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 2 years ago
Can someone please check this; \[\begin{equation*}
f(t)=\cos(t)\big(h(t\pi)h(t)\big)
\end{equation*}\]
\[
\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal
\newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x
\begin{align*}
\intl0\infty{f(t)e^{pt}}t
\end{align*}
\]
 2 years ago
Can someone please check this; \[\begin{equation*} f(t)=\cos(t)\big(h(t\pi)h(t)\big) \end{equation*}\] \[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{pt}}t \end{align*} \]

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UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{pt}}t&=\intl0\infty{\cos(t)\big(h(t\pi)h(t)\big)e^{pt}}t\\ &=\intl0\infty{\cos(t)h(t\pi)e^{pt}}t\intl0\infty{\cos(t)h(t)e^{pt}}t\\ &=\intl\pi\infty{\cos(t)e^{pt}}t\intl0\infty{\cos(t)e^{pt}}t\\ &=\left.\frac{e^{pt}\big(p\cos t+\sin t\big)}{(p)^2+1^2}\right_\pi^\infty\left.\frac{e^{pt}\big(p\cos t+\sin t\big)}{(p)^2+1^2}\right_0^\infty\\ &=\frac{e^{p\pi}\big(p\cos \pi+\sin \pi\big)}{p^2+1}+\frac{\big(p\cos 0+\sin 0\big)}{p^2+1}\\ &=\frac{pe^{p\pi}}{p^2+1}\frac{p}{p^2+1}\\ \\ &=\frac{p\big(1+e^{p\pi }\big)}{p^2+1}\\ \end{align*}\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.0I haven't had my coffee yet, but I don't quite see how you managed to make h(t) vanish

phi
 2 years ago
Best ResponseYou've already chosen the best response.0I think h is the step function

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.0ah, okay, sure, I didn't realize that we could assume anything about h(t)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1isn't that Laplace transform? .. let p = s

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1The unit heaviside step function \[h(ta) =\begin{cases}0&t<a\\ 1&t≥a\end{cases} \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1358356261290:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1just integrate negative of cos(x) e^(px) from 0 to pi

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yes!! multiple ways of representing unit step function.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1think (h(t−π)−h(t)) as a switch (toward negative) that get's on at t=0 and off at t=pi .

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yes that's correct!!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=Integrate+e^%28pt%29+cos%28t%29+from+0+to+pi

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1there's a short formula.for int e^(ax) cos(bx) dx ...also note that you can shift integral for step function.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=Integrate+e^%28at%29+cos%28bt%29

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1358358498070:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1358358737716:dw this should be simpler.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1Ah yeah I can see that now, one integral is simpler than two

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{pt}}t&=\intl0\infty{\cos(t)\big(h(t\pi)h(t)\big)e^{pt}}t\\ &=\intl0\infty{\cos(t)h(t\pi)e^{pt}}t\intl0\infty{\cos(t)h(t)e^{pt}}t\\ &=\intl\pi\infty{\cos(t)e^{pt}}t\intl0\infty{\cos(t)e^{pt}}t\\ &=\intl\pi0{\cos(t)e^{pt}}t\\ &=\left.\frac{e^{pt}\big(\sin (t)+p\cos (t)\big)}{(p)^2+1^2}\right_\pi^0\\ &=\frac{\sin 0+p\cos 0e^{p\pi}\big(\sin \pi+p\cos \pi\big)}{p^2+1^2}\\ &=\frac{p+pe^{p\pi}}{p^2+1^2}\\ &=\frac{p\left(1+e^{p\pi}\right)}{p^2+1^2}\\&\color{red}\checkmark \end{align*} \]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1thanks experimentX

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yw!! you work hard man!!
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