finalsoln

# finalsoln - ME 375 FINAL EXAM Tuesday May 3 2011 Division...

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ME 375 FINAL EXAM Tuesday, May 3, 2011 Division Shaver / Krousgrill/Deng (circle one) SOLUTION Instructions (1) This is a closed book examination, but you are allowed three, single-sided 8.5 × 11 crib sheets. (2) You have two hours to work all five problems on the exam. (3) 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. You must show all of your work to receive any credit. (4) 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. Please write your name in the top right-hand corner of each page. (5) A table of Laplace transform pairs and properties of Laplace transforms is provided at the end of this exam. Problem No. 1 (20%) ________________________ Problem No. 2 (20%) ________________________ Problem No. 3 (20%) ________________________ Problem No. 4 (20%) ________________________ Problem No. 5 (20%) ________________________ TOTAL (***/100%) ________________________

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May 3, 2011 SOLUTION 2 PROBLEM 1 PART A (4%) Consider the electrical circuit shown below. Develop an expression for the impedance Z s ( ) = V i s ( ) I s ( ) . Express your final answer as a single fraction. Solution Z = R + 1 R + 1 Ls + 1 / Cs 1 = R + 1 R + Cs LCs 2 + 1 1 = R + LRCs 2 + R LCs 2 + RCs + 1 = 2 LCs 2 + RCs + 2 LCs 2 + RCs + 1 L A i v i (t) B + C R - D E R
May 3, 2011 SOLUTION 3 PROBLEM 1 (continued) PART B (4%) The system y + 4 y = 8 u t ( ) is given initial conditions of y 0 ( ) = y 0 and an input of u t ( ) = unit step function . Determine the value of y 0 such that the transient portion of the response is ZERO. Solution sY y 0 + 4 Y = 8 s Y = y 0 + 8 / s s + 4 = sy 0 + 8 s s + 4 ( ) = A s + B s + 4 y t ( ) = Ah t ( ) + B 4 t B = s + 4 ( ) Y s ( ) s = 4 = sy 0 + 8 s s = 4 = y 0 2 To eliminate the transient term, we need B = 0 y 0 = 2 Alternate approach: y ss = lim s 0 sY s ( ) = lim s 0 sy 0 + 8 s + 4 = 2 If one starts at the steady-state solution, it will remain at that state. Therefore, need y 0 = 2 to eliminate transient terms in solution.

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May 3, 2011 SOLUTION 4 PROBLEM 1 (continued) PART C (4%) Determine the equivalent spring stiffness for the system shown below. Solution
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finalsoln - ME 375 FINAL EXAM Tuesday May 3 2011 Division...

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