## edr1c 3 years ago what is the period for the fourier expansion of f(t) = { 5, 0<= t <= 2; 0, 2<t<4}

1. edr1c

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

2. TuringTest

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

3. experimentX

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

4. TuringTest

you mean$f(t)=5(1-u_2(t))$?

5. TuringTest

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

6. TuringTest

why 1/5 ?

7. experimentX

sorry ... (t - 2)/2

8. experimentX

|dw:1343833073250:dw|

9. experimentX

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

10. experimentX

|dw:1343833287038:dw|

11. TuringTest

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

12. experimentX

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

13. experimentX

however that was not necessary ... i guess

14. TuringTest

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

15. experimentX

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

16. TuringTest

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

17. experimentX

|dw:1343833717021:dw|

18. experimentX

|dw:1343833858118:dw|

19. TuringTest

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

20. TuringTest

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

21. TuringTest

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

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

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

24. Neemo

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

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

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

27. edr1c

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

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

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.