A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Anas.P

  • one year ago

fourier transform question the first question please as soon as possible

  • This Question is Closed
  1. anas.p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathmate

  2. anas.p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @e.mccormick

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

    Find the fourier Transform of \[e^{-a^{2}x^{2}} ,a>0\]Given \[\int\limits_{-\infty}^{\infty} e^{-t} dt = \sqrt{\pi}\]

  4. anas.p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r @Nnesha

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

    hint: first step, we have to compute this integral: \[\large g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx\]

  6. anas.p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i tried... this method substitution method. integration by parts but it keeps rotating back with nothing reducing. if you could show me a step by step process of solving this i would really be grateful

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

    now, we can write this: \[\Large \begin{gathered} {a^2}{x^2} - i\omega x = {a^2}{x^2} - i\omega x - \frac{{{\omega ^2}}}{{4{a^2}}} + \frac{{{\omega ^2}}}{{4{a^2}}} = \hfill \\ \hfill \\ = {\left( {ax - \frac{{i\omega }}{{2a}}} \right)^2} + \frac{{{\omega ^2}}}{{4{a^2}}} \hfill \\ \end{gathered} \] so we get: \[\Large \begin{gathered} g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx = \hfill \\ \hfill \\ = {e^{ - \frac{{{\omega ^2}}}{{4{a^2}}}}}\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {ax - \frac{{i\omega }}{{2a}}} \right)}^2}}}} dx \hfill \\ \end{gathered} \]

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

    now we have to make this variable change: \[\Large z = ax - \frac{{i\omega }}{{2a}}\]

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

    where z is the new variable

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

    hint: we can use this identity: \[\Large \int_{ - \infty }^{ + \infty } {{e^{ - {z^2}}}} dz = 2\int_0^{ + \infty } {{e^{ - {z^2}}}dz} \]

  11. anas.p
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks man... Really helpful... I had no idea that we had to use the complete the square method. Thanks again.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.