Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

S_Student

  • one year ago

laplace cosat and sinat

  • This Question is Open
  1. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    L(cos at)=\[s/(s ^{2}+a ^{2})\] and L(sin at)=\[a/(s ^{2}+a ^{2})\]

  2. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    if u are interested i can tell u the derivation for these

  3. S_Student
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanx friend...and ya i want to expain me in detail..

  4. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    ok ...

  5. S_Student
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx

  6. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    \[e ^{iat}=\cos(at)+isin(at)\] taking laplace on both sides:: \[L(e ^{iat})=L(\cos (at))+iL(\sin at)\]...............(1) we know that :\[L(e ^{iat})=1/(s-ia)\] now rationalizing(ie multiplying numerator and denominator by (s+ia): \[L(e ^{iat})=(s+ia)/((s+ia)(s-ia))\] \[L(e ^{iat})=(s+ia)/(s ^{2}+a ^{2})\] \[L(e ^{iat})=(s/(s ^{2}+a ^{2}))+i (a/(s ^{2}+a ^{2}))\]................(2) compare 1 and 2::

  7. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    U GET : \[L(\cos at)=s/(s ^{2}+a ^{2})\] and \[L(\sin at)=a/(s ^{2}+a ^{2})\]

  8. S_Student
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is enough for me..

  9. S_Student
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanx dear..

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

    Slight variation ____________ \[\begin{align*} \mathcal L\left\{ \sin({n}t)\right\}(p)&=\int\limits_0^\infty \sin({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak I}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak I}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak I}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak I}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak I}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{n}{{p}^2+{n}^2}\\ \end{align*}\] ____________ \[\begin{align*} \mathcal L\left\{\cos ({n}t)\right\}(p)&=\int\limits_0^\infty \cos({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak R}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak R}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak R}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak R}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak R}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{{p}}{{p}^2+{n}^2}\\ \end{align*}\]

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.