ECE-34: Signals and Systems Summer 23 PROBLEM One period of the DTFS coefficients is given by: X[] = (/3) 2, 8. Solutions - Homewor # 3 a) What is the fundamental period 'N' of the time-domain signal x[n]? b) Using MATLAB, plot X[] for three periods. Plot the magnitude and the phase spectra. c) Find the time-domain signal x[n] (provide x[n] as a function of 'n'). Plot x[n] for three periods. a) The period of is the same as that of. N = 9. b) X[] = (/3) 2, 8. Note that this equation only wors for one period. We have to generate infinite replicas on both sides. clear all; close all; clc = :8; X = (/3).^(2*); % The function as given only wors from to 8 X_3p = [X X X]; % Here, we generate replicas _3p = :26; n = :26; % 3 periods x = 8./(9 - exp(i*2*n*pi/9)); % This function is periodic (N=9), so it % wors for all 'n' figure; subplot (2,,), stem(_3p, abs(x_3p),'.b'); axis ([ 28.2]); set(gca, 'Fontsize',8); xlabel(''); title (' X[], 3 periods'); subplot (2,,2), stem(_3p, angle(x_3p),'.r'); set(gca, 'Fontsize',8); xlabel(''); title ('arg(x[]), 3 periods'); figure; subplot (2,,), stem(n,real(x),'.b'); set(gca, 'Fontsize',8); xlabel('n'); title ('Re(x[n]), 3 periods'); subplot (2,,2), stem(n,imag(x),'.r'); set(gca, 'Fontsize',8); xlabel('n'); title ('Im(x[n]), 3 periods'); X[], 3 periods.8.6.4.2 5 5 2 25 arg(x[]), 3 periods.5 -.5-5 5 2 25 3
ECE-34: Signals and Systems Summer 23 c) Re(x[n]), 3 periods.8.6.4.2 5 5 2 25 3 n. Im(x[n]), 3 periods.5 -.5 -. 5 5 2 25 3 n PROBLEM 2 Identify the appropriate Fourier representation (FT, DTFT, FS, DTFS) for each of the following signals. If the signals are periodic, provide the fundamental period and the fundamental angular frequency a) x[n] = cos((6 /3)n + /3) b) x[n] = exp(j( /4)n) c) x(t) = cos(t/6) d) x(t) = e -t u(-t + 2) e) x(t) = sin(( /5)t) f) x(t) = cos(( /3)t + /5) g) x[n] = [n+2] + [n-4] h) x[n] = (3/8) n u[n-3] Once you identified the appropriate Fourier representation, use the defining equation to obtain the DTFS coefficients, the FS coefficients, the DTFT, or the FT. a) Signal is periodic with DTFS Since the signal is a cosine, we can use the inspection method: Thus, we have the DTFS coefficients for one period ( ):
ECE-34: Signals and Systems Summer 23 b) Signal is periodic with DTFS Since the signal is a complex exponential, we can use the inspection method: Thus, we have the DTFS coefficients for one period ( ): c) Signal is periodic with FS Since the signal is a cosine, we can use the inspection method: Thus, we have the FS coefficients for all : d) Non-periodic FT u(-t+2) The integral diverges, thus the FT representation does not exist. 2 t e) Signal is periodic with FS. Signal is a sine, thus we can use the inspection method: Thus, we have the FS coefficients for all : f) Signal is periodic with FS Since the signal is a cosine, we can use the inspection method: Thus, we have the FS coefficients for all :
ECE-34: Signals and Systems Summer 23 g). Non-periodic signal DTFT h). Non-periodic signal DTFT PROBLEM 3 Use the defining equation for the DTFT to evaluate the frequency-domain representations of the following signals. You must show the procedure. a) x[n] = (3/5) n (u(n-4) - u(n+4)) b) x[n] = b n, b < c) x[n] = 2 [5-3n] d) x[n] = (/4)( [n] + 3 [n-] + 2 [n] + [n-3]) e) x[n] = 2 + e -3n a) -6-5 -4-3 u[n-4] - u[n+4] - 2 3 4 5 6 n - b) c) [n+5] = v[n] -6-5 -4-3 - 2 3 n -6-5 -4-3 [-3n+5] = v[-3n] - 2 3 n
ECE-34: Signals and Systems Summer 23 d) e) tends to infinity. Thus, is undefined PROBLEM 4 Determine the time-domain signals corresponding to the following DTFTs. You must show the procedure. a) X(e j ) = sin(2 ) + jcos(2 ) b) X(e j ) = 3sin(4 ) c) X(e j -j /2 ) = (/2)e d) X(e j ) = cos( ) + sin( /2) a) Thus: b) We use the time-shift property of the DTFT along with the fact that the DTFT of an impulse is. And we determine that: In exercise 4(a), we demonstrate that the DTFT of is Finally: c)
ECE-34: Signals and Systems Summer 23 d) Finally: PROBLEM 5 The following LTI system has an input described by: x[n] h[n] y[n] = x[n]*h[n] x[n] = sin((5 /7)n + /8) The Fourier representation of the impulse response h[n] is given by: H[] = e -, on N-. a) Determine the period 'N' of the signal x[n]. b) Determine the DTFS coefficients X[]. c) Obtain the frequency domain representation Y[] of the output signal y[n]. d) Using MATLAB, plot X[], H[], and Y[] for three periods. Plot the magnitude and the phase spectra. a) b) Since the signal is a sine, we can use the inspection method: Thus, we have the DTFS coefficients for one period ( ):
ECE-34: Signals and Systems Summer 23 c), on 3 is defined for a different range. We can either redefine from -4 to 8, or from to 3. Let's redefine from to 3: One period X[] One period -6-5 -4-3 - 2 3 4 5 6 7 8 9 2 3 4 Now, using the convolution property, we get the DTFS for one period: d) MATLAB: clear all; close all; clc = :3; _3p = :4; % In the period to 3: H =.*exp(-); X(:4) = ; X(6) = (/2i)*exp(i*pi/8); X() = -(/2i)*exp(-i*pi/8); Y(:4) = ; Y(6) = (/2i)*exp(i*pi/8)*4*5*exp(-5); Y() = -(/2i)*exp(-i*pi/8)*4*9*exp(-9); X_3p = [X X X]; % Here, we generate replicas H_3p = [H H H]; % Here, we generate replicas Y_3p = [Y Y Y]; % Here, we generate replicas figure; subplot (2,,), stem(_3p, abs(x_3p),'.b'); axis ([ 45.8]); set(gca, 'Fontsize',8); xlabel(''); title (' X[], 3 periods'); subplot (2,,2), stem(_3p, angle(x_3p),'.r'); set(gca, 'Fontsize',8); xlabel(''); title ('arg(x[]), 3 periods'); figure; subplot (2,,), stem(_3p, abs(h_3p),'.b'); set(gca, 'Fontsize',8); xlabel(''); title (' H[], 3 periods'); subplot (2,,2), stem(_3p, angle(h_3p),'.r'); set(gca, 'Fontsize',8); xlabel(''); title ('arg(h[]), 3 periods');
ECE-34: Signals and Systems Summer 23 figure; subplot (2,,), stem(_3p, abs(y_3p),'.b'); set(gca, 'Fontsize',8); xlabel(''); title (' Y[], 3 periods'); subplot (2,,2), stem(_3p, angle(y_3p),'.r'); axis ([ 45.5 2.5]); set(gca, 'Fontsize',8); xlabel(''); title ('arg(y[]), 3 periods');.8 X[], 3 periods.6.4.2 5 5 2 25 3 35 4 45 2 arg(x[]), 3 periods - 5 5 2 25 3 35 4 45.4 H[], 3 periods.3.2. 5 5 2 25 3 35 4 45 arg(h[]), 3 periods.5 -.5-5 5 2 25 3 35 4 45
ECE-34: Signals and Systems Summer 23.25 Y[], 3 periods.2.5..5 5 5 2 25 3 35 4 45 2 - arg(y[]), 3 periods 5 5 2 25 3 35 4 45 PROBLEM 6 Use the properties of Fourier representation (e.g., time-differentiation, convolution, time-shift, frequencyshift) to find the FT of: Note: '*' denotes convolution. Hint: It might help you that the FT of e -at u(t) is /(a+j ) Differentiation Property of FT: If Then:, then: Convolution Property of FT: Then: Knowledge of a common FT pair: Then: where:, and Also: where:, and Time Shift Property of FT: If Then:, then:
ECE-34: Signals and Systems Summer 23 Finally: PROBLEM 7 Given the following DTFS pair ( = /): Evaluate the time-domain signal y[n] for the following DTFS coefficients Y[]. These DTFS coefficients Y[] happen to have a relationship with the DTFS coefficients X[]. You can use properties of the DTFS. a) Y[] = (/2)(X[-4] + X[+4]) b) Y[] = 3X[] c) Y[] = X[] (*)X[], where (*) denotes periodic convolution. a) We use the frequency shift property: b) Here, we also use the frequency shift property: c) Here, se use the multiplication property: