Lesson%2018 - Lesson18 Challenge17...

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Lesson 18 Lesson 18 Challenge 17 Lesson 18: Convolution Theorem Challenge 18
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  Lesson 18 Challenge 17 Where are the zeros of a 5th-order comb FIR? Response A 5 th  order comb filter has a transfer function given by H(z)=1-z -4 . It follows  that the FIR’s  zeros  are located at those z 0  such that H(z 0 )=0.  They are : z 0  = 1 = 1+j0; H(z=1)=1-1 =0 z 0  = e j π /2 = 0+j1; H(z=j1)=1-(j1) -4 =1-1 =0 z 0  = e j π = -1+j0; H(z=-1)=1-(-1) -4 =1-1 =0 z 0  = e j3 π /2 = 0-j1; H(z=-j1)=1-(-j1) -4 =1-1 =0 The correspond to normalized frequencies of  ω  = 0,  ± π /2, and  π .
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  Lesson 18 Lesson 18 Z-Transform of a time-series x[n] was formally defined. [ ] [ ] [ ] k n k x n x N k - = = δ 0 ( 29 [ ] k N k z k x z X - = = 0 Z
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  Lesson 18 Lesson 18 Z-Transform Properties and Conventions Linearity Delay Theorem Transfer function x[n]= a x 1 [n] + b x 2 [n]   a X 1 (z) + b X 2 (z) = X(z)  ( 29 : ] [ z X z k n x k Z - → - ( 29 [ ] n h z h z H Z k N k k → = - = 0 The values of z which set H(z) to zero (zero gain) are called the filter’s  zeros .
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  Lesson 18 Lesson 18 Example Design a 15 th  order lowpass FIR with a passband cutoff frequency equal to  f 0 = 0.1f s Display the FIR’s zero locations and and frequency response. >> h=fir1(14,0.2); >> zplane(h,1)
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Lesson 18 Lesson 18 There are 14 zeros. There are 8 on the unit circle.
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Lesson%2018 - Lesson18 Challenge17...

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