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Find Fourier transform for: rect(t – 2). cos(500Ł) Select one: O a F(rect(t – 2). cos(500t)] = je+3j[sinc -500 D500 + sinc 2 %3D O b. F(rect(t – 2). cos(500t)] = fe 5(e-300) sinc! O c F[rect(t – 2). cos(500t)] = sinc + sinc- -500 2 +e j(w÷500) sinc500 %3D -502 +498 O d. F(rect(t – 2).cos(500t)] = [e+25(-500), sinc +et3j(w+500), sinc %3D +300 O e F[rect(t – 2). cos(500t)] = e+500je (sinc, + sinc] %3D O f. F(rect(t – 2). cos(500t)] = e j [sinc + sinc +300 O g. F(rect(t – 2). cos(500t)] = 1-500ju "2[sinc, + sinc]

Find Fourier transform for: rect(t – 2). cos(500Ł) Select one: O a F(rect(t – 2). cos(500t)] = je+3j[sinc -500 D500 + sinc 2 %3D O b. F(rect(t – 2). cos(500t)] = fe 5(e-300) sinc! O c F[rect(t – 2). cos(500t)] = sinc + sinc- -500 2 +e j(w÷500) sinc500 %3D -502 +498 O d. F(rect(t – 2).cos(500t)] = [e+25(-500), sinc +et3j(w+500), sinc %3D +300 O e F[rect(t – 2). cos(500t)] = e+500je (sinc, + sinc] %3D O f. F(rect(t – 2). cos(500t)] = e j [sinc + sinc +300 O g. F(rect(t – 2). cos(500t)] = 1-500ju "2[sinc, + sinc]

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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Find Fourier transform for: rect(t – 2). cos (500t) Select one: O a. F(rect(t – 2). cos(500t)) = e+2j" [sinc- ロ-500 + sinc W+500 %3D O b. F(rect(t – 2). cos(500t)] =e 25(w–500) sinc"-00 O . F[rect(t – 2). cos(500t)] = (sinc +e 25(w+500). sinc"t500 2 2 w-502 W+498 + sinc- O d. F(rect(t – 2). cos(500t)] = (e+2j(w–500). . +et2j(w+500) sinc"+500 sinc W-500 O e. F(rect(t – 2). cos(500t)) = e+500ju[sinc", + sinc) w-2 w+2 O f. F(rect(t – 2). cos(500t)] = e-ju [sinc- -2ju -500 + sincs00 0十500 500jw g. F(rect(t – 2). cos(500t)] = 2[sinc+ sinc
Transcribed Image Text:Find Fourier transform for: rect(t – 2). cos (500t) Select one: O a. F(rect(t – 2). cos(500t)) = e+2j" [sinc- ロ-500 + sinc W+500 %3D O b. F(rect(t – 2). cos(500t)] =e 25(w–500) sinc"-00 O . F[rect(t – 2). cos(500t)] = (sinc +e 25(w+500). sinc"t500 2 2 w-502 W+498 + sinc- O d. F(rect(t – 2). cos(500t)] = (e+2j(w–500). . +et2j(w+500) sinc"+500 sinc W-500 O e. F(rect(t – 2). cos(500t)) = e+500ju[sinc", + sinc) w-2 w+2 O f. F(rect(t – 2). cos(500t)] = e-ju [sinc- -2ju -500 + sincs00 0十500 500jw g. F(rect(t – 2). cos(500t)] = 2[sinc+ sinc
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