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Im reading about fourier series, I have the solutions manual for a problem. But i dont understand it

The problem is

$ U(t) = \begin{cases} 1, & \text{0≤t<1} \\[2ex] 0, & \text{1≤t<2} \end{cases}$ $ U(t) = U(t+n*2)$

https://i.stack.imgur.com/n737S.png

The period is 2, so we have the lower limit 0 and upper limit 2. Why do the upper limits change?

mangekyou
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  • Do you understand what happens in the first line (for $a_0$)? It's exactly the same in the other two lines (for $a_n$ and $b_n$), it's just not written out. – Hans Lundmark Nov 12 '19 at 14:06
  • No. I really dont understand it. I understand $\frac2T$ and the limits of integration from 0 to T but not more than that. – mangekyou Nov 12 '19 at 14:15
  • OK, what about $\int_0^2 u(t) , dt = \int_0^1 u(t) , dt + \int_1^2 u(t) , dt$ then? That's a basic property of integrals. – Hans Lundmark Nov 12 '19 at 14:19
  • No, I didn't know about this property. This is the first time I see it. But Iooking at it ,I understand how this property works now. – mangekyou Nov 12 '19 at 14:22
  • Well, that's a bit strange, since you should have come across that long before getting to Fourier series... – Hans Lundmark Nov 12 '19 at 14:25
  • Anyway, once you have that identity, you just substitute the values for your function $u(t)$ on the respective intervals. – Hans Lundmark Nov 12 '19 at 14:25
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    I honestly have never seen it before. But I think I maybe got it. $$\int_0^T u(t),dt= \int_0^2 u(t),dt= \int_0^1 u(t),dt + \int_1^2 u(t),dt$$

    We have $ U(t) = \begin{cases} 1, & \text{0≤t<1} \[2ex] 0, & \text{1≤t<2} \end{cases}$

    $\text{since u(t) = 1 in the interval between 0 and 1. We get the integral } \[2ex]

    $$\int_0^1 1,dt$

    $\text{since u(t) = 0 in the interval between 1 and 2. We get the integral } \[2ex]

    $$\int_1^2 0,dt$

    Is this correct?

     – mangekyou Nov 12 '19 at 14:41
  • Right, that's it! – Hans Lundmark Nov 12 '19 at 14:47
  • Thanks a lot for explaining it. I was really stuck – mangekyou Nov 12 '19 at 14:54
  • One thing Im not completely sure about is why the limits of integration for a and b are also 0,1 – mangekyou Nov 12 '19 at 15:52
  • See my very first comment above. – Hans Lundmark Nov 12 '19 at 17:29
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    I see it now. $$\int_0^1 1cos(nwt) ,dt + \int_1^2 0cos(nwt) ,dt = \int_0^1 cos(nwt) ,dt $$ – mangekyou Nov 12 '19 at 18:41

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