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

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

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

phi Group TitleBest ResponseYou've already chosen the best response.0
I think h is the step function
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
isn't that Laplace transform? .. let p = s
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1358356261290:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
h=H=u=θ
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
just integrate negative of cos(x) e^(px) from 0 to pi
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yes!! multiple ways of representing unit step function.
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yes that's correct!!
 one year ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1358358498070:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1358358737716:dw this should be simpler.
 one year ago

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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
thanks experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yw!! you work hard man!!
 one year ago
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