MAE143A Signals & Systems - Homework 8, Winter 2013 due by the end of class Tuesday March 5, 2013.

Transcription

1 MAE43A Signals & Systems - Homework 8, Winter 3 due by the end of class uesday March 5, 3. Question Measuring frequency responses Before we begin to measure frequency responses, we need a little theory... Oh No! Part (a): Use the cos(a + B) formula to show that cos (θ) = [ + cos(θ)], () sin (θ) = [ cos(θ)]. () And use the sin(a + B) formula to show that: sin(θ) cos(θ) = sin(θ). (3) cos(θ) = cos(θ + θ), he first two answers follow. o show the third relation; = cos(θ) cos(θ) sin(θ)sin(θ), = cos (θ) sin (θ), cos (θ) [ cos (θ)], = [ sin (θ)] sin (θ), cos (θ), = sin (θ). sin(θ) =sin(θ + θ), = sin(θ) cos(θ) + cos(θ) sin(θ), =sin(θ) cos(θ). Part (b): Now consider: the input signal, u(t) = A cos(ωt + φ), its so-called quadrature signal z(t) = A sin(ωt + φ), and the corresponding output signal y(t) =AB cos(ωt + φ + ψ). Clearly, we may write y(t) =AB cos(ωt + φ) cos(ψ) AB sin(ωt + φ)sin(ψ). (4) Use (-3) to show the following relationships. y (t) dt = A B, u (t) dt = A, u(t)y(t) dt = A B z(t)y(t) dt = A B cos(ψ), sin(ψ).

2 Hence, B = = y (t) dt u (t) dt ψ = atan = atan y (t) dt u (t) dt z(t)y(t) dt, z(t)y(t) dt, u(t)y(t) dt u(t)y(t) dt, (5). (6) hese formulæ are highly useful in understanding communications systems and radio. y (t) dt = A B y (t) dt = [ + cos(ωt +φ +ψ)] dt, = A B + [sin(ω +φ +ψ) sin( + φ +ψ)], 4ω A B = A B. he u expression is the same, but now has B =and ψ =. Now consider, using (4), + [sin(ω +φ +ψ) sin( + φ +ψ)] 4ω u(t)y(t) =A B cos (ωt + φ) cos(ψ) A B cos(ωt + φ)sin(ωt + φ)sin(φ), = A B {[ + cos(ωt +φ)] cos(ψ) sin(ωt +φ)sin(ψ)}, u(t)y(t) dt = A B cos(ψ), and taking its as yields the results, because the other terms all tend to zero. Further, z(t)y(t) =A B sin(ωt + φ) cos(ωt + φ) cos(ψ) A B sin (ωt + φ)sin(φ), = A B {[sin(ωt +φ) cos(ψ) [ cos(ωt +φ)] sin(ψ)}, z(t)y(t) dt = A B cos(ψ),., Question matlab and the estimation of frequency response Part (a): We are going to use the formulæ (5-6) to estimate a frequency response in matlab.

3 Download the code trfnest.m from the class website, either ED or oodgeroo. Run the code and explain what the code is doing in terms of the formulæ above. What is happening when the two different plotted curves diverge at the right-hand ends? Part (b): Create the uvec and corresponding yvec from Part (a) for each of three frequencies: ω=.5,.5, 9 radians/sec. Call them u, y, u, y, u3, y3 respectively. Plot each pair against time using, say, plot(t,u,t,y). Zoom in and identify how to measure the magnitude and phase from each plot. You already have the correct answers from Part (a). So indicate on the time plots the quantities needed to calculate gain and phase. Next plot the input and output signals against one another using, say, plot(u,y). How now can the magnitude and phase be measured from this figure, which is called a Lissajous figure? [See: [Hint: Consider that the x signal, u, is A cos(ωt + φ) and the y signal, y, is AB cos(ωt + φ + ψ) as in Part (a) and ask about when these signals cross the axes or reach their extremities.]

4 Problem : Part A: he code is using input and quadrature signals over a range of frequencies as well as the corresponding response signals to find β and φ over a range of frequencies. It does this by numerically integrating the quantities in equations (5) and (6). We will end up with a graph of relative magnitudes and phases. It also uses Matlab s built in command bode in order to compare. Because we are sampling at a finite rate, as we increase the frequency of the input (and correspondingly, the output) there will be a point at which the signal is no longer being accurately represented. We call this point the Nyquist frequency. his is what is happening on the plot. Note: he plot that results from the code given in part A has units of Hertz, not rad/sec. his differs from the label. Part B: We can measure the relative magnitudes in Figure by comparing the height of the peaks of the input and output signals. We can measure the phase by seeing how far the peaks of the input and output are shifted in time and converting by Φ = 8 freq(rad/sec). π he relative amplitudes can be devised from Figure by comparing the maximum values the graphs obtain on the X and Y axis. he phase can be determined by the direction the graphs are plotted (clockwise or counterclockwise) as well as the slope and shape of the graphs.

5 .4 w=.5 rad/sec.5 w=.5 rad/sec. w=9 rad/sec y y y u..5.5 u..5.5 u3 Figure : Plotting the Lissajous figures.8 w=.5 rad/sec u y.8 w=.5 rad/sec u y.8 w=9 rad/sec u3 y Signal Value Signal Value Signal Value ime (S) ime (S) ime (S) Figure : Plotting the response with the three inputs

6 = 6; % time frame in seconds for experiment s =.; % sample time milliseconds 3 t = [:s:]'; % time vector sys = tf([ ],[. ]); % sys with zero at s=,... poles at s=., 6 7 % Part b 8 w=[.5.5 9]; 9 u=cos(w()**pi*t); u=cos(w()**pi*t); u3=cos(w(3)**pi*t); y = lsim(sys,u,t); 3 y = lsim(sys,u,t); 4 y3 = lsim(sys,u3,t); 5 6 figure() 7 subplot(,3,) 8 plot(t,u,t,y) 9 x([,5]) title('w=.5 rad/sec') xlabel ('ime (S)') ylabel ('Signal Value') 3 legend('u','y') 4 subplot(,3,) 5 plot(t,u,t,y) 6 x([,5]) 7 title('w=.5 rad/sec') 8 xlabel ('ime (S)') 9 ylabel ('Signal Value') 3 legend('u','y') 3 subplot(,3,3) 3 plot(t,u3,t,y3) 33 x([,.5]) 34 title('w=9 rad/sec') 35 xlabel ('ime (S)') 36 ylabel ('Signal Value') 37 legend('u3','y3') figure() 4 subplot(,3,) 4 plot(u,y) 4 title('w=.5 rad/sec') 43 xlabel ('u') 44 ylabel ('y') 45 subplot(,3,) 46 plot(u,y) 47 title('w=.5 rad/sec') 48 xlabel ('u') 49 ylabel ('y') 5 subplot(,3,3) 5 plot(u3,y3) 5 title('w=9 rad/sec') 53 xlabel ('u3') 54 ylabel ('y3') 3 Figure 3: Matlab code for ODE function in Problem, Part B