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

edr1c

what is the period for the fourier expansion of f(t) = { 5, 0<= t <= 2; 0, 2<t<4}

  • one year ago
  • one year ago

  • This Question is Closed
  1. edr1c
    Best Response
    You've already chosen the best response.
    Medals 0

    because normally the question i'd encountered has the same interval, eg -2<t<0; 0<t<2 etc.

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

    maybe we can do a shift of the function by substituting u=t-2

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

    |dw:1343832213995:dw| this can be transformed into regular Heaviside function

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

    you mean\[f(t)=5(1-u_2(t))\]?

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

    but that doesn't fix the intervals, which I do think need to be symmetric

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

    why 1/5 ?

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

    sorry ... (t - 2)/2

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

    |dw:1343833073250:dw|

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

    the Fourier expansion of left is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html

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

    |dw:1343833287038:dw|

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

    I still don't see why you need the /2 part

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

    that will make the interval of |dw:1343833578897:dw|

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

    however that was not necessary ... i guess

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

    but why does it have to be from -1 to 1? can't it just be symmetric?

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

    i guess yes i can be ... i thought unit would make easier.

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

    I just did \(x=t-2\)\[A_0=\frac1{2(2)}\int_{-2}^2f(x)dx=\frac14\int_{-2}^05dx=\frac52\]

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

    |dw:1343833717021:dw|

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

    |dw:1343833858118:dw|

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

    \[A_n=\frac12\int_{-2}^2f(x)\cos\left({n\pi x\over2}\right)dx=\frac52\int_{-2}^0\cos\left({n\pi x\over2}\right)dx=\frac5{n\pi}\sin(n\pi)\]

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

    \[B_n=\frac52\int_{-2}^0\sin\left(n\pi x\over2\right)dx=\frac5{n\pi}\left(\cos(n\pi)-1\right)\]

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

    but now this formula will be for f(x)=f(t-2) so I guess this is wrong oh well, been a while since I've done fourier series

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

    If your question is about the period of the function...(piecewise continuous ) then the answer will be , normally ,"4",...because ! a periodic function is entirely defined by its restriction "to one period"... Usually, that's what we do for fourier expansion's exercises !

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

    doesn't anyone know how to plug this into wolf?

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

    so , because "giving answers" is not allowed ! I'll let you do the calculation by yourself ! \[a_n=\frac{2}{4}\int\limits_{0}^{4}f(t)\cos(n \frac{2 \pi}{4}t)dt\] \[b_n=\frac{2}{4}\int\limits_{0}^{4}f(t)\sin(n \frac{2 \pi}{4}t)dt \] To calculate these integrals just "split them" \[\int\limits_{0}^{2}-------+\int\limits_{2}^{4}--------\]

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

    Plot[5/2 + 5*4/(2*Pi)*Sum[Sin[(2 n + 1)*Pi*(x)/2]/(2 n + 1), {n, 0, 10}], {x , 0, 10}]

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

    |dw:1343836422904:dw| the example at wolframmathworld was for 0 to 2L

    • one year ago
  27. edr1c
    Best Response
    You've already chosen the best response.
    Medals 0

    owh so for the expansion it doesnt necessary to be symmetrical? because the formula i'd received in the notes is from -L to L, where L is period/2.. thats the only question im wondering about. if it can be non-symmetrical then everythings is ok already.

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

    if it's symmetric then the calculation is easier ... for eg. if the function is odd ... then you don't have to calculate a_0 and a_n if it's even then you don't have to calculate b_n

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

    the fastest way to get Fourier series is to reduce the function into some known odd or even function ... it would help you to skip lot's of unnecessary steps.

    • 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.