Midterm-I Solution

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Yale University *

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Mechanical Engineering

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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)).
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/’”
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