MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions


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1 MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter, F H (s), with a cutoff frequency of ω H. F H (s) = 1 [(ω H /s) (ω H /s)+1][(ω H /s) (ω H /s)+1]. (1a) Express this filter as a rational function of s. F H (s) = s 4 [s ω H s+ω 2 H ][s ω H s+ω 2 (1b) H ]. 1b. Combining the lowpass and highpass filters, describe how a bandpass filter, F B (s), may be constructed. Put the lowpass and highpass filters in series to make a bandpass filter. (1c) Should you add them or multiply them? why? You should multiply them, because they are being applied in series (just like we multiply D(s) and G(s), which also appear in series). (1d) Compute the single rational transfer function of the necessary bandpass filter, F B (s). F B (s) = F L (s) F H (s) = s 4 [(s/ω L ) (s/ω L )+1][(s/ω L ) (s/ω L )+1][s ω H s+ω 2 H ][s ω H s+ω 2 H ] (1e) 1c. Now specify how you should select ω L and ω H. You may take ω L to be about 3000 Hz and ω H to be about 2000 Hz to form the desired bandpass filter. Actually, since both F L (s) and F H (s) mess up the phase of the signal near their respective cutoff frequencies (that is, no linear filter is ideal...), it is preferable to make ω L a tad higher than this, and ω H a tad lower than this. (1f) 1d. You notice that the mic is about 1 cm closer to the sound source than the speaker, and that the speed of sound is about v = 350 m/s. Assuming a purely sinusoidal tone at f = 2500 Hz [and, thus, a sinusoid with a wavelength of λ = v/ f = 0.14m), about how much phase change will this cause? It will cause about (1/14) 360 = 26 of phase lead. (1g) What kind of compensation can you apply in your circuit to correct for this? You can apply a bit of lag compensation, with D(s) = (s + z)/(s + p) with p < z, taking p/z 3 and pz = 2500 Hz = π rad/sec. This better synchronizes the phase of the output from the speaker with the audio signal it is designed to cancel, which takes a while to travel from the microphone to the speaker. (1h) Question 2. SYSTEMS: Linearizing the dynamics of a pendulum. The nonlinear equations of motion of a pendulum are ml 2 d2 θ dt 2 = u mglsin(θ). 2a. If the pendulum is held at a constant angle θ, calculate the corresponding equilibrium value of the torque u. u = mgl sin(θ) Note also that dθ/dt = 0. (2a)
2 2 MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions 2b. Assuming that the perturbation quantities {θ,u } are small, linearize the full nonlinear equations of motion about the equilibrium condition. (sin identity) (small angle approximation) (equilibrium condition) (rearrange) Compute the linear transfer function from U (s) to Θ (s). dt 2 = (u+u ) mglsin(θ+θ ) dt 2 = (u+u ) mgl{sin(θ)cos(θ )+cos(θ)sin(θ )} dt 2 = (u+u ) mgl{sin(θ) 1+cos(θ) θ ]} ml 2 d2 (θ ) dt 2 = u mglcos(θ) θ [ ] d 2 dt 2 + a 0 θ = b 0 u where a 0 = g l cos(θ), b 0 = 1 ml 2 Θ (s) U (s) = b 0 s 2 + a 0 Assuming m = 1 kg, l = 1 m, and g = 9.8 m/s 2, where are the poles and/or zeros of this transfer function? Two poles at ± a 0 = ± g cos(θ)/l. 2c. If θ = 0: Two zeros at infinity (a.k.a. the resaurant at the end of the universe). u = 0, two poles at ±iω d = ±i 9.8, two zeros at infinity. System is neutrally stable because of the poles on the imaginary axis. The impulse response is a sinusoidal oscillation at frequency ω d that neither grows nor decays. (2b) (2c) (2d) (2e) 2d. If θ = π/4: u = mgl.707, two poles at ±iω d = ±i , two zeros at infinity. System is neutrally stable because of the poles on the imaginary axis. The impulse response is a sinusoidal oscillation at frequency ω d that neither grows nor decays. (2f) 2e. If θ = π/2: u = mgl, two poles at the origin (a double integrator!), two zeros at infinity. System is neutrally stable because of the poles on the imaginary axis. By taking the inverse Laplace transform of Y(s) = G(s) U(s) = (1/s 2 ) 1, the impulse response is seen to be a unit ramp, y(t) = t for t > 0. (2g) 2f. If θ = 3π/4: u = mgl.707, two poles at ± , two zeros at infinity. System is unstable stable because of the pole in the RHP. The impulse response is a diverging exponential (plus a converging exponential). (2h) 2g. If θ = π: u = 0, two poles at ± 9.8, two zeros at infinity. System is unstable stable because of the pole in the RHP. The impulse response is a diverging exponential (plus a converging exponential). (2i)
3 MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions 3 Question 3. (& CONTROL): Root locus and Bode tools in classical control design. Consider the unstable system G(s) = Y(s) U(s) = 1 (s 1)(s+10). 3a. Taking D(s) =, rewrite H(s) = G(s)D(s)/[1 + G(s)D(s)] as a rational function of s: H(s) = +(s 1)(s+10) = s 2 + 9s+( 10) = (s+z + )(s+z ). (3a) Compute the exact pole locations of the closedloop system: z ± = [ 9 ± 81 4( 10)]/2. (3b) 3b. Taking D(s) =, accurately plot the root locus plot: Note that the pair of closedloop poles move in from the openloop poles [that is, from z ± = ( 9 ± 121)/2 = ( 9 ± 11)/2 = { 10,1}] for small, along the real axis to z ± = ( 9 ± 0)/2 = 4.5 (i.e., both at the midpoint between the openloop poles) for = 81/4+10 = 30.25, and then move out along the vertical line directly above and below this midpoint location to z ± = ( 9 ± i 4( 10) 81)/2 for > The resulting plot is thus given below. (3c) c. Is proportional control sufficient to stabilize the system? Compute the minimum value of necessary to stabilize the system. Yes. > 10 moves both poles into the LHP. (3d) 3d. If is selected to achieve the fastest rise time possible while keeping M p 5%, where would the closedloop poles be? What is the natural frequency ω n of these poles? At s = 4.5 ± 4.5i, as marked above, with ζ = and ω n = (3e) [This is achieved with 50.5.] 3e. The openloop system G(s)D(s) has two more poles than zeros. Explain where the other two zeros of this system are, in terms of the intuitive mapping of the complex plane introduced by Riemann. The other two zeros are effectively at infinity in the extended complex plane. When the extended complex plane is mapped onto the Riemann sphere, with the origin mapped to the south pole, the point at infinity is mapped to the north pole (a.k.a. the Restaurant at the End of the Universe). (3f)
4 4 MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions 3f. Sketch the Bode plot corresponding to G(s). If proportional control D(s) = is applied with the gain adjusted to give a crossover frequency ω c = 5 rad/sec, what is the approximate phase margin in the closedloop system? What value of achieves this crossover frequency? Noting the Bode plotting rules for both unstable poles as well as stable poles, the Bode plot is given above. Note that the slope goes from 0 to 1 to 2, and the phase gives a bump akin to that of a lead compensator, with a maximum of 55 of phase lead. To get crossover at ω c = 5 rad/sec, the gain should set as follows: G(iω c )D(iω c ) = iω c 1 iω c + 10 = 1 = Crossover is not quite at the peak of the bump in the phase, resulting in a PM of only about 50. (3g) 3g. Noting the approximate design guides available in the root locus and Bode settings, explain how these two interpretations are consistent. In the root locus plot, we saw that θ = 45 ζ = sin(θ).7, and thus we expected M p 5%. In the Bode plot, we saw that PM = 50, and thus ζ PM/100 =.5; we will thus expected a bit more overshoot with this design (M p 15%). Significantly, note that these design guides are only approximate!! (3h)
5 MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions 5 Question 4. I want my... I want my PID! (or not...) A PID compensator may be written ( D PID (s) = p 1+ 1 ) T I s + T s 2 + s Ds = p T D s where = p T D and, usually, T I > T D. T D + T 1 I T D = (s+z +)(s+z ), s 4a. Assuming T I T D, calculate simple expressions for z + and z, and draw the corresponding Bode plot. z ± = 1 [ 1 ± ] 1 4T D /T I 1 [ ] { 1 1 ±(1 2T D /T I ) =, 1 } 2T D 2T D T I T D The Bode plot of D PID (s) is given below. (4a) b. If there is noise entering the system with very high freqeuncy components, describe the problem with using this type of compensator. High frequencies are amplified a lot in the loop due to the derivative compensation of the PID. We would thus drive our actuators hard (burning them out) while chasing high frequency noise. (4b) 4c. How serious of a problem are delays when closing feedback loops? Use what you know about Pade expansions of e ds to explain clearly what the problem with delays is all about. Delays are a significant problem when closing control loops. Interpreting the Laplace transform e ds of the delay function δ σ (t d) using a Pade approximation, there are poles in the LHP and zeros in the RHP. Recalling that the root locus goes from the openloop poles to the openloop zeros, increasing the feedback gain too far can thus lead to closedloop instability. (4c) 4d. Describe how you can tune three of the compensators developed in class in order to perform more deliberate loop shaping on the Bode plot than is possible with a PID compensator: which would you use where, and why? 0. Select the target ω c based on a rise time (ω c = 1.8/t r ) or bandwidth (ω c = ω BW /1.4) constraint. 1. Apply lead compensation near ω c to give adequate PM to meet the overshoot constraint. 2. Apply lag compensation well below ω c to boost up lowfreqeuncy gain to get good tracking. 3. Apply a lowpass filter well above ω c to provide adequate disturbance rejection. Finally, select such that crossover actually occurs at ω c. (4d)
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