Solucionario parte 4 Matemáticas Avanzadas para Ingeniería - 2da Edición - Glyn James
Short Description
Download Solucionario parte 4 Matemáticas Avanzadas para Ingeniería - 2da Edición - Glyn James...
Description
4 Fourier series Exercises 4.2.9 1(a) 1 a0 = π
0
−π
−πdt +
π
tdt 0 2 π
t π 1 π2 1 0 2 (−πt)−π + −π + =− = = π 2 0 π 2 2 0 π 1 −π cos ntdt + t cos ntdt an = π −π 0 0 t π 1 1 π − sin nt sin nt + 2 cos nt + = π n n n −π 0
2 1 − 2 , n odd = (cos nπ − 1) = πn πn2 0, n even 0 π 1 −π sin ntdt + t sin ntdt bn = π −π 0 t π 0 1 π 1 + − cos nt + 2 sin nt = cos nt π n n n −π 0 3 , n odd 1 = (1 − 2 cos nπ) = n 1 n − , n even n Thus the Fourier expansion of f (t) is 2 1 3 π + sin nt − sin nt − 2 cos nt + 4 πn n n n even n odd n odd ∞ ∞ ∞ sin(2n − 1)t π 2 cos(2n − 1)t sin 2nt i.e. f (t) = − − − +3 2 4 π n=1 (2π − 1) (2n − 1) 2n n=1 n=1 f (t) = −
c Pearson Education Limited 2004
192
Glyn James: Advanced Modern Engineering Mathematics, Third edition
1(b) 0 1 t2 π + πt (t + π)dt = = π 2 2 −π −π 0 0 sin nt cos nt 1 1 (t + π) an = (t + π) cos ntdt = + π −π π n n2 −π
0, n even 1 2 = (1 − cos nπ) = , n odd πn2 πn2 0 cos nt sin nt 1 0 1 1 −(t + π) (t + π) sin ntdt = = − bn = + π −π π n n2 −π n 1 a0 = π
0
Thus the Fourier expansion of f (t) is ∞ 2 1 π sin nt cos nt − f (t) = + 2 4 πn n n=1 n odd ∞ ∞ π 2 cos(2n − 1)t sin nt i.e. f (t) = + − 4 π n=1 (2n − 1)2 n n=1
1(c) From its graph we see that f (t) is an odd function so it has Fourier expansion ∞ bn sin nt f (t) = n=1
with
2 π 2 π t sin ntdt f (t) sin nt = 1− bn = π 0 π 0 π π 1 t 1 2 2 − 1− cos nt − = sin nt = 2 π n π πn nπ 0
Thus the Fourier expansion of f (t) is ∞ 2 sin nt f (t) = π n=1 n
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 1(d) From its graph f (t) is seen to be an even function so its Fourier expansion is ∞ a0 + f (t) = an cos nt 2 n=1 with 2 a0 = π 2 an = π 2 = π
π
0
0
0
π
2 f (t)dt = π
π/2
2 cos tdt = 0
2 f (t) cos ntdt = π
π/2
2 4 π/2 [2 sin t]0 = π π
π/2
2 cos t cos ntdt 0
[cos(n + 1)t + cos(n − 1)t]dt
π/2 2 sin(n + 1)t sin(n − 1)t + = π (n + 1) (n − 1) 0 π 1 π 1 2 sin(n + 1) + sin(n − 1) = π (n + 1) 2 (n − 1) 2 0, n odd 4 1 − , n = 4, 8, 12, . . . = π (n2 − 1) 1 4 , n = 2, 6, 10, . . . 2 π (n − 1) Thus the Fourier expansion of f (t) is ∞ 2 4 (−1)n+1 cos 2nt f (t) = + π π n=1 4n2 − 1
1(e) π 1 t t 4 2 sin cos dt = = 2 π 2 −π π −π π π 1 1 1 1 t an = cos(n + )t + cos(n − )t dt cos cos ntdt = π −π 2 2π −π 2 2 2 1 2 1 2 sin(n + )π + sin(n − )π = 2π (2n + 1) 2 (2n − 1) 2 4 , n = 1, 3, 5, . . . π(4n2 − 1) = 4 − , n = 2, 4, 6, . . . π(4n2 − 1) bn = 0 1 a0 = π
π
c Pearson Education Limited 2004
193
194
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Thus the Fourier expansion of f (t) is ∞ 4 (−1)n+1 cos nt 2 f (t) = + π π n=1 (4n2 − 1)
1(f )
Since f (t) is an even function it has Fourier expansion ∞
a0 an cos nt f (t) = + 2 n=1 2 π 2 π | t | dt = tdt = π a0 = π 0 π 0 π 1 2 π 2 t sin nt + 2 cos nt t cos ntdt = an = π 0 π n n 0
0, n even 2 4 = (cos nπ − 1) = − 2 , n odd πn2 πn Thus the Fourier expansion of f (t) is
with
4 1 π − cos nt 2 π n2 n odd ∞ π 4 cos(2n − 1)t i.e. f (t) = − 2 π n=1 (2n − 1)2 f (t) =
1(g) π 1 π 1 2 t − πt 0 = 0 (2t − π)dt = a0 = π 0 π π π 2 1 1 (2t − π) sin nt + 2 cos nt an = (2t − π) cos ntdt = π 0 π n n 0
4 2 = (cos nπ − 1) = − πn2 , n odd πn2 0, n even π π (2t − π) 2 1 1 − cos nt + 2 sin nt bn = (2t − π) sin ntdt = π 0 π n n 0
0, n odd 1 2 = − (cos nπ + 1) = − , n even n n c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition
195
Thus the Fourier expansion of f (t) is
f (t) =
−
2 4 cos nt + − sin nt 2 πn n n even
odd ∞ ∞ 4 cos(2n − 1)t sin 2nt i.e. f (t) = − − π n=1 (2n − 1)2 n n=1 n
1(h) 1 a0 = π = = an = =
= =
1 bn = π
0
(−t + e )dt + −π
π
t
t
(t + e )dt 0
t2 0 π 1 t2 t t − +e +e + π 2 2 −π 0 2 1 2 π + (eπ − e−π ) = π + sinh π π π 0 π 1 t t (−t + e ) cos ntdt + (t + e ) cos ntdt π −π 0 t t 0 0 1 1 1 − sin nt + 2 cos nt ne sin nt + et cos nt −π + 2 π n n (n + 1) −π π t π 1 t 1 t sin nt + 2 cos nt + 2 ne sin nt + e cos nt 0 + n n (n + 1) 0 2 cos nπ eπ − e−π 2 (−1 + cos nπ) + πn2 π(n2 + 1) 2 2 (cos π − 1) cos nπ sinh π , cos nπ = (−1)n + 2 2 π n (n + 1)
0
t
(−t + e ) sin ntdt + −π
π
t
(t + e ) sin ntdt 0
0 t π 1 1 1 t cos nt − 2 sin nt + − cos nt + 2 sin nt = π n n n n −π 0 t π 2 t e cos nt e sin nt n − + + 2 π +1 n n2 −π n 2n cos nπ(eπ − e−π ) = − cos nπ sinh π, cos nπ = (−1)n =− 2 π(n + 1) π(n2 + 1) c Pearson Education Limited 2004
196
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Thus the Fourier expansion of f (t) is ∞ 2 π 1 (−1)n − 1 (−1)n sinh π + sinh π + cos nt f (t) = + 2 π π n=1 n2 n2 + 1 −
2
∞ 2 n(−1)n sinh π sin nt π n=1 n2 + 1
Since the periodic function f (t) is an even function its Fourier expansion is ∞
a0 + f (t) = an cos nt 2 n=1 with π 1 2 2 3 − (π − t) (π − t) dt = = π2 π 3 3 0 0 π π 2(π − t) 2 2 (π − t)2 2 2 sin nt − an = (π − t) cos ntdt = cos nt − 3 sin nt π 0 π n n2 n 0 4 = 2 n 2 a0 = π
π
2
Thus the Fourier expansion of f (t) is ∞ π2 1 f (t) = cos nt +4 3 n2 n=1
Taking t = π gives 0=
∞ 1 π2 +4 (−1)n 2 3 n n=1
so that ∞ 1 2 (−1)n+1 π = 12 n2 n=1
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 3
Since q(t) is an even function its Fourier expansion is ∞
q(t) = with
a0 + an cos nt 2 n=1
2 π a0 = π 0 2 π an = π 0
Qt dt = Q π π Qt 2Q t 1 cos ntdt = 2 sin nt + 2 cos nt π π n n 0
0, n even 2Q 4Q = 2 2 (cos nπ − 1) = − 2 2 , n odd π n π n Thus the Fourier expansion of q(t) is ∞ 4 cos(2n − 1)t 1 q(t) = Q − 2 2 π n=1 (2n − 1)2
4
1 π 1 10 5 sin tdt = [−5 cos t]π0 = a0 = π 0 π π π π 5 5 sin t cos ntdt = [sin(n + 1)t − sin(n − 1)t]dt an = π 0 2π 0 π cos(n + 1)t cos(n − 1)t 5 − + , n = 1 = 2π (n + 1) (n − 1) 0 1 5 cos nπ cos nπ 1 − − = − + 2π n+1 (n − 1) n+1 n−1
0, n odd, n = 1 5 10 =− (cos nπ + 1) = , n even − π(n2 − 1) π(n2 − 1)
Note that in this case we need to evaluate a1 separately as π 1 π 5 a1 = 5 sin t cos tdt = sin 2tdt = 0 π 0 2π 0 π 5 π 5 bn = sin t sin ntdt = − [cos(n + 1)t − cos(n − 1)t]dt π 0 2π 0 π 5 sin(n + 1)t sin(n − 1)t − , n = 1 =− 2π (n + 1) (n − 1) 0 = 0 , n = 1 c Pearson Education Limited 2004
197
198
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Evaluating b1 separately π 5 π 5 sin t sin tdt = (1 − cos 2t)dt b1 = π 0 2π 0 π 1 5 5 t − sin 2t = = 2π 2 2 0 Thus the Fourier expansion of f (t) is ∞ 5 10 cos 2nt 5 + sin t − π 2 π n=1 4n2 − 1
f (t) =
5 a0 = = an = = =
1 bn = π
0 π 1 2 2 π dt + (t − π) dt π −π 0 1 π 4 1 2 0 3 π t −π + (t − π) = π2 π 3 3 0 0 π 1 2 2 π cos ntdt + (t − π) cos ntdt π −π 0 (t − π)2 π 0 2 1 π2 2(t − π) + cos nt − 3 sin nt sin nt sin nt + π n n n2 n −π 0 2 n2
0
2
π sin ntdt + −π
0
π
2
(t − π) sin ntdt
0 (t − π)2 π (t − π) 2 1 π2 − cos nt cos nt + 2 + − sin nt + 3 cos nt = π n n n2 n −π 0 2 2 π 1 π − + (−1)n = π n n 2 π [1 − (−1)n ] = (−1)n − n πn3 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Thus the Fourier expansion of f (t) is ∞ ∞ 4 sin(2n − 1)t (−1)n 2 2 2 π sin nt − cos nt + f (t) = π + 3 n2 n π n=1 (2n − 1)3 n=1
5(a)
Taking t = 0 gives ∞ 2 π2 + π2 2 = π2 + 2 3 n2 n=1
and hence the required result
5(b)
∞ 1 1 = π2 2 n 6 n=1
Taking t = π gives ∞ 2 π2 + 0 2 (−1)n = π2 + 2 2 3 n n=1
and hence the required result ∞ (−1)n+1 1 2 π = 2 n 12 n=1
6(a)
c Pearson Education Limited 2004
199
200
Glyn James: Advanced Modern Engineering Mathematics, Third edition
6(b)
The Fourier expansion of the even function (a) is given by ∞
f (t) =
a0 + an cos nt 2 n=1
with 2 a0 = π
π/2
π
tdt + 0
(π − t)dt
π/2
π π 2 1 2 π/2 1 t + − (π − t)2 = = π 2 0 2 2 π/2 π/2 π 2 t cos ntdt + (π − t) cos ntdt an = π 0 π/2 π/2 π − t π 1 1 2 t sin nt + 2 cos nt sin nt − 2 cos nt + = π n n n n 0 π/2 1 2 2 nπ − 2 (1 + (−1)n ) = cos 2 π n 2 n n odd 0, 8 = − 2 , n = 2, 6, 10, . . . πn 0, n = 4, 8, 12, . . . Thus the Fourier expansion of f (t) is ∞ 2 cos(4n − 2)t π f (t) = − 4 π n=1 (2n − 1)2
Taking t = 0 where f (t) = 0 gives the required result. c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition
201
7
1 a0 = π
0
π
t (2 − )dt + π
2π
t/πdt π
π t2 2π t2 1 2t − = =3 + π 2π 0 2π π π 2π t 1 t cos ntdt an = (2 − ) cos ntdt + π 0 π π π π t 2π t 1 1 1 2 sin nt − sin nt − sin nt + cos nt + cos nt = π n πn πn2 πn πn2 0 π 2 = 2 2 [1 − (−1)n ] π n
0, n even 4 = , n odd π 2 n2 π 2π t 1 t sin ntdt (2 − ) sin ntdt + bn = π 0 π π π π t 2π 1 2 t 1 1 − cos nt + sin nt + − sin nt = cos nt − cos nt + π n πn πn2 πn πn2 0 π =0 Thus the Fourier expansion of f (t) is ∞ 4 cos(2n − 1)t 3 f (t) = + 2 2 π n=1 (2n − 1)2
Replacing t by t − 12 π gives ∞ 3 4 cos(2n − 1)(t − π) 1 f (t − π) = + 2 2 2 π n=1 (2n − 1)2
c Pearson Education Limited 2004
202
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Since π π 1 cos(2n − 1)(t − π) = cos(2n − 1)t cos(2n − 1) + sin(2n − 1)t sin(2n − 1) 2 2 2 n+1 = (−1) sin(2n − 1)t ∞ 1 3 4 (−1)n+1 sin(2n − 1)t f (t − π) − = 2 2 2 π n=1 (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) =
bn sin
n=1
with 2 bn =
0
nπt
t nπt nπt 2 nπt 2 − t sin sin dt = cos + nπ nπ 0
2 cos nπ =− nπ Thus the Fourier expansion of f (t) is ∞ nπt 2 (−1)n+1 sin f (t) = π n=1 n
9 Since f (t) is an odd function (readily seen from a sketch of its graph) its Fourier expansion is ∞ nπt f (t) = bn sin n=1 with 2 bn =
0
nπt K ( − t) sin tdt
2 K nπt Kt nπt K nπt = − cos + cos − sin nπ nπ (nπ)2 0 2K = nπ c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Thus the Fourier expansion of f (t) is f (t) =
10 1 a0 = 5
∞ nπt 2K 1 sin π n=1 n
5
3dt = 3 0
5 1 15 nπt nπt dt = sin 3 cos =0 5 5 nπ 5 0 0 5 1 15 nπt 1 5 nπt dt = − cos bn = 3 sin 5 0 5 5 nπ 5 0
6 3 [1 − (−1)n ] = nπ , n odd = nπ 0, n even
1 an = 5
5
Thus the Fourier expansion of f (t) is ∞ 6 (2n − 1) 3 1 sin πt f (t) = + 2 π n=1 (2n − 1) 5
11 π/ω A ω 2A − cos ωt A sin ωtdt = = π ω π 0 0 π/ω π/ω Aω Aω sin ωt cos nωtdt = [sin(n + 1)ωt − sin(n − 1)ωt]dt an = π 0 2π 0 π/ω cos(n + 1)ωt cos(n − 1)ωt Aω − + = , n = 1 2π (n + 1)ω (n − 1)ω 0 2 A A 2(−1)n+1 − 2 = [(−1)n+1 − 1] = 2 2 2π n − 1 n −1 π(n − 1)
0, n odd , n = 1 2A = − , n even π(n2 − 1) 2ω a0 = 2π
π/ω
Evaluating a1 separately Aω a1 = π
0
π/ω
A sin ωt cos ωtdt = 2π
π/ω
sin 2ωtdt = 0 0
c Pearson Education Limited 2004
203
204
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Aω bn = π
π/ω
0
Aω sin ωt sin nωtdt = − 2π
0
π/ω
[cos(n + 1)ωt − cos(n − 1)ωt]dt
π/ω Aω sin(n + 1)ωt sin(n − 1)ωt − =− , n = 1 2π (n + 1)ω (n − 1)ω 0
= 0, n = 1 Aω π/ω 2 Aω π/ω A b1 = sin ωtdt = (1 − cos 2ωt)dt = π 0 2π 0 2 Thus the Fourier expansion of f (t) is ∞ π A cos 2nωt 1 + sin ωt − 2 f (t) = π 2 4n2 − 1 n=1
12
Since f (t) is an even function its Fourier expansion is ∞
f (t) =
a0 nπt + an cos 2 T n=1
with 2 a0 = T an = =
2 T
T
0
0 2
T
T 2 1 3 2 t t dt = = T2 T 3 0 3 2 T 2 Tt nπt 2tT 2 2T 3 nπt nπt nπt 2 dt = sin + − t cos cos sin T T nπ T (nπ)2 T (nπ)3 T 0 2
4T (−1)n (nπ)2
Thus the Fourier series expansion of f (t) is ∞ T2 nπt 4T 2 (−1)n f (t) = cos + 2 3 π n=1 n2 T
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 13 2 a0 = T 2 an = T
T
0
T
0
T 2E 1 2 E tdt = 2 t =E T T 2 0 E 2πnt t cos dt T T
T 2πnt 2πnt T 2 2E tT cos =0 sin + = 2 T 2πn T 2πn T 0 2E T 2πnt dt t sin bn = 2 T 0 T T 2E tT 2πnt E 2πnt T 2 = 2 − sin =− cos + T 2πn T 2πn T 0 πn
Thus the Fourier expansion of e(t) is ∞ E1 2πnt E − sin e(t) = 2 π n=1 n T
Exercises 4.3.3 14
Half range Fourier sine series expansion is given by f (t) =
∞
bn sin nt
n=1
with 2 bn = π
0
π
π 1 2 − cos nt 1 sin ntdt = π n 0
2 [(−1)n − 1] nπ
0, n even 4 = , n odd nπ Thus the half range Fourier sine series expansion of f (t) is =−
∞ 4 sin(2n − 1)t f (t) = π n=1 (2n − 1)
Plotting the graphs should cause no problems. c Pearson Education Limited 2004
205
206
Glyn James: Advanced Modern Engineering Mathematics, Third edition
15
Half range Fourier cosine series expansion is given by ∞
a0 + an cos nπt f (t) = 2 n=1 with 2 a0 = 1 an = 2
1
0 1
0
(2t − 1)dt = 0
(2t − 1) cos nπtdt
1 2 (2t − 1) sin nπt + =2 cos nπt nπ (nπ)2 0 4 = [(−1)n − 1] (nπ)2
0, n even 8 = − , n odd (nπ)2 Thus the half range Fourier cosine series expansion of f (t) is ∞ 8 1 f (t) = − 2 cos(2n − 1)πt π n=1 (2n − 1)2
Again plotting the graph should cause no problems.
16(a) a0 = 2
0
an = 2
0
1
1
1 1 4 (1 − t2 )dt = 2 t − t3 = 3 0 3 (1 − t2 ) cos 2nπtdt
1 2t (1 − t2 ) 2 sin 2nπt − =2 cos 2nπt + sin 2nπt 2nπ (2nπ)2 (2nπ)3 0 1 =− (nπ)2 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition bn = 2
1
0
(1 − t2 ) sin 2nπtdt
1 2t (1 − t2 ) 2 cos 2nπt − =2 − sin 2nπt − cos 2nπt 2nπ (2nπ)2 (2nπ)3 0 1 = nπ Thus the full-range Fourier series expansion for f (t) is f (t) = f1 (t) =
16(b)
∞ ∞ 1 1 2 11 − 2 sin 2nπt cos 2nπt + 3 π n=1 n2 π n=1 n
Half range sine series expansion is f2 (t) =
∞
bn sin nπt
n=1
with
bn = 2
0
1
(1 − t2 ) sin nπtdt
(1 − t2 ) cos nπt − =2 − nπ 2 1 =2 − (−1)n + 3 (nπ) nπ 2 , n = nπ 1 4 2 + , n nπ (nπ)3
2t 2 sin nπt − cos nπt (nπ)2 (nπ)3 2 + (nπ)3
1 0
even odd
Thus half range sine series expansion is ∞ ∞ 2 4 1 11 sin 2nπt + + f2 (t) = sin(2n − 1)πt π n=1 n π n=1 (2n − 1) π 2 (2n − 1)3
16(c)
Half range cosine series expansion is ∞
a0 f3 (t) = an cos nπt + 2 n=1 c Pearson Education Limited 2004
207
208
Glyn James: Advanced Modern Engineering Mathematics, Third edition
with
a0 = 2
0
an = 2
0
1
1
(1 − t2 )dt =
4 3
(1 − t2 ) cos nπtdt
1 (1 − t2 ) 2 2t =2 cos nπt + sin nπt sin nπt − nπ (nπ)2 (nπ)3 0 −4(−1)n = (nπ)2 Thus half range cosine series expansion is f3 (t) =
∞ 4 (−1)n+1 2 + 2 cos nπt 3 π n=1 n2
Graphs of the functions f1 (t), f2 (t), f3 (t) for −4 < t < 4 are as follows
17
Fourier cosine series expansion is ∞
a0 + an cos nt f1 (t) = 2 n=1 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition
with
2 π 1 (πt − t2 )dt = π 2 a0 = π 0 3 π 2 an = (πt − t2 ) cos ntdt π 0 π (π − 2t) 2 2 (πt − t2 ) sin nt + cos nt + 3 sin nt = π n n2 n 0 2 = − 2 [1 + (−1)n ]
n 0, n odd 4 = − 2 , n even n
Thus the Fourier cosine series expansion is ∞ 1 2 1 f1 (t) = π − cos 2nt 2 6 n n=1
Fourier sine series expansion is
f2 (t) =
∞
bn sin nt
n=1
with
2 π (πt − t2 ) sin ntdt bn = π 0 π (πt − t2 ) (π − 2t) 2 2 − cos nt + sin nt − 3 cos nt = π n n2 n 0 4 [1 − (−1)n ] = 3 πn
0, n even 8 = , n odd πn3
Thus the Fourier sine series expansion is ∞ 1 8 f2 (t) = sin(2n − 1)t π n=1 (2n − 1)3
c Pearson Education Limited 2004
209
210
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Graphs of the functions f1 (t) and f2 (t) for −2π < t < 2π are:
18
2a x, 0
View more...
Comments
