Question

2. Using a z-transform table, show that a) 2k+14[k – 1] +ek-[k] 2+ ane) b) kyku[k – 1] Hint: Express 1 k-1) in terms of u[k]. c) (2-cos(k)]u[k –1] 2-0.52+0.25 d) k(k-1) (k - 2)24-34[k - m e for m=0,1,2 or 3

Answer 1

Summary- This problem is solved by using the basic concepts of Z-transform. Below are the images uploaded in sequence.

z-transform Solution - Using the Basics of z-transform alk) uzT) XCZ) X (Z) = a(k) zrk KE- some important properties and formulas of z-transform Z -a ak uCk] a ark] . AEK-1 a za 2 , 2 +21 z ko x (Z). Urk-1 Z * 2-1 *Z & (k-ko) u Kuck a – zd X (Z) dZ Kakukk az (1-02-12 - Coswoz 1-2 cow z tz Coswo k uk] sin wok u{k] a 2 Sin Nort 1-2coswort 2²

2x+'u (k-1] te tk-1 ulk] 2k+1 -1 + u(k-1] + ek xet ulk] sku(K-1). 22 & & ek u(k] ek ufk] < -> [atuloy 21 2k- u (k-1) 2 + 2 (: shifting] Z -2 elz- 6 kgk u(k-1) k g K-141 UCK-1) {. By supplying +18-1 at the power (k-1+1) 8K-H'u (K-1) (K-1) z klu (k-1) trk-1 u (K-1) Cor) k. r. rk-lu(K-1) * ulk) u z z-8 oklu (k-1) 22

K. ok- ulk-1) - zd dz 2-8 -2 (2-2) 2 (-1) de - z (2-82 80, the Z-transform for K. 8. ok-lu(K-1) rz rz-82 © [ 2* cos II K UCK-10) let Tr = Wo. ta {cos (A+B) = COSACOS B-SMASMBZ Cos IT & UCK-1) 3 24 cos wo (k-1+1) UCK-1) 2 - (cos Wa CK-1). Coswo - sin Wolk-1) sin wo) & u (K-1) UCK-1) - Sinwolk-1) Sinuo UK-1)/ 2k | coswolk-1) cost Coswok & I - Coswoz-1 1-2 Coswo zl +2=2 (1-Coswozt ) z 1 ((-210swozt + z 2) Cas Wolk-1) =

30 As we know that &* cos wok usk] & I-a coswoz- P 1-22coswo 2-1 + 2 2-2 30 (1)k cos Wo (K-1) Coswo alk-1) (1) K-141 Coswolk-1) costo u (k-1) K-1 coswo(k-1) U (K-1) coswo XI Z-Transform will be 80 (1 - 1 cos Wo Z-1) z 4 x Coswoo T-284 coswo 2-1 +() 22 Si put cos wo = cos TI = cos 180 - Cosso = 1 / 1 I- Cast. Z 1+ 1 z 4

Now, for & 2-k sin wo (K-1) U(K-1) sinub. (1 k-lt sin wo (k-1) ulk-1) sinwo. (ksin wo (K-1) ulk-1). sinw x 1 z-transform. [. ak sin wok ulk) & 1 Lsm wo el (1-2005 wo alto Now, zo transform. - I sin wo zlxz-1 I sing . 1 - 2XLX12'++ 80, on Adding Both the a-transforms we get. 1 - Z7 & 1 7-2 4 0.25 (2-1) 2² - 0.52 +0.25

d. K(K-1)(x-2) 2K-3 ulk-m] for m=0 KCK-1) (K-2) 2-3 u [k] K(K²-3k +2) 2K-3 ulk] (k²_34²+26) 2K-3 u [k] K² 2k-3 u (k] – 36²2K-3 ulk] +2k 2-3 u(x] Kanuck) # xz1 (1-02-172 Kauf < -24 14 21 -Zd. T dz 11-ą)? Q T dz 2 d T20 dz (462037 203-3)**] - zla (2-0) – 2247 I (2-0)3

k² & usk] < – 2(22-2²-220) (2-x) 3 z (x² + 2x) (2-4)3 kausk] « – zdj z ( 2² + 2d ) | -2[co besoin 3(2xz ce+) 1 (2-0)6 af (x+222) – 32 (2²+ 2x) L (2-2) 4 -2 +224 – 3762—37367 (Z-a) 4 KB 2X X FUCK) - 3K? =* XUCH) + 2 *K2KW(K) 8 l 2-frame for PER 12-30)-72) [ 03] on simpily fing this we get - 62 (3-2) 4