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Let x1[n], x2[n] and x3[m] are three sequences, each of length N, whose DFTS are given as X1(k), X2(k) and X3(k) respectively. If X3(k) = X1(k).X,(k), then the expression for X3[m] is CO x[m] = E[m]k[(2-m);] m=0 N-1 x[m]= E[m]x[(2- m)y] m=0 x3[m] = x[m]x,[(2 - m)N] m=0 N-1 *[m] = [m][n –m] m=0

Let x1[n], x2[n] and x3[m] are three sequences, each of length N, whose DFTS are given as X1(k), X2(k) and X3(k) respectively. If X3(k) = X1(k).X,(k), then the expression for X3[m] is CO x[m] = E[m]k[(2-m);] m=0 N-1 x[m]= E[m]x[(2- m)y] m=0 x3[m] = x[m]x,[(2 - m)N] m=0 N-1 *[m] = [m][n –m] m=0

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Let x1[n], x2[n] and x3[m] are three sequences, each of length N, whose DFTS are given as X1(k), X2(k) and X3(k) respectively. If X3(k) = X1(k).X2(k), then the expression for X3[m] is CO *[m] = E[m]k[(2-m);] m=0 N-1 x[m]= E[m]x[(2 - m)y] m=0 x3[m] = E[m]k[(2- m) y] m=0 N-1 x3[m] = [m]2[n-m] m=0
Transcribed Image Text:Let x1[n], x2[n] and x3[m] are three sequences, each of length N, whose DFTS are given as X1(k), X2(k) and X3(k) respectively. If X3(k) = X1(k).X2(k), then the expression for X3[m] is CO *[m] = E[m]k[(2-m);] m=0 N-1 x[m]= E[m]x[(2 - m)y] m=0 x3[m] = E[m]k[(2- m) y] m=0 N-1 x3[m] = [m]2[n-m] m=0
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