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nagle_differential_equations_ISM_Part52

# nagle_differential_equations_ISM_Part52 - Exercises 7.8 30...

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Exercises 7.8 30. Let Y := L { y ( t ) } . The Laplace transform of the left-hand side of the given equation (with the imposed initial conditions) is ( s 2 + 1) Y ( s ) For the right-hand side, one has L k =1 δ ( t - 2 ) ( s ) = k =1 e - 2 kπs . Hence, Y ( s ) = 1 s 2 + 1 k =1 e - 2 kπs . 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 ) = k =1 h k ( t ) := k =1 sin ( t - 2 ) u ( t - 2 ) = k =1 (sin t ) u ( t - 2 ) = (sin t ) k =1 u ( t - 2 ) . Evaluating y ( t ) at, say, t n = ( π/ 2) + 2 for n = 1 , 2 , . . . we see that y ( t n ) = sin π 2 + 2 k =1 u ( t - 2 ) = n k =1 (1) = n → ∞ with t n → ∞ , meaning that the bridge will eventually collapse. 32. By taking the Laplace transform of ay + by + cy = δ ( 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

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Chapter 7 (ii) If the roots of as 2 + bs + c are complex, then they are α ± , where α and β = 0 satisfy H ( s ) = 1 a [( x - α ) 2 + β 2 ] so that h ( t ) = L - 1 { H ( s ) } ( t ) = 1 e αt sin βt , and, again, it is clear that h ( t ) 0 as t → ∞ if and only if the real part α
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nagle_differential_equations_ISM_Part52 - Exercises 7.8 30...

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