6 In the figure below, the input signal x(t) to an LTI system is a periodic signalwith period T and CFSC CH(jw) = 1+ w. Then, the output of the system y(t) can be expressed as a CFS-. The system frequency response is1+kwo2n,expansion as (wo =x(t)y(1)h(t)2nA. y(t) = Lk=-8(t – kT)1+|k|B. y(t) = EC. y(t) = Lk=-0 1+kD. y(t) = Lk=-02ejkwotZk=-001ejkwotejkwot

Answered: Image /qna-images/answer/d20e514b-e180-45e4-92be-e41b86fdb353.jpg

a the figure below, the input signal x(t) to an LTI system is a periodic signal ith period T and CFSC C% 2 The system frequency response is 1+kwo (jw) = 1+ w. Then, the output of the system y(t) can be expressed as a CFS spansion as (wo =7) x(t) Y(1) h(t) 27 A. y(t) = E- B. y(t) = Lk=-0 С. у(t) — 0 D. y(t) = Lk=-0 k=-∞ 1+|k| 8(t – kT) 100 2ejkwot - eikwot Lk=-00 1+k 1 ejkwot

a the figure below, the input signal x(t) to an LTI system is a periodic signal ith period T and CFSC C% 2 The system frequency response is 1+kwo (jw) = 1+ w. Then, the output of the system y(t) can be expressed as a CFS spansion as (wo =7) x(t) Y(1) h(t) 27 A. y(t) = E- B. y(t) = Lk=-0 С. у(t) — 0 D. y(t) = Lk=-0 k=-∞ 1+|k| 8(t – kT) 100 2ejkwot - eikwot Lk=-00 1+k 1 ejkwot

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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A given FM transmitter is modulated with a single sinusoid, the output for no load power is 100W into 50Ω resistive load. Then, the frequency of the modulating signal is carefully adjusted from zero until the modulated carrier encountered its zero amplitude as observed in spectrum analyzer. Under these conditions, determine the: (Use the provided Bessel table) a. the modulation index and the number of significant sideband b. total power on its significant sidebands. c. relative total power of the carrier and each of sideband. Ans. (a) Mf =2.4, N=5; (b) 99.86 W; b Pc=0W, P1=27.04W, P2=18.49W, P3=4W, P4=0.36W, P5=0.04W
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