Question

5. For each of the following h[n] and x[n], characterize the output of the LTI system with impulse response h[n] and input x[n]: a. h[n] = u[n] – u[n – 4] and x[n] = 2^u[n]. b. h[n] = 8[n – 3] and x[n] = en+2. c. h[n] = (-1)" u[n] and x[n] = u[n – 1].

Answer 1

149 hing you for above ata system output is ging; input rin) and impulse response hen) To find output yon) - hin) * xen) a) hin) = uin) - uin-4) and rin) = 2' uin) uin) = 1 170 =o no uln - 4) = 1 n74 na 4 UID-Wa uin) >n -> 'o 3 4 . tun-uin-4) - hin) , Uin) - uin-4) hin) = u(n) - uin-4) from aboo figure it ie :-hin can be represented as Sin) + S(1-1) +811 -2)+sin- 3 As we know sing = 1 no sn-1)= 1 ;n=1 = 0 elowhere O else

t(n) + S(n) = x(n) x in-nd * sin) = xin-nd xon) * Sinnb) = xLong) (a) ylw= hin) + xen) - Toron) + 6(n +1) + sin-23 +8(n-3) ] + ren) = n(n) + n(n+1) + 2 (0-2) + x (0-3) n - +1 uln+1) -21 uln-2) + 3 u(n-3) T 2 CT (On G + 2 +2 sob ont2 yen) - hen) + LIM) hen) = sin-3); xin] = ent yin) - sin-3) + 8 X (n) = xL 1-3) 0-3+2 = 0-1 e = e

Answer 2

(a) The impulse response of the discrete time system is, A[-]= =x[n]-s[n-1] Apply Z-transform on both sides. (z-1) "M X[2] (2-1) Y[z] 2 3 X [2] * 2-1 2-1 The input to the system is, x[n] = 2^u[n] X (2) - ; -2. Determine Y [z]. Y[:)=11 21 2-2 ***=-1)3-2) 12–183-2) [2-2 2-1) 48(2-2) 2-1 15 z - 8 z 2 Apply inverse Z-transform to obtain the output. =- »[n] = k13 2012 (b) The impulse response of the discrete time system is, [n] = m = 0[n–3] Apply Z-transform on both sides.

The input to the system is, x[n] =£**? = e’e" = 7.389e" X (2) = 7.3892 = 7.389 -2.718 Determine 9[z]. Y[z]=zº x [2] ==*(1.389 2.718) Apply inverse Z-transform to obtain the output. y[n] =7.389(2.718)** The impulse response of the discrete time system is, 1.[] = " * = (+1)*7[1] Apply Z-transform on both sides. H[z]=Y[2] 2 17. X[] 2+1 The input to the system is, x[n] =u[n–1] *(z)==" Determine Y[z]. Y[z]=z\x[z] (3+1)=-1) ** 22-1) 21:40] Apply inverse Z-transform to obtain the output. y[n] = 0.5u[n] -0.5(-1)"