The Fourier series coefficients of the periodic signalx(t) = E-o [u(t- k) – u(t - k- 1)]k is evenA. Ck =j2kn1B. Ck =J2kn1C. CR =j2kn-(1-ek/2), k # 0(1-e k/3), k # 0(1-e-kr), k # 0D. Cx =(1-e13km"), k # 0E. C =(1-e 12kr), k # 0%3DJkn1%3Djkn%23

by putting the value of k in given expression and drawing the wave of x(t)

The Fourier series coefficients of the periodic signal x(t) = E-o [u(t - k) – u(t – k- 1)] %3D k is even A. C =(1-e ka/2), k # 0 (1-e k/3), k# 0 2km (1-e k"), k # 0 D. C =(1-e-13kr), k # 0 %3D j2kn В. Ск 12kn %3D С. Ск %3D Jkn

The Fourier series coefficients of the periodic signal x(t) = E-o [u(t - k) – u(t – k- 1)] %3D k is even A. C =(1-e ka/2), k # 0 (1-e k/3), k# 0 2km (1-e k"), k # 0 D. C =(1-e-13kr), k # 0 %3D j2kn В. Ск 12kn %3D С. Ск %3D Jkn

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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Question

100%

The Fourier series coefficients of the periodic signal
x(t) = E-o [u(t- k) – u(t - k- 1)]
k is even
A. Ck =
j2kn
1
B. Ck =
J2kn
1
C. CR =
j2kn
-(1-ek/2), k # 0
(1-e k/3), k # 0
(1-e-kr), k # 0
D. Cx =(1-e13km"), k # 0
E. C =(1-e 12kr), k # 0
%3D
Jkn
1
%3D
jkn
%23

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