Question

5. (4 points) Let x1 [n] be a discrete-time signal defined as 21 [n] = 2e-n/4u[n], n e Z, and u[n] is the unit step sequence. Mark all of the true statements. (xv) x1[-n] = 2er/4uſ-n]. (xvi) O x1[n] is real valued. (xvii) O x1[n] is neither purely odd nor purely even, but it is periodic. (xviii) O x1[-n] = -2er/4u[-n]. 6. (4 points) Similarly, 22[n] = 2e-4nu[n], n e Z. Mark all of the true statements. (xix) O Pæ{rz[n] + 22[-n]} = limn–400 2N+I =-n|22[n]2 +00. (xx) O Poo{x2[n]} = lim N + 2N+1 ER2-\12[n]|2 = 0. (xxi) O u[n] = u[mnfor m > 0, m € Z>0, m is a positive integer. (xxii) O X2[n] = x1(16n). 7. (5 points) Let 23[n] be a discrete-time signal defined as x3[n] = ejn/32 + e-in/32, n e Z. Recall the inverse Euler's relations 2 cos(0) = pj@ +e-j@ and 2j sin(() = ejo – e-jº. Mark all of the true statements. (xxiii) O x3[n] is neither even nor odd. (xxiv) O Ex{13[n]} + . (xxv) O 0<P{<3[n]}< . (xxvi) O x3[n] is real-valued. (xxvii) x3[n] is periodic. 8. (9 points) Plot and label yı[n] = (ej tzn – e-stä")u[n] for n € (-6, 6] on the left grid and y2[n] = y1 (3n), for n € [-6,6], on the right grid.

Answer 1

For all the question, we should remember some points.

If in a signal x(n),

1. If x(-n)= x(n), signal is even.

2. If x(-n)= -x(n), signal is odd.

3. If both 1 and 2 doesn't satisfy. It is neither even nor odd.

Power signal and Energy Signal: If a signal is periodic and it's any value is not infinity. then it is power signal, the Energy of power signal is Infinity and the Power of Energy signal is Zero.

Answers:

For Question 5,6 and 7. True statements are XV, XVI, XVIII, XIX, XX, XXI, XXII, XXIV, XXV, XXVI, XXVII.

Explanations:

( NOW, Givea Zin): 26/4.4in) 210) Qxe-04(0)- 2 R10 - 2014 Now Plotting this on graph we get - 2 évy - (FOI) п (A Now & En] = 2 é (-/ 4 (-2) 1 he/4 -(-)) So this statement in TRUE. (XV) Since (-o to So in so a Hence n is alwas a integen lying o un should be 1, ihn (0, 0) xin) is also a real valued siner. statement is TRUE. (xvii) For checking even and odd Sional, if xn)-TED) then slonal it even. and if x-n) - 2n) then sional inodd. mere, 2m): de 94 ) . Now, e, no 2014u (n) .. and -x, 1nf -2eny (n) – n. (1 Now, here x(n) = xiton) [from ear ( 17 andir) and xen) & nino. Horse Signal is neishu add nor even.

for checking periodicity of the tirrel. x(n) - rint) means sionce repeats itself afty some period but from Fis(1) we can see that sinal alwas decreasing in nature and terras to Cero at in hinily. So 'It is NOT PERIODIC. (6) Given, x 10) = 2e in un). (xix) xn1-n) = 20 noul-m! PER (7) + X2(-))} = power of reason) + Allung]. plotting xan) + 22-1), we get (7) Un) + (-)1: (*) 72042,264 get 12 n even so we can see that sinal w n nahre. means, leondl= |as(n)). Now, for linding its power, in - Efeinst 2 = lim 2Ntl n=-N lim i 20 2 E (zin] NAN No No a nence h statement Rus. ix TRUE. (xx) is x2 (n). we plot only we gent, en rée siout we cannot so its not a periodic sind is power.

(xxi tin un) is a unit impuse tanchon.. its value is I for no 0, 3, 2, ß an n- o! 1 2 3 A 5 now, for olmn) , m>o. let m=2, 4120) Ulan Since we can see the any then tu 2 3 4 na . we can see that it we multiply integer with n (should be positive) we can see same valu one her signal that will be exact like. umi, = positive integer, So, und= almn) for moom Stectement is, TRUE. (xxii) since 2, 10) = 2ê hunt h Now, x, (16n) = 2 é (nx16) 4 (160) 2e Arullon) .:4(160) uin), hence, x, 16h): 2e- anun. T erin), So, the statement in TRUE.

(7) Given, rg (0) = n/2 t eine nez 10, eta Sino = Coso = eie teje 2 (xxiii) Ecoleo 25 then, robin = 2.000 n a 23(-2)= acos (-2). [since coo econd 200ņC : So x2 (n)= xatns.. signal is even .. er hence statement ix FALSE. Gore (xxiv) since signal is periodic with, (641). 32. Be 1/3 ro im (6471). so its power slonalt hence wirts energy should be nence statement on TRUE intinite. - lxxv) since it in periodic and power Signal, its power should be finite. hence of Poolag (n)) <0. NuGH Statemel is correct exvi) yes, it is real valued because for evey value of n we get olarnik) that 14 real. of xxvii) since it is (az) in function cosine, so it in periodic with a period od 64TT. Con que xxiv explained]

For the Question number 8,

We should know that sin x= [e(jx)-e(-jx)]/2j;

so here y1(n) = Sin(Pi*n/6)u(n).

and y2(n) = Sin(Pi*n/2)u(n). { since, u(mn)= u(n), if m is possitive integer).

Now plotting these on graph we can get,

시와 1 1 | 당 당 농 농앉 | Now, f , M (BN) = sin (I x3n) un) hoy sin chemin jum p에) 45- 43 -2 - idix3 4 sim