me375_sp11_hwk04_soln

# me375_sp11_hwk04_soln - ME 375 – Spring 2011 Homework No...

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Unformatted text preview: ME 375 – Spring 2011 Homework No. 4 Due: Friday, February 11 Problem No. 1 Part A: (20%) For each of the following systems, determine the transfer function G s ( ) = Y s ( ) / U s ( ) and calculate the poles of this transfer function. a) 5 y + 7 y = 15 u t ( ) b) y + 10 y + 21 y = 4 u t ( ) c) y + 14 y + 58 y = 6 u t ( ) + 4 u t ( ) d) x + 6 x − 8 y = u t ( ) and y + 6 y + 8 x = u t ( ) [Since we are looking for the transfer function (forced response), zero ICs will be used in the following Laplace transforms] Part a): L 5 y + 7 y [ ] = 15 L u t ( ) ⎡ ⎣ ⎤ ⎦ ⇒ 5 s + 7 ( ) Y s ( ) = 15 U s ( ) ⇒ G s ( ) = Y s ( ) U s ( ) = 15 5 s + 7 = N s ( ) D s ( ) D s ( ) = 5 s + 7 = ⇒ p 1 = − 7 / 5 Part b): L y + 10 y + 21 y [ ] = 4 L u t ( ) ⎡ ⎣ ⎤ ⎦ ⇒ s 2 + 10 s + 21 ( ) Y s ( ) = 4 U s ( ) ⇒ G s ( ) = Y s ( ) U s ( ) = 4 s 2 + 10 s + 21 = N s ( ) D s ( ) D s ( ) = s 2 + 10 s + 21 = ⇒ p 1,2 = − 10 ± 10 2 − 4 ( ) 21 ( ) 2 = − 10 ± 4 2 = − 3, − 7 Part c): L y + 14 y + 58 y [ ] = 6 L u t ( ) ⎡ ⎣ ⎤ ⎦ + 4 L u t ( ) ⎡ ⎣ ⎤ ⎦ ⇒ s 2 + 14 s + 58 ( ) Y s ( ) = 6 s + 4 ( ) U s ( ) ⇒ G s ( ) = Y s ( ) U s ( ) = 6 s + 4 s 2 + 14 s + 58 = N s ( ) D s ( ) D s ( ) = s 2 + 14 s + 58 = ⇒ p 1,2 = − 14 ± 14 2 − 4 ( ) 58 ( ) 2 = − 14 ± 6 j 2 = − 7 ± 3 j Part d): L x + 6 x − 8 y [ ] = L u t ( ) ⎡ ⎣ ⎤ ⎦ ⇒ s + 6 ( ) X s ( ) − 8 Y s ( ) = sU s ( ) ⇒ X s ( ) = 8 Y s ( ) + sU s ( ) s + 6 L y + 6 y + 8 x [ ] = L u t ( ) ⎡ ⎣ ⎤ ⎦ ⇒ s + 6 ( ) Y s ( ) + 8 X s ( ) = U s ( ) ⇒ s + 6 ( ) Y s ( ) + 8 8 Y s ( ) + sU s ( ) s + 6 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = U s ( ) ⇒ s + 6 ( ) 2 + 64 ⎡ ⎣ ⎤ ⎦ Y s ( ) = s + 6 ( ) − 8 s ⎡ ⎣ ⎤ ⎦ U s ( ) ⇒ G s ( ) = Y s ( ) U s ( ) = − 7 s + 6 s 2 + 12 s + 100 = N s ( ) D s ( ) D s ( ) = s 2 + 12 s + 100 = ⇒...
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## This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.

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me375_sp11_hwk04_soln - ME 375 – Spring 2011 Homework No...

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