Lesson_140 - Lesson14 Challenge13...

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Lesson 14 Lesson 14 Challenge 13 Lesson 14:  FIR Frequency response Challenge 14:
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  Lesson 14 Challenge 13 Compare the frequency responses of two L=6 order LTI FIRs shown below: h1={1,1,1,1,1,0} h2={0,1,1,1,1,1} Compare them in terms of magnitude and phase frequency responses.
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  Lesson 14 Challenge 13 Computing the frequency response H 1 (e j ω ) = 1 +  e -j ω  + e -j2 ω  +  e -j3 ω  + e -j4 ω  =   = e -j2 ω  ( e j2 ω  +  e j ω  +1 + e -j ω  + e -j2 ω =  e -j2 ω   ( 1+2cos( ω )+2cos(2 ω )); - π≤ ω < π H 2 (e j ω ) = e -j ω  + e -j2 ω  + e -j3 ω  + e -j4 ω  + e -j5 ω  =   = e -j3 ω  ( e j2 ω  +  e j ω  +1 + e -j ω  + e -j2 ω =  e -j3 ω   ( 1+2cos( ω )+2cos(2 ω )); - π≤ ω < π
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  Lesson 14 Challenge 13 Comparing the magnitude frequency responses one determines that they are  equal. |H 1 (e j ω )| = |1 +  e -j ω  + e -j2 ω  +  e -j3 ω  + e -j4 ω  |=   = |e -j2 ω  ( e j2 ω  +  e j ω  +1 + e -j ω  + e -j2 ω )|  =   |e -j2 ω   | | ( 1+2cos( ω )+2cos(2 ω ))| =  | ( 1+2cos( ω )+2cos(2 ω ))|; - π≤ ω < π |H 2 (e j ω )| = |e -j ω  + e -j2 ω  + e -j3 ω  + e -j4 ω  + e -j5 ω  |=   |e -j3 ω   | |( e j2 ω  +  e j ω  +1 + e -j ω  + e -j2 ω )|  =  |( 1+2cos( ω )+2cos(2 ω ))|; - π≤ ω < π
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  Lesson 14 Challenge 13 The phase response of the first filter is   H 1 (e j ω )=-2 ω  and for the second    H 2 (e j ω )=-3 ω . H 1 (e j ω ) =   e -j2 ω  +    ( 1+2cos( ω )+2cos(2 ω ))    e -j2 ω  = -j2 ω ; - π≤ ω < π  H 2 (e j ω ) =   e -j3 ω  +   ( e j2 ω  +  e j ω  +1 + e -j ω  + e -j2 ω  e -j3 ω   = -j3 ω ; - π≤ ω < π
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  Lesson 14
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Lesson_140 - Lesson14 Challenge13...

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