2025-L14su10

2025-L14su10 - ECE2025 Summer 2010 Lectures 14 and15 IIR...

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1 6/28/2010 1 ECE2025 Summer 2010 Lectures 14 and15 IIR Filters 30 Jun 10 07 Jul 10 6/28/2010 3 ANNOUNCEMENTS ± HW #6 due July 6 and 7 ± Monday recitations (Taylor) turn it in at the beginning of lab on Wednesday July 7. ± June 30-July 1: ± Turn in Lab #5 ± Do Lab #6 6/28/2010 4 SECOND QUIZ ± Quiz #2, in lecture, Friday, July 2 ± 10% of final grade ± You can use the full 1:20-2:30 time ± Review session tomorrow — Klaus 2447 6:30 pm ± Emphasis: ± HWs #3 through #5 ± Lectures #6 through #12 ± No z-transforms, ± Closed book, closed notes, except: ± One 8.5”X11” help sheet allowed, handwritten , OK to write on both sides 6/28/2010 5 READING ASSIGNMENTS ± This Lecture: ± Chapter 8, Sects. 8-1 to 8-6 ± Other Reading: ± Recitation, next lecture: all of Ch. 8
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2 6/28/2010 6 LECTURE OBJECTIVES ± Show how to compute the output y[n] ± FIRST-ORDER CASE (N=1) ± z-transform: Impulse Response h[n] ÅÆ H(z) y [ n ] = a l y [ n l ] l = 1 N + b k x [ n k ] k = 0 M ± INFINITE IMPULSE RESPONSE FILTERS ± Define IIR DIGITAL Filters ± Have FEEDBACK FEEDBACK : use PREVIOUS OUTPUTS 6/28/2010 7 THREE DOMAINS Z-TRANSFORM-DOMAIN POLYNOMIALS: H(z) FREQ-DOMAIN k j M k k j e b e H ω ˆ 0 ˆ ) ( = = TIME-DOMAIN = = M k k k n x b n y 0 ] [ ] [ } { k b 6/28/2010 8 Quick Review: Delay by n d IMPULSE RESPONSE SYSTEM FUNCTION d n z z H = ) ( ] [ ] [ d n n x n y = FREQUENCY RESPONSE ] [ ] [ d n n n h = δ d n j j e e H ˆ ˆ ) ( = 6/28/2010 9 y [ n ] = 1 L x [ n k ] k = 0 L 1 Quick Review: L-pt Averager IMPULSE RESPONSE h [ n ] = 1 L [ n k ] k = 0 L 1 SYSTEM FUNCTION = = 1 0 1 ) ( L n n L z z H
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3 6/28/2010 10 LOGICAL THREAD ± FIND the IMPULSE RESPONSE, h[n] ± INFINITELY LONG ± IIR IIR Filters ± EXPLOIT THREE DOMAINS: ± Show Relationship for IIR: h [ n ] H ( z ) H ( e j ˆ ω ) H ( z ) = h [ n ] z n n = 0 6/28/2010 11 ONE FEEDBACK TERM ± CAUSALITY ± NOT USING FUTURE OUTPUTS or INPUTS y [ n ] = a 1 y [ n 1] + b 0 x [ n ] + b 1 x [ n 1] FIR PART of the FILTER FEED-FORWARD PREVIOUS FEEDBACK ± ADD PREVIOUS OUTPUTS 6/28/2010 12 FILTER COEFFICIENTS ± ADD PREVIOUS OUTPUTS ± MATLAB ± yy = filter([3,-2],[1,-0.8],xx) y [ n ] = 0.8 y [ n 1] + 3 x [ n ] 2 x [ n SIGN CHANGE FEEDBACK COEFFICIENT 6/28/2010 13 COMPUTE OUTPUT
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4 6/28/2010 14 COMPUTE y[n] ± FEEDBACK DIFFERENCE EQUATION: y [ n ] = 0.8 y [ n 1] + 5 x [ n ] y [0] = 0.8 y [ 1] + 5 x [0] ± NEED y[-1] to get started 6/28/2010 15 AT REST CONDITION ± y[n] = 0, for n<0 ± BECAUSE x[n] = 0, for n<0 6/28/2010 16 COMPUTE y[0] ± THIS STARTS THE RECURSION: ± SAME with MORE FEEDBACK TERMS y [ n ] = a 1 y [ n 1] + a 2 y [ n 2] + b k x [ n k ] k = 0 2 6/28/2010 17 COMPUTE MORE y[n] ± CONTINUE THE RECURSION:
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5 6/28/2010 18 PLOT y[n] 6/28/2010 19 y [ n ] = a 1 y [ n 1] + b 0 x [ n ] IMPULSE RESPONSE u [ n ] = 1, for n 0 h h δ ] [ ) ( ] [ 1 0 n u a b n h n = 6/28/2010 20 IMPULSE RESPONSE ± DIFFERENCE EQUATION: ± Find h[n] ± CONVOLUTION in TIME-DOMAIN h [ n ] y [ n ] = h [ n ] x [ n ] x [ n ] IMPULSE RESPONSE y [ n ] = 0.8 y [ n 1] + 3 x [ n ] h [ n ] = 3(0.8) n u [ n ] LTI SYSTEM 6/28/2010 21 PLOT IMPULSE RESPONSE h [ n ] = b 0 ( a 1 ) n u [ n ] = 3(0.8) n u [ n ]
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6 6/28/2010 22 Infinite-Length Signal: h[n] ± POLYNOMIAL Representation ± SIMPLIFY the SUMMATION H ( z ) = h [ n ] z n n =−∞ APPLIES to Any SIGNAL = −∞ = = = 0 1 0 1 0 ] [ ) ( ) ( n n n n n n z a b z n u a b z H 6/28/2010 23
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2025-L14su10 - ECE2025 Summer 2010 Lectures 14 and15 IIR...

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