Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x) = x/2 sin nx (2) f(x)=-x 25 sin nx (3) f(x)= sinx sin x (4) f(x) = cosx COS X (5) f(x)=xsin x -1+ COS nX (6) f(x) = xcos x 2n(-1) (n-1)(n+1) %3D sin(nx) 0. 7) f(x) = 0
Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x) = x/2 sin nx (2) f(x)=-x 25 sin nx (3) f(x)= sinx sin x (4) f(x) = cosx COS X (5) f(x)=xsin x -1+ COS nX (6) f(x) = xcos x 2n(-1) (n-1)(n+1) %3D sin(nx) 0. 7) f(x) = 0
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![Exercise (3-2): find Fourier series on [0,2n]
Question
Answer
(1) f(x) = x/2
sin nx
(2) f(x)=-x
2 sin nx
(3) f(x)= sin x
sin x
(4) f(x) = cosx
COS X
(5) f(x)=xsin x
-1+
COS nX
(6) f(x)=xcos x
2n(-1)
%3D
sin(nx)
(n-1)(n+1)
0.
7) f(x) =
0<xくて
1
l-(-1)"
sin nx
1 <x<2T
-1
0<x<7
-1+(-1)"
8) f(x) = {
1
sin nx
11
T<x<27
3
(9) f(x) = {
sin nx
12 <x<2n
(10) f(x)=
-R/4
ーてくx<0
Σ
1 -1+(-1)"
sin nx
0<x<A
0<x<T
37
(-1)" -1
sin nx
(11) f(x) = {
COS nx-
4
T T<x<2n
0<x<T
s-1)" -cos nx +
オーX
sin nx
|(12) f(x)=
0.
0<xく元
(13) f(x)= {
27-x
COS nx
T<x< 27](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2721001d-a6db-4704-8d36-161a2b964ec1%2F4298760f-5342-4e0b-bc07-c1766c9dcc34%2Fpc75b2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise (3-2): find Fourier series on [0,2n]
Question
Answer
(1) f(x) = x/2
sin nx
(2) f(x)=-x
2 sin nx
(3) f(x)= sin x
sin x
(4) f(x) = cosx
COS X
(5) f(x)=xsin x
-1+
COS nX
(6) f(x)=xcos x
2n(-1)
%3D
sin(nx)
(n-1)(n+1)
0.
7) f(x) =
0<xくて
1
l-(-1)"
sin nx
1 <x<2T
-1
0<x<7
-1+(-1)"
8) f(x) = {
1
sin nx
11
T<x<27
3
(9) f(x) = {
sin nx
12 <x<2n
(10) f(x)=
-R/4
ーてくx<0
Σ
1 -1+(-1)"
sin nx
0<x<A
0<x<T
37
(-1)" -1
sin nx
(11) f(x) = {
COS nx-
4
T T<x<2n
0<x<T
s-1)" -cos nx +
オーX
sin nx
|(12) f(x)=
0.
0<xく元
(13) f(x)= {
27-x
COS nx
T<x< 27
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