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LAPLACE TRANSFORM
Mohammed R. asked • 05/05/17
Find Laplace Transform of e-t Cos2t
How do i find laplace transform the equation
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Calvin H. answered • 05/05/17
TUTOR 4.9 (15)
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You can apply the definition L{f(t)} = Integral (from 0 to infinity) of e^(-st)f(t)dt
or apply the form on the Laplace Transform for e^(at)cos(bt):
L{e^(at)cos(bt)} = (s-a)/[(s-a)^2+b^2]
The above form could be found by applying a translation by a to the Laplace Transform of cos(bt):
L{e^(at)f(t)} = F(s-a)
since
Integral (from 0 to infinity) e^(at)e^(-st)f(t)dt
= Integral (from 0 to infinity) e^[(s-a)t]f(t)dt
= F(s-a) for a > s
L{cos(bt)} = Integral (0 to infinity) e^(-st)cos(bt)dt
= -1/s{[e^(-st)cos(bt) | (0 to infinity) + b Integral (0 to infinity) e^(-st)sin(bt)dt}
= -1/s{-1 + b (-1/s)[e^(-st)sin(bt) | (0 to infinity) - b Integral (0 to infinity) e^(-st)cos(bt)dt]
= 1/s - (b/s)^2 Integral (0 to infinity) e^(-st)cos(bt)dt
=> Integral (0 to infinity) e^(-st)cos(bt)dt = (1/s)/[1+(b/s)^2]
= (1/s)/[(s^2+b^2)/s^2] = s/(s^2+b^2)
So L{cos(bt)} = s/(s^2+b^2) = F(s)