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final2006

# final2006 - Problem 1(8 points The time-delay rule allows...

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Problem 1. (8 points) The time-delay rule allows computation of Laplace transforms of many signals of engineering interesting. As an illustration, you are asked to compute the Laplace Transform of the following functions, which appear as the output of basic rectifier circuits. (a) (4 points) The function f ( t ) = max { sin( t ) , 0 } 1( t ), i.e., the function that is equal to sin( t ) whenever it is positive and zero otherwise, as shown in the left figure. Hint : Notice that you can write the equation f ( t ) f ( t π ) = sin( t )1( t ). Apply to both sides of this equation the Laplace Transform and solve for F ( s ). (b) (4 points) The function g ( t ) = abs(sin( t )1( t ), i.e., the absolute value of the function sin( t )1( t ) as shown in the right figure. Hint : As in (a), try to write g ( t ) as the sum of f ( t ) and a properly modified version of f ( t ), then take Laplace Transform and use the solution of (a) to find the solution for G ( s ). 0 0 0.5 1 1.5 2 t π 2 π 3 π 4 π 5 π f(t) 0 0 0.5 1 1.5 2 t t π 2 π 3 π 4 π 5 π g(t) Solution: (a) We start from the equation f ( t ) f ( t π ) = sin( t )1( t ) . Applying the Laplace Transform to both sides, we get F ( s ) e πs F ( s ) = 1 s 2 + 1 . Solving for F ( s ): F ( s ) = 1 1 e πs 1 s 2 + 1 . 1

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(b) It is clear that g ( t ) can be written as g ( t ) = f ( t ) + f ( t π ) . Applying the Laplace Transform to both sides, we get G ( s ) = F ( s ) + e πs F ( s ) . Solving for G ( s ): G ( s ) = ( 1 + e πs ) F ( s ) . Plugging F ( s ) from (a) into the above equation, G ( s ) = 1 + e πs 1 e πs 1 s 2 + 1 . Note (not required in the exam) : using the fact that tanh ( π 2 s ) = 1 e - πs 1+e - πs , we can write G ( s ) = 1 tanh ( π 2 s ) 1 s 2 + 1 . That’s the way it’s shown in tables. Some people tried to solve the problem using infinite series (the hard way, in fact nobody succeeded). For this, note that f ( t ) = sin( t )1( t ) + sin( t π )1( t π ) + sin( t 2 π )1( t 2 π ) + ... = i =0 sin( t )1( t ) . Then, F ( s ) = i =0 e iπs s 2 + 1 = 1 s 2 + 1 i =0 ( e πs ) i = 1 s 2 + 1 1 1 e πs , using the sum of the geometric series. Similarly, f ( t ) = sin( t )1( t ) + 2 sin( t π )1( t π ) + 2 sin( t 2 π )1( t 2 π ) + ... = sin( t )1( t ) + 2 i =1 sin( t )1( t ) . 2
Then, F ( s ) = 1 s 2 + 1 + 2 i =1 e iπs s 2 + 1 = 1 s 2 + 1 + 2 s 2 + 1 i =1 ( e πs ) i = 1 s 2 + 1 + 2e πs s 2 + 1 i =0 ( e πs ) i = 1 s 2 + 1 1 + e πs 1 e πs . 3

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Problem 2. (8 points) The Laplace Transform is not only useful for solving ODE’s and in control design, but can also be used to easily compute diﬃcult improper integrals that sometimes appear in applications. This problem illustrates how to do it.
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final2006 - Problem 1(8 points The time-delay rule allows...

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