Question

3.(10 points) Two linear time invariant (L TI) systems with impulse response hy(k) and h2(k) are connected in cascade as shown in Figure 1. Let x(k) be the input, yı(k) be the output of the first LTI, and yz(k) be the output of the second LTI. Let h;(k) = k(0.5)* u(k), hz(k)= ku(k), and x(k) = (0.7)* u(k). Use z-transform to (a) find yı(k). (b) find y2(k). x(k) yi(k) h;(k) h2(k) ya(k)

Answer 1

8.3). two linear time ervaricent (LTI) systems with impulse response holk) and halk) are connected in cascade as a(1) Y(K) halk) Y(K) hick) • a given values hilk) = * (.5julk) 3 halk)= Kulk) and alk) = (67) Kulk) we know Vick ) =alk) ☆hilk) Convolution in time domain is multiplication in z-domain or frequency domain ? Y, CZ) = x(2).Hi(z) z-transform of K (a)kulk) - we know = azt (1-azulya And (akulk) has L z-transtorm i-azt - Hicz)= .sz (1--52-1) And X(z) = 1 + (7) 2 :! Y, 12) os 1-1.772" 4 { isal |-.52-1)2 have to we To find in verse z-transform use partial fraction

Y,(7) - P + !-(-7) Z' Qa't R 1-05 z1j2 .si p(1--52"}}+60='+R )(1--7€") (1-072-1) (1 -.5214 (1-0727) (1-15 21/2 .sé! = P(-252*+1-2')+Q2--7822+R--PRE! Compare coefficient of 21, 2² and constant .25P-:70 =0 B .25p -(i) 07 PAR=0 RE-P-lii) -p+8.7R = 5 -ci il Put Eq. (i) and eg. (ii) in Eq. (iii) -p+ 25 P to7p =.5 "P=8.75 8 = 3.125 and R=-8075 Y, 12)- 8.75 + 3.125z_8.75 (1.521/2 1-.7z1 How Evaluate Equation in such form 8.75 •5zzi ) Y, (Z) - 8.75 |-.721 +3.1251.57) .5 (l-isz-1)2) - (1--Sz-!? 07 Y, (2) take inverse Z-transform :: V,(K)= 8.75 1-7)*ulk) +6-25 kl-s}uCK)-14-5 (k+Jes)* ückt!)

zt 2 2را 2 - )) or •szt (4--221) el (1-z-1)2 (28+R? (1-7*2) (1-2)2 b) halk)=kulk) Z-trana form of halk) is H2(Z) = we know Y₂(k) Y2(k)=Yock) *hak) Y2(2)= Y; (2). H2L2) Y212) 11-157-1)? partial tractions for enverse Z-transform (1-oselja . Ti-Z1j2 :.sz? = All--se")*(1-z"){+(8='tr )11-177")(1+z")? +(52'+T}(1--72"}(1--57" for (P) put a = t = 10 - (ii) hl=+=2 + UztR tszitt -1 Y₂(z ) = A To73 - (1) put kl 5 -5(40)2= P(1--5x19)*(1-1012 'P=68.05 for I and R ::29+RS-s -(lii s and T put =(5+1)(1--7) (12:5)? S+T = 6.667 -civ) Now put žta in Eq. (i) we get (v) 2 =) For 5 . PERE T= put z = 3 in Ceq.(i) - 44.1 2-9.09 p + 128 +4R +755 +.257 (vi)

from eg. (VI and (ii) R+T=-68.05 and Ta-68.05-R - (vi) 28+ s =-5+6.67 +68.05 S = 69.717-28 2.-4.13-9,09868.05 +128 +46-5-28) +-45(69.717-28) +o25 (-68.05 +5+208) il-4.1 = 38-602 :: Q = 199.3 --403-6 R =-5-2x18 s = 69.717 -28 =-329 T = (-68.05 +5 +208) = 335055 + (-329)2' +335-55 (1-212 + 199.32-40316 [1-521)2 ''Y2 () 68.05 |--72-1 simpler form as Evaluate this Equation in 403.6 | .szz (1--SZ? 10:25:02) – 403-4(+522" + 199.3/1521 5 Yz1Z) -68.05 l-etzt - 32gz + 335-55(72') T1-71)2 (1-2172 Take inverse z-transform Yield Y(k)= 68.05(-7)ku(k) + 398.6K (5)* ulk) -807-2 (k+1)(-*(K+1)-329 kulk) +335.55(k+1) u(k+)