solucionario parte 4 matemáticas avanzadas para ingeniería - 2da edición - glyn james

7/30/2019 Solucionario parte 4 Matemticas Avanzadas para Ingeniera - 2da Edicin - Glyn James

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4

Fourier series

Exercises 4.2.9

1(a)

a0 =1

0dt +

0

tdt

=1

(t)0 +

t22

0

=

1

2 +

2

2

=

2

an =1

0 cos ntdt +

0

t cos ntdt

=1

nsin nt

0

+

t

nsin nt +

1

n2cos nt

0

= 1n2

(cos n 1) = 2

n2, n odd

0, n even

bn =1

0 sin ntdt +

0

t sin ntdt

=1

n

cos nt0

+ t

ncos nt +

1

n2sin nt

0

=1

n(1 2cos n) =

3

n, n odd

1

n

, n even

Thus the Fourier expansion of f(t) is

f(t) = 4

+n odd

2n2

cos nt +

n odd

3

nsin nt

n even

1

nsin nt

i.e. f(t) = 4 2

n=1

cos(2n 1)t(2 1)2 + 3

n=1

sin(2n 1)t(2n 1)

n=1

sin2nt

2n

cPearson Education Limited 2004


2/76

192 Glyn James: Advanced Modern Engineering Mathematics, Third edition

1(b)

a0 =1

0

(t + )dt =1

t2

2+ t

0

=

2

an =1

0

(t + )cos ntdt =1

(t + )

sin nt

n+

cos nt

n2

0

=1

n2(1 cos n) =

0, n even

2

n2, n odd

bn = 1

0

(t + )sin ntdt = 1

(t + ) cos ntn

+ sin ntn2

0

= 1n


f(t) =

4+

n odd

2

n2cos nt

n=1

1

nsin nt

i.e. f(t) =

4+

2

n=1

cos(2n 1)t(2n

1)2

n=1

sin nt

n

1(c) From its graph we see that f(t) is an odd function so it has Fourier

expansion

f(t) =n=1

bn sin nt

with

bn = 2

0

f(t)sin nt = 2

0

1 t

sin ntdt

=2

1

n

1 t

cos nt 1

n2sin nt

0

=2

n


f(t) =2

n=1

sin nt

n



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Glyn James: Advanced Modern Engineering Mathematics, Third edition 193

1(d) From its graph f(t) is seen to be an even function so its Fourier

expansion is

f(t) =a02

+n=1

an cos nt

with

a0 =2

0

f(t)dt =2

/20

2cos tdt =2

[2sin t]

/20 =

4

an =2

0

f(t)cos ntdt =2

/20

2cos t cos ntdt

=2

/20

[cos(n + 1)t + cos(n 1)t]dt

=2

sin(n + 1)t

(n + 1)+

sin(n 1)t(n 1)

/20

=2

1

(n + 1)sin(n + 1)

2+

1

(n 1) sin(n 1)

2

=

0, n odd

4

1

(n2 1) , n = 4, 8, 12, . . .4

1

(n2 1), n = 2, 6, 10, . . .


f(t) =2

+

4

n=1

(1)n+1 cos2nt4n2 1

1(e)

a0 =

1

cos

t

2 dt =

1

2sin

t

2 =

4

an =1

cost

2cos ntdt =

1

2

cos(n +

1

2)t + cos(n 1

2)t

dt

=2

2

2

(2n + 1)sin(n +

1

2) +

2

(2n 1) sin(n1

2)

=

4

(4n2 1) , n = 1, 3, 5, . . .

4(4n2 1) , n = 2, 4, 6, . . .

bn = 0



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f(t) =2

+

4

n=1

(1)n+1 cos nt(4n2 1)

1(f) Since f(t) is an even function it has Fourier expansion

f(t) =a02

+n=1

an cos nt

with

a0 =2

0

| t | dt = 2

0

tdt =

an =2

0

t cos ntdt =2

t

nsin nt +

1

n2cos nt

0

=2

n2(cos n 1) =

0, n even

4n2

, n odd


f(t) = 2 4

n odd

1n2

cos nt

i.e. f(t) =

2 4

n=1

cos(2n 1)t(2n 1)2

1(g)

a0 =1

0

(2t

)dt =1

t2 t

0

= 0

an =1

0

(2t )cos ntdt = 1

(2t )

nsin nt +

2

n2cos nt

0

=2

n2(cos n 1) =

4

n2, n odd

0, n even

bn =1

0

(2t )sin ntdt = 1

(2t )

ncos nt +

2

n2sin nt

0

=

1

n

(cos n + 1) = 0, n odd

2

n , n even



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f(t) =n odd

4n2

cos nt +n even

2n

sin nt

i.e. f(t) = 4

n=1

cos(2n 1)t(2n 1)2

n=1

sin2nt

n

1(h)

a0 =1

0

(t + et)dt +0

(t + et)dt

=1

t22

+ et0

+ t2

2+ et

0

=1

2 + (e e) = + 2

sinh

an =

1

0(t + e

t

)cos ntdt +0 (t + e

t

)cos ntdt

=1

t

nsin nt +

1

n2cos nt

0

+1

(n2 + 1)

net sin nt + et cos nt

0

+

t

nsin nt +

1

n2cos nt

0

+1

(n2 + 1)

net sin nt + et cos nt

0

=2

n2(1 + cos n) + 2cos n

(n2 + 1)

e e2

=

2

(cos 1)

n2+

cos n

(n2 + 1)sinh , cos n = (1)

n

bn =1

0

(t + et)sin ntdt +0

(t + et)sin ntdt

=1

tn

cos nt 1n2

sin nt0

+ t

ncos nt +

1

n2sin nt

0

+n2

2 + 1

et cos nt

n+

et sin nt

n2

= n

(n2 + 1) cos n(e

e

) = 2n

(n2 + 1) cos n sinh , cos n = (1)n



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f(t) =

2+

1

sinh

+

2

n=1

(1)n 1

n2+

(1)n sinh n2 + 1

cos nt

2

n=1

n(1)nn2 + 1

sinh sin nt

2 Since the periodic function f(t) is an even function its Fourier expansion is

f(t) =a02

+n=1

an cos nt

with

a0 =2

0

( t)2dt = 2

1

3( t)3

0

=2

32

an =2

0

( t)2 cos ntdt = 2

( t)2

nsin nt 2( t)

n2cos nt 2

n3sin nt

0

=4

n2


f(t) =2

3+ 4

n=1

1

n2cos nt

Taking t = gives

0 =2

3+ 4

n=1

1

n2(1)n

so that

1

122 =

n=1(1)n+1

n2



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3 Since q(t) is an even function its Fourier expansion is

q(t) =a02

+

n=1

an cos nt

with

a0 =2

0

Qt

dt = Q

an =2

0

Qt

cos ntdt =

2Q

2

t

nsin nt +

1

n2cos nt

0

= 2Q2n2

(cos n 1) = 0, n even 4Q2n2

, n odd

Thus the Fourier expansion of q(t) is

q(t) = Q

1

2 4

2

n=1

cos(2n 1)t(2n 1)2

4

a0 =

1

0 5sin tdt =

1

[5cos t]0 =

10

an =5

0

sin t cos ntdt =5

2

0

[sin(n + 1)t sin(n 1)t]dt

=5

2

cos(n + 1)t

(n + 1)+

cos(n 1)t(n 1)

0

, n = 1

=5

2

cos nn + 1

cos n(n 1)

1n + 1

+1

n 1

=

5

(n2

1)(cos n + 1) =

0, n odd, n = 1

10

(n2 1), n even

Note that in this case we need to evaluate a1 separately as

a1 =1

0

5sin t cos tdt =5

2

0

sin2tdt = 0

bn =5

0

sin t sin ntdt = 52

0

[cos(n + 1)t cos(n 1)t]dt

= 52

sin(n + 1)t

(n + 1) sin(n 1)t

(n 1)0

, n = 1

= 0 , n = 1c

Pearson Education Limited 2004


8/76


Evaluating b1 separately

b1 =5

0

sin t sin tdt =5

2

0

(1 cos2t)dt

=5

2

t 1

2sin2t

0

=5

2


f(t) =5

+

5

2sin t 10

n=1

cos2nt

4n2 1

5

a0 =1

0

2dt +

0

(t )2dt=

1

2t0 +

13

(t )30

=

4

32

an =1

0

2 cos ntdt +

0

(t )2 cos ntdt

=1

2n

sin nt0

+ (t )2

nsin nt +

2(t )n2

cos nt 2n3

sin nt0

=2

n2

bn =1

0

2 sin ntdt +

0

(t )2 sin ntdt

=1

2n

cos nt0

+ (t )2

ncos nt + 2

(t )n2

sin nt +2

n3cos nt

0

=1

2n

+2

n(1)n

=

n(1)n 2

n3[1 (1)n]



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f(t) =2

32 +

n=1

2

n2cos nt +

(1)nn

sin nt

4

n=1

sin(2n 1)t(2n 1)3

5(a) Taking t = 0 gives

2 + 2

2=

2

32 +

n=1

2

n2

and hence the required resultn=1

1

n2=

1

62

5(b) Taking t = gives

2 + 0

2

=2

3

2 +

n=1

2

n2

(

1)n

and hence the required result

n=1

(1)n+1n2

=1

122

6(a)



10/76


6(b)

The Fourier expansion of the even function (a) is given by

f(t) =a02

+n=1

an cos nt

with

a0 =2

/20

tdt +

/2

( t)dt

= 2

12 t2/20

+12 ( t)2

/2= 2

an =2

/20

t cos ntdt +

/2

( t)cos ntdt

=2

tn

sin nt +1

n2cos nt

/20

+ t

nsin nt 1

n2cos nt

/2

=2

2

n2cos

n

2 1

n2(1 + (1)n)

=

0, n odd

8n2

, n = 2, 6, 10, . . .

0, n = 4, 8, 12, . . .


f(t) =

4 2

n=1

cos(4n 2)t(2n 1)2

Taking t = 0 where f(t) = 0 gives the required result.



11/76


7

a0 =1

0

(2 t

)dt +

2

t/dt

=1

2t t

2

2

0

+ t2

2

2

= 3

an =1

0

(2 t

)cos ntdt +

2

t

cos ntdt

=1

2n

sin nt tn

sin nt 1n2

cos nt0

+ t

nsin nt +

1

n2cos nt

2

=2

2n2 [1 (1)n

]

=

0, n even

4

2n2, n odd

bn =1

0

(2 t

)sin ntdt +

2

t

sin ntdt

=1

2n

cos nt +t

ncos nt 1

n2sin nt

0

+ t

ncos nt +

1

n2sin nt

2

= 0


f(t) =3

2+

4

2

n=1

cos(2n 1)t(2n 1)2

Replacing t by t 12 gives

f(t

1

2

) =3

2

+4

2

n=1

cos(2n 1)(t )(2n 1)2



12/76


Since

cos(2n 1)(t 12

) = cos(2n 1)t cos(2n 1) 2

+ sin(2n 1)t sin(2n 1) 2

= (1)n+1 sin(2n 1)t

f(t 12

) 32

=4

2

n=1

(1)n+1 sin(2n 1)t(2n 1)2

The corresponding odd function is readily recognised from the graph of f(t).

Exercises 4.2.11

8 Since f(t) is an odd function the Fourier expansion is

f(t) =n=1

bn sinnt

with

bn =2

0

t sinnt

dt =2

t

n

cosnt

+ n2

sinnt

0

= 2n

cos n


f(t) =2

n=1

(1)n+1n

sinnt

9 Since f(t) is an odd function (readily seen from a sketch of its graph) its

Fourier expansion is

f(t) =

n=1

bn sinnt

with

bn =2

0

K

( t)sin nt

tdt

=2

K

ncos

nt

+

Kt

ncos

nt

K

(n)2sin

nt

0

=

2K

n



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f(t) =2K

n=1

1

nsin

nt

10

a0 =1

5

50

3dt = 3

an =1

5 5

0

3cosnt

5dt =

1

515

nsin

nt

5 5

0

= 0

bn =1

5

50

3sinnt

5dt =

1

5

15

ncos

nt

5

50

=3

n[1 (1)n] =

6

n, n odd

0, n even


f(t) =3

2+

6

n=11

(2n

1)

sin(2n 1)

5t

11

a0 =2

2

/0

A sin tdt =

A

cos t

/0

=2A

an =A

/0

sin t cos ntdt =A

2

/0

[sin(n + 1)t sin(n 1)t]dt

=A

2

cos(n + 1)t

(n + 1)+

cos(n 1)t(n 1)

/

0

, n= 1

=A

2

2(1)n+1

n2 1 2

n2 1

=A

(n2 1) [(1)n+1 1]

=

0, n odd , n = 1 2A

(n2 1) , n even

Evaluating a1 separately

a1 =

A

/0 sin t cos tdt =

A

2/0 sin2tdt = 0



14/76


bn = A

/

0

sin t sin ntdt = A2

/

0

[cos(n + 1)t cos(n 1)t]dt

= A2

sin(n + 1)t

(n + 1) sin(n 1)t

(n 1)/0

, n = 1

= 0, n = 1

b1 =A

/0

sin2 tdt =A

2

/0

(1 cos2t)dt = A2


f(t) =A

1 +

2sin t 2

n=1

cos2nt

4n2 1

12 Since f(t) is an even function its Fourier expansion is

f(t) =a02

+

n=1

an cosnt

T

with

a0 =2

T

T0

t2dt =2

T

1

3t3T0

=2

3T2

an =2

TT

0

t2 cosnt

Tdt =

2

TT t2

nsin

nt

T+

2tT2

(n)2cos

nt

T 2T3

(n)3sin

nt

TT

0

=4T2

(n)2(1)n

Thus the Fourier series expansion of f(t) is

f(t) =T2

3+

4T2

2

n=1

(1)nn2

cosnt

T



15/76


13

a0 = 2T

T0

ET

tdt = 2ET2

12

t2T0

= E

an =2

T

T0

E

Tt cos

2nt

Tdt

=2E

T2

tT

2nsin

2nt

T+ T

2n

2cos

2nt

T

T0

= 0

bn =2E

T2

T0

t sin2nt

Tdt

= 2ET2 tT2n cos 2ntT + T2n

2

sin 2ntTT0

= EnThus the Fourier expansion of e(t) is

e(t) =E

2 E

n=1

1

nsin

2nt

T

Exercises 4.3.3

14 Half range Fourier sine series expansion is given by

f(t) =n=1

bn sin nt

with

bn =2

0

1sin ntdt =2

1

ncos nt

0

=

2

n

[(

1)n

1]

=

0, n even

4

n, n odd

Thus the half range Fourier sine series expansion of f(t) is

f(t) =4

n=1

sin(2n 1)t(2n 1)

Plotting the graphs should cause no problems.



16/76


15 Half range Fourier cosine series expansion is given by

f(t) =a02

+

n=1

an cos nt

with

a0 =2

1

10

(2t 1)dt = 0

an = 210

(2t 1)cos ntdt

= 2

(2t 1)

nsin nt +

2

(n)2cos nt

10

=4

(n)2[(1)n 1]

=

0, n even

8(n)2

, n odd

Thus the half range Fourier cosine series expansion of f(t) is

f(t) = 82

n=1

1

(2n 1)2 cos(2n 1)t

Again plotting the graph should cause no problems.

16(a)

a0 = 2

10

(1 t2)dt = 2t 13

t310

=4

3

an = 2

10

(1 t2)cos2ntdt

= 2

(1 t2)

2nsin2nt 2t

(2n)2cos2nt +

2

(2n)3sin2nt

10

= 1

(n)2



17/76


bn = 210 (1 t

2

)sin2ntdt

= 2

(1 t

2)

2ncos2nt 2t

(2n)2sin2nt 2

(2n)3cos2nt

10

=1

n

Thus the full-range Fourier series expansion for f(t) is

f(t) = f1(t) =2

3 1

2

n=1

1

n2cos2nt +

1

n=1

1

nsin2nt

16(b) Half range sine series expansion is

f2(t) =n=1

bn sin nt

with

bn = 210 (1 t

2

)sin ntdt

= 2

(1 t

2)

ncos nt 2t

(n)2sin nt 2

(n)3cos nt

10

= 2

2

(n)3(1)n + 1

n+

2

(n)3

=

2

n, n even

2

1

n+

4

(n)3 , n odd

Thus half range sine series expansion is

f2(t) =1

n=1

1

nsin2nt +

2

n=1

1

(2n 1) +4

2(2n 1)3

sin(2n 1)t

16(c) Half range cosine series expansion is

f3(t) =a0

2

+

n=1

an cos nt



18/76


with

a0 = 2

1

0

(1 t2)dt = 43

an = 2

10

(1 t2)cos ntdt

= 2

(1 t2)

nsin nt 2t

(n)2cos nt +

2

(n)3sin nt

10

=4(1)n

(n)2

Thus half range cosine series expansion is

f3(t) =2

3+

4

2

n=1

(1)n+1n2

cos nt

Graphs of the functions f1(t), f2(t), f3(t) for 4 < t < 4 are as follows

17 Fourier cosine series expansion is

f1(t) =a0

2

+

n=1

an cos nt



19/76


witha0 =

2

0

(t t2)dt = 13

2

an =2

0

(t t2)cos ntdt

=2

(t t2)

nsin nt +

( 2t)n2

cos nt +2

n3sin nt

0

= 2n2

[1 + (1)n]

= 0, n odd 4

n2, n even

Thus the Fourier cosine series expansion is

f1(t) =1

62

n=1

1

n2cos2nt

Fourier sine series expansion is

f2(t) =

n=1

bn sin nt

with

bn =2

0

(t t2)sin ntdt

=2

(t t

2)

ncos nt +

( 2t)n2

sin nt 2n3

cos nt

0

= 4n3 [1 (1)n]

=

0, n even

8

n3, n odd

Thus the Fourier sine series expansion is

f2(t) =8

n=1

1

(2n 1)3 sin(2n 1)t



20/76


Graphs of the functions f1(t) and f2(t) for 2 < t < 2 are:

18

f(x) =2a

x , 0 < x

solucionario parte 4 matemáticas avanzadas para ingeniería - 2da edición - glyn james

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