Midterm-I Solution

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Oct 30, 2023

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PLACE NAME OR INITIALS HERE: 1 Exam Number: ME 132 Midterm I Monday, October 2, 2023, 01:10 PM to 2:00 PM HONOR PLEDGE: Copy (NOW) and SIGN (after the exam is completed): I have neither given nor received aid on this exam, nor have I observed a violation of the Engineering Honor Code. SIGNATURE (Sign after the exam is completed) OLUT 10/ FULL NAME (PRINTED) STUDENT ID RULES: CLOSED BOOK . CLOSED CLASS NOTES . CLOSED HOMEWORK . CLOSED SLIDES . ONE SHEET OF NOTE PAPER (DOUBLE-SIDED) . CALCULATOR ALLOWED. O U A W The maximum possible score is 60. To maximize your score on this exam, read the questions carefully and write legibly. For those problems that allow partial credit, show your work clearly in this booklet.

PLACE NAME OR INITIALS HERE: Multiple-Choice / True-False Section of the Exam. Supporting reasons not required, There are five questions, each worth five (5) points. 1. L (T or F): The steady-state response of a stable linear time-invariant system, due to a sinusoidal input, depends on the initial condition. 2._ £ (Tor F): One of the advantages of open-loop control is its ability to correct for errors due to disturbances. . Which of these statements is true for the input-output system 3 +1? = u. (circle ALL that apply). (a) System is Memoryless. @System is Causal. @System is Time-invariant (d) System is Linear 4. A step input u(t) = u,,t > 0 is fed into a first-order system with initial condition y(0) = 8. The step response is y(t) = 2e~2* + 6. Which of the following are possible? (circle ALL th at apply). Remember that the step response for first-order systems is of the form y(t) = (yo - %uss)e-_at + guss. (a) a=—2,b=3,u, = —4. ABPa =2,b=6,u,, = 2. C@a= 2,b= —3,u,, = —d. (d)Y 6 =2, b=86, 5y =4 5. Consider the parallel connection of two stable first-order systems L, £, shown below. ...........................

PLACE NAME OR INITIALS HERE: 3 Suppose the frequency response functions of the systems Xy, Z; are G, (w), G2(w) respectively. Let G(w) denote the frequency response function of the parallel system, i.e., the FRF from u — y. Then (circle ALL that apply) (@) G(w) = G1(w)Ga(w)- YG(w) = G1(w) + Ga(w). @The parallel system is bounded-input bounded-output stable. Input u(t) = sin(wt) produces the following steady-state output: Yes () = |G (w)] sin(wt + £G1 (W) + |Ga(w)| sin(wt + £G2(w)).

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w2 e BTt R 6 £ 0),0,8) & bk N 41, wlo) v G o s = beth s -é—mwzgoéfié/ﬂow. W’MW}&WJ/ e Uiy =)+ l). pu 8780 7 W'MW/W/’”

PLACE NAME OR INITIALS HERE: ° 6. (1x10 = 10 points) Consider SpaceX’s rocket booster landing problem. For this system, mix and Igaiwh the fOI_IOWing- Enter your answers (i.e. the letters (a) through (j)) in the boxes provided. nly one option goes in each box. No supporting reasons are required. Reference: Time Delay: § Bandwidth: Controller: ;\ Noise: . 4 Plant: O Sensor: & Actuator: b Disturbancezc- Uncertainty: e d (a) The rocket booster. (b) The gimballed rocket engine that provides vectored thrust. (c) Aerodynamic forces acting on the rocket booster. (d) Structural changes in the rocket booster due to atmospheric re-entry. (e) Position of SpaceX’s landing pad. (f) Maximum rate of changing the orientation of the rocket engine. (g) Radar signal to provide position of the rocket booster. (h) Guidance and navigation device that processes measurements and commands and adjusts the rocket engine thrust and orientation. ‘ (i) Effect of radar signal transmitted from ground station to the rocket and reflected back, and transmission of computed position from the ground station to the rocket for onboard control. (j) Radar reflections due to obstacles in the environment and other aircraft.

PLACE NAME OR INITIALS HERE: 7. (3+3+4+45 = 15 points) In the following figure, three systems are shown: the (open—loop) controller K, the (open-loop) plant P, and the closed-loop system. For each of the systems, their respective unit-step responses (i.e., responses to a step input with unit magnitude, us; = 1) are also shown. Assume any initial conditions are zero. | | o ] ‘ Input Output e - K —> 04 | ol . ] ol L L L A R W S S 0 1 2 3 4 5 6 7 8 9 10 Time (s) (a) (b) 10 T 7 T S T T 1 8 = - e o - E 6 Input Output ——r P > [ 2 " 0 . s L 0 1 2 3 4 5 6 T 8 9 10 Time (s) (c) (d) 1 T o8 Input + > Ougzut e | K P > oal o e . _ 02 0 . 0 1 2 3 4 5 6 & 8 9 10 Time (s) (e) (£ a) For the three systems (the controller, the plant, and the closed-loop system), based on the above step responses, identify which of the systems are static and which are first-order systems. Lombeeller K 2 Zevolh— 05w P‘ QM*- 'P Fres t»_ O'B'JCY . Fi‘\'&{' —ovdex KP delop i

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: pLACE NAME OR INITIALS HERE: 7 b) For the first-order systems, specify which ones are stable and why. i.e., suppose the static systems c) Identify the system parameters for each of the three systems, bu; identify the were written as y = bu and the first-order systems were written as y + ay = parameters b or a,b for each system. - L= 0-5 = o= Vi )> ;—:_,_—-"""' :&'_‘/‘L > b=a” -

PLACE NAME OR INITIALS HERE: d) Suppose one of the first-order systems was y + 10y = 100u and one of the static systems was Y = 10u. Draw the bode plots for these two systems on the given figures. (Recall that the bode plots tell you what is the amplification / attenuation of the output as well as the additional phase of the output sinusoid when then system is excited by a sinusoid at a particular frequency.) w0 BodeiPlal “ Rode Pt : §; 20 g 20 — [} g'm I 3 é’.zo - i -40 * i : -40 4 + - 107 10° 10’ 102 103 107! 10° 10! 102 10° 90 —— 980 §°45 4 g.ss 1 ;-E, 0 . § 0 il S I P N -9‘1)0“ 1~(-)° 1t;' 102 g " ‘ ‘ 10° 10 10 10’ 102 10° Frequency - log scale ~ Frequency - log scale (a) First-Order System 7 + 10y = 100u (b) Static System y = 10u - < =10 54 =60) > b2 #;low '6\(0)] = 20 dA 0 (5 U= & & wt L 0.~ 10 = |0 T @ > (}f i _lo tov ol = \61(03\ _ Ju(w) = S - 20 d8 AR VR L, o

£ NAME OR INITIALS HERE: 9 : 2 p[,AC " 8. (1x4 + 2 + 4 = 10 points) A plant, with input i i \ ) put u, disturbance d and output y 1s overned b y(t) = 2y(t) + 3u(t) + d(t). Pt 55 Y (a) Is the process stable? o= "R = WM We' (b) Consider a proportional-control strategy u(t) = Kir(t)+ K, (r(t) = y(t))- Determine the closed- loop differential equation relating the variables (y, 7, d). j-ay = afkn g h — kgt [t d o yrlyDy= Greawdn 4 (c) As a function of Ko, what is the steady-state gain from d — y in the closed-loop system? R Obng 3k (d) As a function of K1, Ko, what is the steady-state gain from r — y in the closed-loop system? 556\ =, 3[%1‘3& ™y 3k ~a (e) For what values of Ky and K3 is the closed-loop system stable? sbe /A & %A K, & &

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PLACE NAME OR INITIALS HERE: 10 (f) Choose K3, K3 so that the closed-loop system is stable and steady-state gain from r — y equals 1 (for perfect tracking), and steady-state gain from d — y equals 0.1 (for disturbance attenuation). — | = Slpny= 01 P 3 ; > 3&:\1 =>.