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SIGNALS AND SYSTEMS QUIZ 1 ECE, Quizzes of Signals and Systems

International Institute of Information TechnologySignals and Systems

SIGNALS AND SYSTEMS QUIZ 1 ECE SUNJECT-ELECTRONICS ECE COURSE-SIGNALS AND SYSTEMS YEAR-2025 PROFESOR-RAJESH.K Course Overview: This course introduces the fundamental concepts and mathematical tools used to analyze and process signals and systems. It lays the foundation for further study in areas like communications, control systems, signal processing, and electronics. Course Objectives: Understand different types of signals (continuous-time and discrete-time) and their properties. Analyze linear time-invariant (LTI) systems using convolution and system properties. Apply Fourier series, Fourier transform, Laplace transform, and Z-transform for signal and system analysis. Explore sampling theory and its implications in digital signal processing.

Typology: Quizzes

2024/2025

Available from 06/23/2025

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Name: Roll No: Marks: INDIAN INSTITUTE OF INFORMATION TECHNOLOGY KOTTAYAM Department of Electronics and Communication Engineering Indian Institute of IEC 122 SIGNALS & SYSTEMS Information Technology i7T- January 202 Kottayam Quiz I - January 2025 Time: 15 minutes Semester II Max marks: 10 . . 1, n=0,1.2.3.... 1. Consider the signal x[n] = sin(2zn)u[n], where u[n] = . 0, otherwise The period of this signal x[n] is (A) 4 (B) 3 (C) 2 PF! Ans; Fay n=O, ><(nJ=0 n=O : x[n) = sin(21.0) ufo] = Ox! =O xfoj n=i1 + xfn] = Sin (mt) ufi] = Ox! =O N=L! xo] = sinQam.2) uk] = Ox 1 =O . : “3-2-4 22345 2. A periodic signal x(t) of period T, is given by lL. |th Hi t Te Tr TU 7 A To Tt dic component of x(t) =4/ xe@ dt = £| at = L(t)” = 1-07) (ovevoge value) Tr en ey Tt OT Te jos 3. A system with input x(/) and output y(/) is defined by the input-output relation 2 y(t) = | x(t)dr The system will be (A) causal, time-invariant and unstable (C) non-causal, time-invariant and unstable (B) causal, time-invariant and stable er non-causal, time-variant and unstable gis future Cime Aes: Cawralty: Put t22/ a= fxcdde present = time Output signal depends on futuve values of input signal. Non-cavsat 7 Time Lavorionce + Deloyed xlt-to—y lt) = fx (t-te —D Input ' —% bout —2(t-t.) Outpe (= ylt-t) = [ xmdt —® Delceal 4: d f In QO, pet t-t,=r T— - 0% + A> ~ data=da Tn lt A Wt -t 72t-to _, © becomes y, = f xajyda —@ — 0 aro yay Time Vag ing Stability: Pat x(t)2u(t) (Bounded Input] =2t gO = fumar ~~ +00 #00 As t—-- yor = Jug@de = Jodt = 00 [Unbounded | ’ We = ° Unstable eo Total Enevgy of x,[)5 £y = Zac's 0-250-540-7574 Hors trostpo2st £y.,= 2:75 joules To tal Energy of x,[n) ; Ex, _ = |x Pell Pp 07s OS*4O.2S HO HORS Os EO-7e 4/* h=-0 &, = 3 “75 joules t a Ex, > Ex, 1 too 2 6. The integral tn in Pe /25(1 — 2t) dt is equal to 2n 1 els (B) els (C) —ne7 2 (D) 1 Nal 8V20 Av2n V2n Ans: §(t-2t) = S$ @(L-4) = 4 §G-*) [> Scat) = 450] = 1 §-t) [--&@ is an even function] <3 = fo 2 1 [ee th § (28) at = al tes $G-b)dt 6g —<& Aa Oxley §CE-Es) - Ewe _te® ~ 8\ar 7. The signal energy of the continuous-time signal x(t) = [(r- Du(t — 1)] — [@ — Dut - 2)] - [@ — 3)u(t - 3)] + [(t -— Fur - 4)] is Ans, The signal x (+) may be plotted as shown below: a(t) 1 + °° bo Energs E= [ixaldt = FO, MOO, YO _ SF joules Sa 3 3 3 Sa nt i{ =), i(—|n 8. A discrete-time signal x[”] =e ‘ BT ye ql is down-sampled to get the signal x z[m] such that a(n] =x[4n]. The fundamental period of the down-sampled signal xg[n] is Ans; j(S8)an J Zan < [>] = x(4n] =e +e J? ) n gan x, 0] = e +¢€ _—4 — Fundomental | ——— _ N, <2 Pert od No N,= 32 i f N, = LEM (N,,N,) = LOM (3,2) = 6 samples 9. Consider the discrete-time signal x[7] = u[—n + 5] — u[n + 3], where 1, n>0 my) co The smallest n for which x[n] =Qis__~3 Ans: The signal x fn] meoy be plotted as shawn below: ubars] L | | 1 etnJz 0 From “5-43-21 O12 34 5 n ose a -3 to % u(n+3] a) = n 4 < 4 -8 7-7 -G°5 “4 -32-2-1 “3-2-1 O12 345 n 10. The mean square value of the given periodic waveform f(f) is | nAC) Ans: Mean square value of £¢t) CMsv)