Question

A discrete-time LTI system with a cumpling frequency of 8kHz is shown in Figure 4.1. The rectangular boxes with the label z provide one sample period delay to their input signals. The circular components are adders or subtractors. The triangular components provide linear gain factors of or by, where i is 0, 1 or 2 a) Derive the system transfer function H(z). b) Find the difference equation relating the output y[n) and input x[n). c) Given that the gain values are: 1,-1, ,--2/12, b. - 1 4--18/12, 4,-0.92 Find the poles and zeros of this system. d) Sketch the frequency response of this system. 4 A Figure THE END

Answer 1

From the figure above; x, [n] = x[n] + a₂ [n-2 ] - 4_IM) = , En] + 9, 42 [n-1] - 2- Transform of Gi); X2(+) = x,(z) + a, z' X(7) Z » X2(2) = X, (2) in) 7- Staroform of (i); X; (3) = X(z) + az X2 (2) X(z) + azt X, (Z) Scanned with (Frem egen in l-a, E'

-a, z x,27) = (-1-4, 7) x(2) ) y [nj = bo. *2 [n] + b, 2₂ [n-1]+ b2 22 [n-2] Iking 7- Tearsform; Y(7)= b. X2 (2) + 6, 7" Xa(z)+ bę z ? X2 (7) (bo + b 2 + b2 22 ) X2 (7) IN [From Ov) wa) (a type ') xo) - Sed

a) Systen tranfer function H1Z) = x 12 H(Z) = but b, 2' + b2 2 2 t-a, zl - azz-2 B) Y(2) = HL) X(7) 9 (1-a, 7'- 0 2-2 ) (7) = (6+'46,72)x(2) Y(z) – a, z-" V(z) – az ? Y12) = b, XC7)+6,7%) 1 tb. z 2 X(2) Taking inverse z- Transform; y[n] = a, y In-] 234142] bax[n] +b, 2 [m-nl] + b2 x [n-2] Equation (vi) is the difference equation relating y [n]. & x[n] Scanned with CamScanner

a= -18/52 , az=0.92 . (7) = 1 - 27' + z-? Itu z' -0.97+? Zeros of this system are given by the solutions of the equation 1 -2 Pobs of this system are giver by the solutions of the equation: > z’+ 19 2-0.9=0 => z= -12 V ) + 480"9 X09 12 - 0.506, -1.779

:: Zelor of the system = + (1+j), F (17) Poles ..... = 0.506, -1779 it te Z-'-0"97-2 for getting frequency response, z= eit (st digital frequency CV .: Hrei) = 1-2 ejor + ejar 1+ 18 errogejad ) ile evle-0.96MY Now, e a cose + jsin 2 a Heim) = 1 - 52 (cosch-jsin) + cos (22) -j sin (2) 1+ 18 (cose- j sum) 0.9 (cosart-jsur 2) + 18 = (1-v2cost2 + ca 212 ) +j (18 sin. 2 - sis 28) (1 + Is cose -09 022) + j (-18 sinh +009 since CS Scanned with CamScanne

Thejm) = (-12 con 0 t co 2-2) + (pe cin 2 - singh) VC+1- 091603202 + ( a la sin D +0°9 sin zgje +18 co 12+ cos2 (2 In 3-0, (Hein) = 0.426 / (from enterian (Hein) Their is symmetric about origir as it is an even function Hej2) LH leder) > Odd function Lule) → 0° as 2 0 & a