x(t) = 9sin (20t) cos (200nt))%3DThe Fourier transform of the given signal isjAx(f) - (#) (5(S – 90) + 8 (f + 110) – 8 (f – 110) – 6 (f + 90))The numerical values of the constants are A=and B=

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x(t) = 9sin (20t) cos (200nt)) %3D The Fourier transform of the given signal is x(f) = (#) (5(f – 90) + 5 (f + 110) – 8 (f – 110) – 6 (f + 90)) jA The numerical values of the constants are A= and B=

x(t) = 9sin (20t) cos (200nt)) %3D The Fourier transform of the given signal is x(f) = (#) (5(f – 90) + 5 (f + 110) – 8 (f – 110) – 6 (f + 90)) jA The numerical values of the constants are A= and B=

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

x(t) = 9sin (20t) cos (200nt))
%3D
The Fourier transform of the given signal is
jA
x(f) - (#) (5(S – 90) + 8 (f + 110) – 8 (f – 110) – 6 (f + 90))
The numerical values of the constants are A=
and B=

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