Question

2. Short-answer questions on various topics (20 marks total) In each case, clearly explain the reasoning behind your answer. a. (3 marks) A causal LTI system has an impulse response h[n], having an even component he[n] and an odd component ho[n]. The portion of he[n] for n > 0 is he[n] = 0.5(0.32)", n > 0. Determine the complete description of he[n], ho[n], and h[n]. b. (3 marks) A system has the transfer function z– 7z+6 H(2) = 22 -0.12 -0.2 (-1)(2+3)(2-2) (z - 0.5) (z +0.4) Is this system causal? Is this system stable? (2 marks) A D-T signal has the characteristic that x[n] = 0 for n <No where N, is some integer constant. The bilateral Z-Transform of this sequence is X(z) = En=-x[n]z-". What is the lowest value No can take for the Region of Convergence of X(z) to include z = ? d. (2 marks) A system's input-output relationship is described as y(t) = 3cos (6)x(t – 2). Can you or can you not use the system's impulse response to determine the output of the system as y(t) = h(t) *x(t)? e. (2 marks) A system's input-output relationship is described as y(t) = (3x(t – 2). Can you or can you not use the system's impulse response to determine the output of the system as y(t) = h(t) *x(t)? For questions f. through j. consider the system y[n] = x[n – k] f. (1 mark) Does this system have memory? g. (1 mark) Is this system invertible? h. (1 mark) Is this system causal? i. (1 mark) Is this system BIBO stable? j. (2 marks) Is this system time-invariant? k. (2 marks) Is this system linear?

Answer 1

Solutions d) Given equation is : ( not possible) | 9 ) = 2 ok (1)x(-). No we cannot we the system impule response to determine output of the systems as y Ct) n(t) x ult)" be cause we cant tend yle) = b[t) x xk) by wing former tram foron

so that's why I hate it is not or laplace possible en M e) Given that (Yes we can find emput output relation yht) = (3x (+-2)] Here the Answer is yes because we can we system impulse in y(t) = n(t) x x(t) i lets see by whing Jowner tram forms yle) = (a (4-2) a(t) cosy) = -x (4-2) 4. (W) = rejew xi (w) 4 (6) : -ei zw inverse founder transformi ht) =-816-2) for x(+-2) so, y) = (t-) . h(t) = 8(+-2) henu we can find yft=ht) x (t)

Casual) b) frörn that. ( Not - WAY = 25-72+6 342012-0-2, G-4 (243)2-2) Tz-o's) (Z +04) for a Casual function Order of be. Denominator should be greater theis numerator here . denominator = 2 numerator a3. Casual So et es not h) Eivers (it is not casual) 96) = { *(n*] k=5 ure here y(0) = x(n+8) + x (n+4) ---- +2(6-5) Therefore output is depending on part Present and future inputs that's why it is not casual

Eliven ( Yes A Er stable) yron(n++) K -5 you = n(n+s) 4u (n+y) +--- tuins) To get ning put ? (n) = s(m) and ghl= n(n) so then h (n) = ( T+2 4 8 ( n +) + ... + C (n- or h(n) = 1, VI !}. ... To be stable e in(t) = finite value na so ų n(A) = 11t . ll times=11( trite value) n= - . . n= -f hence of es stable f) Given (yes, it is having memory) K -5 . . or y 6) = a (n75-) +x (n+y) ---(-6) so output gol is depending on pant input also this why it is having memory un

g) Gliver yn = {x (n+c) (ites inventable) 40) K=.5. lt . .. @) - 5 () y) = { $(n+) it u(n)=8(n-1) so yin) = final -k) k=-5 so we will get unique output for every different inputs so it is investable

of Given. you os x (nt) ( 1168 tiene un cant) B in the vanant y(meno) > y (n_no) y (n_no) = { x(nn hot) y (n , ho) = {n(n-hork) k=-5 Time Variant ZS * Eleven 462 x {n+) by simplyting Hit Ka-5 : yi(n) + 42 (n) = 4(n) for y, (n); Put all - fx i nate) for y= (n) , put i 6) = x2 (n) os y oth > Ex=(+) kas : for y=(n) put a(n) := x1(h) + x2 (n)

so yen2 = { * (nt) +*2 (nu) kaos Y36) = Ye (G) + 4 = (um) nime at es lineas

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