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edr1c

  • 2 years ago

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

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  1. edr1c
    • 2 years ago
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    because normally the question i'd encountered has the same interval, eg -2<t<0; 0<t<2 etc.

  2. TuringTest
    • 2 years ago
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    maybe we can do a shift of the function by substituting u=t-2

  3. experimentX
    • 2 years ago
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    |dw:1343832213995:dw| this can be transformed into regular Heaviside function

  4. TuringTest
    • 2 years ago
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    you mean\[f(t)=5(1-u_2(t))\]?

  5. TuringTest
    • 2 years ago
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    but that doesn't fix the intervals, which I do think need to be symmetric

  6. TuringTest
    • 2 years ago
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    why 1/5 ?

  7. experimentX
    • 2 years ago
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    sorry ... (t - 2)/2

  8. experimentX
    • 2 years ago
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    |dw:1343833073250:dw|

  9. experimentX
    • 2 years ago
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    the Fourier expansion of left is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html

  10. experimentX
    • 2 years ago
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    |dw:1343833287038:dw|

  11. TuringTest
    • 2 years ago
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    I still don't see why you need the /2 part

  12. experimentX
    • 2 years ago
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    that will make the interval of |dw:1343833578897:dw|

  13. experimentX
    • 2 years ago
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    however that was not necessary ... i guess

  14. TuringTest
    • 2 years ago
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    but why does it have to be from -1 to 1? can't it just be symmetric?

  15. experimentX
    • 2 years ago
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    i guess yes i can be ... i thought unit would make easier.

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

  17. experimentX
    • 2 years ago
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    |dw:1343833717021:dw|

  18. experimentX
    • 2 years ago
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    |dw:1343833858118:dw|

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

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

  21. TuringTest
    • 2 years ago
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    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

  22. Neemo
    • 2 years ago
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    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 !

  23. experimentX
    • 2 years ago
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    doesn't anyone know how to plug this into wolf?

  24. Neemo
    • 2 years ago
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    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" \[\int\limits_{0}^{2}-------+\int\limits_{2}^{4}--------\]

  25. experimentX
    • 2 years ago
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    Plot[5/2 + 5*4/(2*Pi)*Sum[Sin[(2 n + 1)*Pi*(x)/2]/(2 n + 1), {n, 0, 10}], {x , 0, 10}]

  26. experimentX
    • 2 years ago
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    |dw:1343836422904:dw| the example at wolframmathworld was for 0 to 2L

  27. edr1c
    • 2 years ago
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    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.

  28. experimentX
    • 2 years ago
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    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

  29. experimentX
    • 2 years ago
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    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.

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