How to calculate f (t) = cos at, t >= 0 by the definition of Laplace?
- az_lenderLv 76 years agoFavorite Answer
F(s) = integral from t = 0 to infinity of
[cos(at)] * e^(-st) dt
Integrate by parts, letting
u = cos(at) and dv = e^(-st) dt
du = -a sin(at) dt and v = (-1/s) e^(-st)
The integral is
uv - integral of v du
= [cos(at) * (-1/s)e^(-st)] +
integral of (-a/s) [sin(at)*e^(-st)] dt
The first term, evaluated at t = infinity and zero,
comes out to just 1/s.
The second term must be integrated by parts a second time,
using u2 = a sin(at) and dv2 = (-1/s)e^(-st) dt, so
du2 = a^2 cos(at) and v2 = (-1/s^2)e^(-st)
Then the original integral must be
1/s - (a/s^2)*sin(at)e^(-st) -
(a^2/s^2) integral of [ (cos at) e^(-st) ] dt
ADD the integral from the RHS to the original integral on the LHS, obtaining:
[1 + (a^2/s^2)] * F(s) =
1/s + [(a/s^2)*(sin at)*e^(-st)]
but note that the last term will be zero BOTH at t = infinity and t = 0.
So you have
F(s) = (1/s) / (1 + a^2/s^2) ]
= s / (s^2 + a^2)
and luckily this coincides with the expected Laplace transform of cos(at).
- RickLv 76 years ago
1. Compute the Laplace transform of f(t).
2. Now use the initial value theorem. See: http://en.wikipedia.org/wiki/Initial_value_theorem
- Anonymous6 years ago
Step by step please.