Laplace transform

[logistic_guy]

Derive the Laplace transform for the following time functions:

\(\displaystyle \bold{a.} \ u(t)\)
\(\displaystyle \bold{b.} \ tu(t)\)
\(\displaystyle \bold{c.} \sin \omega t \ u(t)\)
\(\displaystyle \bold{d.} \cos \omega t \ u(t)\)

 

[khansaheb]

Derive the Laplace transform for the following time functions:

\(\displaystyle \bold{a.} \ u(t)\)
\(\displaystyle \bold{b.} \ tu(t)\)
\(\displaystyle \bold{c.} \sin \omega t \ u(t)\)
\(\displaystyle \bold{d.} \cos \omega t \ u(t)\)

show us your effort/s to solve this problem.

 

[logistic_guy]

\(\displaystyle \bold{a}.\)

\(\displaystyle \mathcal{L}\{u(t)\} = \int_{0}^{\infty} u(t) e^{-st} \ dt = \int_{0}^{\infty} e^{-st} \ dt = -\frac{1}{s}e^{-st}\bigg|_{0}^{\infty} = -\frac{1}{s}(0 - 1) = \frac{1}{s}\)

 

[logistic_guy]

\(\displaystyle \bold{b}.\)

\(\displaystyle \mathcal{L}\{tu(t)\} = \int_{0}^{\infty} tu(t) e^{-st} \ dt = \int_{0}^{\infty} te^{-st} \ dt\)

\(\displaystyle u = t\)
\(\displaystyle du = dt\)
\(\displaystyle dv = e^{-st} \ dt\)
\(\displaystyle v = \frac{1}{-s} e^{-st}\)

\(\displaystyle \int_{0}^{\infty} te^{-st} \ dt = -\frac{t}{s}e^{-st}\bigg |_{0}^{\infty} + \frac{1}{s} \int_{0}^{\infty}e^{-st} \ dt\)


\(\displaystyle = -\frac{t}{s}\left(\lim_{t \rightarrow \infty}\frac{t}{e^{st}} - 0\right) - \frac{1}{s^2}e^{-st}\bigg |_{0}^{\infty} \)


\(\displaystyle = -\frac{t}{s^2}\left(\lim_{t \rightarrow \infty}\frac{1}{e^{st}}\right) - \frac{1}{s^2}\left(\lim_{t \rightarrow \infty}e^{-st} - 1\right)\)


\(\displaystyle = -\frac{t}{s^2}\left(0\right) - \frac{1}{s^2}\left(0 - 1\right) = \frac{1}{s^2}\)

 

[logistic_guy]

\(\displaystyle \bold{c.}\)

\(\displaystyle \mathcal{L}\{\sin \omega t \ u(t)\} = \int_{0}^{\infty} \sin \omega t \ u(t) e^{-st} \ dt = \int_{0}^{\infty} \sin \omega t \ e^{-st} \ dt\)


\(\displaystyle \sin \omega t = \frac{e^{j\omega t} - e^{-j\omega t}}{2j}\)


\(\displaystyle \int_{0}^{\infty} \sin \omega t \ e^{-st} \ dt = \frac{1}{2j}\int_{0}^{\infty} e^{(j\omega - s)t} \ dt - \frac{1}{2j}\int_{0}^{\infty} e^{-(j\omega + s)t} \ dt\)


\(\displaystyle = \frac{1}{2j}\left(-\frac{1}{j\omega - s} - \frac{1}{j\omega + s}\right) = \frac{\omega}{s^2 - j^2\omega^2} = \frac{\omega}{s^2 + \omega^2}\)

We could have solved the integral by integration by parts if we wanted.

 

[logistic_guy]

\(\displaystyle \bold{d.}\)

\(\displaystyle \mathcal{L}\{\cos \omega t \ u(t)\} = \int_{0}^{\infty} \cos \omega t \ u(t) e^{-st} \ dt = \int_{0}^{\infty} \cos \omega t \ e^{-st} \ dt\)

Let us this time solve this integral by integration by parts.

\(\displaystyle u = \cos \omega t\)
\(\displaystyle du = -\omega\sin \omega t \ dt\)
\(\displaystyle dv = e^{-st} \ dt\)
\(\displaystyle v = \frac{1}{-s}e^{-st}\)

\(\displaystyle \int_{0}^{\infty} \cos \omega t \ e^{-st} \ dt = \frac{1}{-s}e^{-st}\cos \omega t\bigg |_{0}^{\infty} - \frac{\omega}{s}\int_{0}^{\infty} \sin \omega t e^{-st} \ dt\)


\(\displaystyle = \frac{1}{-s}\left(0 - 1\right) - \frac{\omega}{s}\int_{0}^{\infty} \sin \omega t e^{-st} \ dt = \frac{1}{s} - \frac{\omega}{s}\int_{0}^{\infty} \sin \omega t e^{-st} \ dt\)

\(\displaystyle u = \sin \omega t\)
\(\displaystyle du = \omega\cos \omega t \ dt\)
\(\displaystyle dv = e^{-st} \ dt\)
\(\displaystyle v = \frac{1}{-s}e^{-st}\)

\(\displaystyle \frac{1}{s} - \frac{\omega}{s}\int_{0}^{\infty} \sin \omega t e^{-st} \ dt = \frac{1}{s} + \frac{\omega}{s^2}e^{-st}\sin \omega t \bigg |_{0}^{\infty} - \frac{\omega^2}{s^2}\int_{0}^{\infty} e^{-st}\cos \omega t \ dt\)


\(\displaystyle = \frac{1}{s} + \frac{\omega}{s^2}(0 - 0) - \frac{\omega^2}{s^2}\int_{0}^{\infty} e^{-st}\cos \omega t \ dt = \frac{1}{s} - \frac{\omega^2}{s^2}\int_{0}^{\infty} e^{-st}\cos \omega t \ dt\)


\(\displaystyle \left(1 + \frac{\omega^2}{s^2}\right)\int_{0}^{\infty} \cos \omega t \ e^{-st} \ dt = \frac{1}{s}\)


\(\displaystyle \int_{0}^{\infty} \cos \omega t \ e^{-st} \ dt = \frac{1}{s\left(1 + \frac{\omega^2}{s^2}\right)} = \frac{1}{s\left(\frac{s^2}{s^2} + \frac{\omega^2}{s^2}\right)} = \frac{s^2}{s\left(s^2 + \omega^2\right)} = \frac{s}{s^2 + \omega^2}\)