Exam 3 (Version 1) - Solution , Sp'12

# In this problem the input is a unit step response and

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Unformatted text preview: ned as, () Using partial fractions, we rewrite this as: () ( ) ( ) ( ) Thus, ( ) ( ) . Therefore, upon inspection, we see that: 8 Upon substitution, () ( * Re-arranging the expression gives us: () ( ( * ) ( ( * ) From the Laplace transform tables, we see that: * ( )+ () ( ) ( )) Thus, the error signal is: () () () ( ( ) Since the value of the trigonometric functions oscillate between see that the dominant term is as and as we . Thus, the steady state error is . (d) The transfer function is now, () () Thus, () 9 Taking partial fractions, we have: () Thus, ( ) ( ) Hence, Therefore, () Re-arranging, () ( ( * ) ( ( * ) From the Laplace transform tables, we see that: * ( )+ () ( ) 10...
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## This note was uploaded on 08/03/2012 for the course IE 474 taught by Professor Srinivasan,c during the Spring '08 term at Purdue.

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