Question

Given 0.2 E(z) (z - 0.2)2(z2 0.6065) a) Use Partial fraction expansion to find the inverse, e(k) b) Use the power series method of inversion to find the first 7 samples of e(k) and verify with the answer from part (a)

Answer 1

GIven O.2 E (z) = O.60G5) Le 티디) - . 2 + CzT D (Z-0-2)2 (2+0 6065) (-0-2) (z-o.2)2 (2t0.6065) Shere 22 (a.21 A=o 0.1914 240 6065 (zio 06s72 8- O 2 O3094 Z40-606 ZEO 2 Ta solve Fer canel D Lets pluq ih value of A & B then sımputy 2 CztD 30 94 -2)2 0-60o 6s) 0-60 65) (Z-o.2)2 Muttiply by (z -v-2)2 ( z40.6065) both sidlecwe ger

(5909-2) +608.0 + (5909.0+2) (7. oz)+)6)0 (zo2) (Cz+D) O. 2 Ta king z= cue have 15311. Go65 to.3094 x 1-6065 2- 64(CtD) C+D= 0 6798 ----0 Taking zeo we get 0232 + o(876 t o.04 D 0 2713 D= plugging value of D--o 2713 in ed2 O we ger S161- O19 14 o1915z- 0.23) Now O-30 94 E(z= (z-o-2) (z-0-2)2 (z+0.6065 Ta lany In Ver se Z-hansfom ( we get K-I = 10.19(4 (0.2) uc .54t Ch-1) (0.2) uik eik) +So1915 Sin (.478t) 378 780 (23tbo) s 2) z023. Z Sin (0'7782) ロ478 (9778zto96 Reumangeng term cohare Cos o ,k-I 547 (k- ek) 1914 (or2) K-1 SinAO .246) (0.778)^gincn) (0-478) Sin() G 2.46 foruta ase a ucn-!) bz Sinta = bh sincan) z2-2Cosa ( bz)+ b1 bsinca) 7-2C0s a Cbz sinta (n)

O 2 E (z) = (z2)2.60 65) (2) 3 20606S) zo42t o0 4 O 2 E () to6465 Z.0-47 o.242G Z 0.0 24 3 4 2 Z E (z) we nave Inverse 7-haunsfovm Talng e (k) O22 (K-4)- o982(k5) + urlo 2( K-61 o 042 (eF)4 veify 02461 (077) 'sin 1914(u. 2) + o To elk=1) using part a) elk-1) 1914 o~19 (4 - e (k-) ohich hold ue foun part (b) a lso elk)kel as (-elk)= o2Z ( K-4) -g.082(k-s o102(K-6)+ e od 2 (k- all existfor k74 fer kea) elkeT)zo EDMI NOTE 6 PR

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