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# INTRO TO DIG SIG PRO E C E 467

Clemson

GPA 3.84

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This 8 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 82 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 LECTURE10 Finding residue at a pole of order gt 1 Examgle Fz has pole of order 4 atfzpo b b b b Thean az p 1 2 3 4 quot 2 ltzpor z pogt3 2 por for O lt 2 p0 lt r1 due to the existence of Laurent Series Define ltIgtz z p04 Fz DZ Zanz pon4 quot39b1zquotpo3 b z po2 b32quotpob4 n0 1912 n 4X2 p0 gt63 3b1z po2 2b2 2 quot p0 ha I Zan n0 2 Zan n0 n4n3z p0n2 32b1zp02b2 13quot2 i an n 4n 3n 2z p0n1 3 2b1 n0 lt1quotz i 2b d b Im Szggoq Z 3 1 an 1 290 In general the residuebl of Fz at a pole of order M at z p0 is 1 dM M r m F M 1iz391gtrgodz T JZ p 2 V z ECE T467 667 Introduction to Digital Signal Processing Examgle Fz l 2 12z 2 Res of Fz at 22 limz 2Fz 1 1 z2 2 12 Res of Fz at 21 um umW nm391 1 z ndz 2 2 H1 2 2 2412 2 Now use above method toevaluate inverse z transform Given Fz with ROCz gt 2 212Z 2 2 rightsided signal 1 Zn 1 12 27 W2 2 12z 2 and enclose orlgln 1 11 1 z 512712 Resgm at poles mSIde C 2 gt 2 For n 21 only poles inside C are at 21 22 Therefore for n 21 01 Zn 1 at 21 Res at 22 z 1gt2ltz 2 pow2 fn Res ECE T467 667 Introduction to Digital Signal Processing At 211 n 1 n2 n 1 mdz quotmz 2n 1z z 1 z gt1 dz 2 2 2 1 Z 22 lt 1ltn 1gtlt11 n 39quot 1 A1222 n 1 lim 2 2 2 1 z gt2 z Therefore for n 21 fn 2 quot n For nlt1 recall Fz Lf 222 t 1gf0ig 1 2 g 22 J Since ROC lslzl gt 2 all For large 2 as z gt oo fn must be 0 for nlt0 these terms 9 0 otherwise these terms would blow up a z gt co gt imam f0 umvo 2 2 1 z2 Overall result fn 2 1 njum 1 ECE T467 667 Introduction to Digital Signal Processing Notethat f12 1O f221 2O f32231 f423 44 etc in general if the ROC of Fz has the form lzl gt a Then fn must be 0 for all nltO and f0 lim Fz Zv Also If the ROC of Fz has the form al lt z lt oo then fn 0 for a finite no of negative n values and lim Fz co z wo Examgie 4 Let F1z Fz 24 Z 21222 New R00 2 lt z lt co Recall 24Fz lt gt fn 4 gt f1n fn 4 gt f1 l f3 1 4 z and llm F1z 7 gt oo Z o v Z I ECE T467 667 Introduction to Digital Signal Processing Examgle z Fz R0012 lt la 2 a n I fnj Z z 1 1 2 dzquot 2174 C z a 2njCz a 144 For n 2 0 no poles inside C gt fn O for n 2 O For nlt0 poles QLQ present inside C at 20 2 fn 22 O for nlt0 f 1 Re 3 at 20 zz a 1 1 llm z aOZ a a 1 f 2Res 2 at 20 2 z a d 1 1 1 Im Im z gtOdz Z a z gt0 2 432 a2 Simiiarly f 3 1 a39nd 4 I a x a 5 39 ECE T 467 667 Introduction to Digital Signal Processing General method for nding fn when Fz has a ROC of the form lzl lt la Start with fn Fzzquotquot dz n zltr1 Let p gtzl and dz1 2 dp Z p P Let 2 m on C with rltr1 2 p 19 5 on the new contour C in pplane r As angle of 2 on C goes from 0 a 2n counterclockwise angle of p on 0 goes from 27 gt O clockwise n1 1 1 1 1 3111 33 quot F quot H 27 91110 1 DZ 3 lt r1 Reversing directions of integral cancels the sign 1 1 fn F Ep 1dp E Useful when Fz has 3 ROC J 1 inside a circle or between two circles 391 of finite radius l gt fn ZRes Fp quot1 at poles inside 039 pl gtrl where ROC of 1 Fz is z lt M ECE T467 667 Introduction to Digital Signal Processing Apply to previous example Fz E E a ROC l2 lt Ia p39quot 1 at poles inside 0 kv J I No poles inside 6 for nlt1 1 pole at inside 039 a For nlt1 1 1 1 fn Ilm quotquot1 t plp a1 app a a 1 p ajp 1 n1 n a im1 a ns1 f T m E p a a a D We already showed that xn O for n 2 O by working in the z plane for the previous Fz and the same ROC ECE T 467 667 Introduction to Digital Signal Processing Overall result 39 fn a u n 1 B for n gt 1 Go back to general form of Fz Fz f2z2 f 1z fO igltgl all fn must be 0 for n 21 Since ROC includes 20 f0 mm 0

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