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anonymous
 4 years ago
what is the period for the fourier expansion of
f(t) = { 5, 0<= t <= 2; 0, 2<t<4}
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because normally the question i'd encountered has the same interval, eg 2<t<0; 0<t<2 etc.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0maybe we can do a shift of the function by substituting u=t2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343832213995:dw this can be transformed into regular Heaviside function

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0you mean\[f(t)=5(1u_2(t))\]?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0but that doesn't fix the intervals, which I do think need to be symmetric

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0sorry ... (t  2)/2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343833073250:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0the Fourier expansion of left is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343833287038:dw

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I still don't see why you need the /2 part

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0that will make the interval of dw:1343833578897:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0however that was not necessary ... i guess

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0but why does it have to be from 1 to 1? can't it just be symmetric?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i guess yes i can be ... i thought unit would make easier.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I just did \(x=t2\)\[A_0=\frac1{2(2)}\int_{2}^2f(x)dx=\frac14\int_{2}^05dx=\frac52\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343833717021:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343833858118:dw

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[A_n=\frac12\int_{2}^2f(x)\cos\left({n\pi x\over2}\right)dx=\frac52\int_{2}^0\cos\left({n\pi x\over2}\right)dx=\frac5{n\pi}\sin(n\pi)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[B_n=\frac52\int_{2}^0\sin\left(n\pi x\over2\right)dx=\frac5{n\pi}\left(\cos(n\pi)1\right)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0but now this formula will be for f(x)=f(t2) so I guess this is wrong oh well, been a while since I've done fourier series

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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
 4 years ago
Best ResponseYou've already chosen the best response.0doesn't anyone know how to plug this into wolf?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so , 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
 4 years ago
Best ResponseYou've already chosen the best response.0Plot[5/2 + 5*4/(2*Pi)*Sum[Sin[(2 n + 1)*Pi*(x)/2]/(2 n + 1), {n, 0, 10}], {x , 0, 10}]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343836422904:dw the example at wolframmathworld was for 0 to 2L

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0owh 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 nonsymmetrical then everythings is ok already.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0the 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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