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Q If the energy of a continuous-time signal x(t) is E and the energy of the signal 2x(2t — 1) iscE,thenc is 2% (rounded off to 1 decimal place). Co Ans: 2(y>E = fixes rae v0 , go 2x(2t-)>£F “= flax@t-dl aes 4 flxcee-a)| Ut to -» det 2t-1=T t—> -% > T—»-00 YWdtsdt t—> % 1 t—>0% dt= at 2 * 0 e= flO ae = 2 f bets) lux = 26 a é E’. cE c=u2 Q. ‘The period of the discrete-time signal x[n] described by the cquation below is N= 48 (Round off to the nearest integer). [n] =1+ 3sin(— +2) Ssi T TT. x[n] = sin{( 1 +5 sin(S n D Ans [n] = 1 + 3si (= +32) ssin@ m) x[n} = Sin | nN +} — osIn(> Nn -— 8 4 3 4 onan aH N, =16 samples N,=6 somples N=LEM(N,N2) = LOMCI6,6) = 48 samples «@) Consider the system as shown below x(t) V(t) where y(t) = x(e‘). The system is (A) linear and causal. (BY linear and non-causal. (C) non-linear and causal. (D) non-linear and non-causal. Ans: Linearity Cos ality x, (4) y(t) = 5 (e) Put t=o0 mt) > ¥ (= x,Cet) yl =* (e) = (1) + t nn al (ax, (t)+ bx lO)]—s y, ) = arale J+ bx fer) NON CAUSAL a,(= ay, +by, ) we LINEAR Q. Consider a continuous-time system with input x(t) and output y(t) given by y(t) = x(0) cos(t) This system is (A) linear and time-invariant (B) non-linear and time-invariant (Oy linear and time-varying (D) non-linear and time-varying Ans: GH= cos (t) x(t) —* CVEAR a TIME-VARYING (coefficient of xlt)is a tenctim of t) (2) The period of the signal x() = 8sin(0 8rt+ 1) We (A) 0.47 s (B) 0.87 s (C) 1.25 s DP} 2.5 s Ans : We = O-8t Te = 2 = 2% ~ 25sec Wo o-3T @ Given a sequence.x{m], to generate the sequence y[n]=x[3—4n], which one of the following procedures would be correct? (A) First delay x[n] by 3 samples to generate z,[n], then pick every 4" sample of z,[7] to generate z,[n], and then finally time reverse z,[7] to obtain y[n] (PF First advance xn] by 3 samples to generate z,[m]. then pick every 4" sample of z,[1] to generate z,[{m], and then finally time reverse z,[n] to obtain y[n] (C) First pick every fourth sample of x[n] to generate v,[n], time-reverse y,[n] to obtain v,[n] and finally advance v,[n] by 3 samples to obtain y[n] (D) First pick every fourth sample of x[n] to generate v,[n], time-reverse v,[n] to obtain v,[n] and finally delay v,[m] by 3 samples to obtain y[n] Ans: [1] = xJn+3) —» x[-n+3] = x[-4n+3] =— vance - a COR) z Take every 44 * sample zt] — 2c [n+ 3] —» x[4n+3)- —s» x£4n+3]) @OY Take ever won ath Cylrt) avn q comple rep (Q) A system with input x(t) and output y(t) is defined by the input-output relation -2 y(t) = f x(x)dr The system will be (A) causal, time-invariant and unstable (C) non-causal, time-invariant and unstable fy fewer : Pat gC) = frcoae ———> Non-causal preant vo (B) causal, time-invariant and stable Ans non-causal, time-variant and unstable : . u Time —vaviant “Uns table” laive 2 (+) = ult)) Q. Let y(t) =x(4t), where x(t) is a continuous-time periodic signal with fundamental period of 100 s. The fundamental period of y(t) is 25° s (rounded off to the nearest integer). Ans x(t) —> Ty =100sec Pf P=xlat) oe ‘_ Te — 25sec yay [compress by 4 Q. The output of a system y(C) is related to its input x(t) according to the relation y(t) = x(t) sin(2mt). This system is Br Linear and time-variant (B) | Non-linear and time-invariant (C) | Linear and time-invariant (D) | Non-linear and time-variant Q The signal sin(V2706) is (A) | periodic with period T= V2 wef not periodic (C) | periodic with period T= 27 (D) | periodic with period T= 417? Ans: Sin ((27e+7D) = sin(fzre )—O No value of To satisfies OQ, except %B=0 " Mot periodic “ Q. Consider a system with input x(t) and output y(t) related as follows dy n= fe"x(0| Which one of the following statements is TRUE? (A) The system is nonlinear (B) The system is time-invariant [ey The system is stable (D) The system has memory Ans: Lineayi ty + x, 4)—> got) niet [e*x.co] my = Ae [eta.to] [ax,(4)4 bx,(t)]—> y,() = a [e*(axtt+ bat) _ ad fe Ste]+ bd. [e*x,co/ Gs) = ay, (+ by, Linear Q) A causal and stable LTI system with impulse response h(t) produces an output y(t) for an input signal x(t). A signal x(0.5t) is applied to another causal and stable LTI system with impulse response h(0.5t). The resulting output is YF — 2y(0.5t) (B) 4y(0.5t) (C) _0.25y(2t) (D) 0.25 y(0.25¢) Ans: x(t) > y@= x(#) * hC) x(at) PO (xt)= 2 eo) y(t) oct) xh x05 => yA =x(osxh(ost) = 4— gest) = ayle-s 4) jos] © For a unit step input u[n], a discrete-time LTI system produces an output signal (26[n + 1] + 6[n] + 6[n — 1] ). Let y[n] be the output of the system for an input ((3)" u(n]). The value of y[0] is > > Ms lr] —a inp > sir Co tep sponte) S[n)= 28(neQ + SiS[-) a sree, hla] = s(n] sln-i) =(25[n+J+ Sin+ $(n-i) - (28l)+ S(r-i]+ §[r-2)) s[o) shri} hfn) = 25[9*')- Sin] - §[n-2] “Ola gi7 ylrd= rn] ¥ hh] = adr] x [2stord-& [a-Si] x{rpk${h-N3] =2c(n-nJ gio) anorg xb) 0-3 Put n=o0 vs =2 xf1] - x{oJ—>{-2] zr] =(4) "ulr] i y lo] =2(4)-O-O [ \" Vs —— -1-1 216 1 2 ” ¥ [o]J= 9 (Q) The output y[n] of a discrete-time system for an input x[n] is n| = max |x[k]|. yn] = _max_|x[K]| The unit impulse response of the system is (A) 0 for all n. (8) 1 for all n. or unit step signal u[n]. (0) unit impulse signal 5[n]. gf = 3B => 2+hblJ=3 hfij= 3-2 =1 g PJ = 4 SS 14 2A +h) = 4 1+ 2004+ Af) = h2] =4-2-1=4 yi] = hli+ 2h2] = 14+20)=3 yl) = hb) =41 . 10 y [3] + yl4] = 10(3) +1 =3] Q. Two sequences x,[n] and x2[n] have the same energy. Suppose x,[n] = @ 0.5" u[n]. where a is a 1 1 2 g) IPP 1 positive real number and u[n] is the unit step sequence. Assume xo[n] =| V1.5 forn=0,1 0 otherwise. Then the value of @ is fas: a [n)= & (4)'ae9) Same Enengy x, [n} = er 2n=0o,! : o 3; otherwise 20 . b,2 Epes 2 |xQvur- Z| ‘ hz -60 n=-00 nz0 nzo zy 2 3 00 ce B= [1+ > Ge 8] = 4 Sy 7 _ 4 ao 4 = S/*[al/* es) + (y= 2—2@ N=-% ve nei Given, £,.> Ey, => eaal *. 3 ae = 33 ne Q. The result of the convolution x(—t) * 5(—t — to) is (A) x(t + to) (B) x(t — to) (C) x(-t + to) Aes x(—t — to) Ans: x(t)x (t-te) = x (t-te) Prt. = (-t) x &(- t-te) = x(-t-t,) tz-ty Q. The impulse response of an LTI system can be obtained by (A) differentiating the unit ramp response {BF differentiating the unit step response (C) integrating the unit ramp response (D) integrating the unit step response ha&= a slt) Q. A continuous, linear time-invariant filter has an impulse response h(t) described by _(3 forO Nonzero in n-k e[-3,9) ~3kErn-F nN-JskenNr3 a ae A 8 (@) Let _fl, O3n59 _/l, OSn=N alm] {5 elsewhere and Ala] = {g elsewhere where N =< 9 is an integer. Determine the value of N, given that y[n] = x[n] *hLa] and yl4] = 5, yl14] = 0. q ns: ylnta x[o] tbo) = Sx hbo = So 21H Abr A} k = 700 kzo yo] = xfophbe)+ xf] hi] + x12] hfn-d) +--+ 2[Y) Alo] —D Given: gll= a pat nad in: gla} = 2¢Lo] M4) + fy hL3) + =L2)n02) + 2L 3] A1) + M4] Ale) gk [hla] +t Als} Al) +b olj+ thle] => Nz— "nonzero" Nz NZ pe we we Gy és) c=!) G@)») c=) NEG ia] = xf(sJh@) +x [EJ4l(Q +x{7J hla + x(3j Al6J+addhb&) =O Given) glia] = xD C + xl6]al@ +xl7snld + 8 Olt 4 i Oo 19 1 oO i oe ar) CAN] eC; neo bhj=15,0S7S4 N24 (Q) For the systems described by the following equations, with the input x(1) and output y(t), determine which are invertible and which are noninvertible. For the invertible systems, find the input-output relationship of the inverse system. t @ xio= ff xorae (b) yf) =x"(), x(0) real, n integer ) wy) dx(t) © yYO= =, (d) y(@) =x@Gr—6) (c) y(t) = coslx()] () xy =e, x(t) real t Ans: (a) jg = fxende — Integrator -% x(t) = a go —> Difleventiatew (lnverse sz stem) (bd ye = fe(t))"_» nt power of x(t) x(t)= xy ye — vin yoot of g® (inievse <5 stem) When n is even, x(t) and —2x(t) produce same output. J The system is non invertible. © (t)= A xt) dDifferentiatov I ae 209 = f y(tdt—» Lntegrator ( (averse gO =A cae+s)=2 ne. oulp' .!, Non- invertible By (6) = x(3t-6) = x [3(¢-2)] x= y [s+ 2) = x(t) ©) gt) = cosfxct)] x(t) = cof [gle] Cinverse< A DT signal 2e(n) , Where x[nj= 0 for N A signal alt) , where x(thz0 for tro Moti causal 5eq,nene—> A DT signal x{n], wheve r[eJ2o for N>0 Noncausal signal—> A signal act) that has nonzero yalues in both pesitive anol negative time . Noncausal sequence —> A DT signa | x(n] thet has nonzero values in both positive and negative time Ans: Lineay convoalutio : h g—tjae gl] abl --—- 7 - Alr]= Lun a\gel gl) gh: a (Hftgle gl) 4gld-- -- nif bebe} t G)i¢gel +90 tah) : ginl= 4 gel, Lge) +901, 5 9 +p gereged,—f de Given yfel=1—> gel=t gi ot = $90) + gl =i argd=2 gitl-o (@) The following are the impulse responses of discrete-time LTI systems. Determine whether each system is causal and/or stable. Justify your answers. (a) Alm} = (5)"uln} {b) A[n] = (0.8)"u[n + 2] (c) Ala] = G)"ul-a] (d) Afn] = (5)"u[3 — n] (e) A{n] = (—4)"u[n} + (L.01)"u[n ~ 1] (f) An] = (—4)"u[n] + (L.01)"uf1 — 1) (g) Aln] = n(Z)"uln — 1] Ans: A Dr system with impulse response Ath) ss causal ié Alfn]a90 5 n< Stable — (f) hie) = te 7a) dni) re'}+be'] at Ah @*(1-t)=0 at l-t=9o t=1 Causal Stable gH Home work (Q) Evaluate the step response for the LTI systems rep- resenved by the following impulse responses: (a) b[n] = (—1/2)"u[n] (b) bin] = 8[n] — d[n — 2] (c) bln] = (- 1)"{u[n + 2] — u[n — 3]} (d) bin] = nu[n] (e) b(t) = ell (f) b(t) = 6 (t) (g) h(t) = (1/4) (u(t) — u(t — 4)) (h) b(t) = u(t) 
