Question

Question 4. (20 points) Compute the DTFT of the discrete-time signals, 1) x[n] = n(0.5)"u[n]. (opt) 2) x[n] = n(0.5)”cos(4n)u[n]. (opt) 3) x[n] = (0.5)" cos(4n)u[n]. (7pt)

Answer 1

o 1 x [n] = n(0.5 Mucol The DIFT of x[n] is X(W) = ²n (0.5)n ejun n3-00 = n(ouseowth The above expression is an arithmetic geometric programing for an infinite length the sum of arithmetic geometric progression is where a = initial term r=common ratio d = difference n (0.5 e solo so FO Since the first term is zero so Consider *wan Coben Хр(о<be x (w) = : 0.5 esw lo. sojāt +0.5euw - (-asgow)

2] x [n] = n (OS) cos(4n) u [n] @ rcoso - edº tego X[n] = 760.5" (e in the sky] an (o.si esant poen] an (o.5. 219)? +65.2:4yn {n.co.Setya x n coseto ] for n (o. Sapago DTFT of n.co.5x pj4jn is 7,0) = în Coistely gwn a in corso slewyn - - 3(4-0) nad The above expression is an arithmetic Progression for an infinite length the sum of arithmetic geometric Progression is Q = initial term + Tre r= common ratio d= difference

Coste(4-0) ) x(w) e n (0.5 x picu-w) yn 0.5 e5(4-w) 0.5 €ö€4-w) 1-0.5 21(4-w) (1-0.5 Quic Grass? for noise 349 1 eet 0.5 e 3 (4-0) X(W) = Pos zö(4-w) * (T-0. seolarci)! . 2. 1-O . xW)(xy (60) + x2 (w) ] fois eu (4-6) (1-005 poches to +0.58(4-0) +0.568(40) 1-0.5 06(4-0) 12 1-0. 5cm 4.0.5 244-w) 1-0.5 @jcrat?

3 x [n] = (0.5) cos un u(n) to Cose - eso tēso _ o compare ea o & @ 4n = coso i-e 0= 40 Cosun= esante 340 2 substitute ea ☺ in ea o x(0) = (0.5lb cosan a(n) - toissi porta ques ) . te a 2 fons en su (os con lucens It is in the form of a una Here a = Cong pua)"+ Co. gejala carn) .. we know that DT FT of alcu(n) y 6 i-arvo

6 ea for o ea apply so + | و = (دی) x منی ( مانع 6 . 5 تم ( ع .6 - 1)" - مانم لماذج 0 . 5-1 ا لماذم ال كما في 0 . 5 -| |