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ECE 513 - DIGITAL SIGNAL PROCESSINGCranos M. WilliamsAssigned: November 11, 2019HOMEWORK 9 Solution, DUE NOVEMBER 20, 20191.Frequency Representation of Up-sampling and Down-sampling: (15 Pts)The discretetime Fourier transform (DTFT) of a signalx[n] is shown in Figure 1a. The responseis zero forB≤ |ω| ≤π.↓Mx[n]y1[n]↑Lx[n]y2[n](a)(c)(b)ωB-B|X(ω)|1Figure 1: a.)DTFT ofx[n], b.)Down-sampling block diagram, c.)Up-sampling blockdiagram(a) Referring to Figure 1b, draw the DTFT|Y1(ω)|of the output signal for a down-sampling factor of M = 2 for the following cases. Your plots should range from-π≤ω≤π. Label all relevant frequencies.i. B =π5ii. B =π2iii. B =3π4iv. B =πFigures 2 shows the output of down-sampled signals.1
ECE 513 - Homework 9 Solution2(b) Referring to Figure 1c, draw the DTFT|Y2(ω)|of the output signal for the followcases. Your plots should range from-π≤ω≤π. Label all relevant frequencies.
ECE 513 - Homework 9 Solution3(c) Use the MATLAB functionfir2to produce a 101 sample sequencex[n] whosemagnitude response matches the response in Figure 1a.Usingyour ownup-sampling and down-sampling routines, test the results above. Did your drawingsmatch the results MATLAB gave? If they didn’t, explain why not. Turn in yourdrawings and the MATLAB plots.
ECE 513 - Homework 9 Solution4
ECE 513 - Homework 9 Solution5-1-0.500.518.104.22.168.81Normalized Frequency|X(ω)|Original Signal: B = π/5-1-0.500.522.214.171.124.81Normalized Frequency|Y1(ω)|Original Signal Downsampled by 2Figure 4: Output withB=π/5-1-0.500.5126.96.36.199.81Normalized Frequency|X(ω)|Original Signal: B = π/2-1-0.5