## 辅导案例-ME 37500

• July 5, 2020

ME 37500 Midterm Exam February 14th, 2017 6:30 – 7:30 pm CL50 224 Name: ________________________________ HWID: Section: Deng / Shelton / Roll (circle one) Instructions (1) TURN OFF your cell phones and tablets. If you are using a smart watch, please also turn it off and put it in your bags. (2) Write your name and HWID at the top of each page. (3) This is a closed book examination, but you are allowed one single-sided 8.5”×11” hand written crib sheet. (4) Only ME approve exam calculator: TI-30XIIs can be used (5) You have one hour to work all four problems on the exam. (6) If you use extra pages, be sure to write your NAME and HW ID on the top of each extra page. Ensure that your pages are in the correct order and re-staple the packet together. (7) Any work on the back of the page do not count and will not be graded. (8) You must show all of your work to receive any credit. Clearly mark your final answer. (9) Use the solution procedure we have discussed: what are you given, what are you asked to find, what are your assumptions, what is your solution, does your solution make sense. (10) You must write neatly and should use a logical format to solve the problems. You are encouraged to really “think” about the problems before you start to solve them. (11) A table of Laplace transform pairs and properties of Laplace transform is attached at the end of this exam set. Problem 1 (15 points) ________________________ Problem 2 (20 points) ________________________ Problem 3 (40 points) ________________________ Problem 4 (25 points) ________________________ TOTAL (***/100 points) ________________________ Name: HWID: Page 2 of 8 PROBLEM 1 (15 points) (A) For the closed-loop system shown below, find the characteristic equation. (5 pts.) The transfer function is () = () () = ()1+() = 1+1+� 1 + �� + � = (+)(+)(+)+. Thus, the characteristic equation is ( + )( + ) + = 0. Alternatively, 2 + ( + ) + ( + ) = 0. (B) Find the zeros of transfer function (). (5 pts.) () = (3 + )(2 + )( + )2(2 + + ) The zeros are located at = 0, = − 3 , and = ±√. (C) You are working with a feedback system that automatically maintains its two closed-loop poles at = −2 ± 2, regardless of the number of system zeros. Assume there are currently no zeros, and you must introduce a single zero into this system. If you want to minimize overshoot when a unit step is applied at the reference input, should you place the zero at = −1, = −2, or = −3? Why? (5 pts.) Since the complex conjugate closed-loop poles are fixed, we desire the fastest possible zero to minimize step response overshoot. Therefore, we choose = −3. Name: HWID: Page 3 of 8 PROBLEM 2 (20 points) (A) Consider the following closed-loop transfer function: () = ( + 2) 4 + 53 + 452 + 196 Based on the Hurwitz Criterion, is this system unstable? Why or why not? (5 pts.) According to the Hurwitz Criterion, it is a necessary condition of stability that all coefficients in the characteristic polynomial (transfer function denominator) be of the same sign, and non-zero. Since the term is missing, the system is indeed unstable. (B) Consider the following closed-loop transfer function: () = ( − 1) 4 + 23 + 32 + 4 + Use a Routh array to determine the values of that make the closed-loop system stable. (15 pts.) Characteristic equation: () = 4 + 23 + 32 + 4 + = 0. Constructing a Routh array: 4: 1 3 3: 2 4 2: 1 1: 4 − 2 0: For closed-loop stability, all entries in the first column must be of the same sign. Thus, 0 < < 2. Name: HWID: Page 4 of 8 PROBLEM 3 (40 points) Consider the following closed-loop system, in which is a non-negative integer ( = 0,1, 2, 3, … ) Let () = 0 for this non-unity feedback system, we have () = ()1 + ()(), and () = () − () = 0 − () = − ()1 + ()(). (A) What is the minimum integer that causes the output () to completely reject a unit-step disturbance ()? (15 pts.) () = 25 + 251 + � 1� (2) � 25 + 25� = 25 −1 −1( + 25) + 252 () = = − = −25−1−1( + 25) + 252 To reject a step input, we need a Type 1 system, which requires at least 1 differentiator in the numerator of (). Thus, ≥ 2. Alternatively, using the final value theorem, with () = 1 , () = lim→0 ⋅ 1 ⋅ 25−1−1( + 25) + 252 = lim→0 25−1−1( + 25) + 252 Since 0−1 = ∞, and 00 = 1, we need 0≥1 = 0 in the numerator as → 0 to drive the steady-state output to zero. Hence, ≥ 2. () () () − + () + + () () 1 2 25 + 25 Name: HWID: Page 5 of 8 PROBLEM 3 (continued) (B) If = 1, () = 1 , and () = 0, what is the sensitivity function for the steady-state output to changes in parameter ? (15 pts.) Given that, () = ()1 + ()() = � 25 + 25�1 + �1� �21 � � 25 + 25�() = 25( + 25) + 252 (), ⇒ () = 25 + 25(1 + 2)(), therefore, = lim→0 2525(1 + 2) 1 = 11 + 2. The sensitivity of to variation in is then, = ⋅ = ⋅ 1 + 21 ⋅ (1 + 2)(0)− 1(2)(1 + 2)2 = − 22(1 + 2) (C) Let represent the sensitivity of steady-state output to feedback gain . Assume = 0.2 and = − 1.6 when = 2. If increases to 2.1, what is the change in ? (10 pts.) It is given that at a nominal value of = 2, the sensitivity is =2 = −1.6 and the nominal steady-state output is = 15 = 0.2. We can alternately express the sensitivity as = limΔ→0 �Δ ��Δ � . If moves from 2 to 2.1, the denominator is Δ = 0.1 2 = 0.05. Since Δ is small, Δ ≈ ⋅ �Δ �. Thus, Δ ≈ (0.2) �−85� (0.05) = −0.016. Name: HWID: Page 6 of 8 PROBLEM 4 (25 points) Consider the following unity feedback system: where 1( ) ( 1)( 2) G s s s = + + a) Let ( ) as bC s cs d + = + , determine the coefficients a , b , c , and d using the direct pole placement method to locate the roots such that the desired closed-loop characteristic polynomial is 2 3 2( 6)( 4 20) 10 44 120s s s s s s+ + + = + + + . (10 pts.) 2 3 2 3 2 ( 1)( 2)( ) 1( ) ( 6)( 4 20) (3 ) (2 3 ) (2 ) 10 44 120 1 3 10 7 2 3 44 21 2 120 106 G C G C CLD D N N D s s cs d as b s s s cs c d s c d a s d b s s s c c d d c d a a d b b + = + + + + + = + + + + + + + + + + = + + + = + = ⇒ = + + = ⇒ = + = ⇒ = Therefore, the controller is: 21 106( ) 7 sC s s + = + Name: HWID: Page 7 of 8 PROBLEM 4 (continued) b) It is desired that the controller cancel one of the poles of the plant at −1. Use the direct pole placement method to design a controller ( )C s that satisfies this condition while putting the closed-loop system poles at 1 1s = − and 2,3 2 4s j= − ± . Hint: You need to determine the controller form (different from part (a)) and its coefficients. (15 pts.) Let ( 1)( ) C C s NC s D + = 2 2 ( 1)( 2) 1( 1) ( 1)( 4 20) ( 2) 4 20 G C G C CL C C C C D D N N D s s D s N s s s s D N s s + = + + + + = + + + + + = + + Note that the controller has to be proper, and the orders of the polynomials on the left and right side of the equation should match. let ( 1) ( 1)( ) C C s N s aC s D bs c + + = = + 2 2 2 ( 2)( ) 4 20 (2 ) (2 ) 4 20 1 2 4 2 2 20 16 s bs c a s s bs b c s c a s s b b c c c a a + + + = + + + + + + = + + = + = ⇒ = + = ⇒ = Therefore, the controller is: 16( 1)( ) 2 sC s s + = + Name: HWID: Page 8 of 8 Laplace Transform Pairs and Properties ( ( )F s is the Laplace transform of ( )f t . Note: ( ) 0, 0f t t= ∀ < ) ( ), 0f t t ≥ ( )F s Comments ( )tδ 1 Unit Impulse 1 1 s Unit Step t 2 1 s Unit Ramp 21 2 t 3 1 s Unit Parabolic te α− 1 s α+ Exponential tte α− ( )sin tω 2 2s ω ω+ Sinusoidal ( )cos tω 2 2 s s ω+ ( )sinte tα ω− ( ) (0 )sF s f −− First Derivative ( ) ( )nf t nth Derivative ( )df t T− ( )dsTe F s− Time delay ( )te f tα− ( )F s α+ Modulation/translation n tt e α− ( ) 1 ! n n s α ++ (0 ) lim ( )sf sF s + →∞= Initial Value Theorem ( )f t converges to a constant value as t →∞ 0( ) lim ( )sf sF s→∞ = Final Value Theorem

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