ENGR 2422 Engineering Mathematics 2

Faculty of Engineering and Applied Science
2005 Winter

Laplace Transform Example (Second Shift Theorem)

Question

Find the Laplace transform of

f(t) = (t^2 - 4t e^(-2t)) H(t-3)

Solution

In order to use the second shift theorem, the function multiplying H(t – 3) must be re-expressed as a function of (t – 3), not t.

t^2 - 4t e^(-2t) = (t – 3 + 3)^2
– 4(t – 3 + 3) e^–2(t – 3 + 3)
= (t – 3)^2 + 6(t – 3) + 9
– 4(t – 3) e^–2(t – 3)e^(–6)
– 12 e^–2(t – 3)e^(–6)
But if
g(t – 3) = (t – 3)^2 + 6(t – 3) + 9
– 4(t – 3) e^–2(t – 3)e^(–6)
– 12 e^–2(t – 3)e^(–6)
then
g(t) = t^2 + 6t + 9
– 4t e^(–2t) e^(–6)
– 12 e^(–2t) e^(–6)
One version of the second shift theorem, applied to this situation, states that
$L{g(t–3) H(t–3)} = L{g(t)} e^(–3s)$
$L{f(t)} = e^(–3s) L{ t^2 + 6t + 9 – 4t e^(–2t) e^(–6) – 12 e^(–2t) e^(–6)}$
The first shift theorem appears next:
$L{– (4t + 12) e^(–2t) e^(–6)} = L{– (4t + 12) e^(–6)} evaluated at s -> s+2$
The Laplace transform of the original f (t) then follows: