By Paulo S. R. Diniz, Eduardo A. B. da Silva, Sergio L. Netto

This new, fully-revised version covers all of the significant subject matters of electronic sign processing (DSP) layout and research in one, all-inclusive quantity, interweaving thought with real-world examples and layout trade-offs. development at the luck of the unique, this variation contains new fabric on random sign processing, a brand new bankruptcy on spectral estimation, enormously increased insurance of filter out banks and wavelets, and new fabric at the answer of distinction equations. extra steps in mathematical derivations cause them to more uncomplicated to stick with, and a major new characteristic is the homemade part on the finish of every bankruptcy, the place readers get hands-on event of fixing functional sign processing difficulties in a variety of MATLAB experiments. With a hundred and twenty labored examples, 20 case reports, and virtually four hundred homework routines, the booklet is vital studying for someone taking DSP classes. Its designated combination of concept and real-world sensible examples additionally makes it a terrific reference for practitioners.

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**Example text**

N=0: n=1: 4 4 + d2 + 0 = 0 −→ d2 = − 9 9 8 8 12 − d2 − d3 = 0 −→ d3 = − d2 = 9 9 9 Thus, for n ≤ 0, the complete solution is y(n) = 12 4 n 4 n n 2 − (−1) + n (−1) . 9 9 9 (d) We have to solve the following difference equation 5 y(n) − y(n − 1) + y(n − 2) = (−1)n u(n) = cos (πn) u(n), 6 where y(−2) = y(−1) = 0. Using operator notation, the former equation becomes 5 1 − D−1 + D−2 {y(n)} = cos (πn) u(n). 6 The homogeneous equation is 5 yh (n) − yh (n − 1) + yh (n − 2) = 0. 6 Then, the characteristic polynomial equation from which we derive the homogeneous solution is 5 ρ2 − ρ + 1 = 0.

Thus, the steady-state response (response obtained when n is “very large”) is given by √ 8 π 3 3 π y(n) = sin n − cos n . 91 3 91 3 30 CHAPTER 1. 18 (a) y (n) = sin (ωn − 2ω) u (n − 2) + sin (ωn − ω) u (n − 1) + sin (ωn) u (n) Steady-state ⇒ n ≥ 2. 7. 19a. (b) 32 CHAPTER 1. 8. 19b. 9. 19c. 5 2<∞ = = The system is BIBO stable. 5n |h (n)| = n=−∞ n=0 The above sum is unbounded, and so the system is unstable. 1 10 9 The above sum is unbounded, and so the system is unstable. 34 CHAPTER 1. DISCRETE-TIME SYSTEMS (d) 0 ∞ |h (n)| = 2−n , l = −n n=−∞ n=−∞ ∞ l = 2 l=0 The above sum is unbounded, and so the system is unstable.

13 N M ai y (n − i) = i=0 bl x (n − l) l=0 M y (n) = N bl x (n − l) − l=0 ai y (n − i) i=1 25 We suppose the input signal x (n) is causal, so that x (n) = 0, ∀n < 0. Thus, N y (n) = − ai y (n − i), ∀n < 0 i=1 Considering that the N independent auxiliary conditions are set to zero, we conclude that y (n) = 0, ∀n < 0. Then, y (0) = b0 x (0) y (1) = b0 x (1) + b1 x (0) − a1 y (0) = b0 x (1) + b1 x (0) − a1 b0 x (0) .. If x (n) = kx1 (n) + x2 (n), and remembering that y1 (n) = y2 (n) = 0, ∀n < 0, y (0) = kb0 x1 (0) + b0 x2 (n) = ky1 (0) + y2 (0) y (1) = b0 [kx1 (1) + x2 (1)] + b1 [kx1 (0) + x2 (0)] − a1 [kb0 x1 (0) + b0 x2 (n)] y (1) = kb0 x1 (1) + kb1 x1 (0) − ka1 y1 (0) + b0 x2 (1) + b1 x2 (0) − ka1 y2 (0) ..