Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x)=x/2 sin nx (2) f(x)= -x sin nx (3) f(x)=sin x sin x (4) f(x)= cos x COS X 2 -COS nx n -1 (5) f(x)%=xsin.x -1+ 6) f(x)=xcos x 2n(-1)" sin(nx) H(n-1)(n +1) 0
Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x)=x/2 sin nx (2) f(x)= -x sin nx (3) f(x)=sin x sin x (4) f(x)= cos x COS X 2 -COS nx n -1 (5) f(x)%=xsin.x -1+ 6) f(x)=xcos x 2n(-1)" sin(nx) H(n-1)(n +1) 0
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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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
sin nx
(3) f(x)=sin x
sin x
(4) f(x) = cos x
COS X
(5) f(x)=xsin x
-1+
-COS nx
n -1
6) f(x)=xcos x
2n(-1)"
sin(nx)
(n-1)(n +1)
0<x<A
-i-CD sin nx
1-(-1)"
(7) f(x) =
1 T<x< 2n
2ペ-1+(-1)"
sin nx
0<x<T
(8) f(x) =-
1
元くx<2元
0<xくだ
3 1-(-1) sin nx
1.
(9) f(x) =-
|2 くx<2元
2
-7/4
(10) f(x)=
- T<x<0
1-1+(-1)"
sin nx
0<x<A
0<xくて
37
-1)" -1
sin nx
(11) f(x) =
COS nx -
T Tくx<2元
0<x<A
sin nx
(12) f(x) =-
COS NX +
てくx<2元
4
(13) f(x) =
0<x<T
(-1)" -
COS nx
27-x
A<x< 27
IT -n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F314c7c98-2bcb-4e6f-b8c3-33e3cefec393%2Fd202bdbb-e0d8-4974-b76e-43b538e85ee3%2Fq711z1_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
sin nx
(3) f(x)=sin x
sin x
(4) f(x) = cos x
COS X
(5) f(x)=xsin x
-1+
-COS nx
n -1
6) f(x)=xcos x
2n(-1)"
sin(nx)
(n-1)(n +1)
0<x<A
-i-CD sin nx
1-(-1)"
(7) f(x) =
1 T<x< 2n
2ペ-1+(-1)"
sin nx
0<x<T
(8) f(x) =-
1
元くx<2元
0<xくだ
3 1-(-1) sin nx
1.
(9) f(x) =-
|2 くx<2元
2
-7/4
(10) f(x)=
- T<x<0
1-1+(-1)"
sin nx
0<x<A
0<xくて
37
-1)" -1
sin nx
(11) f(x) =
COS nx -
T Tくx<2元
0<x<A
sin nx
(12) f(x) =-
COS NX +
てくx<2元
4
(13) f(x) =
0<x<T
(-1)" -
COS nx
27-x
A<x< 27
IT -n
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