Solutions_Assignment3_EEE505_Fall2023

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© A. Papandreou-Suppappola, Arizona State University EEE505, Fall 2023 Assignment 3 Solutions 1. [3 pts] (a) Compute the Wigner distribution (WD) of the signal x 1 p t q = 2cos p 2 πf 1 t q ; show all steps. (b) Obtain the WD of x 2 p t q = 2cos p 400 π p t ´ 5 qq e j 2 π p 50 q t ; do not compute it using the WD definition. (c) Does the WD of x 1 p t q in part (a) have any cross terms (CTs)? If so, provide the following: (i) CTs location in the time-frequency domain (ii) CTs direction of oscillation in time and rate of repetition per second (or oscillation frequency) (iii) CTs direction of oscillation in frequency and rate of repetition per Hz (or oscillation period). (d) Using x 1 p t q from part (a), consider the signal x 3 p t q = x 1 p t q ` u p t q ´ u p t ´ 2 q ˘ , with f 1 = 200 Hz. (i) Selecting a sampling frequency to avoid aliasing in the WD, use M ATLAB to simulate and plot the WD of x 3 p t q ; provide your code and plot (with correctly labeled axis). (ii) How does your plot compare to your answers in parts (a) and (c)? (e) Consider the signal x 4 p t q = 2cos p π 400 t q ` 2cos p π 600 t q , @ t . (i) How many CTs does the WD of x 4 p t q have? (ii) Provide the locations of the CTs in the time-frequency (TF) domain. Solution (a) We derived the WD of a complex sinusoid in class as x p t q “ e j 2 πf 0 t WD ÝÝÑ W x p t, f q “ δ p f ´ f 0 q . To show it W x p t, f q “ 8 ż ´8 x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ e ´ j 2 πτf d τ “ 8 ż ´8 e j 2 πf 0 p t ` τ 2 q e ´ j 2 πf 0 p t ´ τ 2 q e ´ j 2 πτf d τ “ 8 ż ´8 e ´ j 2 πτ p f ´ f 0 q d τ “ δ p f ´ f 0 q As cos p 2 πf 0 t q = 0 . 5 p e j 2 πf 0 t ` e ´ j 2 πf 0 t q is the sum of two complex sinusoids, its WD is the sum of the two auto WDs and one cross term (see pg 1 of Notes 3e for the derivation of the cross term). x p t q “ e j 2 πf 1 t ` e j 2 πf 2 t WD ÝÝÑ W x p t, f q “ δ p f ´ f 1 q ` δ p f ´ f 2 q ` 2 δ ´ f ´ f 1 ` f 2 2 ¯ cos p 2 πt p f 1 ´ f 2 qq Using this result, the WD of x a p t q is x a p t q “ 2cos p 2 πf 1 t q “ e j 2 πf 1 t ` e ´ j 2 πf 1 t WD ÝÝÑ W x a p t, f q “ δ p f ´ f 1 q ` δ p f ` f 1 q ` 2 δ p f q cos p 2 πt p 2 f 1 qq (b) We note that x b p t q = 2cos p 400 π p t ´ 5 qq e j 2 π p 50 q t is the same as x a p t q with f 1 = 200 Hz and shifted in time by 5 s and in frequency by 50 Hz. As the WD preserves both time and frequency shifts, then W x b p t, f q “ W x a p t ´ 5 , f ´ 50 q “ δ p f ´ f 1 ´ 50 q ` δ p f ` f 1 ´ 50 q ` 2 δ p f ´ 50 q cos p 2 π p t ´ 5 qp 2 f 1 qq “ δ p f ´ 250 q ` δ p f ` 150 q ` 2 δ p f ´ 50 q cos p 2 π p 400 qp t ´ 5 qq

© A. Papandreou-Suppappola, Arizona State University (c) The WD of x a p t q has one cross term that (i) is located at f = 0 (along the time axis); (ii) from the term cos p 2 πt p 2 f 1 qq , it oscillates in time with oscillation frequency 2 f 1 ; and it does not oscillate in frequency (d) (i) The M ATLAB code to generate and plot the WD of x a p t q is provided below. Note that the sampling frequency was chosen to be f s = 1000 which satisfies f max = 200 ď f s { 4 to avoid aliasing in the WD. f1=200;fs=1000;Ts=1/fs;Td=2;N=(Td/Ts)+1;t=linspace(0,Td,N); % Signal parameters x=cos(2 * pi * f1 * t).’; % Cosine signal % Using TFR toolbox [Wx,tt,ff]=tfrwv(x); % WD tt=tt * Ts;ff=ff * fs; % (t, f) in (s, Hz), 0 to fs/2, CTs at f=0 (difficult to see) % To plot WD from -fs/4 Hz to fs/4 Hz top=ceil(N/2);topn=top-1; Awdx=Wx([top:end 1:topn],:);freq=linspace(-fs/4,fs/4,N); xa=hilbert(x);Wxa=tfrwv(xa); % WD, analytic signal, 0 to fs/2, no CTs % Using SP toolbox [Wxx, fff, ttt]=wvd(x,fs); % appears to automatically take analytic signal % Plots of WD figure(1);subplot(411); imagesc(tt, ff, abs(Wx));axis xy; clim([-50 50]);colormap(jet);colorbar; % Change limits to better view WD title(’WD (0 to f_s/2) of cos(2 pi (200) t),f_s=1,000’); xlabel(’time, s’); ylabel(’frequency, Hz’); subplot(412);imagesc(tt, freq, abs(Awdx)); axis xy; clim([-50 50]); colormap(jet); colorbar; title(’WD (-f_s/4 to f_s/4) of cos(2 pi (200) t), f_s=1,000’); xlabel(’time, s’); ylabel(’frequency, Hz’); subplot(413); imagesc(tt, ff, abs(Wxa)); axis xy; clim([0 100]); colormap(jet); colorbar; title(’WD (0 to f_s/2) of analytic of cos(2 pi (200) t), f_s=1,000’); xlabel(’time, s’); ylabel(’frequency, Hz’); subplot(414);imagesc(ttt, fff, abs(Wxx)); axis xy; clim([0 100]); colormap(jet); colorbar; title(’WD SP toolbox, appears analytic cos(2 pi (200) t), f_s=1,000’); xlabel(’time, s’); ylabel(’frequency, Hz’); Figure 1(a) shows the WD of x c p t q in period p 0 , f s { 2 q . As expected, there are two auto terms that appear at the correct frequencies, f 1 = 200 Hz and ´ f 1 = ´ 200 = ´ 200 ` f s { 2 = ´ 200 ` 500 = 300 Hz. Figure 1(b) shows the same WD but within p´ f s { 4 , f s { 4 q . Although the CT is present at f = 0 , between the two original frequencies 200 Hz and -200 Hz, it is harder to see in Figure 1(a). The CT location and its geometry are better observed in Figure 2(a) from 0 to 20 Hz, Figure 2(a) from 480 to 500 Hz, and in Figure 2(c), -10 to 10 Hz.

© A. Papandreou-Suppappola, Arizona State University WD (0 to f s /2) of cos(2 pi (200) t), f s =1,000 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (a) time, s 0 100 200 300 400 500 frequency, Hz -50 0 50 WD (-f s /4 to f s /4) of cos(2 pi (200) t), f s =1,000 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (b) time, s -200 -100 0 100 200 frequency, Hz -50 0 50 Figure 1: WD of x c p t q using the TFR toolbox function tfrwvd for frequencies (a) 0 to f s { 2 ; (b) ´ f s { 4 to f s { 4 . Figure 3(a) shows the WD of the analytic signal of x c p t q ; as expected, there are no negative frequencies. All three figures were obtained using the TFR toolbox function tfrwvd . Figure 3(b) was obtained using the signal processing toolbox function wvd.m and the real signal x c p t q ; even though the analytic signal is not used, it appears that the function computes the WD of the analytic signal. WD of cos(2 pi (200) t) from 0 to 20 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (a) time, s 0 5 10 15 20 frequency, Hz -200 -100 0 100 WD of cos(2 pi (200) t) from 480 to 500 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (b) time, s 480 485 490 495 500 frequency, Hz -200 -100 0 100 WD of cos(2 pi (200) t) from -10 to 10 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (c) time, s -10 -5 0 5 10 frequency, Hz -200 -100 0 100 Figure 2: WD in Figure 1(a) (a) 0 to 20 Hz; (b) 480 to f s { 2 and (c) -10 to 10 Hz. (d) (ii) As expected from parts (a) and (b), the WD plot in Figure 1 shows two auto terms at 200 and -200 Hz and one cross term at 0 Hz.

© A. Papandreou-Suppappola, Arizona State University WD (0 to f s /2) of analytic of cos(2 pi (200) t), f s =1,000 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (a) time, s 0 100 200 300 400 frequency, Hz 0 20 40 60 80 100 WD SP toolbox, appears analytic cos(2 pi (200) t), f s =1,000 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (b) time, s 0 100 200 300 400 500 frequency, Hz 0 20 40 60 80 100 Figure 3: (a) WD of the analytic signal of x c p t q using the TFR toolbox, frequency 0 to f s { 2 ; (b) WD of x c p t q using the SP toolbox function wvd.m , frequency 0 to f s { 2 . (e) The signal x 4 p t q = 2cos p π 400 t q ` 2cos p π 600 t q , @ t can be written as e j 2 π 200 t ` e ´ j 2 π 200 t ` e j 2 π 300 t ` e ´ j 2 π 300 t (i) As there are N = 4 distinct components, then the number of CTs is given by N p N ´ 1 q{ 2 = 6 (ii) The CT locations are half way between any two from the set of frequencies f 1 = 200 , ´ f 1 = ´ 200 , f 2 = 300 , ´ f 2 = 300 Hz. 300 and -300 Hz. Thus, the CT locations occur, @ t , at frequencies f 1 ` p´ f 1 q 2 “ 0 Hz f 2 ` p´ f 2 q 2 “ 0 Hz f 1 ` f 2 2 “ 200 ` 300 2 “ 250 Hz f 1 ´ f 2 2 “ 200 ´ 300 2 “ ´ 50 Hz f 2 ´ f 1 2 “ 300 ´ 200 2 “ 50 Hz ´ f 1 ´ f 2 2 “ ´ 200 ´ 300 2 “ ´ 250 Hz

© A. Papandreou-Suppappola, Arizona State University 2. [2 pts] Demonstrate that the WD satisfies the unitarity property given below; show all steps of the derivation. ˇ ˇ ˇ ˇ ż x p t q y ˚ p t q dt ˇ ˇ ˇ ˇ 2 “ ż ż WD x p t, f q WD y p t, f q dt df Solution To show unitarity, replace the definition of WD 8 ż ´8 8 ż ´8 WD x p t, f q WD y p t, f q d t d f “ 8 ż ´8 8 ż ´8 ” x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ e ´ j 2 πfτ d τ ı” y ´ t ` η 2 ¯ y ˚ ´ t ´ η 2 ¯ e ´ j 2 πfη d η ı d t d f “ 8 ż ´8 8 ż ´8 8 ż ´8 8 ż ´8 x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ y ´ t ` η 2 ¯ y ˚ ´ t ´ η 2 ¯ e ´ j 2 πfτ e ´ j 2 πfη d t d f d τ d η “ 8 ż ´8 8 ż ´8 8 ż ´8 x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ y ´ t ` η 2 ¯ y ˚ ´ t ´ η 2 ¯ « 8 ż ´8 e ´ j 2 πf p τ ` η q d f ff d t d τ d η “ 8 ż ´8 8 ż ´8 8 ż ´8 x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ y ´ t ` η 2 ¯ y ˚ ´ t ´ η 2 ¯ δ p τ ` η q d t d τ d η 8 ż ´8 8 ż ´8 x ´ t ` τ 2 ¯ x ˚ ´ t ´ τ 2 ¯ y ´ t ´ τ 2 ¯ y ˚ ´ t ` τ 2 ¯ d t d τ 8 ż ´8 8 ż ´8 « x ´ t ` τ 2 ¯ y ˚ ´ t ` τ 2 ¯ ff« x ´ t ´ τ 2 ¯ y ˚ ´ t ´ τ 2 ¯ ff ˚ d t d τ Let t 1 = t ` τ { 2 and t 2 = t ´ τ { 2 ; then dt 1 dt 2 = dt dτ , and 8 ż ´8 8 ż ´8 WD x p t, f q WD y p t, f q dt df “ ˜ 8 ż ´8 x p t 1 q y ˚ p t 1 q d t 1 ¸˜ 8 ż ´8 x p t 2 q y ˚ p t 2 q d t 2 ¸ ˚ “ ˜ 8 ż ´8 x p t q y ˚ p t q d t ¸˜ 8 ż ´8 x p t q y ˚ p t q d t ¸ ˚ “ ˇ ˇ ˇ ˇ ˇ 8 ż ´8 x p t q y ˚ p t q d t ˇ ˇ ˇ ˇ ˇ 2

© A. Papandreou-Suppappola, Arizona State University 3. [3 pts] The linear frequency-modulated (LFM) signal is often used in radar as its parameters can be selected to provide high resolution in both range and range-rate. These parameters are obtained from estimates of the received signal’s multipath (time shift) and Doppler (frequency shift). Consider the LFM signal s p t q = g p t q e j 2 π p α { 2 q t 2 , where α = 40 , 000 Hz 2 , effective duration T d = 1 s and g p t q is a Gaussian window. The received (noiseless) signal from two transmissions is r p t q “ r 1 p t q ` r 2 p t q “ g p t ´ 0 . 2 q e jπ p 20 , 000 t ` α p t ´ 0 . 2 q 2 q ` g p t ´ 0 . 4 q e jπ p 20 , 000 t ` α p t ´ 0 . 4 q 2 q , t P p 0 , 2 q (a) What is the minimum sampling frequency f s needed to avoid aliasing when computing the WD of r p t q ? (b) Assuming that the (non-aliased) WD of s p t q is very highly-localized in TF, and thus appears as a line with starting TF point (0, 0) and ending TF point (1 s, 40 kHz), express the WD of r p t q in terms of the WD of s p t q ; do not compute the WD of s p t q . (c) How many auto-terms (ATs) and cross-terms (CTs) does the WD of r p t q have? (d) (i) Sketch the WD of r p t q in TF, with correctly labeled axis. (ii) Provide the location of the starting and ending TF points for each AT and each CT. (iii) Provide the direction and frequency of oscillation for each CT. Solution (a) The LFM signal is given by s p t q = g p t q e jπα t 2 = g p t q e j 2 π 20 k t 2 , t P p 0 , 1 q , where 20 k = 20000 Hz. The received signal can also be written in terms of TF shifted versions of the transmit signal, r p t q “ r 1 p t q ` r 2 p t q “ s p t ´ 0 . 2 q e j 2 π 10 k t ` s p t ´ 0 . 4 q e j 2 π 10 k t , t P p 0 , 2 q To avoid aliasing in the WD, the maximum frequency f max of r p t q must satisfy f max ď f s { 4 . However, Since t P p 0 , 2 q , the Fourier transform of r p t q is is zero for ´ f s { 2 to 0; as a result, we can select f s using f max ď f s { 2 . To find f max , we use the instantaneous frequency (IF) of r 1 p t q and r 2 p t q . Specifically, using the IF of the transmit signal s p t q , IF s p t q = α t = 40 k t , t P p 0 , 1 q , IF r 1 p t q “ 10 k ` α p t ´ 0 . 2 q “ 10 k ` 40 k p t ´ 0 . 2 q , t P p 0 . 2 , 1 . 2 q IF r 2 p t q “ 10 k ` α p t ´ 0 . 4 q “ 10 k ` 40 k p t ´ 0 . 4 q , t P p 0 . 4 , 1 . 4 q the maximum frequency of r 1 p t q is IF r 1 p 1 . 2 q = 50 kHz and the maximum frequency of r 2 p t q is IF r 2 p 1 . 4 q = 50 kHz Thus, to avoid aliasing, we select f s ě 100 kHz (b) Denoting the WD of s p t q as W s p t, f q , then W r p t, f q “ W r 1 p t, f q ` W r 2 p t, f q ` W r 1 r 2 p t, f q ` W r 2 r 1 p t, f q “ W s p t ´ 0 . 2 , f ´ 10 k q ` W s p t ´ 0 . 4 , f ´ 10 k q ` 2 Re t W r 1 r 2 p t, f qu where we used W r 2 r 1 p t, f q = W ˚ r 2 r 1 p t, f q . With two components, r p t q has L = 2 ATs and M = 1 CT. We can express the CTs in terms of W s p t q as (see Notes 3d) C r 1 r 2 p t, f q “ 2 Re t W r 1 r 2 p t, f qu “ 2 W s p t ´ t 12 , f ´ f 12 q cos ´ 2 π p ν 12 t ´ τ 12 f q ` 2 πf 12 τ 12 ¯

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