problemset8_Soln - The University of Texas at Austin...

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The University of Texas at Austin Department of Electrical and Computer Engineering EE362K: Introduction to Automatic Control—Fall 2009 Solutions to Problem Set Eight C. Caramanis November 18, 2009. 1. The solution to the given differential equation is x ( t ) = x (0) e at + u ( t ) * e at where * means convolution. Taking the Laplace transform of both sides gives X ( s ) = x (0) s - a + U ( s ) s - a One may note that the above relation can be readily found from the Laplace transform of the original differential equation. To compute the transfer function, set x (0) = 0 to get G xu = X ( s ) U ( s ) = 1 s - a s = a is a pole of this transfer function. u ( t ) = e at or equivalently U ( s ) = 1 s - a gives X ( s ) = x (0) s - a + 1 ( s - a ) 2 Taking the inverse Laplace transform of both sides gives x ( t ) = L - 1 { x (0) s - a + 1 ( s - a ) 2 } = x (0) e at + te at 2. The transfer function of the system is given by T ( s ) = C ( sI - A ) - 1 B + D = 1 J t s 2 - mgl 3. (a) Setting u ( t ) = 0 gives y ( n ) + a 1 y ( n - 1) + ... + a n y = 0 Substituting y ( t ) = e λt in the above equation gives ( λ n + a 1 λ n - 1
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This note was uploaded on 02/10/2010 for the course EE 362K taught by Professor Friedrich during the Fall '08 term at University of Texas at Austin.

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problemset8_Soln - The University of Texas at Austin...

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