Continuous and discrete time signals and systems by Mrinal Mandal, Amir Asif

By Mrinal Mandal, Amir Asif

This textbook provides an advent to the elemental recommendations of continuous-time (CT) and discrete-time (DT) signs and structures, treating them individually in a pedagogical and self-contained demeanour. Emphasis is at the easy sign processing rules, with underlying thoughts illustrated utilizing useful examples from sign processing, multimedia communications, and bioinformatics. Following introductory chapters, the textual content is separated into elements. half I covers the theories, recommendations, and purposes of CT indications and platforms and half II discusses those subject matters for DT, in order that the 2 may be taught independently or jointly. Accompanying the publication is a CD-ROM containing MATLAB code, audio clips, photos, interactive courses, and sign animations. With over three hundred illustrations, 285 labored examples and 385 homework difficulties, this textbook is a perfect creation to the topic for undergraduates in electric and machine engineering. extra assets, together with recommendations for teachers, can be found on-line at www.cambridge.org/9780521854559.

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4, where the values of the odd-numbered samples of q2 [k], highlighted with the gray background, are obtained by taking the average of the values of the two neighboring P1: RPU/XXX P2: RPU/XXX CUUK852-Mandal & Asif QC: RPU/XXX May 25, 2007 T1: RPU 18:7 44 Part I Introduction to signals and systems Fig. 27. 5k] of signal x[k], where unknown sample values are interpolated. 2 0 −10 −8 −6 −2 −4 0 2 4 6 k 10 8 samples at k and k − 1 obtained from x[k]. The waveform for q2 [k] is plotted in Fig. 27. 3 Time inversion The time inversion (also known as time reversal or reflection) operation reflects the input signal about the vertical axis (t = 0).

49) P1: RPU/XXX P2: RPU/XXX CUUK852-Mandal & Asif QC: RPU/XXX May 25, 2007 34 T1: RPU 18:7 Part I Introduction to signals and systems Solution 5 − jt 5 − jt δ(t) = 2 7+t 7 + t2 (ii) Using Eq. 46) yields (i) Using Eq. 46) yields ∞ t=0 δ(t) = ∞ (t + 5)δ(t − 2)dt = −∞ 5 δ(t). 7 ∞ [(t + 5)]t=2 δ(t − 2)dt = 7 −∞ δ(t − 2)dt. −∞ Since the integral computes the area enclosed by the unit step function, which is one, we obtain ∞ ∞ (t + 5)δ(t − 2)dt = 7 −∞ δ(t − 2)dt = 7. −∞ (iii) Using Eq. 5π +2 δ(ω − 5)dω.

29) and it is plotted in Fig. 12(d). 30) 0 t =0  −1 t < 0. The CT sign function sgn(t) is plotted in Fig. 12(e). Note that the operation sgn(·) can be used to output the sign of the input argument. The DT signum P1: RPU/XXX P2: RPU/XXX CUUK852-Mandal & Asif QC: RPU/XXX May 25, 2007 T1: RPU 18:7 26 Part I Introduction to signals and systems x(t) = u(t) 1 x[k] = u[k] 1 t k 0 0 (a) (b) ( ) t x(t) = rect t 1 −t 2 t t 2 0 ( k x[k] = rect 2N + 1 1 (c) ) k −N 0 N (d) x[k] = sgn(k) x(t) = sgn(t) 1 1 t 0 k 0 −1 −1 (e) ( f) r(t) = tu(t) r[k] = ku[k] slope = 1 t k 0 0 (g) (h) x(t) = sin(w0t) x[k] = sin(W0k) 1 1 t 2p w0 0 2p W0 0 (j) (i) 1 x(t) = sinc(w0t) x[k] = sinc(W0k) 1 t 1 −w 0 (k) k 0 k 0 1 w0 (l) Fig.

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