Question

Consider the difference equation y n]-(a+2) yln - 1] +2ayln - 2] = 3n] +6ax[n - 2] where a0,2 is some real constant 1. (10pt) Find the homogenous (=complementary) solution(s) to the equation 2. (10pt) Find the impulse response of the LTI system described by the equation 3. (5pt) For which values of a (if any) is the system stable?

Answer 1

Soln:- Given difference equation : 0 y[n] –(2+2) Y(n-3.+2a y tn-2] = ?[1] + Ga x(n-2] For homogenous solution, – +2) 9h-J+ 2a 9nz] o We assume the solution to be expo- nential yh(n) = 1h Substituting in egn, we get tn - (@+2) 11+ 2a 112-0 ori 19-24 12-6+2)/ +22) = 0 Roots of ²(a+2) Atza cire 12 (1+2 + (a+2)2 40X2

= (+2) } /a44+44 - 8 a . 2 X = (2+2) + ✓ a?+4-4a 2 X = (2+2) + s (0-2)2 @+2) + (2-2) 2 Tz (@+2) +62-2) (2+2) - (2-2) Az a, General solution yhan) = Cid,"+ in Yu(n) ? Cila)"+ 512M (10

From difference ern, we put ☺ no or, yo -(0+2) YED) +29 442)=0 or yoz (+2)YEV - 2a Yee) (in) similarly, putting na yo -(0+2) y®) + 2a YH) = 0 or you – Q+2) ((a+2) YEV)– 2a yer)) +2a 464 zo from cinj ori yo - (Q+2)? YED) +20 (+2) y-2) - tza 441) zo or yuz yev { (at2)+2x} + :- 4+2 ( 2a (atz)) iv

Also, putting nzo in vii) @ Yn oz :C1+5 V - putting nal Yu (z ca+26 il Equating (w with cili Ci+9 = (2+2) y EU -2 a yea Nii) Equsting & i) with aw 61a +26 = you/ a?+4 +2a) + -ye92044a) Qili viixa actas = (1+2a) YED) – 2a2 y42) solving vill) and (ix) simultaneously,

67(2-a)= yw (4) + y<2) (-42) or, az 4 y 6) - 4a 4 (2) (2-0) Putting G in ii), we get ci- you (19+2 - ) + Y-3) (-2a + 4 2 ) C 4-( 44a + Y+z) (-4/6 + 29²786 ) Giz yeu le ) + you () 10 . 12-9)

Hence the homoge nous or comple mentany solution is Yum, c, (a)"+ G. (2) Титания Flyveno + yees (o ta] an 4 4+1) - 4a yer) e (2-a) b Impulse response, Given, yin) – (0+2)yht] +2a y h2] z an] + 69 xfn-2] - Evalue ting Z transform>

YZ) -@+2)2+ Y(z) +20 2-2 yceso 3 X 2) + 6 a 2-2 X(Z) a+2) ori. Y.?) (1 - - 2 - 1 + 2q 2-2) = X(z) ( 1+6024) or year or 2 Y(z) 1 X(2) at 6a2-2 1- (2+2) 2-1+2azz = ₂² + 6a z?- (9+2)2tza = 1+ (a+2)2 +49 Z²(a +22 729 Y(Z) = 1+ (2+2)2+44 X2 (2-2) (2-a)

NA AVA Y 2 = 1 + 1 X(?) a²+60 + TZ-a)' 6a44 (2-2) HZz 1+ (@+69), (62+4) (2-a) (Z-2) hmz sn]+(a?46a) Z-W)" + Gare (6a+4)2-1(2) - Impulse response (o system is stable if all poles Jie inside unit circle and ROC includes unit circle Poles are at a and 2 271. For no value of a will the system be stable