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Proof this Laplace transform L(f(t)u(t-a)) = e^-as L(f(t+a))

Solution

The usual L.T. formula for translation on the t-axis is given by

L(u(t a)f(t a) ) = e ^(as)*F(s), where F(s) = L (f(t)) , a > 0

Applying this equation one can write:

L (u(t a)g(t a) ) = e ^as(L g(t)) ;

This says that to get the factor on the right side involving g, we should replace t a by t in the function g(t a) on the left, and then take its Laplace transform. Apply this procedure to the function f(t), written in the form f(t) = f((t a) + a); we get (“replacing t a by t and then taking the Laplace Transform”)

L (u(t a)f((t a) + a)) = e ^as*L (f(t + a) )

Hence proved.