Consider the non-periodic signal x(t) shown below where t1 =1 andt2 = 3.x(t)1-t2t1t2-1The Fourier transform of the signal is given by:x (w) = 2 sin (dw) – sin(30)]a)O b) X(w) = w? [2 sin (3w) – 3 sin(w)]Oc)x (w) = sin (3u) – 3 sin (w)]- 3 sin ( wd)X(w) = 3 sin (3w) – sin (w)|sin ( w0ª x(x) = 3 [2sin(324) – 3sin (2) ]n(2)]2 sin ( 3w- 3 sin ( 2w

We use differentiation in time domain property to find the Fourier transform of the given signal.

Consider the non-periodic signal x(t) shown below where t1 =1 and t2 = 3. x(t) 1 -t2 -t1 t1 t2 1 The Fourier transform of the signal is given by: x (w) = [2 sin(dw) – žsin (3)] a) sin(3w b) X(w) = w² [2 sin(3w) – 3 sin (w)] - x (w) = [sin (3«) O c) - 3 sin (w O d) x (w) = 3 [3 sin(3w) – sin (w)] O e) x (w) = 2 sin (3w) – 3 sin (20)] u) – 3 sin ( 2w 2 sin ( 3w

Consider the non-periodic signal x(t) shown below where t1 =1 and t2 = 3. x(t) 1 -t2 -t1 t1 t2 1 The Fourier transform of the signal is given by: x (w) = [2 sin(dw) – žsin (3)] a) sin(3w b) X(w) = w² [2 sin(3w) – 3 sin (w)] - x (w) = [sin (3«) O c) - 3 sin (w O d) x (w) = 3 [3 sin(3w) – sin (w)] O e) x (w) = 2 sin (3w) – 3 sin (20)] u) – 3 sin ( 2w 2 sin ( 3w

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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