anonymous
  • anonymous
what is the period for the fourier expansion of f(t) = { 5, 0<= t <= 2; 0, 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
because normally the question i'd encountered has the same interval, eg -2
TuringTest
  • TuringTest
maybe we can do a shift of the function by substituting u=t-2
experimentX
  • experimentX
|dw:1343832213995:dw| this can be transformed into regular Heaviside function

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TuringTest
  • TuringTest
you mean\[f(t)=5(1-u_2(t))\]?
TuringTest
  • TuringTest
but that doesn't fix the intervals, which I do think need to be symmetric
TuringTest
  • TuringTest
why 1/5 ?
experimentX
  • experimentX
sorry ... (t - 2)/2
experimentX
  • experimentX
|dw:1343833073250:dw|
experimentX
  • experimentX
the Fourier expansion of left is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html
experimentX
  • experimentX
|dw:1343833287038:dw|
TuringTest
  • TuringTest
I still don't see why you need the /2 part
experimentX
  • experimentX
that will make the interval of |dw:1343833578897:dw|
experimentX
  • experimentX
however that was not necessary ... i guess
TuringTest
  • TuringTest
but why does it have to be from -1 to 1? can't it just be symmetric?
experimentX
  • experimentX
i guess yes i can be ... i thought unit would make easier.
TuringTest
  • TuringTest
I just did \(x=t-2\)\[A_0=\frac1{2(2)}\int_{-2}^2f(x)dx=\frac14\int_{-2}^05dx=\frac52\]
experimentX
  • experimentX
|dw:1343833717021:dw|
experimentX
  • experimentX
|dw:1343833858118:dw|
TuringTest
  • TuringTest
\[A_n=\frac12\int_{-2}^2f(x)\cos\left({n\pi x\over2}\right)dx=\frac52\int_{-2}^0\cos\left({n\pi x\over2}\right)dx=\frac5{n\pi}\sin(n\pi)\]
TuringTest
  • TuringTest
\[B_n=\frac52\int_{-2}^0\sin\left(n\pi x\over2\right)dx=\frac5{n\pi}\left(\cos(n\pi)-1\right)\]
TuringTest
  • TuringTest
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
anonymous
  • anonymous
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 !
experimentX
  • experimentX
doesn't anyone know how to plug this into wolf?
anonymous
  • anonymous
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" \[\int\limits_{0}^{2}-------+\int\limits_{2}^{4}--------\]
experimentX
  • experimentX
Plot[5/2 + 5*4/(2*Pi)*Sum[Sin[(2 n + 1)*Pi*(x)/2]/(2 n + 1), {n, 0, 10}], {x , 0, 10}]
experimentX
  • experimentX
|dw:1343836422904:dw| the example at wolframmathworld was for 0 to 2L
anonymous
  • anonymous
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.
experimentX
  • experimentX
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
experimentX
  • experimentX
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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