Lesson_100

# Lesson_100 - x[k]={1. 0.5, 0.25, 0.125, … };...

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EEL 3135: Signals and Systems Dr. Fred J. Taylor, Professor Lesson Title: Convolution Lesson Number: 10 (Section 5-3) Challenge 10 Problem P5-6 in the text asks you to consider using an L=4 sample moving average filer to process a signal x[k]=(0.5) k u[k], where u[k] is a unit step function beginning at k=0. You are asked to determine the output y[k] for k [-5,10]. The textbook notes that the sum shown below, can be expressed in close form: 1 : 1 ) 1 ( < - - = + = a if a a a a N M N M k k To illustrate, 1 : 1 1 ... 1 1 1 ) 1 ( 0 1 0 0 1 0 0 3 2 < - - = - = - = + + + + = - = + = + = = + = = = a if a a a a a a a a a a a a a N k k N k k N k N k k k k k k k --------------------- For a=0.5, the L=4 MA linear convolution yields:
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Unformatted text preview: x[k]={1. 0.5, 0.25, 0.125, … }; h=0.25{1,1,1,1} y[-5]= y[-4]=y[-3]=y[-2]= y[-1] = 0 y[0]=0.25; y[1]=0.375; y[2]=0.438; y[3]=0.469; <= maximum output y[4]=0.234; y[5]=0.117; y[6]=0.059; y[7]=0.029; y[8]=0.015; … For a=0.5, and an MA of order of L, an integer, L>1, what would the maximum output be? At sample index k=L-1, the output is maximum and equal to: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ) ( ) ( 1 1 5 . 1 ) 5 . ( 5 . 1 5 . 5 . 4 1 5 . 4 1 4 1 L L L k k L k k a-=--= = ∑ ∑-=-= 1...
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## This note was uploaded on 08/21/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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