Chapter 17, Solution 1. 2 periodic ω...

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

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

Chapter 17, Solution 1. 2 periodic ω...

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