429-2009-HW4key

# 429-2009-HW4key - Homework 4 Answer Key By David Simcha...

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Homework 4 Answer Key By: David Simcha Problem 1: We’re asked to prove that: ( 1 ) f ( s ) g ( s )= ± 0 e - st ± t 0 g ( τ ) f ( t - τ ) dτdt Note: I’m using τ here instead of t’ because, in a calculus context, t’ looks too much like taking a derivative of t. We also make the the following assumption, which is justiFed by the fact that f(t) is implicitly multiplied by u(t) when working in Laplace space, meaning that f(t) = 0 for t < 0: ( 2 ) ± t g ( τ ) f ( t - τ ) =0 Therefore, we can write that: ( 3 ) ± t 0 g ( τ ) f ( t - τ ) = ± 0 g ( τ ) f ( t - τ ) We then make the following changes of variables: ( 4 ) τ = τ ( 5 ) t = u + τ ( 6 ) du = dt Substituting these changes of variables, as well as (3), we can rewrite (1) as: ( 7 ) ± 0 ± 0 e ( - s )( u + τ ) f ( u ) g ( τ ) dτdu We now expand the exponential term as follows: ( 8 ) e - s ( u + τ ) = e - su e - s τ Using this, (7) can now be rewritten as the product of two single integrals, since terms with only u can be treated as constants when integrated w.r.t. tau and vice-versa: ( 9 ) ± 0 g ( τ ) e - τ s ± 0 e - f ( u ) du This is, by deFnition of a Laplace transform, ˜ f ( s g ( s ). Q.E.D. 1

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Problem 2: To derive the inverse Laplace transform of a repeated pole, we Frst take the inverse Laplace transform of the following expression: ( 1 x ( s )= 1 ( s + α )( s + ± + α ) Using ”standard” results derived from the lecture on contour integration this evaluates to: ( 2 ) x ( t e - α t ± - e - ( ± + α ) t ± Combining terms of the r.h.s. and placing them over a common denominator, we get: ( 3 ) e - - e - ± t - ± The derivative of the denominator w.r.t. ± is trivially 1. Taking the derivative of the numerator w.r.t. ± , we get: ( 4 ) d ( e - - e - - ) = te - - Evaluating at ± = 0 and simplifying we get: ( 5 ) x ( t - αt Now for a second repeated pole, we do essentially the same thing, but with two epsilon variables. Starting with the following expression: ( 6 x ( s 1 ( s + α s + α + ± 1 s + α + ± 2 ) And again taking the inverse Laplace transform by standard methods: ( 7 ) x ( t e - - ± 2 t ± 2 ( ± 2 - ± 1 ) - e - - ± 1 t ± 1 ( ± 2 - ± 1 ) + e - ± 1 ± 2 We again need to combine terms and place them over a common denominator to apply L’Hopital’s rule: ( 8 ) ± 1 e - ( α + ± 2 ) t - ± 2 e - ( α + ± 1 ) t + ± 2
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429-2009-HW4key - Homework 4 Answer Key By David Simcha...

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