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14.5: Shifting Theorem

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    5505

    • This is a very useful theorem, and one that is almost trivial to prove. (Try it!) It is

    \[\textbf{L}\left(e^{-at} y(t) \right) = \bar{y}(s+a).\]

    For example, from the table, we have \(\textbf{L}(t) = 1/s^2\). The shifting theorem tells us that \(\textbf{L}\left(te^{-at} \right) = 1/(s+a)^2\). I'm sure you will now want to expand your table even more. Or you may want to go the other way, and cut down the table a bit! After all, you know that \(\textbf{L}(1) = 1/s\). The shifting theorem, then, tells you that \(\textbf{L}(e^{at}) = 1/(s-a)\), so that entry in the table is superfluous! Note that you can use the theorem to deduce either direct or inverse transforms.