Lesson%2011 - Lesson11 Challenge10 Lesson11 FIRarchitecture...

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Lesson 11 Challenge 10 Lesson 11 FIR architecture Challenge 11
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    Challenge 10 Problem P5-6 in the text asks you to consider using an L=4 sample moving  average filter to process the 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 as: 1 : 1 ) 1 ( < - - = + = a if a a a a N M N M k k
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    Challenge 10 Sidebar derivation 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 division long proof a a a a a N k k N k k N k N k k k k k k k
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    Challenge 10 -5                     0                                             10  L=4 MA filter, x[k]= (0.5) k u[k] 0.25 1.0 0.469
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    Challenge 10 For a=0.5, the L=4 MA linear convolution yields: 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;   (@L-1) y[3]=0.469;   <= maximum output  (@L) y[4]=0.234;  y[5]=0.117;  y[6]=0.059;  y[7]=0.029;  y[8]=0.015;  …
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    Challenge 10 In general, for an L th  order MA filter the maximum filter output to an input  signal x[k]=(0.5) k  u[k] occurs at sample index (L-1) (after L samples), and it  corresponds to:  ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ) ( ) ( 0 1 0 1 0 5 . 0 1 ) 5 . 0 ( 5 . 0 1 5 . 0 5 . 0 4 1 5 . 0 4 1 4 1 L L L k k L k k a - = - - = = - = - = Textbook formula.
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    Lesson 11 FIR Architecture [ ] [ ] = - = M k k k n x h n y 0
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    Lesson 11 FIRs are fundamentally important. How are they realized?  =  architecture
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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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Lesson%2011 - Lesson11 Challenge10 Lesson11 FIRarchitecture...

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