ECE 421 Introduction to Signal Processing Project 1 - Solutions

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Wilfrid Robert Stokes
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1 1. (10 credits) Given, xx oooo (tt) = cos (2ππFF cc tt) xx iiii (tt) = cos (2ππππππ) ECE 421 Introduction to Signal Processing Project 1 - Solutions Dror Baron, Spring 2017 The AM modulated output satisfies, xx ffff1 (tt) = xx oooo (tt)xx iiii (tt) = cos(2ππff cc tt) cos(2ππππππ) = 1 2 (cos(2ππ(ff cc + FF)tt) + cos(2ππ(ff cc FF)tt)) 2. (10 credits) The AM demodulation output is xx ffff2 (tt) = xx oooo (tt) xx ffff1 (tt) = cos(2ππff cc tt) [ 1 2 (cos(2ππ(ff cc + FF)tt) + cos(2ππ(ff cc FF)tt))] = 1 4 (cos(2ππ(2ff cc + FF)tt) + cos(2ππ(2ff cc FF)tt) + cos(2ππππππ) + cos(2ππ( FF)tt)) = 1 4 (cos(2ππ(2ff cc + FF)tt) + cos(2ππ(2ff cc FF)tt)) + 1 cos(2ππππππ) 2 Note that the cosines with frequencies (2FF cc + FF) and (2FF cc FF)are not in the baseband, and these frequencies will be removed by the low pass filter after frequency mixing at the analog demodulator. 3. (16 credits) The objective is to find the lowest sampling frequency,, such that the entire modulated frequency band is shifted into the baseband [0, 20 kkkkkk]. The sampling frequency for the base band signal is = 40kkkkkk. Any lower sampling frequency will cause folding of the base band. Thus The carrier frequency, FF cc = (FF ll+ff h ), should become 0 Hz in the base band for correct mapping 2 by downsampling. This implies that FF cc = h, where h ZZ +, ZZ + is the set of positive integers. Let h be the largest integer multiple such that h FF cc. Then h = FF cc, where is the floor operation. The minimum sampling frequency is = FF cc h

4. (10 credits) The minimum sampling frequencies are 10 a. = 40 kkkkkk b. = 43.33 kkkkkk 5. (12 credits) Given, = 1200 kkkkkk. 12 [FF ll, FF h ] = [100 kkkkkk, 140 kkkkkk] = 40 kkkkkk.

2 4. (10 credits) The minimum sampling frequencies are 10 a. = 40 kkkkkk b. = kkkkkk 5. (12 credits) Given, = 1200 kkkkkk. 12 [FF ll, FF h ] = [100 kkkkkk, 140 kkkkkk] = 40 kkkkkk. From we have mm = nn. xx ffff1 mm = xx ffff1 nn, The above equation can be rewritten as nn = mmmm = mm Hence the downsampling factor is kk =. Substituting the corresponding values, k = (24 credits) The perceived frequencies are a. Frequency of ss 1 (tt) = 15 kkkkkk (See Figure 1). b. Frequency of ss 2 (tt) = 5 kkkkkk (See Figure 2). c. Frequency of ss 3 (tt) = 5 kkkkkk (See Figure 3). Figures: Figure 1: Original and downsampled signals of ss 1 (tt)

parameters fband = 120*10^3; fas = 1200*10^3; fs = 40000; DSF = fas/fs; % (a) First signal A_1

3 Figure 2: Original and downsampled signals of ss 2 (tt) MATLAB code for part 6: clc; clear all; close all; format long; Figure 3: Original and downsampled signals of ss 3 (tt) % signal parameters fband = 120*10^3; fas = 1200*10^3; fs = 40000; DSF = fas/fs; % (a) First signal A_1 = 1; % carrier frequency % analog sampling frequency % audio sampling frequency % Down Sampling Factor

4 f_1 = 135*10^3; % (b) Second signal A_2 = 1; f_2 = 125*10^3; % (c) Third signal A_3 = 1; f_3 = 115*10^3; % time range t_end = *5; t = (1/fas:1/fas:t_end); t_downsample = (DSF/fas:DSF/fas:t_end); % Downsampling (demodulation) % (a) First signal ana_sig = A_1*cos(2*pi*f_1.*t);%signal title('sinusoidal signal, s_1(t)'); legend('s_1(t)','downsampled s_1(t)','location','northeast'); axis([0 3*10^ ]); % (b) Second signal ana_sig = A_2*cos(2*pi*f_2.*t);%signal title('sinusoidal signal, s_2(t)'); legend('s_2(t)','downsampled s_2(t)','location','northeast'); axis([0 5*10^ ]); % (c) Third signal ana_sig = A_3*cos(2*pi*f_3.*t);%signal title('sinusoidal signal, s_3(t)'); legend('s_3(t)','downsampled s_3(t)','location','northeast'); axis([0 5*10^ ]);

5 7. (18 credits) a) Linear or Non-linear I. Suppose that x(n) is a real valued signal. Then, IIII{xx(nn)} = 0. Now consider, jjjj(nn) as the input signal. It is observed that output will be xx(nn), because the imaginary part is x(n). As a result, the input is multiplied by jj, but the output is not. Thus, this system has not the Homogeneity property and it is nonlinear. II. This system is not linear because it is does not have Homogeneity. For instance, if xx(nn) = 1, then yy(nn) = 3. If xx 1 (nn) = 2xx(nn) = 2 then yy 1 (nn) = 4. However, if the system was linear, then the output for xx 1 (nn) = 2xx(nn) should be 2yy(nn) = 6, and it is not. So it is non-linear. As a fast check for all systems, every linear system should have the zero input zero output property (ZIZO), where a zero input is mapped to a zero output. In our case, the system does not satisfy ZIZO. b) TI or TV I. If xx 1 (nn) is the input, then yy 1 (nn) = xx 1 ( nn). But for an input xx 2 (nn) = xx 1 (nn nn 0 ), yy 2 (nn) = xx 2 ( nn)= xx 1 (nn nn 0 ) = xx 1 ( nn + nn 0 ) yy 1 (nn nn 0 ). Thus, it is time-variant. II. For this system, as an intuitive solution, we can say that shifting the input by one time step shifts the output by two steps, and so it is time-variant. c) BIBO or not BIBO i. If xx(nn) = 1, then yy(nn) =, and the system is not BIBO. ii. If xx(nn) = jj 1+nn 2, then yy(nn) = eenn2 +1, and the system is not BIBO.

b) discrete-time iv) aperiodic (finite energy)

b) discrete-time iv) aperiodic (finite energy) EE 464 Frequency Analysis of Signals and Systems Fall 2018 Read Text, Chapter. Study suggestion: Use Matlab to plot several of the signals and their DTFT in the examples to follow. Some types of signal