solhwset2

solhwset2 - Solutions to Homework Set 2 MAE143B Summer II...

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Solutions to Homework Set 2 - MAE143B, Summer II, 2009 1.1 First, we multiply denominator and numerator of G ( s )by 1 m and k to reach the desired form G ( s )= 1 k k m s 2 + d m s + k m We can write d m as 2 q k m 1 2 d mk = d m This way, by de±ning K = 1 k , ω n = q k m and β = d 2 mk we can obtain the desired form G ( s K ω 2 n s 2 +2 βω n s + ω 2 n 1.2 As we learned in the lecture, impulse response of the G ( s )is y ( t G ( t ). We will simply take the inverse Laplace of G ( s ) by completing squares, G ( s K ω 2 n ( s + n ) 2 + ω 2 n (1 β 2 ) We need to take inverse Laplace of this expression. If we look at the Table A.2 at the appendix A of the book, we see the following transformation: L 1 { b ( s + a ) 2 + b 2 } = e at sin( bt ) By writing G ( s )as G ( s n p 1 β 2 ω n p 1 β 2 ( s + n ) 2 + ω 2 n (1 β 2 ) Now consider, b = ω n p 1 β 2 and a = n . We can reach G ( t ) G ( t n p 1 β 2 e n t sin( ω n p 1 β 2 t ) The frequency of the oscillation is ω d = ω n p 1 β 2 1.3 If β 1, then G ( s ) will not have imaginary poles anymore. It will have real distinct poles; s 1 = n + ω n p β 2 1 ,s 2 = n ω n p β 2 1 Then, we can write G ( s )a s G ( s C 1 s s 1 + C 2 s s 2 ,mean s G ( t ) will be only exponential (no sinusoid), G ( t C 1 e s 1 t + C 2 e s 2 t 1

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1.4 Because poles of T.F are on the left hand plane, we know that G ( s ) is stable, then we can use Final Value Theorem, lim t →∞ y ( t ) = lim s 0 sy ( s ) = lim s 0 s 1 s 2 + ms + k | {z } G ( s ) 1 s |{z} input = 1 k 1.5 We know; u ( s )= K p [ r ( s ) y ( s )] and y ( s G ( s
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solhwset2 - Solutions to Homework Set 2 MAE143B Summer II...

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