find f(t) for L{f(t)}={((1-e^s)(1+e^2s))/s^2}?

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  • find f(t) for L{f(t)}={((1-e^s)(1+e^2s))/s^2}?


Answer #1 | 23/12 2013 19:59
I'll write it as L{f(t)} = 1/s^2 + e^(2s)/s^2 - e^s/s^2 - e^(3s)/s^2 In the Laplace Transform table in Pauls Online Math Notes, formula #27 says that e^(-cs) F(s) is the the Laplace transform of u(t-c) f(t-c), where u(t-c) is the unit step function at t=c. And formula #3 implies that 1/s^2 is the transform of f(t) = t. So in your problem, f(t) seems to be t + (t+2)u(t+2) - (t+1)u(t+1) - (t+3)u(t+3) It's not obvious what the meaning of the f values on the negative domain might be, since the Laplace transform operates on f(t) only from 0 to +infinity. The expression I wrote for f(t) is uniformly equal to -1 on the interval (0,infinity).

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