Download Digital Signal Processing: A Computer-Based Approach by Sanjit Mitra PDF

By Sanjit Mitra

"Digital sign Processing: A Computer-Based method" is meant for a two-semester direction on electronic sign processing for seniors or first-year graduate scholars. in accordance with person suggestions, a few new issues were additional to the second one variation, whereas a few extra themes from the 1st version were got rid of. the writer has taken nice care to arrange the chapters extra logically via reordering the sections inside chapters. extra worked-out examples have additionally been integrated. The ebook comprises greater than 500 difficulties and a hundred and fifty MATLAB exercises.New issues within the moment version comprise: finite-dimensional discrete-time structures, correlation of signs, inverse structures, procedure identity, matched filter out, layout of analog and IIR electronic highpass, bandpass and bandstop filters, extra on FIR filters, spectral research of random indications and sparse antenna array layout.

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5) + = 2 or α = – . 5) + , n ≥ 0. 06y[n − 2] = 2 n µ[n] with y[–1] = 1 and y[–2] = 0. 06y c [n − 2] = 0 . 2 . 2) n . For the particular solution we choose y p [n] = β(2) n . 06)2 n− 2 = 2 n µ[n]. 9662 . 2) n + 200 n 2 . 0356. 9662(2) n , for n ≥ 0. 06y[n − 2] = x[n] − 2 x[n − 1] with x[n] = 2 n µ[n] , and y[–1] = 1 and y[–2] = 0. 06y[n − 2] = 2 n µ[n] − 2 (2 n −1 ) µ[n −1] = δ[ n]. The solution of this 28 equation is thus the complementary solution with the constants determined from the given initial conditions y[–1] = 1 and y[–2] = 0.

Now, the DTFT of g1 [−n] is given by G1 (e − jω ) . Hence, the DTFT of g 3 [n] is given by G 3 (e jω ) = e − j3ω G1(e − jω ) + e − j4ω G1(e jω ). (d) g 4 [n] = g1 [n] + g 1[−(n − 7)]. Hence, the DTFT of g 4 [n] is given by G 4 (e jω ) = G1(e jω ) + e− j7ω G1(e − jω ). , ∞  ∞  ∞  ∞  y[n] e − jωn =  x 1[n] e− jωn   x 2 [n] e − jωn   x 3 [n] e− jωn   n= −∞   n =−∞   n =−∞  n =−∞ ∑ ∑ ∑ ∑  ∞  ∞  ∞       (a) Therefore, setting ω = 0 we get y[n] =  x1 [n]   x 2 [n]   x 3[n]  .

168b) we get P˜ [k] = k =0 Hence ˜p[n] = N −1 ∑ p˜[n]e− j2πkn / N = 1. n= 0 1 N N−1 ∑ e− j2πkn / N . 37 X ω=2πk /N ∞ ∑ = X(e j2πk /N ) = x[n] e− j2πkn / N , – ∞ < k < ∞. n= −∞ ˜ [k + l N] = X(e j2π( k+ l N) / N ) = X(e j2πk / N e j2πl ) = X(e j2πk / N ) = X ˜ [k]. (a) X 1 (b) x˜ [n] = N = N−1 ∑ k =0 N−1  ∞  ˜ [k]e j2πkn / N = 1 ∑  ∑ x[l ]e − j2πkl /N  e j2πkn / N X N k =0  l = −∞   N −1 − j2πkr  1 N−1 ∞ 1 ∞ j2πk(n− l ) /N ˜ x[l] e . Let l = n + r ⋅ N . Then x [n] = x[n + r ⋅ N] ∑ ∑ ∑  ∑e .

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