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Given the above function I have to show that the Fourier series is :

$$f(t)=\frac{1}{2}+\frac{6}{\pi}(\cos\omega_0t -\frac{1}{3}\cos3\omega_0t+\frac{1}{5}\cos5\omega_0t-\frac{1}{7}\cos7\omega_0t+\cdots)$$

Can I get help on how to start solving this?

2 Answers 2

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Step number 1 is to fins the periodicity of the function. By looking at the graph, you can see that the period is $T$. You also notice that the function is even, so only cosine terms will contribute to the Fourier series.
Step number 2 involves writing out explicitly what your function looks like (not the series). You should only write it for one period. You can choose any start point $t$ and write how it should look in the interval $[t, t+T]$. If you start at $0$, you need to describe 3 intervals $([0,T/4], [T/4,3T/4], [3T/4,T])$. The same thing if you go from $-T/2$ to $T/2$. In this case, your best bet is to go from $-T/4$ to $3T/4$, since you have only two intervals to think of. If $-T/4\le t\lt T/4$ then $f(t)=2$. For $T/4\le t\lt3T/4$, $f(t)=-1$
Now you write $f(t)$ as a Fourier series containing only cosine terms: $$f(t)=a_0/2+\sum_{n=1}^\infty a_n\cos\left(\frac{2\pi}{T} n t\right)$$ You can identify $\omega_0=\frac{2\pi}{T}$. The coefficients $a_n$ can be found from $$a_n=\frac{2}{T}\int_{t_0}^{t_0+T}f(t)\cos(n\omega_0 t)dt$$ And you should choose $t_0$ to be $-T/4$
For example $$a_0=\frac{2}{T}\int_{-T/4}^{3T/4}f(t)dt=\frac{2}{T}\int_{-T/4}^{T/4}f(t)dt+\frac{2}{T}\int_{T/4}^{3T/4}f(t)dt=\\=\frac{2}{T}\frac{T}{2}(2-1)=1$$ This way the first term in the series is $a_0/2=1/2$

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    I did this for $a_n$ but i dont know how to get to $\frac{6}{\pi}$ $$a_n=\frac{4}{T}\int_{0}^{T/2} f(2) \cos n\omega_0dt+\frac{4}{T}\int_{T/4}^{T/2} f(-1) \cos n\omega_0dt$$ $$=\frac{4}{T}\left[\frac{\sin n\omega_0t}{n\omega_0}\right]_{0}^{T/4}+\frac{4}{T}\left[\frac{-\sin n\omega_0t}{n\omega_0}\right]_{T/4}^{T/2}$$ $$=\frac{4}{T}\left[\frac{\sin n\frac{\pi}{2}}{n\omega_0}-0\right ]+\frac{4}{T}\left[ -\frac{\sin n\pi}{n\omega_0}+\frac{\sin n\frac{\pi}{2}}{n\omega_0}\right ]$$ – 2017-02-14
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    Look at your integration limits and at my integration limits. You need to integrate the first integral from $-T/4$ to $T/4$ and the second from $T/4$ to $3T/4$. You also forgot to plug in the values of $2$ and $-1$ – 2017-02-14
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The function is even so $b_n=0$ for all $n$. Then \begin{eqnarray} a_0 &=& \dfrac{1}{L}\int_{-L}^{L}f(x)dx\\ &=& \dfrac{1}{L}\int_{-\frac12T}^{\frac12T}f(x)dx\\ &=& \dfrac{2}{L}\int_{0}^{\frac12T}f(x)dx\\ &=& \dfrac{2}{T}\int_{0}^{\frac14T}2\,dx+\dfrac{2}{T}\int_{\frac14T}^{\frac12T}-1\,dx\\ &=& 1-\frac12\\ &=& \frac12 \end{eqnarray} and \begin{eqnarray} a_n&=& \dfrac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx\\ &=& \dfrac{2}{T}\int_{-\frac12T}^{\frac12T}f(x)\cos\frac{2n\pi x}{T}dx\\ &=& \dfrac{4}{T}\int_{0}^{\frac14T}2\cos\frac{2n\pi x}{T}dx+\dfrac{4}{T}\int_{\frac14T}^{\frac12T}-\cos\frac{2n\pi x}{T}dx\\ &=& \dfrac{4}{T}(2)\frac{T}{2n\pi}\sin\frac{2n\pi x}{T}\Big|_{0}^{\frac14T}+\dfrac{4}{T}(-1)\frac{T}{2n\pi}\sin\frac{2n\pi x}{T}\Big|_{\frac14T}^{\frac12T}\\ &=& \frac{4}{n\pi}\sin\frac{n\pi}{2}-\frac{2}{n\pi}(\sin n\pi-\sin\frac{n\pi}{2})\\ &=& \frac{6}{n\pi}\sin\frac{n\pi}{2} \end{eqnarray} with $\omega_0=\dfrac{2\pi}{T}$ we have \begin{eqnarray} f(x)&=& \frac12+\sum_{n=1}^{\infty}\frac{6}{n\pi}\sin\frac{n\pi}{2}\cos\frac{2n\pi x}{T}\\ &=& \frac12+\sum_{n=1}^{\infty}\frac{6}{n\pi}\sin\frac{n\pi}{2}\cos n\omega_0x\\ &=& \frac12+\frac{6}{\pi}\left(\cos\omega_0x-\frac{1}{3}\cos3\omega_0x+\frac{1}{5}\cos5\omega_0x+\cdots\right) \end{eqnarray}