Function is:
$v(t)=4$ for $0< t< \frac\pi2$
$v(t)=-4$ for $-\frac{\pi}{2}< t< 0$
$v(t)=0$ for $-\pi< t< -\frac \pi2$ and $\frac\pi2< t< \pi$
I solved this and got :
$v(t)=\frac8\pi (\sin(t) + \sin(2t) + \frac1{3\sin(3t)} + \frac1{5\sin(5t)} + \frac1{3\sin(6t)} + \frac1{7\sin(7t)}+...)$ and it is correct.
but then it asks me for this:
It can be shown that the fourier transform of $v(t)$ is given by: $v(u)=(-\frac{4i}{u \pi})(1-\cos(u \pi^2))$
a) By representig $v(u)$ as a complex number of the form $p-iq$, evaluate $p$ and $q$ and at $q$ values of $u=\frac{n}{2\pi}$, for integer values of $n$ in the range $0$ to $7$. hence sketch $q$ over the interval from $u=0$ to $u=\frac7{2\pi}$
b) By comparing $p$ and $q$ found in part a) with the values of the coefficients ($a_m$ and $b_m$) (the ones I find at the beginning!) explain the relationship between fourier transform of a periodic function and the coefficients of its fourier series.
I have no idea :( please help