By Sanjit Mitra
"Digital sign Processing: A Computer-Based process" is meant for a two-semester direction on electronic sign processing for seniors or first-year graduate scholars. in line with consumer suggestions, a couple of new themes were additional to the second one version, whereas a few extra subject matters from the 1st variation were got rid of. the writer has taken nice care to arrange the chapters extra logically by way of reordering the sections inside of chapters. extra worked-out examples have additionally been incorporated. The publication comprises greater than 500 difficulties and a hundred and fifty MATLAB exercises.New themes within the moment variation contain: finite-dimensional discrete-time structures, correlation of indications, inverse structures, method id, matched clear out, layout of analog and IIR electronic highpass, bandpass and bandstop filters, extra on FIR filters, spectral research of random signs and sparse antenna array layout.
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Extra info for Digital Signal Processing: A Computer-Based Approach
Example text
4 Now, y˜ [0] = ∑ ˜x[r]h˜[0 − r] = x˜[0]h˜[0] + x˜[1] h˜[4] + ˜x[2] ˜h[3] + ˜x[3] ˜h[2] + x˜[4]h˜ [1] = –8. r=0 3 ∑ ˜x[r]h˜[1 − r] = ˜x[0]h˜ [1] + x˜[1]h˜[0] + x˜[2] h˜[3] + x˜[3] h˜[2] = 3. Similarly y˜ [1] = r=0 Continuing the process we can show that y˜ [2] = −15, y˜ [3] = 16, and y˜ [4] = −8. N-1 can be accumulated to ψ˜ k [n], where 0 ≤ k ≤ N – 1. Hence in this case the Fourier series representation involves only N complex exponential sequences. Let x˜ [n] = 1 N N −1 ∑ N−1 ∑ X˜ [k]e j2πkn / N k =0 x˜ [n] e − j2πrn / N = n= 0 1 N then N −1 N−1 ∑∑ n= 0 k =0 ˜ [k]e j2π(k −r )n / N = 1 X N N −1 N−1 ∑ ˜ [k] X k =0 ∑ e j2π(k − r)n / N .
N= −∞ ∞ (b) X(e jπ ) = (c) (d) π ∑ x[n](−1)n = 2 + 4 +1 + 5 + 3 − 2 − 4 +3 = 14. n= −∞ ∫−πX(e jω )dω = 2πx[0] = −4π. ∞ π jω 2 X(e ) dω = 2π ∑ x[n] 2 = 168π. ∫−π (Using Parseval's relation) n= −∞ (e) π 2 dX(e jω ) dω = 2π −π dω ∫ ∞ ∑ n ⋅ x[n] 2 = 1238π. 27 Let G1 (e jω ) denote the DTFT of g1 [n]. (b) g 2 [n] = g1[n] + g1 [n − 4]. Hence, the DTFT of g 2 [n] is given by G 2 (e jω ) = G1 (e jω ) + e− j4ω G1(e jω ) = (1+ e − j4ω )G1(e jω ). (c) g 3 [n] = g 1[−(n −3)]+ g 1[n − 4]. Now, the DTFT of g1 [−n] is given by G1 (e − jω ) .
X*[<ý–n >N]} Using linearity and results of part (b) we get 1 {X[k] 2 + X*[k]} = Re{X[k]}. – x*[< –n >N]}. Again using results of part (b) and linearity we get 1 {X[k] 2 –X*[k]} = j Im {X[k]}.
Digital Signal Processing: A Computer-Based Approach by Sanjit Mitra
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