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Unformatted text preview: 4.2 1 Solution to problem 4.2 a) A firstorder transfer function has the form: ( ) 1 K G s s = + Where K is the steadystate gain, and is the time constant. In our case, the gain is: 5 K = b) The time constant is: 10 = c) Remember the Final Value Theorem (FVT), from Chapter 3, which states that: lim ( ) lim ( ) t s y t s Y s = The Laplace transform of the output  ( ) Y s can be calculated using the transfer function ( ) G s , which relates the input and the output: ( ) 5 2 ( ) ( ) ( ) ( ) ( ) 10 1 Y s G s Y s G s U s U s s s = = = + Now we are ready to apply the FVT, however we must first check whether the conditions for the applicability of the theorem are satisfied. Recall from the solution of Exercise 3.9, that the theorem can only be applied when the poles of ( ) sY s have negative real parts. In this case, 5 2 10 ( ) 10 1 10 1 s Y s s s s s = = + + has only one pole, which is negative ( 1 10 p =  ). Therefore the FVT is applicable, and yields:...
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 Spring '09
 RafiqulGani

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