test301-2-f09 - The force or input u is applied to mass m 2...

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A-AE 301 exam two, fall 2009. Please put your best answer clearly written on the exam in a box. Two page of notes and no calculators. NAME: 1
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Problem 1. Solve the following difierential equation where the initial conditions are all zero _ y + y = 5 cos(2 t ) : y ( t ) = 2
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Problem 2. Compute e At for A = 2 4 ¡ 2 2 ¡ 4 2 3 5 : e At = 3
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Problem 3. Compute the transfer function for the following state space system 2 4 _ x 1 _ x 2 3 5 = 2 4 ¡ 2 ¡ 1 1 0 3 5 2 4 x 1 x 2 3 5 + 2 4 2 0 3 5 u y = h 1 1 i x 3a) The transfer function is G ( s ) = Y ( s ) U ( s ) = 3b) The impulse response g ( t ) for this system is given by g ( t ) = 4
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Problem 4. Consider the transfer function Y ( s ) U ( s ) = G ( s ) = 8476 s 3 + 274 s 2 ¡ 37628 s + 4 ( s 2 + 9789 s + 1)( s 2 + 321 s + 2) Assume that the input u ( t ) = 12. Then flnd lim t !1 y ( t ) = 5
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Problem 5. Construct a circuit consisting of a resistor, capacitor and inductor whose transfer function is given by G ( s ) = 2 s 2 2 s 2 + s + 1 : 6
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Problem 6. Design a circuit in Simulink using integrators, summers and gains for the transfer function G ( s ) = 3 s + 1 s 2 + s + 4 : 7
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Problem 7. Find the transfer function for the state space system: _ x = 2 6 6 6 6 6 6 4 0 1 0 0 0 0 1 0 0 0 0 1 ¡ 10 ¡ 20 ¡ 30 ¡ 40 3 7 7 7 7 7 7 5 x + 2 6 6 6 6 6 6 4 0 0 0 1 3 7 7 7 7 7 7 5 u y = h 1 2 3 0 i x Y ( s ) U ( s ) = G ( s ) = 8
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Figure Problem 8 Problem 8. Consider the mass spring damper system in Figure 8. The force or input
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Unformatted text preview: The force or input u is applied to mass m 2 . The distance from the ±oor to m 2 is x 2 , while the distance from the ±oor to m 1 is x 1 . Moreover, m 1 = m 2 = 1 ; k 1 = k 2 = 1 and c = 1 : Find the transfer function from u to the output y = x 1 . Y ( s ) U ( s ) = G ( s ) = 9 i u y Z 1 Z 2 Figure Problem 9 Problem 9. Find the transfer function with the input u and output voltage y for the circuit given in the Figure for Problem 9 where Z 1 is a resistor R 1 = 1 and a capacitor C = 1 in parallel, while Z 2 consists of a single resistor R 2 = 1. 10 Laplace transform pairs g ( t ) G ( s ) = R 1 e ¡ st g ( t ) dt – ( t ) 1 1 1 =s t n n ! =s n +1 e at 1 = ( s ¡ a ) cos( !t ) s= ( s 2 + ! 2 ) sin( !t ) != ( s 2 + ! 2 ) e at t n n ! = ( s ¡ a ) n +1 e at cos( !t ) ( s ¡ a ) = (( s ¡ a ) 2 + ! 2 ) e at sin( !t ) != (( s ¡ a ) 2 + ! 2 ) 11...
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