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edr1c
Group Title
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
edr1c Group Title
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

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edr1c Group TitleBest ResponseYou've already chosen the best response.0
because normally the question i'd encountered has the same interval, eg 2<t<0; 0<t<2 etc.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
maybe we can do a shift of the function by substituting u=t2
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343832213995:dw this can be transformed into regular Heaviside function
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
you mean\[f(t)=5(1u_2(t))\]?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
but that doesn't fix the intervals, which I do think need to be symmetric
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
why 1/5 ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
sorry ... (t  2)/2
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343833073250:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the Fourier expansion of left is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343833287038:dw
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I still don't see why you need the /2 part
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
that will make the interval of dw:1343833578897:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
however that was not necessary ... i guess
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
but why does it have to be from 1 to 1? can't it just be symmetric?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
i guess yes i can be ... i thought unit would make easier.
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343833717021:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343833858118:dw
 one year ago

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
but 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
 one year ago

Neemo Group TitleBest ResponseYou've already chosen the best response.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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
doesn't anyone know how to plug this into wolf?
 one year ago

Neemo Group TitleBest ResponseYou've already chosen the best response.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

experimentX Group TitleBest ResponseYou've already chosen the best response.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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1343836422904:dw the example at wolframmathworld was for 0 to 2L
 one year ago

edr1c Group TitleBest ResponseYou've already chosen the best response.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 nonsymmetrical then everythings is ok already.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.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

experimentX Group TitleBest ResponseYou've already chosen the best response.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
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