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# INTRO TO DIG SIG PRO E C E 467

Clemson

GPA 3.84

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This 12 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 20 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 39 LECTURE 11 EXAMPLE 1 HI 1 ltz Exz with ROC l lt zIlt 2 zquot dz tun L 4 1 ch1ltzlt z 1X2 9 2 2 2 2 1 2 39 Res of 271139 2 m at poles inside C z z For n21 only pole inside C is at z Therefore for n21 ECE 39l 467 667 Introduction to Digital Signal Processing For nlt1 use pplane approach 1 1 fn F D39quot dp and p l L 3 E H 2 2 ltlp lt 2 3 p k w J nk 394 lt 2 n1 P 1113 p 2 p 2 n1 P ZResF Ej 1 2p j WW poie at pole at p2 2 gt fn 2 Res at poles inside 0 223 lt ipl lt 2 3 For n51 only pole inside C is at Therefore for n51 wn1 3 1 fn 210 31g1g23pj p ECE T467 667 Introduction to Digital Signal Processing zj n J n quot39 quot n1 2 n S1 1 9 3 2 2 Combining results ltngtlnquot m lmm 1snsoo oeSnSO 001m 9 2 ODIN Note We could also written the combined answer as n i n1 fn un 2 u n 1 2SnSon SnS1 Other methods for finding the Inverse ztransform Long Division Partia Fraction Expansion Long Division Consider Xz RoC z gt a z a 2 2 3 1az a z a32 zalz z a a a a22 1 a 39z391 azz 1 aaz392 aazi2 ECE T467667 Introduction to Digital Signal Processing Since Xz 2 xnz n we see that x01 x1a x2a2 x3a3 n We see that the qeneral result is xn aquotun Now consider Xz 2 with R00 l4 ltla Due to shape of ROC xn must be leftsided Try to perform long division in a different manner gt x 1 i x 2 13 x 3 13 In general xn a u n 1 ECE T467667 Introduction to Digital Signal Processing EXAMPLE 1 28 I xz1z8 28 1 ROC zgt1 1 2quot8 z3916 z 24 z8 1 z8 z8 1 1 128 28 2 8 16 16 1 224 1n8mm012 xn 16 124 11 0 othern X EXAMPLE 8 Xz Z 8 1 ROC 2 lt1 8 16 24 Z Z Z 28 Z16 x8 1 Z16 x 16 1 216 224 X 24 1 224 1 n8mm123 gt xn 0 other n ECE l 467 667 Introduction to Digital Signal Processing Partial Fraction Expansion Method Examgle Xz 721 R00 z gt1 32 42 1 z 22 Factor as Xz 1 z 1 1 Z l 3 ROC all z 2l gt1 zl gt 15 gt xn 5n un 2n un General Method of Partial Fraction Exgansion To be seen Factorization of Xz gig yields a sum of terms of the form SEEP 9 when order of Nz lt order of Dz However typical 2 a 39 table entries Table 22 have terms of the form Z z ak ie numerator 2 instead of numerator const Therefore the following algorithm is used Xz a0 a1za222 aNzN 1 Form39 A 2 M 2 b0 jrb1zb2z sz 2 Skip if NltM Perform long division until remainder polynomial has order lt M ECE T467 667 Introduction to Digital Signal Processing Examgle A 2 2 1 Xz ltzltoo z z v J 4 8 nos Stegi Xz 23z Z 22 z 4 Stegz 3 2I 22 zlz3 2 4 8 23 221 z i E22Zzlt order2 4 8 3 2 9 3 2 2 4 16 32 23 z 3 e order1 16 32 Therefore we can write XZ 3 23Z3 Z1L 2 Z 4 22 Z1 4 8 ECE T467 667 Introduction to Digital Signal Processing Back to the general algorithm Step 3 Entering step 3 we can write XZ N M N M i 39 T quot012 M 1 do d1z dM1z D0 bz szM Define Es Pz 0 if Step 2Vwas skipped EXAMPLE Assume Pz has one k1th order pole and other poles are 15 order poles Then A A A M A 912 1 kt 1 L4 11 i 2quotZr ZZr Z Zr ik1Z Z Y J k the order pole at zzr st 1 order poles at zzk1zk3 zM To find the A J for the 1 order poles Al z z z jk1 k2 M 6 22 I To find the A1 for the k h order poles k j 1 T jjagZTlt izrzrk 1 zil 1quot 2 k ZZr i ECE T467 667 Introduction to Digital Signal Processing EXAMPLE Given I z wz WZ pole of order 1 at 221 where Pz has 39 39 pole of order 3 at zb Also order of Vz lt order of Wz 4 Using results of Step 3 Pzcan be expressed as A13 A12 Au A1 2 b3 z b2 zb z a Pz A1z a 1 zl as Per 9 za Now consider z b3 Pz A13 A12z b AMP by M232 2 b3 PZ A13 2 b as per with k3 j32 1 do ltz b3quotvltz1 A 13 33 dzo ECE T467667 Introduction to Digital Signal Processing Also agile b3 z A12 A112z bgt z b Egg Some fn and b3 z A12 2 b as per with k3 i2 3 Am gillz b Mall b Finally aggkz b3 I z 2A z b 5E Some other fn And 8922 2 M31142 2A11 z b as per 9 with k3 j1 1 d2 3 An Era243 PZ Steg 4 Back to general algorithm MPY each term obtained in step 3 by z to get Xz 10 ECE T467 667 Introduction to Digital Signal Processing 39 Step 5 Find 3 for each term of Step 4 using Tables whenever possible EXAMPLE Apply steps 3 4 5 to previous Example Aquot Recall coming out of Step 2 3231 E zz 4 3s21 Z 4 z2Z 4 8 Yz 51 1 5 31 z z 2 4 From 3222 s 16 5 2 2 21 2 4 11 ECE T467 667 Introduction to Digital Signal Processing and E z3 A2 z1 Tm 16 32 4 1 16 Z 2 i 1 1 z z 4 4 Xz 5 1 17 1 gt z 1 1 z 2 Z 16 Z 2 4 Steg 4 MPY by z gtXz22 z Z 1 Z 4 2 1 1621 2 z 4 Step 5 Termbytermfinal 3 E E l quotI 12 l quot xn5n2 48n122 un 164 un 12

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