Exam 1 Fall 2020 AERE 331 Take-Home Due 9/24(F) 2pm via Canvas pdf Name__________________________ 1 [NOTE: All work, answers and plots must be placed directly beneath the problem part to receive credit. Matlab code that supports the same should be placed in the Appendix unless stated otherwise. I will not go to the Appendix to search for work.] PROBLEM 1(30pts) A space vehicle will use pulsed propulsion to move along the x-axis to a position x2 over a time t2 . A forward pulse force F0 is applied over the first t0 seconds, and a retarding pulse force − F0 is applied over the last t0 seconds. The force f (t ) , vehicle position x(t ) (both theoretical and simulated), and theoretical vehicle velocity v(t ) are shown below for x2 = 10 , t2 = 7 sec , and F0 = 1 . Theoretical Position & Velocity; Simulated Position Force Profile 10 1 10 position(theory) 9 0.8 9 velocity(theory) 8 0.4 7 7 0.2 6 6 5 5 4 4 3 3 2 2 1 1 Force Position 0 -0.2 -0.4 -0.6 -0.8 -1 0 1 2 3 4 5 6 7 0 Velocity Position(sim) 8 0.6 0 0 1 2 Time (sec) 3 4 5 6 7 Time Figure 1 Force profile (left), and position and velocity profiles (right). (a)(8pts) Denote the vehicle mass as m. (i) Begin with Newton’s law , ∑f = mv to arrive at a mathematical expression for the velocity profile v(t ) as function of the parameters m, F0 , x2 , t0 , and t2 . (ii) Substitute the given values for the parameters (assume m = 1 ) into your general expression to validate it in relation to the given velocity profile. [Yes, this will require computing integrals.] Solution: (b)(4pts) It should be clear that the only force acting on the space vehicle is f (t ) . From Newton’s law arrive at the transfer functions Gv ( s ) = V ( s ) / F ( s ) and Gx ( s ) = X ( s ) / F ( s ) . Keep m general. Solution: 2 (c)(6pts) For the general force profile given in (a) use a picture to show that f = (t ) F0 1(t ) − 1(t − t0 ) − 1( t − (t2 − t0 ) )  . Solution: Figure 1(c) Picture to show that the shape of the force profile in (a) is f = (t ) F0 1(t ) − 1(t − t0 ) − 1( t − (t2 − t0 ) )  . (d)(6pts) For any y (t ) ↔ Y ( s ) we have y (t − ∆) ↔ e −∆sY ( s ) . Hence, in relation to (c) we have: ) F ( s= ∆ 1  1 Q( s ) . Treat Q( s ) as a F0 1 − e − t0 s − e − (t2 −t0 ) s =   s s ( ) ‘transfer function’. (i) Compute the step response of this ‘transfer function’ over time T=0:.001:7 to arrive at a plot of f (t ) , using F0 = 1 , t0 = 2 , and t2 = 7 . (ii) Validate f (t ) by comparing it to the force profile shown in (a). Solution: [Include your Matlab code HERE.] Figure 1(d) Transfer function-based force profile. (e)(6pts) (i) Use your results from (b) and (d) in the ‘lsim’ command to arrive at overlaid plots of v(t ) and x(t ) . (ii) Discuss how these plots compare to those in (a) Solution: [See code @ 1(e).] Figure 1(e) Simulation-based plots of v(t ) and x(t ) . Remark. The position equation can be shown to be:  0.5(v0 / t0 )t 2 for 0 ≤ t < t0  x(t )  v0 (t − t0 ) + x(t0 ) for t0 ≤ t < t2 − t0  2 2   0.5(v0 / t0 ) (t2 − t0 ) − t  + (v0t2 / t0 ) [t − (t2 − t0 ) ] + x(t2 − t0 ) for t2 − t0 ≤ t2 I refrained from addressing them, being well aware of the care that must be taken in arriving at them. Grading would be a headache! From them, one can arrive at the valuable result: t2= t0 + x2 / v0 . PROBLEM 2(40pts) A transport aircraft at altitude of 10,000 ft and speed of 487 ft/s has the yaw rate/rudder position transfer function: G (s) = ∆ r (s) = δ r (s) 3 0.73s 3 + 1.43s 2 + 0.30 s + 0.24 . Using the method of Partial Fraction Expansion s + 2.21s 3 + 1.78s 2 + 2.13s + 0.07 4 (PFE) we can express G ( s ) as the sum of the roll, Dutch roll, and spiral mode transfer functions: 0.008722 0.6065s + 0.02975 0.1148 + 2 + =Gr ( s ) + Gdr ( s ) + Gs ( s ) . s + 1.858 s + 0.3182 s + 1.115 s + 0.03378 (a)(4pts) Use the ‘roots’ command (twice) to verify that the poles of G ( s ) are consistent with those of the three G (s) = component transfer functions. Solution: [Include your code HERE.] (b)(10pts) (i) Compute the time constants of the three modes. (ii) Compute the damping ratio and damped natural frequency of the Dutch roll mode. Solution: [Show all work HERE.] (c)(8pts) The block diagram for control of these dynamics is shown below. (i) Use ‘rlocus’ and the data cursor to show that (i) ζ ≅ 0.9 and Gc ( s=) K ≅ 2.21 , and (ii) that the three closed loop poles for this value are: pr ≅ −1.93 roll; pdr ≅ −0.6 ± i 0.3 Dutch roll ; psp ≅ −0.66 spiral. rc (s) − Gc (s ) G (s) r (s) Solution: [See code @ 2(c).] Figure 2(c) Root locus & data cursor information. (d)(7pts) Compute the percent reduction in the time constant for each mode, and the percent increase in the damping ratio associated with the underdamped mode. Solution: [Show all work HERE.] 4 (e)(6pts) For Gc ( s) = 2.21 the CL transfer function, scaled to have unity static gain, is: 3 2 ∆ r (s)  0.60  1.61s + 3.16 s + 0.66 s + 0.53 . W= (s) = rc ( s )   4 3 2  0.53  s + 3.82 s + 4.94 s + 2.79 s + 0.60 (i): Give the value for c, such that cG ( s ) has unity static gain. (ii) Overlay the step responses for the scaled original and closed loop transfer functions. (iii) Quantify the most obvious improvement resulting from feedback control. Solution: [See code @ 2(e).] Figure 2(e) Step responses for scaled G ( s) and W ( s) . (f)(5pts) In (e) you should have observed that W ( s) has a large spike very near to t = 0 . Investigate the zeros of W ( s) to identify the cause of the spike. Solution: 5 PROBLEM 3(30pts) A motor to be used for a solar array angular positioning has 10 . In this problem you will design a unity feedback command s ( s + 0.5) control system having a CL pole at s0 =−2 + i 2 . G p (s) = (a)(8pts) Beginning with a sketch, arrive at the ‘defect angle’ of this pole, per the root locus angle criterion. Then use this angle to place the controller zero. Solution: [Show all work HERE.] Figure 3(a) Sketch to find defect angle. (b)(3pts) Regardless of your answer in (a), assume here that Gc = ( s ) K ( s + 2.3) . From a root locus plot for the CL system use the data cursor to find the value of K. Solution: [See code @ 3(b).] Figure 3(b) CL root locus with data cursor information. (c)(8pts) Now, instead of a PD controller, consider the lead compensator Gc ( s) = K ( s + 2) . (i) Find the required value for β s+β as you did in (a). [Note: If you think a bit, you might find that a sketch is not needed.] (ii) Find the value for K as you did in (a). (iii) Use the data cursor to find the value of the third CL pole. Solution: [See code @ 3(c).] Figure 3(c) Root locus plot. 5( s + 2) . (b ) (d)(6pts) Assume that G= 0.35( s + 2.3) and Gc( c ) ( s) = c (s) s + 16 (i) Overlay plots of the CL step response for each controller. (ii) Briefly discuss how they compare. Solution: [See code @ 3(d).] Figure 3(d) CL step responses. The required total controller power is related to the area beneath the magnitude curves. Clearly, the PD controller requires much more power than the lead controller. Use the ‘integral’ command to quantify the ratio of the controller powers over the interval [10−1 ,103 ] rad/sec. Solution: [Include your code HERE.] 40 Magnitude (dB) viewed as magnitude or power: 10 log10 M 2 (ω ) = 20 log10 M (ω ) . 60 20 0 90 Gcb Gcc Phase (deg) (e)(5pts) The Bode plots for the two controllers are shown at right. The magnitude in dB is M (ω ) dB = 20 log10 M (ω ) . The controller power is given by the square of the magnitude, and in dB is defined as P(ω ) dB = 10log10 P(ω ) . Hence, the vertical axis in the figure can be 6 Controller Bode Plots 45 0 10 -1 10 0 10 1 10 2 10 3 Frequency (rad/s) Figure 3(d) Bode plots for each controller. Appendix Matlab Code [NOTE: This code should support the answers/ plots given beneath a given problem part. I will not search through this code to find answers.] %exam1.m 9/13/21 %PROBLEM 1 %(d): T=0:.001:7; s=tf('s'); figure(10) plot(T,f) title('TF-Based Force Profile') xlabel('Time (sec)') grid %(e): figure(11) yyaxis left plot(T,x) ylabel('Position') yyaxis right plot(T,v) ylabel('Velocity') title('Simulation-based Position and Velocity') grid %================================================ %PROBLEM 2 Np=[0.73 1.43 0.30 0.24]; Dp=[1 2.21 1.78 2.13 0.07]; G=tf(Np,Dp); %--- Partial Fraction Expansion--[num,den]=tfdata(G,'v'); [r,p,k]=residue(num,den); s=tf('s'); Gr=r(1)/(s-p(1)); G21=r(2)/(s-p(2)); G22=conj(G21); Gdr=G21+G22; Gsp=r(4)/(s-p(4)); %#################################### %(c): figure(20) grid %(d): Gc=2.21; W=(0.6/0.53)*tf([1.61 3.16 0.66 0.53],[1 3.82 4.94 2.79 0.60]); figure(21) title('Step Responses for Scaled G(s) and W(s)') legend('G(s)','W(s)') grid %(f): %===================================== %PROBLEM 3 %(b): %(c): %(d): %(e): 7