ENGR 2422 Engineering Mathematics 2
Laplace transform of cos ωt

Four different methods for obtaining the Laplace transform of the cosine function are presented here:

directly, from the definition of the Laplace transform
via the exponential function
using the Maclaurin series expansion
via the derivative of the sine function

Direct Derivation

Let f (t) = cos ωt and F (s) = { cos ωt }
$F(s) = Integral_0^oo { exp(-st) cos t } dt$
Using integration by parts,
Integration by parts, tabular method
F(s) = [exp(-st)/w^2 (w sin wt - s cos wt)] - (s/w)^2 F(s)
(1 + (s/w)^2) F(s) = 0 - (-s/w^2)
F(s) = (s/w^2)*(w^2 / (s^2 + w^2))
F(s) = s/(s^2 + w^2)

Complex Exponential Form

First note that $cos x = (e^x + e^(-x)) / 2 and L{ e^(at) } = 1 / (s-a)$ .
F(s) = L { (exp(jwt) + exp(-jwt))/2 }
= ( 1/(s-jw) + 1/(s+jw) ) / 2
F(s) = (s+jw + s-jw) / [2(s^2 - (jw)^2)]
F(s) = s/(s^2 + w^2)

Maclaurin Series

The Maclaurin series for the cosine function is
cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ...
Also using the linearity property of Laplace transforms:
$L{cos wt} = Sum{(-1)^n w^(2n) L{ t^(2n) } / (2n)!$
$= Sum{(-w^2/s^2)^n} / s$
which is a geometric series, first term a = 1/s and common ratio r = –(ω/s)².
$L{cos wt} = (1/s) / (1 + w^2/s^2)$
F(s) = s/(s^2 + w^2)

Derivative Method

The rule for the Laplace transform of a derivative is needed here:
L{ f'(t) } = s L{ f(t) } - f(0)
We also need to know in advance the Laplace transform of the sine function.
L{ cos wt } = L { (sin wt / w)' }
= s L{ (sin wt)/w } - 0
F(s) = s/(s^2 + w^2)

[Index of Brief Notes]

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Created 2004 03 13 and last modified 2004 03 13 by Dr. G.H. George

ENGR 2422 Engineering Mathematics 2 Laplace transform of cos ωt

Direct Derivation

Complex Exponential Form

Maclaurin Series

Derivative Method

ENGR 2422 Engineering Mathematics 2
Laplace transform of cos ωt