sol5.09 - ECE521 Linear Systems Fall 2009 Homework 5...

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Unformatted text preview: ECE521 Linear Systems Fall 2009 Homework 5 Solutions Problem 1: Consider the system ˙ x = parenleftbigg- 3 5- 3 parenrightbigg x + parenleftbigg 1 parenrightbigg · 1 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright u x (0) = parenleftbigg 1 parenrightbigg y = [1 2] · x 1. Find the transfer function G ( s ) . 2. We know that y ( t ) = Ce tA x (0) + integraldisplay t Ce A ( t- τ ) Bu ( τ ) dτ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright y (0) ( t ) Using Laplace transform, find the output y (0) ( t ) when the initial state is zero. (Use Laplace transform tables). 3. Having computed y (0) ( t ) find y ( t ) . We have G ( s ) = C ( sI- A )- 1 B = [1 2] parenleftbigg 1 s +3 5 ( s +3) 2 1 s +3 parenrightbiggparenleftbigg 1 parenrightbigg = 5 ( s + 3) 2 + 2 s + 3 = 2 s + 11 ( s + 3) 2 . We know that Y ( s ) = G ( s ) U ( s ) where Y ( s ) and U ( s ) are Laplace transforms of y (0) ( t ) and u ( t ), respectively. As L (1) = 1 s we obtain Y ( s ) = 2 ( s + 3) 2 + 11 s ( s + 3) 2 Partial fraction expension of the second term give 11 s ( s + 3) 2 = 11 9 s- 11 9( s + 3)- 11 3( s + 3) 2 Thus Y ( s ) =- 5 3 · 1 ( s + 3) 2- 11 9 · 1 s + 3 + 11 9 · 1 s 2 The inverse Laplace transform gives y (0) ( t ) as y (0) ( t ) = (- 5 3 te- 3 t- 11 9 e- 3 t + 11 9 ) u step ( t ) where u step ( t ) is the unit step function. Now we calculate the zero input solution. For that we notice that A = parenleftbigg- 3 5- 3 parenrightbigg = parenleftbigg- 3- 3 parenrightbigg + parenleftbigg 0 5 0 0 parenrightbigg = A 1 + A 2 and that the two matrices on the right commute. Henceand that the two matrices on the right commute....
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sol5.09 - ECE521 Linear Systems Fall 2009 Homework 5...

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