Exam (elaborations) TEST BANK FOR Statistical and Adaptive Signal Processing By Dimitris G. Manolakis, Dimitris Manolakis, Vinay K. Ingle, Stephen M. Kogon (Solution Manual)

Sampling frequency Fs = 100 sam/sec (a) Continuous-time signal xc(t) = 2 cos(40πt + π/3) has frequency of 20 Hz. Hence x(n) = xc (t)|t=n/Fs = 2 cos
40πn 100 + π/3  which implies that ω0 = 40π 100 = 2π 5 . (b) Steady-state response yc,ss(t): Given that h(n) = 0.8nu(n), the frequency response function is H(e jω) = 1 1 − 0.8e−jω Since ω0 = 2π/5, the system response at ω0 is H(e jω) = 1 1 − 0.8e−j2π/5 = 0.9343 e−j0.2517π Hence yss(n) = 2(0.9343) cos(2πn/5 + π/3 − 0.2517π), or yc,ss(t) = 1. 868 6 cos(40πt + 0.585π) (c) Any xc(t) that has the same digital frequency ω0 after sampling and the same phase shift as above will have the same steady state response. Since Fs = 100 sam/sec, the two other frequencies are 120 and 220 Hz. 2.2 The discrete-time signal is x(n) = A cos(ω0n)wR(n) where wR(n) is an N-point rectangular window. (a) The DTFT of x(n) is determined as X(e jω) = F [A cos(ω0n)wR(n)] = (A/2) F  e jω0 wR(n)  + (A/2) F  e−jω0 wR(n)  (1) Using the DTFT of wR(n) as F [wR(n)] = N−1 n=0 e−jωn = e−jω(N−1)/2 sin(ωN/2) sin(ω/2) (2) and the fact that complex exponential causes a translation in the frequency domain (1) can be written after a fair amount of algebra and trigonometry as X(e jω) = XR(e jω) + j XI(e jω) 1 2 Statistical and Adaptive Signal Processing - Solution Manual −4 −2 0 2 4 −5 0 5 10 15 20 32−point DTFT (Real) −4 −2 0 2 4 −15 −10 −5 0 5 10 15 32−point DTFT (Imaginary) Figure 2.2bc: Real and Imaginary DTFT and DFT Plots(ω0 = π/4) where XR(e jω) = A 2 cos[(ω − ω0)(N − 1)/2] sin[(ω − ω0)N/2] sin[(ω − ω0)/2] + A 2 cos[(ω + ω0)(N − 1)/2] sin{[ω − (2π − ω0)]N/2} sin{[ω − (2π − ω0)]/2} (3) and XR(e jω) = −A 2 sin[(ω − ω0)(N − 1)/2] sin[(ω − ω0)N/2] sin[(ω − ω0)/2] − A 2 sin[(ω + ω0)(N − 1)/2] sin{[ω − (2π − ω0)]N/2} sin{[ω − (2π − ω0)]/2} (4) (b) N = 32 and ω0 = π/4. The DTFT plots are shown in Figure 2.2bc. (c) The DFT samples are shown in Figure 2.2bc. (d) N = 32 and ω0 = 1.1π/4. The plots are shown in Figure 2.2d. The added spectrum for the second case above (ω0 = 1.1π/4) is a result of the periodic extension of the DFT. For a 32-point sequence, the end of each extension does not line up with the beginning of the next extension. This results in sharp edges in the periodic extension, and added frequencies in the spectrum. 2.3 The sequence is x(n) = cos(πn/4), 0 ≤ n ≤ 15. (a) The 16-point DFT is shown in the top-left plot of Figure 2.3. (b) The 32-point DFT is shown in the top-right plot of Figure 2.3. (c) The 64-point DFT is shown in the bottom plot of Figure 2.3. (d) The zero padding results in a lower frequency sampling interval. Hence there are more terms in the DFT representation. The shape of the DTFT continues to fill in as N increases from 16 to 64. 2.4 x(n) = {1, 2, 3, 4, 3, 2, 1}; h(n) = {−1, 0, 1}