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Find the energy in the signal x[n] = 2(0.12)", n2 0.

Find the energy in the signal x[n] = 2(0.12)", n2 0.

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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1) Find the energy in the signal x[n] = 2(0.12)", n20. %3D 2) Given a periodic signal of x[n] = 2cos(2 m/4) whose period is N = 4. Find its average value and signal power. 3) Given a periodic signal of x[n] = 9e24 whose period is N=4. Find its signal power. 4) Sketch each signal and its even and odd parts. (1) x[n] = 8(0.5)' u[n] (ii) x[n]= u[n] (iii) x[n]=1+ u[n] (iv) x[n] = u[n]-u[n – 4] (v) (vi) x[n]= {6,4,2, 2}
Transcribed Image Text:1) Find the energy in the signal x[n] = 2(0.12)", n20. %3D 2) Given a periodic signal of x[n] = 2cos(2 m/4) whose period is N = 4. Find its average value and signal power. 3) Given a periodic signal of x[n] = 9e24 whose period is N=4. Find its signal power. 4) Sketch each signal and its even and odd parts. (1) x[n] = 8(0.5)' u[n] (ii) x[n]= u[n] (iii) x[n]=1+ u[n] (iv) x[n] = u[n]-u[n – 4] (v) (vi) x[n]= {6,4,2, 2}
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