nt50j3 e

2n33−

Chapter 17, Solution 58. cn = (an – jbn)/2, (–1) = cos(nπ), ωo = 2π/T = 1

cn = [(cos(nπ) – 1)/(2πn2)] – j cos(nπ)/(2n)

Thus

f(t) = ∑+π

π

−π

−π jnt2 e

n2)ncos(j

n21)ncos(

4

Chapter 17, Solution 59. For f(t), T = 2π, ωo = 2π/T = 1. ao = DC component = (1xπ + 0)/2π = 0.5

For h(t), T = 2, ωo = 2π/T = π. ao = (3x1 – 2x1)/2 = 0.5

Thus by replacing ωo = 1 with ωo = π and multiplying the magnitude by five, we obtain

h(t) = ∑−1 ∞

≠−∞=

π+

π+0n

n

t)1n2(j

)1n2(e5j

2

Chapter 17, Solution 60. From Problem 16.17, ao = 0 = an, bn = [2/(nπ)][1 – 2 cos(nπ)], co = 0 cn = (an – jbn)/2 = [j/(nπ)][2 cos(nπ) – 1], n ≠ 0. Chapter 17, Solution 61.

(a) ωo = 1.

f(t) = ao + ∑ φ−ω )tncos(A non

= 6 + 4cos(t + 50°) + 2cos(2t + 35°)

+ cos(3t + 25°) + 0.5cos(4t + 20°) = 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°)

– 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°) + 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°)

= 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t)

– 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t)

(b) frms = ∑∞

=

+1n

2n

2o A

21a

frms

2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 frms = 6.828 Chapter 17, Solution 62.

(a) s3141.0202TT/220o =π

=→π==ω

(b) f ...)90t40cos(1.5)90t20cos(43)tncos(Aa)t( o

1n

onono +++++=φ+ω+= ∑

∞

=

....t80sin8.1t60sin7.2t40sin1.5t20sin43)t(f −−−−−= Chapter 17, Solution 63.

This is an even function.

T = 3, ωo = 2π/3, bn = 0.

f(t) = 5.1t1,2

1t0,1<<<<

ao =

+= ∫∫∫

5.1

1

1

0

2/T

0dt2dt1

32dt)t(f

T2 = (2/3)[1 + 1] = 4/3

an =

π+π=ω ∫∫∫

5.1

1

1

0

2/T

0 o dt)3/tn2cos(2dt)3/tn2cos(134dt)tncos()t(f

T4

=

π

π+

π

π

5.1

1

1

0 3tn2sin

n26

3tn2sin

n23

34

= [–2/(nπ)]sin(2nπ/3)

f2(t) = ∑∞

=

π

π

π−

1n 3tn2cos

3n3sin

n12

34

ao = 4/3 = 1.3333, ωo = 2π/3, an = –[2/(nπ)]sin(2nπt/3)

An =

π

π=+

3n2sin

n2b2

n2na

A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103

The amplitude spectra are shown below. 1.333

0.1103 0.1378

0

0.275

0.551

4321n

5 0

An


The amplitude and phase spectra are shown below.

An 3.183 2.122 1.591 0.4244 0 2π 4π 6π ω nφ 0 2π 4π 6π ω

-180o

Chapter 17, Solution 65. an = 20/(n2π2), bn = –3/(nπ), ωn = 2n

An = 22442n n

9n400b

π+

π=+2

na

= 22n44.441

n3

π+

π, n = 1, 3, 5, 7, 9, etc.

n An 1 2.24 3 0.39 5 0.208 7 0.143 9 0.109

φn = tan–1(bn/an) = tan–1[–3/(nπ)][n2π2/20] = tan–1(–nx0.4712)

n φn 1 –25.23° 3 –54.73° 5 –67° 7 –73.14° 9 –76.74° ∞ –90°

0 10

φn

ωn

14 62 18

–25.23°

–54.73°–67°

–76.74° –73.14°

–60°

–30°

–90°

2.24

An

ωn

0.0.143 0.109 0.208

0.39

18 14 10 6 2 0

Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 5.099510E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 3.184E+00 1.000E+00 1.782E+00 0.000E+00 2 1.000E+00 1.593E+00 5.002E-01 3.564E+00 1.782E+00 3 1.500E+00 1.063E+00 3.338E-01 5.347E+00 3.564E+00 4 2.000E+00 7.978E-01 2.506E-01 7.129E+00 5.347E+00 5 2.500E+00 6.392E-01 2.008E-01 8.911E+00 7.129E+00 6 3.000E+00 5.336E-01 1.676E-01 1.069E+01 8.911E+00 7 3.500E+00 4.583E-01 1.440E-01 1.248E+01 1.069E+01 8 4.000E+00 4.020E-01 1.263E-01 1.426E+01 1.248E+01 9 4.500E+00 3.583E-01 1.126E-01 1.604E+01 1.426E+01 TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT

hapter 17, Solution 67.

he Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. After simulation, the output file includes the following Fourier components,

C T

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

PHASE (DEG)

.000E+00 -8.996E+01 0.000E+00

NO (HZ) COMPONENT COMPONENT (DEG)

1 1.667E-01 2.432E+00 12 3.334E-01 6.576E-04 2.705E-04 -8.932E+01 6.467E-01 3 5.001E-01 5.403E-01 2.222E-01 9.011E+01 1.801E+02 4 6.668E-01 3.343E-04 1.375E-04 9.134E+01 1.813E+02 5 8.335E-01 9.716E-02 3.996E-02 -8.982E+01 1.433E-01 6 1.000E+00 7.481E-06 3.076E-06 -9.000E+01 -3.581E-02 7 1.167E+00 4.968E-02 2.043E-02 -8.975E+01 2.173E-01 8 1.334E+00 1.613E-04 6.634E-05 -8.722E+01 2.748E+00 9 1.500E+00 6.002E-02 2.468E-02 9.032E+01 1.803E+02

TAL HARMON DISTORT N = 2.280 E+01 PERC T TO IC IO 065 EN

Chapter 17, Solution 68. The schematic is shown below. We set the final time = 6T=12s and the center frequency = 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the output file as shown below.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 1.990000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.273E+00 1.000E+00 9.000E-01 0.000E+00 2 1.000E+00 6.367E-01 5.001E-01 -1.782E+02 1.791E+02 3 1.500E+00 4.246E-01 3.334E-01 2.700E+00 1.800E+00 4 2.000E+00 3.185E-01 2.502E-01 -1.764E+02 -1.773E+02 5 2.500E+00 2.549E-01 2.002E-01 4.500E+00 3.600E+00 6 3.000E+00 2.125E-01 1.669E-01 -1.746E+0 -1.755E+02 7 3.500E+00 1.823E-01 1.431E-01 6.300E+00 5.400E+00 8 4.000E+00 1.596E-01 1.253E-01 -1.728E+02 -1.737E+02 9 4.500E+00 1.419E-01 1.115E-01 8.100E+00 7.200E+00 Chapter 17, Solution 69.

The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 2 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 3 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 4 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 5 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 6 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 7 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 8 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 9 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00 TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT

Chapter 17, Solution 70. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00 2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 -4.928E+01 3 1.500E+00 2.111E-01 1.973E-01 -3.985E+01 -4.990E+01 4 2.000E+00 1.247E-01 1.166E-01 -5.870E+01 -6.874E+01 5 2.500E+00 8.538E-02 7.980E-02 -5.680E+01 -6.685E+01 6 3.000E+00 6.139E-02 5.738E-02 -6.563E+01 -7.567E+01 7 3.500E+00 4.743E-02 4.433E-02 -6.520E+01 -7.524E+01 8 4.000E+00 3.711E-02 3.469E-02 -7.222E+01 -8.226E+01 9 4.500E+00 2.997E-02 2.802E-02 -7.088E+01 -8.092E+01 TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT

Chapter 17, Solution 71. The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box. After simulation, the output waveform is as shown. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1) DC COMPONENT = 8.374999E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00 2 1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01 3 1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01 4 2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01 5 2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01 6 3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01 7 3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01 8 4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01 9 4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01 TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT Chapter 17, Solution 72. T = 5, ωo = 2π/T = 2π/5 f(t) is an odd function. ao = 0 = an

bn = ∫ ∫ π=ω2/T

0

10

0o dt)tn4.0sin(1054dt)tnsin()t(f

T4

= 1

0

)nt4.0cos(n2

5x8π

π− = )]n4.0cos(1[

n20

π−π

f(t) = ∑∞

=

ππ−π 1n

)tn4.0sin()]n4.0cos(1[n120


p = ∑+R

V21

RV 2

n2DC

= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW


(a) An = 2n

2n ba + , φ = tan–1(bn/an)

A1 = 22 86 + = 10, φ1 = tan–1(6/8) = 36.87° A2 = 22 43 + = 5, φ2 = tan–1(3/4) = 36.87°

i(t) = 4 + 10cos(100πt – 36.87°) – 5cos(200πt – 36.87°) A

(b) p = ∑+ RI5.0RI 2n

2DC

= 2[42 +0.5(102 + 52)] = 157 W Chapter 17, Solution 75. The lowpass filter is shown below. R + + vs C vo - -

∑∞

=ω

πτ+

τ=

1nos tncos

Tnsin

n1

TA2

TAv

T/n2n,VRCj1

1V

Cj1R

Cj1

V onsn

s

n

no π=ω=ω

ω+=

ω+

ω=

For n=0, (dc component), TAVsoV τ

== (1)

For the nth harmonic,

o

n122

n2o 90

Tnsin

nTA2

RCtanCR1

1V −∠πτ

•ω∠ω+

=−

When n=1, 22

2o

CRT

41

1T

nsinTA2|V

π+

•πτ

=| (2)

From (1) and (2),

4222

222

10x09.39.30CRT

41

CRT

41

110

sinTA2x50

TA

=τ

=π

+→π

+

π=

τ

mF 59.2410x4

10x09.3x1010R2

TC10CRT

413

4251022

2=

π=

π=→=

π+

−


vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence

vo(t) = ∑∞

=

ππ

+1k

)tnsin(n1205 , n = 2k – 1

T = 2, ωo = 2π/T = 2π, ωn = nωo = 2nπ

jωnL = j2nπ; Z = R||10 = 10R/(10 + R) Vo = ZVs/(Z + j2nπ) = [10R/(10R + j2nπ(10 + R))]Vs

Vo = s2222

1

V)R10(n4R100

)R10)(R5/n(tanR+π+

+π−∠ −10

Vs = [20/(nπ)]∠0°

The source current Is is

Is = )R10(n2jR10

n20)R10(

n2jR10

R10V

n2jZV ss

+π+π

+=

π++

=π+

= 2222

1

)R10(n4R100

)R10)(3/n(tann20)R10(

+π+

+π−∠π

+ −

ps = VDCIDC + ∑ φ−θ )cos(IV21

nnsnsn

For the DC case, L acts like a short-circuit.

Is = R10

)R10(5

R10R

510

+=

+

, Vs = 5 = Vo

++π+

+

π+

π

+

+π+

+π

+

π

++

=

−

−

222

122

222

12

s

)R10(16R100

)R10(5

2tancos)R10(10

)R10(4R100

)R10(5

tancos)R10(20

21

R10)R10(25p

ps = ∑∞

=

+1n

onDC

RV

21

RV

=

+

+π++

+π+

+ 222222 )R10(10R100R100

)R10(4R100R100

21

R25

We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals.

R10

)R10(25x107

R25 +

=

100 = 70 + 7R which leads to R = 30/7 = 4.286 Ω


(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence ωo = 2.

T = 2π/ωo = 2π/2 = π

(b) The average value is the DC component = –2

(c) Vrms = ∑∞

=

++1n

2n

2no )ba(

21a

)136810(21)2( 2222222

rms +++++−=V = 121.5

Vrms = 11.02 V


(a) p = ∑ ∑+=+R

VR

VR

V21

RV 2

rms,n2DC

2n

2DC

= 0 + (402/5) + (202/5) + (102/5) = 420 W

(b) 5% increase = (5/100)420 = 21

pDC = 21 W = 105R21VtoleadswhichR

V 2DC

2DC ==

VDC = 10.25 V

Chapter 17, Solution 79. From Table 17.3, it is evident that an = 0,

bn = 4A/[π(2n – 1)], A = 10.

A Fortran program to calculate bn is shown below. The result is also shown.

C FOR PROBLEM 17.79 DIMENSION B(20) A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) – 1.) PRINT *, N, B(N) 10 CONTINUE STOP END

n bn 1 12.731 2 4.243 3 2.546 4 1.8187 5 1.414 6 1.1573 7 0.9793 8 0.8487 9 0.7498 10 0.67

Chapter 17, Solution 80. From Problem 17.55, cn = [1 + e–jnπ]/[2π(1 – n2)]

This is calculated using the Fortran program shown below. The results are also shown.

C FOR PROBLEM 17.80 COMPLEX X, C(0:20) PIE = 3.1415927

A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N))) PRINT *, N, C(N)

10 CONTINUE STOP END

n cn 0 0.3188 + j0 1 0 2 –0.1061 + j0 3 0 4 –0.2121x10–1 + j0 5 0 6 –0.9095x10–2 + j0 7 0 8 –0.5052x10–2 + j0 9 0 10 –0.3215x10–2 + j0

Chapter 17, Solution 81. (a)

A

3T 2TT0

f(t) = ∑∞

=

ω−ππ 1n

o2 )tncos(1n4

1A4A−

2

The total average power is pavg = Frms

2R = Frms2 since R = 1 ohm.

Pavg = Frms2 = ∫

T

0

2 dt)t(fT1 = 0.5A2

(b) From the Fourier series above |co| = 2A/2, |cn| = 4A/[π(4n2 – 1)]

n ωo |cn| 2|cn|2 % power 0 0 2A/π 4A2/(π2) 81.1% 1 2ωo 2A/(3π) 8A2/(9π2) 18.01% 2 4ωo 2A/(15π) 2A2/(225π2) 0.72% 3 6ωo 2A/(35π) 8A2/(1225π2) 0.13% 4 8ωo 2A/(63π) 8A2/(3969π2) 0.04%

(c) 81.1% (d) 0.72% Chapter 17, Solution 82.

P = ∑∞

=1n

2n

2DC

RV

21

R+

V

Assuming V is an amplitude-phase form of Fourier series. But

|An| = 2|Cn|, co = ao |An|2 = 4|Cn|2

Hence,

P = ∑∞

=1n

2n

2o

Rc

R+ 2

c

Alternatively,

P = R

V

2rms

where

∑∑∑∞

−∞=

∞

=

∞

=

=+=+=n

2n

1n

2n

2o

1n

2n

2o

2rms cc2cA

21aV

= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)

= 100 + 2x94.57 = 289.14

P = 289.14/4 = 72.3 W

Chapter 17, Solution 1. 2 periodic ω πω · Hence f(t) = sin(n t) n 10 1 n 1 π π + ∑ ∞ = 5...

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Transcript of Chapter 17, Solution 1. 2 periodic ω πω · Hence f(t) = sin(n t) n 10 1 n 1 π π + ∑ ∞ = 5...