Question

3. Problem 3 [Impulse). One of the most important digital signals is the so-called unit impulse sequence, which is a discrete time function whose sample is equal to zero for all values of the time index n except n = 0, where it has unity value, that is, Ji, n=0, 8[n] = 0, n+0. Page 6 (a) (2 points) Draw a graphical representation of the signal 8[n]. The horizontal axis should indicate the time-index value n = ..., -2,-1,0,1,2,... and the vertical axis the corresponding signal value. Also, plot the delayed signal 84[n] = b[n - 4). Unit impulse sequences are useful because any discrete signal can be written as a weighted lin- ear combination of shifted impulse sequences. The time-shifting operation is used to shift the sequence [n] by a fixed integer value l € Z to generate another sequence de[n] defined as follows: de[n] = b[n- 4, Vn e Z. The following three parts show the claim above. Let s[n] be an arbitrary discrete-time signal. (b) (2 points) Prove that s[n]8[n - 4 = s[(8[n - 4, for all integers n e Z and l € Z.

Answer 1

Pooblem 3 (Impulse) @ Sin] - sen] J 2, no 1on S[n] -3 -2 -1 0 1 2 3 4 Su [n] = 8{n-4)= { 1, n=4 To nth PS4Ln] -2 -1 0 1 2 3 4 5

) S[n] st.) = s(e) sin son) NW o 1 2 3 4 5 6 sm-) 1 --- 1 n 3 -2 -1 of - 1 2 3 4 5 6 9m) sm-2)= s[2] S(1-2) --- -2 - 1 0 1 2 3 4 These toe, sm) [n-e) = S(I) sin-e] sin - Estry Sinl] = se sing] lez DEZ - sm --. +8{n+20+ 8(n+1) + S[m] +8[n-]+${n-2) + --- han nh(n) = 1 & NEZ -3 -2 -1 0 1 2 3 4 The defose, sin sin hen) = sin)

sen) S[n-e] = s[l] s[n-l] #nez & lez 3(- Es[4]8[ne] Let S[n] be on cobifacay discoele-time signal lez sisset) st 4 -3 -- 2 123456h - 1 senje sin+2) +28 (1+1)+38[1]+ 41811-1]+38[n-2)+28(n-3]+8[n-4 - S(-2)(n-(-2) + S(-2)(n-(-2)] + S(0) S(1-0]+S(1)&(n-1) +S(2) S(1-2) +53) S(n-3) +SCUSCH-4 Therefcoe s slejscn-e) - s sinsin-] Sin]: a e EZ lez

1 nyo OM o nco non 4 -3 -2 - 1 2 3 1 0 nu. {n}=W[-] -4 -3 -2 -1 0 1 2 3 4 min- u[m]-win-5) -2 -1 0 1 2 3 4 5 6 puin-5) -2 2 1 2 3 1 ton 4 5 5 -2 -1 0 1 2 3 4 5 6 7