Question

QUESTION 2 (20 MARKS) (a) A continuous time signal x(t) = 3e2tu(-t) is an input to a Linear Time Invariant system of which the impulse response h(t) is shown as h(t) = { .. 12, -osts-2 elsewhere Compute the output y(t) of the system above using convolution in time domain for all values of time t. [8 marks) (b) The impulse response h[n] of an LTI system is given as a[n] = 4(0.6)”u[n] Determine if the system is stable. [3 marks) (c) Given a causal LTI system described by y[n] – vln – 1] + X «In – 2] = 2x[n – 1] Determine the impulse response h[n] of this system. You are NOT ALLOWED to use any transform methods (assume initial rest). [9 marks)

Answer 1

لن Given 2+ 201+ = 3€ (-+) and hit) 2 -oots-2 - 244-2 otherwise he Knowe that output reci'll be g(t) = htact) - Se h(t) alt-t) dt 20 2(t-1) f 241-t-2) 3e if- d+T)'d t Ef 2 2/ t-t) 2.3e dł t 2 2 et-zł at r- zł = 6 e dł S6 e e -2 A glt) ge : नया-2) re ēxt] y lt) Betleet & 2 (4+2) din y (t) = 3-ge 27 be - 2

(2) 6) Given Impulse Response of ahhts system is n h {n} = 4/006 ulo) condition for stability :- If impulse response is absolutely summable S ई/01 र il; a 200 n=coo SE 4.(0.6) u{n n N . 4 $ (016 neo 4 x 1 1-0.6 sto Satisfies the Condition for stability TR

so system is stable © y[n] - 4 + 3_ y[n-2) = 2X[ns] -0 2 We then 2 know that that if if a[n] = {[n] output of y (n] = b[n] Impulse Response from egno. put x[n] = S[n] by [n] = h En] so, = b[2]=i[n-1]+[n-1= 2 3/2-1] 5[n] - h[n 1] -3 h (n-2] +28(2-1] ZD Given that system is causal, so, * hen] =.o f n=-1, -2, -3; and so 1 for n=o for all value of n

to. [n] - [n] 3 hn-21 +28Cn-17 20 for M=0, 10JE 오 네 3 V-2) +28-1 =0+O4O 5 hj= for, n= J E 도시이 3 HJ+ 2810) - 20 오x0.3 kp + 2xl는 2 5 220 Al=2 foy. n=2 *h127 = 월-M0J+280T=_x2-0to 20 h12)=요 16 5 for n=3, hE 농1-옮T+2812) 20 32 T 놓 X solfo 3 x2 7 0 - 20 5 ha] - 0 - 98 50, hIn= 0, 2, 1.6, 2.98,