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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 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)
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) 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
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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