Question

If f(t) satisfies those conditions f(t) is even function which has the property of f(t) = f(-t) Laplace transform of f(t) is

How to prove this? Laplace transform

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Answer #1

a) e st f(at) dt L(f(at))

Consider 2 cases: a>0

We substitute du at u dt = and our equation becomes

L(f(at)) ef(u) du a

But as est f(t) dt F(s) we have e st/a f(t) dt F(s/a)

So that F L(f(at))

Now the case a<0 in which case we substitute b a so that

est fbt) dt L(f(at)) est f(at) dt

Using even property, we get L(f(at))=\int_{0}^{\infty}e^{-st}f(bt)\,dt=\frac{1}{b}F\left ( \frac{s}{b} \right )

That is, for a<0 we have L(f(at))=\frac{1}{(-a)}F\left ( \frac{s}{(-a)} \right )

Combining these we get L(f(at))=\frac{1}{|a|}F\left ( \frac{s}{|a|} \right )

b) If we have an odd function, we get:

a>0 then F L(f(at)) as the equation is unaffected

If a<0 then we get

est fbt) dt L(f(at)) est f(at) dt

Which is L(f(at))=-\int_{0}^{\infty}e^{-st}f(-bt)\,dt and so

L(f(at)) -F -a) (-a)

So that L(f(at))=\text{sgn}(a)\frac{1}{|a|}F\left(\frac{s}{|a|} \right )

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