Welcome to our community

Be a part of something great, join today!

Problem of the Week #20 - August 13th, 2012

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thanks to those who participated in last week's POTW!! Here's this week's problem!

-----

Problem: Suppose that $\displaystyle g(t) = \int_0^t f(\tau)\,d\tau$. If $G(s)$ and $F(s)$ are the Laplace transforms of $g(t)$ and $f(t)$ respectively, show that $G(s) = \dfrac{F(s)}{s}$.

Recall that $\displaystyle F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st}f(t)\,dt$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
This week's problem was correctly answered by BAdhi and Sudharaka.

Sudharaka's solution:

\[g(t) = \int_0^t f(\tau)\,d\tau\]

\begin{eqnarray}


G(s)&=&\mathcal{L}\{g(t)\}\\


&=&\int_0^{\infty} e^{-st}\int_0^t f(\tau)\,d\tau\,dt\\


&=&\int_0^{\infty}\int_{0}^{t}e^{-st}f(\tau)\,d\tau\,dt


\end{eqnarray}


Note that, \(0<\tau<t<\infty\). Changing the order of integration we get,


\begin{eqnarray}


G(s)&=&\int_0^{\infty}\int_{\tau}^{\infty}e^{-st}f(\tau)\,dt\,d\tau\\


&=&\int_0^{\infty}f(\tau)\int_{\tau}^{\infty}e^{-st}\,dt\,d\tau\\


&=&\int_0^{\infty}f(\tau)\frac{e^{-s\tau}}{s}\,d\tau\\


&=&\frac{1}{s}\int_0^{\infty}e^{-s\tau}f(\tau)\,d\tau\\


\therefore G(s)=\frac{F(s)}{s}


\end{eqnarray}

BAdhi's solution:

Let's consider the laplace transform of $g(t)$

$$G(s)=\mathcal{L}\{g(t)\}=\int_0^\infty e^{-st}g(t)\, dt$$


since, $g(t)=\displaystyle \int_0^t f(\tau )\, d\tau$


$$\begin{align*}
\mathcal{L}\{g(t)\}&=\int \limits_0^\infty e^{-st}g(t)\, dt \\ &=\int \limits_0^\infty e^{-st}\left[ \int_0^t f(\tau )\, d\tau \right] dt \\ &=\int \limits_{0}^{\infty}\int \limits_{0}^{t} e^{-st}f(\tau )\, d\tau dt\\ \end{align*}$$


By switching variables, the boundaries of $t$ and $\tau$ changes as,


$0<\tau<\infty$ and $\,\tau<t<\infty$


then,


$$
\mathcal{L}\{g(t)\}=\int \limits_0^\infty \int \limits_\tau^\infty e^{-st}f(\tau )\, dtd\tau $$


since $f(\tau )$ is independent on $t$,


$$\begin{align*}
\mathcal{L}\{g(t)\}&=\int \limits_{0}^{\infty}f(\tau )\left[ \int \limits_\tau^\infty e^{-st}\,dt \right] d\tau \\ &=\int \limits_0^\infty f(\tau ) \left[ \frac{e^{-st}}{-s}\right] _\tau^\infty \,d\tau \\ &=\int \limits_0^\infty f(\tau ) \left[ \frac{0-e^{-s\tau }}{-s}\right] \,d\tau (when s>0) \\&=\frac{\displaystyle \int \limits_0^\infty f(\tau ) e^{-s\tau }\,d\tau }{s}\\ &=\frac{\displaystyle \int \limits_0^\infty f(t) e^{-st }\,dt }{s}\\ &=\frac{F(s)}{s} \end{align*}$$


$$\therefore \; G(s)=\frac{F(s)}{s}$$ for $s>0$
 
Status
Not open for further replies.