Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  2. whpalmer4 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  3. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I think h is the step function

    • one year ago
  4. whpalmer4 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ah, okay, sure, I didn't realize that we could assume anything about h(t)

    • one year ago
  5. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    isn't that Laplace transform? .. let p = s

    • one year ago
  6. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The unit heaviside step function \[h(t-a) =\begin{cases}0&t<a\\ 1&t≥a\end{cases} \]

    • one year ago
  7. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1358356261290:dw|

    • one year ago
  8. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    h=H=u=θ

    • one year ago
  9. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    just integrate negative of cos(x) e^(-px) from 0 to pi

    • one year ago
  10. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes!! multiple ways of representing unit step function.

    • one year ago
  11. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    • one year ago
    1 Attachment
  12. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  13. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes that's correct!!

    • one year ago
  14. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    http://www.wolframalpha.com/input/?i=Integrate+e^%28-pt%29+cos%28t%29+from+0+to+pi

    • one year ago
  15. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  16. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    http://www.wolframalpha.com/input/?i=Integrate+e^%28at%29+cos%28bt%29

    • one year ago
  17. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1358358498070:dw|

    • one year ago
  18. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1358358737716:dw| this should be simpler.

    • one year ago
  19. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ah yeah I can see that now, one integral is simpler than two

    • one year ago
  20. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\ &=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\ &=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\ &=\intl\pi0{\cos(t)e^{-pt}}t\\ &=\left.\frac{e^{-pt}\big(\sin (t)+p\cos (t)\big)}{(-p)^2+1^2}\right|_\pi^0\\ &=\frac{\sin 0+p\cos 0-e^{-p\pi}\big(\sin \pi+p\cos \pi\big)}{p^2+1^2}\\ &=\frac{p+pe^{-p\pi}}{p^2+1^2}\\ &=\frac{p\left(1+e^{-p\pi}\right)}{p^2+1^2}\\&\color{red}\checkmark \end{align*} \]

    • one year ago
  21. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks experimentX

    • one year ago
  22. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yw!! you work hard man!!

    • one year ago
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.