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Unformatted text preview: Exercises 7.8 30. Let Y := L{ y ( t ) } . The Laplace transform of the lefthand side of the given equation (with the imposed initial conditions) is ( s 2 + 1) Y ( s ) For the righthand side, one has L X k =1 ( t 2 k ) ( s ) = X k =1 e 2 ks . Hence, Y ( s ) = 1 s 2 + 1 X k =1 e 2 ks . Taking the inverse Laplace transform in this equation yields the following sum y ( t ) of the series of impulse response functions h k ( t ): y ( t ) = X k =1 h k ( t ) := X k =1 sin ( t 2 k ) u ( t 2 k ) = X k =1 (sin t ) u ( t 2 k ) = (sin t ) X k =1 u ( t 2 k ) . Evaluating y ( t ) at, say, t n = ( / 2) + 2 n for n = 1 , 2 ,... we see that y ( t n ) = h sin 2 + 2 n i X k =1 u ( t 2 k ) = n X k =1 (1) = n with t n , meaning that the bridge will eventually collapse. 32. By taking the Laplace transform of ay 00 + by + cy = ( t ) , y (0) = y (0) = 0 , and solving for Y := L{ y } , we find that the transfer function is given by H ( s ) = 1 as 2 + bs + c . We consider the following possibilities. (i) If the roots of the polynomial as 2 + bs + c are real and distinct, say r 1 , r 2 , then H ( s ) = 1 a ( s r 1 )( s r 2 ) = 1 a ( r 1 r 2 ) 1 s r 1 1 s r 2 . Thus, h ( t ) = L 1 { H ( s ) } ( t ) = 1 a ( r 1 r 2 ) ( e r 1 t e r 2 t ) and, clearly, h ( t ) has zero limit as t if and only if r 1 and r 2 are negative. 251 Chapter 7 (ii) If the roots of as 2 + bs + c are complex, then they are i , where and 6 = 0 satisfy H ( s ) = 1 a [( x ) 2 + 2 ] so that h ( t ) = L 1 { H ( s ) } ( t ) = 1 a...
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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