Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Exercise (3-2): find Fourier series on [0,2n]
Question
Answer
(1) f(x)= x/2
sin nx
2.
(2) f(x) = -x
2 sin nx
(3) f(x)=sinx
sin x
(4) f(x) = cos x
COS X
(5) f(x)= xsin x
2
COS nX
-1+
n -1
5 2n(-1)*
(n-1)(n +1)
(6) f(x)=xcos x
sin(nx)
0
(7) f(x) =
1 1-(-1)"
sin nx
0<xくて
1 T<x< 2T
2.
-1
0<xくT
2-1+(-1)"
sin nx
(8) f(x) =
1
元くX<2元
11
[1
(9) f(x)=
0<x<T
3
sin nx
2
2 <x<2T
-n/4
(10) f(x)={
5-1+(-1)"
ーズくx<0
1
sin nx
0<xくだ
0<x<7
37
sin nx
(11) f(x) =
COS NX -
4
|TTくX<2元
0<x<A
(-1)" -1
sin nx
12) f(x)={
COS nX +
元くx<2元
0<xくだ
(-1)"-1
(13) f(x)={
27 - x
COS NX
Transcribed Image Text:Exercise (3-2): find Fourier series on [0,2n] Question Answer (1) f(x)= x/2 sin nx 2. (2) f(x) = -x 2 sin nx (3) f(x)=sinx sin x (4) f(x) = cos x COS X (5) f(x)= xsin x 2 COS nX -1+ n -1 5 2n(-1)* (n-1)(n +1) (6) f(x)=xcos x sin(nx) 0 (7) f(x) = 1 1-(-1)" sin nx 0<xくて 1 T<x< 2T 2. -1 0<xくT 2-1+(-1)" sin nx (8) f(x) = 1 元くX<2元 11 [1 (9) f(x)= 0<x<T 3 sin nx 2 2 <x<2T -n/4 (10) f(x)={ 5-1+(-1)" ーズくx<0 1 sin nx 0<xくだ 0<x<7 37 sin nx (11) f(x) = COS NX - 4 |TTくX<2元 0<x<A (-1)" -1 sin nx 12) f(x)={ COS nX + 元くx<2元 0<xくだ (-1)"-1 (13) f(x)={ 27 - x COS NX
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