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# Lesson%2015 - Lesson 15 Challenge 14 Lesson 15 FIR...

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Lesson 15 Challenge 14 Lesson 15: FIR Frequency Response Special cases Case study MATLAB design Challenge 15

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Challenge 14 Challenge Two N+1=11 order comb filters, having an impulse response  h[k]={1,0,0,0,0,0,0,0,0,0,0,1} and frequency response H(e j ω )=2(cos( ω N/2)e -j ω N/2 are cascaded.  What is the frequency response H 0 (e j ω ) of the cascaded filter? Response Treat the 1 st  impulse response as an input signal having a given frequency  response: H(e j ω )=| H(e j ω )| H(e j ω )=(2(cos(10 ω /2)) -10 ω /2=(2(cos(5 ω )) -5 ω Then  H 0 (e j ω )=| H(e j ω )| 2 2H(e j ω )=4cos(5 ω ) 2 -10 ω
Lesson 15 The frequency response function is a steady-state metric since it is based on an input signal which exists for all time. ( 29 ( 29 ( 29 ω φ j j N m m j i j e e H e h e H = = - = - 1 0 [ ] n j j e Ae n x =

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Lesson 15 The frequency response function can be graphically interpreted in terms of the magnitude and phase  response.  Common frameworks: Magnitude response Linear-linear Log (dB) – linear Log (db) – log (octave)  ( 29 ( 29 ( 29 ω φ j j N m m j i j e e H e h e H = = - = - 1 0 Phase response Linear – linear Linear – log (octave)
Lesson 15 Phase response was noted to be problematic Phase unwrapping FIRs = Linear phase or non-linear phase? Group delay N th  order  linear phase  FIR    τ =(N-1)/2  (developed in earlier lessons) π 0 Unwrapped phase Wrapped phase ( 29 ( 29 ω φ τ d d - =

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Lesson 15 Delay systems y[n]=x[n-n 0   H(e j ω ) = e -j ω n 0   unit magnitude with phase - ω n 0   Discrete-time equations   y[n]=a 0 x[n]+ a 1 x[n-1]+ a 2 x[n-2]+ . ..   Impulse response x[n]= δ [n] is h[n]={a 0 , a 1 , a 2 , … }  -   moral
Lesson 15 Difference equation y[n] = a 0 x[n] – a no x[n-n 0 ] . .. If a

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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%2015 - Lesson 15 Challenge 14 Lesson 15 FIR...

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