Review Problem 3 solution

Review Problem 3 solution - + + + + . Equating the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
3. What is the response of x to a unit step in ( ) u t ? (This is the step response, ( ) h t .) For this problem take ( ) 0 0 x = and ( ) 0 0 x = d . The Laplace Transform Table shows the pair 1 , 0 ! n at t e t n - and ( ) 1 1 n s a + + . If 0 n = and 0 a = the pair degenerates to a unit step and 1 s . From problem 1, ( ) ( ) ( ) 0 0 2 2 2 2 5 2 5 U s s x x X s s s s s + + = + + + + + d . Substituting the zero value initial conditions and noting that the input is ( ) 1 U s s = gives the step response, in frequency space, ( ) ( ) ( ) 2 1 1 2 5 X s H s s s s = = + + . Partial fraction expansion is next required. Assuming the expansion, ( ) 2 2 1 1 2 5 2 5 a bs c s s s s s s + = + + + + + and multiplying the equation by the denominator of the left hand side of the expression gives ( ) ( ) ( ) ( ) 2 2 1 2 5 2 5 a s s bs c s a b s a c s a = + + + + =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + + + + . Equating the coefficients of the polynomial in s gives: 1/5 a = , 1/5 b = -and 2/ 5 c = -. Thus, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 .2 1 .1 2 .2 .2 .4 .2 2 5 1 2 s s X s H s s s s s s-+---= = + = + + + + + . In addition to the Laplace Transform pair noted above, note the transform pair cos at e t ω-and ( ) 2 2 s a s a + + + , and the pair sin at e t-and ( ) 2 2 s a + + . Using the equivalences 1 a = and 2 = , gives ( ) ( ) .2 .2 cos2 .1 sin 2 t t x t h t e t e t--= =--. Or, ( ) ( ) ( ) .2 .2cos2 .1sin 2 t x t h t e t t-= =-+ ....
View Full Document

This note was uploaded on 02/29/2008 for the course ME 242 taught by Professor Perreira during the Spring '08 term at Lehigh University .

Ask a homework question - tutors are online