Laplace transform of the cos function | 9math

Laplace transform of the cos function

Posted December 8th, 2007 by Structure

Laplace transform of $f(t)=\cos t\:h(t)$ is defined by

$\mathcal{L}(\cos t\:h(t))(s)=\int_{0}^{+\infty}\cos t e^{-st}dt$

We can notice that $\cos t=Re(\cos t+i\sin t) =Re(e^{it}$
Then for real s we have

$\mathcal{L}(\cos t\:h(t))(s)=Re(\mathcal{L}(e^{it}\:h(t))(s)=Re\frac{1}{s-i}=Re\frac{s+i}{s^2+1}=\frac{s}{s^2+1}$

So

$\mathcal{L}(\cos t\:h(t))(s)=\frac{s}{s^2+1}$

We have also

$\mathcal{L}(\sin t\:h(t))(s)=\frac{1}{s^2+1}$

Another method is the following
Function $f:R\rightarrow R$ , $f(t)=\cos t\:h(t)$ verifies differential equation

$x"(t)+x(t)=0\;\:x(0+)=1,\:\:x'(0+)=0$

$\mathcal{L}(x(t)(s)=X(s)$

$\mathcal{L}(x'(t)(s)=sX(s)-1$

$\mathcal{L}(x"\;(t)(s)=s^2X(s)-s$

We have

$s^2X(s)-s+X(s)=0$

so

$X(s)=\mathcal{L}(\cos t\:h(t))(s)=\frac{s}{s^2+1}$

As $\mathcal{L}(t f(t))(s)=-\mathcal{L}'( f(t))(s)$ we have also

$\mathcal{L}(t\cos t\:h(t))(s)=-\frac{d}{ds}\frac{s}{s^2+1}=\frac{s^2-1}{(s^2+1)^2}$

Function $f:R\rightarrow R$ , $f(t)=\cos \omega t\:h(t)$ verifies differential equation

$x"(t)+ \omega^2 x(t)=0\;\:x(0+)=1,\:\:x'(0+)=0$

$\mathcal{L}(x(t)(s)=X(s)$

$\mathcal{L}(x'(t)(s)=sX(s)-1$

$\mathcal{L}(x"\;(t)(s)=s^2X(s)-s$

We have

$s^2X(s)-s+ \omega^2 X(s)=0$

so

$X(s)=\mathcal{L}(\cos \omega t\:h(t))(s)=\frac{s}{s^2+ \omega^2}$

and also

$\mathcal{L}(t\cos \omega t\:h(t))(s)=\frac{s^2-\omega^2}{(s^2+ \omega^2)^2}$

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