A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 3 years ago
Can someone please check this; \[\begin{equation*}
f(t)=\cos(t)\big(h(t\pi)h(t)\big)
\end{equation*}\]
\[
\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal
\newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x
\begin{align*}
\intl0\infty{f(t)e^{pt}}t
\end{align*}
\]
UnkleRhaukus
 3 years ago
Can someone please check this; \[\begin{equation*} f(t)=\cos(t)\big(h(t\pi)h(t)\big) \end{equation*}\] \[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{pt}}t \end{align*} \]

This Question is Closed

UnkleRhaukus
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0I think h is the step function

whpalmer4
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1isn't that Laplace transform? .. let p = s

UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358356261290:dw

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

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

experimentX
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1yes that's correct!!

experimentX
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=Integrate+e^%28at%29+cos%28bt%29

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

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

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

UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1thanks experimentX

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yw!! you work hard man!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.