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### Practice Final Problem 8 solution

Course: ME 242, Spring 2008
School: Lehigh
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Word Count: 579

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1999)Find (Spring the total response of the system x + 5 x = u (t ) , x ( 0 ) = -2 , u (t ) for the input shown to the right. 1 Note that G ( s ) = and that the homogeneous solution is x ( t ) = Ae -5t . s+5 0 1 Next write out an analytical form of the input function, u ( t ) = 10us - 5ur ( t - 1) + 5ur ( t - 3) . Alternatively note that the following inputs occur, 10 2 3 t u = {10; 0 t 1,10 - 5 ( t - 1)...

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1999)Find (Spring the total response of the system x + 5 x = u (t ) , x ( 0 ) = -2 , u (t ) for the input shown to the right. 1 Note that G ( s ) = and that the homogeneous solution is x ( t ) = Ae -5t . s+5 0 1 Next write out an analytical form of the input function, u ( t ) = 10us - 5ur ( t - 1) + 5ur ( t - 3) . Alternatively note that the following inputs occur, 10 2 3 t u = {10; 0 t 1,10 - 5 ( t - 1) ;1 t 3, 0; 3 t} . As the Laplace Transform of a unit step is 1/ s , for the first time region x ( s ) = G ( s ) u ( s ) = 10i 1 1 b a i = 10 + . Solving for a and b by s+5 s s s +5 multiplying both sides of the equation by the denominator of the left hand side gives 1 = a ( s + 5 ) + bs . Equating coefficients of like powers of s gives 1 = 5a and 0 = a + b . Thus, a = 1/ 5 and b = - a = -1/ 5 . Therefore x ( s ) = 10 1 1 -5t - and x p ( t ) = 2 (1 - e ) . The total 5 s s+5 solution in this time region becomes, x ( t ) = Ae-5t + 2 (1 - e-5t ) = 2 + Be-5t , where one unknown coefficient has been replaced by another unknown coefficient. The coefficient is found from the initial condition, x ( 0 ) = -2 = 2 + Be-5i 0 = 2 + B , or B = -4 . Thus x ( t ) = 2 - 4e -5t ; 0 t 1 . (Alternatively for this time region take the Laplace transform of the differential equation to yield U ( s) + x0 sX ( s ) - x0 + 5 X ( s ) = U ( s) . Next, rewrite this expression as X ( s ) = and then substitute s+5 10 / s - 2 10 - 2s a b = = + both x0 = x ( 0 ) = -2 and U ( s ) = 10 / s to obtain X ( s ) = . Partial s+5 s ( s + 5) s s + 5 fraction expansion solution methods first yield 10 - 2s = a ( s + 5) + bs = ( a + b ) + s 5a . Thus a = 2 , 2 4 - . Thus the total solution is x ( t ) = 2 - 4e -5t ; 0 t 1 . This is a much s s+5 more efficient approach.) At t = 1 , x ( t = 1) = 2 - 4e -5i1 = x1 , where x1 is now a known number. For a unit ramp at time b = -4 or X ( s ) = 1 1 a b c i 2 = + 2 + . Solving for a , b and c by multiplying both s+5 s s +5 s s sides of the equation by the denominator of the left hand side gives 1 = as ( s + 5 ) + b ( s + 5 ) + cs 2 . zero x ( s ) = G ( s ) u ( s ) = Equating coefficients of like powers of s gives 1 = 5b , 0 = 5a + b and 0 = a + c . Solving gives b = 1/ 5 , a = -1/ 25 and c = 1/ 25 . Thus a unit ramp at zero time has a response in s space of 1 -1 5 1 1 -5t x (s) = + 2 + or in t space has a particular response of x p ( t ) = ( -1 + 5t + e ) . 25 s s s+5 25 But the actual input is 5 unit ramps and it occurs at t = 1 . Thus the total solution for this region is 5 -5 t -1 -5 t -1 x ( t ) = Ae ( ) - -1 + 5 ( t - 1) + e ( ) ;1 t 3 . To determine A note the "initial condition", 25 5 -5 1-1 x ( t = 1) = x1 , so x1 = Ae0 - -1 + 5 (1 - 1) + e ( ) = A . Thus 25 Name_________________________ Page 1 of 2 12/8/2005 5 -5( t -1) -1 + 5 ( t - 1) + e ;1 t 3 . At t = 3 , 25 5 1 -5 3-1 x3 = x1e -10 - -1 + 5 ( 3 - 1) + e ( ) = x1e-10 - 9 + e -10 a known number. 25 5 Finally for t 3 the input is null and thus so is the particular solution. Thus the total solution -5 t - 3 becomes x ( t ) = Ae ( ) . Solving for the unknown coefficient yields x ( t ) = x1e -5( t -1) - A = x3 . x ( t ) = x3e -5( t - 3) ;3 t . Name_________________________ Page 2 of 2 12/8/2005
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