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3.8-1 Show that the autocorrelation function of g(t) = C cos (27t fot + 00) is given by Rg(t) (C2/2) cos 27for, and the corresponding PSD is Sg(f) = (C2/4)[8f – fo) + 8f +fo)I. Hence, show that for a signal y (t) given by y(1) = Co+ C, cos (n27 fot + On) n=1

3.8-1 Show that the autocorrelation function of g(t) = C cos (27t fot + 00) is given by Rg(t) (C2/2) cos 27for, and the corresponding PSD is Sg(f) = (C2/4)[8f – fo) + 8f +fo)I. Hence, show that for a signal y (t) given by y(1) = Co+ C, cos (n27 fot + On) n=1

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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3.8-1

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3.8-1 Show that the autocorrelation function of g(t) = C cos (27 fot + 00) is given by Rg(t) = (C2/2) cos 27for, and the corresponding PSD is Sg(f) = (C²/4)[8(f – fo) + 8f + fo)l- Hence, show that for a signal y (t) given by y(t) = Co + Cn cos (n2T fot + On) n=1
Transcribed Image Text:3.8-1 Show that the autocorrelation function of g(t) = C cos (27 fot + 00) is given by Rg(t) = (C2/2) cos 27for, and the corresponding PSD is Sg(f) = (C²/4)[8(f – fo) + 8f + fo)l- Hence, show that for a signal y (t) given by y(t) = Co + Cn cos (n2T fot + On) n=1
Problems 139 the autocorrelation function and the PSD are given by Ry(t) = Co² + Cn² cos n2n for n=1 Syf) = Co²5(f)+ n²l8V – nfo) + ôf + nfo)] n=1 Hint: Show thatif g (?) = 81 (t)+82(t), then Rg(t) = Rg, (t)+Rg2 (1 )+Rg,82(t)+Rg28, (t), where Rg182(t) = lims→(1/T) T12 81 (1)82(1 + t) dt. If g1 (1) and g2(t) represent any two of the infinite terms in y(t), then show that Rg182 (T) = Rg281 (1) = 0. To show this, use the fact that the area under any sinusoid over a very large time interval is at most equal to the area of the half-cycle of the sinusoid.
Transcribed Image Text:Problems 139 the autocorrelation function and the PSD are given by Ry(t) = Co² + Cn² cos n2n for n=1 Syf) = Co²5(f)+ n²l8V – nfo) + ôf + nfo)] n=1 Hint: Show thatif g (?) = 81 (t)+82(t), then Rg(t) = Rg, (t)+Rg2 (1 )+Rg,82(t)+Rg28, (t), where Rg182(t) = lims→(1/T) T12 81 (1)82(1 + t) dt. If g1 (1) and g2(t) represent any two of the infinite terms in y(t), then show that Rg182 (T) = Rg281 (1) = 0. To show this, use the fact that the area under any sinusoid over a very large time interval is at most equal to the area of the half-cycle of the sinusoid.
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