Question

Q1-20 points) a) Find the transform of the following signals and plot the ROC. 1 x(n)=(-0,357u(n-4)+(0.25?u(n+2) IL- x(n)=-cu-n-1) b) Find the Inverse Z-transform of: z(2-1.5) for (2-0.33)(2-0.5) ROC: z>0.5

Answer 1

Solution - I) 200))= (-0.35)" 4cm-47 +0.25) 4 (n+2) Taking z-toms poem on the both sides weget X(z) = Ż4 3 [(-0.35}"4] +2° Z ((0.25)**] 23 + 23 Z X(z) = 24 2 +3? z+0.35 Z-0.25 Z+0.35 2-0.25 and SULL h = 4 h = -2 ; 4197-4) un+2) quini Right handed Causel fignas → Roc z=20:35 left handed Cousel poles. Signal- → Roc of X(z) containt the combination of the both left honded ond light honded Signal. Hence Roc does not content omny poles. 1317 -0.35 omd (2) >0.25 2= +0:25 F0.3s</21 20.25 Ans.

X(n) we con 1) Ans! -an u(-ne) By applying properties of z-toomsfosm write z Frooms form of the signal xm is given as X(2)= where Roc (2)</al - 1-937 b) Lowense 2 - Toomsoosm of F(2)= 2(2-es) (2-0-33) (2-0-5) & Roc =12170.5 B f122 (2-4.5) A t - Z Z-0.33 2-08 -0:17 (Z-0-33) (205) (Ź-1:5) 0-337-1-17 A= lim (2-0.33) (2-0-33) (2-05) A- (0.33-1-5)/-0.14 -6.882 Z 0.33 (-1) 0:12 = -2.94 (2-105) 2-0-33)(2-0-3) B- lim (2-0.5) Z+0.5 B=5088 Hence f(2)_ 2-105 6-882 5088 2 (2-0-33) (2-0.5) 2-0.33 .2-0.5 f(2)=6.8822 5.882 2-0.33 2-0.5

f(z) 6.8822 5.882 Z-0.33 2-0 Tacing invesse z loomsbosm (with Roc (zbot Hence 6-8822 would equal to zero, Z-0:33 fm) - 5-88 (0's)Mu(m Angs