Solution_to_problem_6.2 - 1 1 1 2 0.5 p--= = = -There are...

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6.2 – 1 Solution to problem 6.2 a) The standard form is: 1 2 ( 1) ( ) ( 1)( 1) s a K s G s e s s θ τ - + = + + Thus we need to rearrange 5 2( 0.5) ( ) ( 2)(2 1) s s G s e s s - + = + + in order to get terms of the form ( ) 1 s + , as follows: 5 0.5(2 1) ( ) 2 2(0.5 1)(2 1) s s G s e s s - + = + + Now the transfer is in standard form, but we can simplify the gain and perform a pole-zero cancellation, since there is a common term in the numerator and denominator: 5 0.5 ( ) (0.5 1) s G s e s - = + b ) We have obtained a first-order plus time delay (FOPTD) transfer function of the form: 1 ( ) 1 s K G s e s - = + The gain is 0.5 K = . The pole is
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Unformatted text preview: 1 1 1 2 0.5 p--= = = -There are no zeroes due to the previous pole-zero cancellation. c ) The 1/1 Pade approximation of the time delay is: 5 5 1 2 5 1 2 s s e s-- + The transfer function is now: ( ) 5 5 0.5 1 0.5 2 ( ) 5 0.5 1 0.5 1 1 2 s s G s e s s s- -+ = + + + Since it is already in the standard form, we can calculate easily: -Gain: 0.5 K =-Poles: 1 2 2 2 5 p p-= -=-Zero: 1 2 5 z =...
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This note was uploaded on 11/20/2009 for the course CHME DTU-abroad taught by Professor Rafiqulgani during the Spring '09 term at Rensselaer Polytechnic Institute.

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