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Problem#4: Sketch the amplitude (magnitude) spectrum and phase spectrum of the periodic signals: x(t) = 2 cos(xt) + sin(2rt) Note: you can use the FT tables. Answer: TABLE 3.1 Short Table of Fourier Transforms g(t) GỰ) g(t) GỰ) a+j2af a>0 10 sin 2rfof j0.5 [8(f +fo) – 8(f – fo)] 2 u(-t) a >0 11 (t) a – j2af 2a 2af 2 3 e-al a² + (2xf)² a>0 12 sgnt 2af 4 te-#u(t) a >0 (a +j2nf)2 n! n() z sinc (7fr) 5 reu(t) a >0 n() sine () (a +j2¤f)*+ 18 2B sinc (27 Br) п 2B 6 8(1) 19 7 1 SF) Sif – fo) 20 B sinc² (7Bt) A cos 2n fot 0.5 [8(f +fo) + 8f – fo)] 21 E 8(t – nT) fo E- 8f – nfo) fo = }

Problem#4: Sketch the amplitude (magnitude) spectrum and phase spectrum of the periodic signals: x(t) = 2 cos(xt) + sin(2rt) Note: you can use the FT tables. Answer: TABLE 3.1 Short Table of Fourier Transforms g(t) GỰ) g(t) GỰ) a+j2af a>0 10 sin 2rfof j0.5 [8(f +fo) – 8(f – fo)] 2 u(-t) a >0 11 (t) a – j2af 2a 2af 2 3 e-al a² + (2xf)² a>0 12 sgnt 2af 4 te-#u(t) a >0 (a +j2nf)2 n! n() z sinc (7fr) 5 reu(t) a >0 n() sine () (a +j2¤f)*+ 18 2B sinc (27 Br) п 2B 6 8(1) 19 7 1 SF) Sif – fo) 20 B sinc² (7Bt) A cos 2n fot 0.5 [8(f +fo) + 8f – fo)] 21 E 8(t – nT) fo E- 8f – nfo) fo = }

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
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Problem#4: Sketch the amplitude (magnitude) spectrum and phase spectrum of the periodic signals: x(t) = 2 cos(xt) + sin(2nt) Note: you can use the FT tables. Answer: TABLE 3.1 Short Table of Fourier Transforms g(1) GF) g(1) GỰ) I u(t) a >0 10 sin 2nfof jo.5 [8(f +fo) – 8f – fo)] a+j2af 2 e u(-t) a>0 I1 u(t) 1 a - j2nf 2a 3 e-all a >0 a² + (2xf)² 12 sgn? 4 te-u(t) a >0 (a +j2nf)? n! (4) 17 n z sinc (æft) 5 re-au(t) a >0 (a +j2nf)n+ 18 2B sinc (27 Bt) 6 8(1) 2B 19 A Sf) sinc2 Sf – fo) 20 B sinc? (7Bt) A 2B cos 2nfot 0.5 [8(f +fo) + 8(f – fo)] 21 E- 8(t – nT') fo E- 8f – nfo) fo = + Ln=-00
Transcribed Image Text:Problem#4: Sketch the amplitude (magnitude) spectrum and phase spectrum of the periodic signals: x(t) = 2 cos(xt) + sin(2nt) Note: you can use the FT tables. Answer: TABLE 3.1 Short Table of Fourier Transforms g(1) GF) g(1) GỰ) I u(t) a >0 10 sin 2nfof jo.5 [8(f +fo) – 8f – fo)] a+j2af 2 e u(-t) a>0 I1 u(t) 1 a - j2nf 2a 3 e-all a >0 a² + (2xf)² 12 sgn? 4 te-u(t) a >0 (a +j2nf)? n! (4) 17 n z sinc (æft) 5 re-au(t) a >0 (a +j2nf)n+ 18 2B sinc (27 Bt) 6 8(1) 2B 19 A Sf) sinc2 Sf – fo) 20 B sinc? (7Bt) A 2B cos 2nfot 0.5 [8(f +fo) + 8(f – fo)] 21 E- 8(t – nT') fo E- 8f – nfo) fo = + Ln=-00
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