Assignment 4 Solutions Continuous-Time Fourier Transform

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1 Assignment 4 Solutions Continuous-Time Fourier Transform ECE 3 Signals and Systems II Version 1.01 Spring Properties of complex numbers. Let c 1 α 1 + jβ 1 and c α + jβ be two complex numbers. a. What is the magnitude of c 1? Express your answer in terms of α 1 and β 1. c 1 α 1 + β 1. b. Show that (c 1 + c ) c 1 + c. (c 1 + c ) ([α 1 + jβ 1 ]+[α + jβ ]) ([α 1 + α ]+j[β 1 + β ]) [α 1 + α ] j[β 1 + β ] [α 1 jβ 1 ]+[α jβ ] c 1 + c c. Show that (c 1 c ) c 1 c. (c 1 c ) ( c 1 e j c1 c e j c) ( c 1 c e ) j( c1+ c ) c 1 c e j( c1+ c ) c 1 e j c1 c e j c c 1c. Frequency concepts. Explain in your own words why there is no natural interpretation of high-frequency in continuous-time. Compare and contrast with the discrete-time case. In discrete time the highest perceivable distinct frequency is π radians/sample. Sinusoidal signals that oscillate at a high frequency than this are indistinguishable from sinusoids with 1

2 lower frequencies. This is due to the fact that discrete-time sinusoids (complex and real) are not distinct, e j(ω+π)n e jωn. However, continuous-time sinusoids (real and complex) are distinct and there is no upper limit to the highest perceivable frequency, in general. Thus, the notions of low-, mid-, and high- frequencies depends on the application when considering continuous-time signals and systems, unlike discrete-time signals and systems. 3. Relationship to LTI systems. a. Solve for the transfer function, H(s), for a first-order highpass filter with a passband gain of 5 and a cutoff frequency of 500 Hz. s H(s) 5 b. Plot the impulse response of this system. First, we must convert H(s) to a proper rational function so that we can solve for the impulse response. s H(s) 5 5 ( 1000π ) π h(t) 5δ(t) 5000πe 1000πt u(t) Impulse 5δ(t) notshown Impulse Response h(t) Time (s) x 10 4 c. What range of frequencies does this filter amplify? Must solve for the frequency where the gain is 1 to identify. Note that if the gain is 1,

3 the magnitude square gain is also 1. H(jω) 1 5 jω jω + ω c ω 5 ω + ωc ω + ωc 5ω ω c 4ω ω ω c π rad/s Therefore all frequencies above f 10.1 Hz are amplified. d. What range of frequencies does this filter attenuate? All frequencies below 10.1 Hz e. Use MATLAB to create the Bode plot of this system. Magnitude $ H(j omega) $ (db) Phase (degrees) Frequency (rad/s) Note that this is consistent with the previous two answers. f. Suppose a sinusoid x(t) 10 cos(1000πt +30 ) is applied to the system. What is the output? y(t) 10 H(j1000π) cos(1000πt H(j1000π)) 10(3.536) cos(1000πt ) cos(1000πt +75 ) 3

4 g. Repeat the previous question for x(t) 10 cos(100πt +45 ). y(t) 10 H(j100π) cos(100πt H(j100π)) 10(0.4975) cos(100πt ) cos(100πt ) h. Repeat the previous question for x(t) 10 sin(100πt 135 ). y(t) 10 H(j100π) sin(100πt H(j100π)) 10(0.4975) sin(100πt ) sin(100πt ) function [] LtiSystemsExample; cutofffrequency 1000*pi; % Cutoff frequency timerange [-0.e-3 e-3]; % Time range frequencyrange cutofffrequency*[ ]; frequenciesevaluate [100*pi 1000*pi]; % Frequencies to evaluate at (rad/s) % Preprocessing t linspace(timerange(1),timerange(),1000); % Create 1000 points over time range u (t>0); wc cutofffrequency; h 5*wc*exp(-wc*t).*u; w logspace(log10(frequencyrange(1)),log10(frequencyrange()),1000); sys tf([5 0],[1 wc]); [magnitude,phase] bode(sys,w); magnitude squeeze(magnitude); phase squeeze(phase); % Unit step function % Plot of the Signal figure; FigureSet(1,5); h plot(t,h); set(h, LineWidth,1.5); xlim(timerange); ylim([0 5*wc*1.05]); FigureLatex; xlabel( Time (s) ); ylabel( Impulse Response $h(t)$ ); title( Impulse $5\delta(t)$ not shown ); box off; AxisLines; AxisSet(8); print( LtiSystems-ImpulseResponse, -depsc ); % Plot of the Bode Plot 4

5 figure; FigureSet(1,5); FigureLatex; subplot(,1,1); h semilogx(w,0*log10(magnitude), r ); set(h, LineWidth,1.5); xlim(frequencyrange); ylim([-30 0]); ylabel( Magnitude $ H(j omega) $ (db) ); box off; AxisLines; subplot(,1,); h semilogx(w,squeeze(phase), r ); set(h, LineWidth,1.5); xlim(frequencyrange); ylim([0 100]); xlabel( Frequency (rad/s) ); ylabel( Phase (degrees) ); box off; AxisSet(8); print( LtiSystems-BodePlot, -depsc ); % Evaluate at Specified Frequencies for c11:length(frequenciesevaluate) [magnitude,phase] bode(sys,frequenciesevaluate(c1)); fprintf( Frequency:%6.1f pi rad/s Magnitude:%5.3f Phase:%5.f degrees\n,frequenciesevaluate(c1)/pi,abs(magnitude),phase end 4. Properties. a. What properties does the CTFT have for all real signals? Complex conjugate symmetry, X( jω)x (jω). b. Real, even signals? Real-valued, X(jω)Re{X(jω)} and even symmetry, X( jω)x(jω). c. Real, even signals with finite energy? Real-valued, X(jω)Re{X(jω)}; even symmetry, X( jω)x(jω); and finite value for all frequencies X(jω) <. d. Real, odd signals with finite power? Imaginary-valued, X(jω) Im{X(jω)}; odd symmetry, X( jω) X(jω); and finite value for all frequencies X(jω) <. e. Real, periodic signals? Complex conjugate symmetry, X( jω) X (jω) and infinite at integer multiples of the fundamental frequency X(jω o l) (an impulse train). 5. Transform examples. Find the Fourier transforms of the following signals. Which of these signals have Fourier transforms that converge? Which of these signals have Fourier transforms that are real? imaginary? a. x(t) cos(1000t). 5

6 x(t) 1 [ e j1000t +e j1000t] e jωot FT πδ(ω ω o ) α 1 x 1 (t)+α x (t) FT α 1 X 1 (jω)+α X (jω) x(t) FT πδ(ω ω o )+πδ(ω + ω o ) This CTFT did not converge (the CTFT is infinite at some frequencies). This is consistent with the knowledge that the signal is a power signal. The signal is even, so the transform is real. b. x(t) 3δ(t). X(jω) 3e jω0 3 x(t)e jωt dt 3δ(t)e jωt dt The CTFT did converge (is finite at all frequencies). transform is real. The signal is even and the c. x(t) 13 cos(100t) 7 sin(500t) cos(100t) FT πδ(ω 100) + πδ(ω + 100) sin(500t) FT jπ [δ(ω + 500) δ(ω + 500)] X(jω)13π [δ(ω + 100) + δ(ω 100)] j7π [δ(ω + 500) δ(ω + 500)] The CTFT did not converge, is not real, and is not imaginary. d. x(t) p 50 (t). 6

7 X(jω) x(t)e jωt dt e jωt dt jω e jωt [ e jω50 e jω50] jω j sin(50ω) jω sin(50ω) ω The CTFT did converge and is real. This is consistent with the knowledge that the signal has finite energy and is even, respectively. 7

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