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Unformatted text preview: Name_________________________ Page 1 of 5 12/8/2005 (Spring 2000) A linear model characterized by the transfer function 2 ( ) 2 2 S G S S S = + + . with zero initial conditions, is excited by the following excitations. Determine the forced responses. a. 3 ) ( = t u . Note that there are three ways to solve this problem. The easiest way is to use Laplace transforms. The transform of the input is ( ) 3 U s s = and thus, because of the zero valued initial conditions ( ) ( ) ( ) ( ) 2 2 2 3 1 3 ( ) 2 2 1 1 X s U s G s s s s = = = + + + + . Using inverse Laplace transforms this gives ( ) 3 sin t x t e t = . ASIDE: The remaining two methods either lead to erroneous solutions or are very tedious. They both start of the same way. In both the homogeneous solution is required. As 2 2 2 s s + + = , one finds the 1 d ω = and 1 n ςω = . The homogeneous solution is ( ) ( ) 1 2 1 2 cos sin cos sin n t t h d d x e C t C t e C t C t ςω ω ω = + = + . The forced part of the solution can be found by in one of two ways. Using operator notation the forced portion is given by ( ) 3* *1 2 t p x u G e = = = . This is obviously incorrect! WHY? THE SYSTEM EQUATION TRANSFORMS THE EFFECTIVE INPUT FROM A STEP TO AN IMPULSE. THUS AN IMPULSE RESPONSE IS WHAT IS NEEDED. Let us look at the correct third approach; start by writing out the differential equation, ( ) 2 2 du t x x x dt + + = dd d . Substituting the input, ( ) ( ) 3 step u t u t = yields ( ) 2 2 3 x x x t δ + + = dd d , where ( ) t δ is the unit impulse. At this point one could take the Laplace Transform of this expression and then utilize the inverse Laplace transform to get the correct answer as in the first method. Alternatively recall how we handled impulse responses before we knew anything about Laplace.. One way to do it is to utilize the homogeneous solution for shifted initial conditions. To determine the shift integrate the differential equation over the time of the impulse....
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 Spring '08
 Perreira

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