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by: Eloy Ferry


Eloy Ferry
GPA 3.84

John Gowdy

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John Gowdy
Class Notes
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This 11 page Class Notes was uploaded by Eloy Ferry on Saturday September 26, 2015. The Class Notes belongs to E C E 467 at Clemson University taught by John Gowdy in Fall. Since its upload, it has received 43 views. For similar materials see /class/214305/e-c-e-467-clemson-university in ELECTRICAL AND COMPUTER ENGINEERING at Clemson University.

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Date Created: 09/26/15
ECE 467667 Introduction to Digital Signal Processing LECTURE 18 One way to show aliasing 7t Aliased response curve for E S s 11 Further analvsis of impulse invariance manninq Rewrite as 1 21t HltZileST jQ THS lksmjg2 where QTm This is a special caseszj of the general cases o in Hz 12Hsj2 nk and QTro 39 T k sojn T467667 Introducticm to Digital Signal Processing Z1 y 21 EEC 2 x a QT 1 1 39 0391 2 n I 1 TIE Each segment of length 2m on the line 0 201 maps one time around the circle of radius e I1T in zplane A 6 zplane J Odb ECE T467667 Introduction to Digital Signal Processing lf His gtooat sk 5k j k then Hz has a pole at z eskT since Hz 26 oo v2 skT value of Hs at other points on the line sok lHZissk 00 We saw this earlier when checking Condition 1 If Hs has a zero at ssk then Hz does no usually have a zero at z eskT since Hleesk1 O value of Hs at other points on the line sok 1HssS it Design Examgle Low pass filter with no ripplegt Use Butterworth filter 02 31t k2 15db 1 db 15 db Y I 211 311 1c 21 gt 21 031 21r since QT co D 39 usin T1 T 467667 Introduction to Digital Signal Processing For analog Butterworth filter 1010 1 log 5 1010 1 Now using n6 solve for no using the passband equation We need the extra performancequot due to roundingup to get n in the stopband to counteract the aliasing inherent in the impulse invariance method 2 h c e 7032 l 2n 39 101 1T Get Hs by mapping the Bth order normalized filter from Table 31b using the LP gtLP mapping Hs 82 51765 1132 53 1st 19318s1 S slt 7032 1209 l 2364osu4945X 2u99453n494sx 243584sa4945 T467667 Introduction toDigiml Signal Processing quot 39 Partial Fraction Exnansion or case of 3 pairs of complex conjugate poles Hs1209 k1 k1 k2 k2 k3 k3 Let s1gjb gt poles at s g i jb sgcjd s c t jd 3ejf s e i if k k 39 Consrder 1 where 31 g 1b k1 k139 k1 k s gib s gib s g ib s gJb k1s g jb k139s g jb egg k1 gk1 k1j jbk1 k139 s gzb2 sg2b2 s2Rek 2Rek z 1 g 121 1 1vgt2bmk1 9Yb Therefore k1 k1 MsN 8 81 sws s g2b2 M 2Rek1 39 where N 49 Rek1 2bImk1 M9 2b1mk1 and s1gjb ECE T467667 Introductirm to Digital Signai Processing Therefore Hs can be written as Ms N Ps Q Rs T 1209 s g2b2 s c2d2 se2f2 where M 2Rek1 P2Rek2 R2Rek3 and N g M Q quotCP quot T 399R Summarizefor n6 1 Find gbcdef by factoring the 3 2nd order terms from the table 2 Find k1 k2 k3 using lims sis for i123 8 Find MNPQRT from For current example Hs is obtained as Hs Us I vLs we 185583 18453 9952 9890 2871s 2856 s 67932 18192 s 49722 4972 s 18202 67922 Table entries from class handout using T1 sa 39 zzeze acoso s a2 m2 z2 22s l cosm e 2a D zequota since 2 a 2 a 23 Sa2co z 229 coswe ECE T467667 Introduction to Digital Signal Processing Write Us as 1 8558 s9943 39 5e7932 18192 The denominator has the form 5 a2 co2 where a6793 and 921819 Should be quotccquot for other table entry 2 218858 2793 2 31250 2 3e793 1819 5 15793 1819 18858 s6793 8558 3150 1819 same 1819 same Now use and to get digital filter 22 ze mg 3 cos1819 U 2 218558 Z2 229 13793 COS1819 e 233793 6793 ze sm1819 32137 22 22equot396793 cos1819 9465795 2871 44662 1 1 1 29712 1 159492 2 Overall Hz UzVzWz where 4 4 1 18558 453042 Vz 2 13911 1 5 Z 392 and W65 1 2 1 106912 36992 1 99722 2570z ECE T467667 Introduction to Digital Signal Processing Another potential method for transforming analog filters to digital filters method of Apmoximatino Derivatives in Differential Equation for Analoo Filter Start with analog filter having desired properties Hls YS gadk 8k XS ick Sk Corresponding differential equation dkvlttgt dk2gts 2 k0 dtk k0 dtk Approximate the first derivative as the first backward difference dyt dt 3 Ynquot n 1 g V1yn isnT Now approximate the second derivative as d2yt dt t E w when W yin ViiYn 1Yn n 1 Yn quot 1 1 2 T ytn 2yn 1yn2 T2 in general the kth derivative is approximated as d39tk t ML 5 vm W vm Vlk39lily i 8 ECE T46766 7 Introductirm to Digital SignaiPri Jccssing in the zdomain 3V1yn yz1 Z i T altvlt2gtryltnmvltzi222 Yz1rz 1 T Eivltkgtiyltniiivltzi quotz T Summarize Approximation of differential equation N m Ecka ynT EdekxnT k0 k0 Expressed in the zdomain this becomes N 1 z 1 k m 11 27 k0 T k0 T ECE T467667 Intrmlucticm to Digital Signal Processing I 1 24 Check the mapping 3 to see whether Condition 1 and Condition 2 are satisfied The mapping can be written sT 1 z 391 1 1 sT z 1sT z Substituting 3 o jQ 1 1 2 1 Z and z 1o1r2tr 1 GT rm 1 I 1 oT2 m2 lf olt0 then 1 csT2 gt1zgtlz2 lt1andzllt1 LVJ LH spl39ane Condition 391 is this satisfied j Stable causal analog filters will map to stable causal digital filters Check for Condition 2 When 00 so that sj 2 the mapping can be written as 1 1 1 jQT 39 Z T 2 2 FOTHjm 1 19 1QT W lz z 1 w 1 QZT s plane does not map to unit circle in zplane 2 1 unless 90 gt Condition 2 riotsatisfied in axis in 10 T467667 Introducticm to Digital Signal Processing Further investigate this mapping for the case of s 19 Z1j 2T 1522T2 when 20 21 1jeT 1 ET when Qze z 162 T2 162T 2 J162 T21 1 eT When QE Z w 39 162 T2 162 T2 When 900 20 When 91 39 1 j 1 1 11 2 2 quot 200 1 Q T J 9quotquot 2 Q 0 Qz e LH splane maps to inside smail circle Q l T 11


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