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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.
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>. Institute of Information Technology Kottayam Time: 20 minutes INDIAN INSTITUTE OF INFORMATION TECHNOLOGY KOTTAYAM Department of Electronics and Communication Engineering IEC 122 SIGNALS & SYSTEMS Quiz V - April 2025 (A) Semester II Max marks: 1C 1. The ROC (region of convergence) of the z-transform of a discrete-time signal is represented by the shaded region in the z-plane. z-transform is represented by A. im Unit cirele Cc. Im ni Ans: 2e(n) = aint If the signal x[m] = (2.0)!"!,c0 < n < +00, then the ROC of its =plone as —— Re aA Re 2, LE LEED Pre ee HEE GLO: im ‘Unit circle =plane (ROC does not exist) <4 No =-trensform/Roc for x>1 2. A discrete-time signal x[n] = 6[n — 3] + 26[”a — 5] has z-transform X(z). If Y(z) = X(—z) is the z-transform of another signal y[n], then (A) yln] = x(n] (B) y[n] =«[-n] Ans: Ler yin] = -x[n] (D) y[n] =—x[-n] X[nJ= §[n-3] +2§(n-5] —2 > Kez) = 274227 Y¥(2) = XX gf) -3 =. - - C2 +2¢2)% = -Z?-227 ~ §£n-3]- 2${n-5] = -x(v] 3. If the impulse response of a discrete-time system is h[n] = —5"u[—n — 1], then the system function H(z) is ‘equal to (A (Cc and the system is unstable or > = 5 and the system is stable (D) = Ans: h[n) = -Suf-n-i] <2 Ho) - 2 : |z[< 5° z-5 71 Roc includes unit civcle Iz[=1 Hence sy stem is stable 4. A sequence x[n] with the z-transform X(z) = 744 <2 — 2z¢ +2 - 3z-‘ is applied as an input to a linear, time-invariant system with the impulse response h[7] = 26[n — 3] where 1, n=0 6[n] = | . 0, otherwise The output at n = 4 is (A) -6 BY zer0 (C) 2 (D) -4 Ans: 42 aed XQ) aot 2242-32 } ez) = X(2) Hez) = (24 z*- 2242-32") 227 Hezy= 223 -2 . Yos= 22 saz tystey z3_¢77 ~ yCn] = 28 (n4i} + 2 5(n-1)—- 4§&-2)+ 4 § -9 -6 & (n-77 n=2 n=3 n=7 yl1=0 5. Letx[n] = (-3] u(n)— (-3) u(—n— 1). The Region of Convergence (ROC) of the z-transform of x(n) is 1 A> 5 SH er kl> 4 | (B) |z| < 3 (D) does not exist Ans: x[n] = (-5) won) ~(-3) u(-n- 1) —2 +5 XQ) = ar + a 7 3 —u“— = OC (z)>4 bzeL foc 4 ssjL«lzl< L 3 —277! 6. The discrete-time transfer function ———— is 1-057 (A) non-minimum phase and unstable (C) minimum phase and stable (B) minimum phase and unstable I non-minimum phase and stable Ans: ~— Hcez)= 1-22! -» Zero: ==2 L-osz'! Pole + zZ=z0-5 Zeve z=2 fs outside Jzj=1. .* System is non-mirimum Phose. Pole z=o0°S inside unit civele [2z/=L covvesponds to stable system 1 10. The z-transform of the sequence x[n] is given by X(z) = Coen with the region of convergence |e] > 2. Then x[2] is “ Ans: Qube aoa = Pala "uf n)<2> 2) = x(n) = (2)"4(9] x(2)U{n} Xe) aa Using matrix methed of convolution (2)” a @|1 4 * C01 = {44,12 32, ---] 2 & 2 =x(2zJ=12