I'm trying to find the expression of the fourier series (using complex numbers) for the following odd squarewave function with period T:
$V(t) = V_0$ for $0<t<T/2$
$V(t) = -V_0$ for $T/2<t<T$
I've come this far :
$$V(t)=\frac{2V_0}{\pi*j}\sum_{n=-\infty,n=odd}^\infty \frac{1}{n}e^{jn\omega t}$$
The book where I got this exercise from gives the answer as:
$$\frac{4V_0}{\pi}\sum_{n=-\infty,n=odd}^\infty \frac{sin n\omega t}{n}$$
My question is how to go from my expression to the book's answer.
Here's what I've tried:
I noticed the $j$ was gone on both sides and the $2$ became a $4$, so I thought about splitting $\frac{2V_0}{\pi*j}$ into $\frac{2}{j}$ and $\frac{V_0}{\pi}$ and then squaring $\frac{2}{j}$ to get $-4$ which would give $-\frac{4V_0}{\pi}$.
Then, $e^{jn\omega t}$ can be written as $cos(n\omega t)+jsin(n\omega t)$ and $cos(n\omega t)$ disappears because we are dealing with an odd function.
The book also suggests using the properties: $sin(\theta)=-sin(-\theta)$ and $cos(\theta)=cos(-\theta)$ which, if I understand correctly, let us write the complex number as $cos(-n\omega t)-jsin(-n\omega t)$. In that case, we would be left with $-jsin(-n\omega t)$.
However, no matter how I try to simplify the expression after that, I'm always left with the $j$ and a minus sign.
Thanks for the help @Steven Clark. I figured it out with some more help from a math professor.
I finally have some time to share the full steps here for anyone interested:
$$V(t)=\frac{2V_0}{\pi*j}\sum_{n=-\infty,n=odd}^\infty \frac{1}{n}e^{jn\omega t}$$
$$V(t)=\frac{2V_0}{\pi}\frac{1}{j}\sum_{n=-\infty,n=odd}^\infty \frac{1}{n}cos(n\omega_0t)+j\frac{1}{n}sin(n\omega_0t)$$
$$V(t)=\frac{2V_0}{\pi}\sum_{n=-\infty,n=odd}^\infty -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)$$
$$V(t)=\frac{2V_0}{\pi}[\sum_{n=-\infty,n=odd}^{-1} -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)+\sum_{n=1,n=odd}^{\infty} -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
Substituting n for -n in the first sum:
$$V(t)=\frac{2V_0}{\pi}[\sum_{n=1,n=odd}^{\infty} -\frac{j}{-n}cos(-n\omega_0t)+\frac{1}{-n}sin(-n\omega_0t)+\sum_{n=1,n=odd}^{\infty} -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
$$V(t)=\frac{2V_0}{\pi}[\sum_{n=1,n=odd}^{\infty} \frac{j}{n}cos(-n\omega_0t)-\frac{1}{n}sin(-n\omega_0t)+\sum_{n=1,n=odd}^{\infty} -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
Now, using sin(θ)=−sin(−θ) and cos(θ)=cos(−θ):
$$V(t)=\frac{2V_0}{\pi}[\sum_{n=1,n=odd}^{\infty} \frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)+\sum_{n=1,n=odd}^{\infty} -\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
$$V(t)=\frac{2V_0}{\pi}\sum_{n=1,n=odd}^{\infty} [\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)-\frac{j}{n}cos(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
$$V(t)=\frac{2V_0}{\pi}\sum_{n=1,n=odd}^{\infty} [\frac{1}{n}sin(n\omega_0t)+\frac{1}{n}sin(n\omega_0t)]$$
$$V(t)=\frac{2V_0}{\pi}\sum_{n=1,n=odd}^{\infty} [\frac{2}{n}sin(n\omega_0t)]$$
$$V(t)=\frac{2V_0}{\pi}*2\sum_{n=1,n=odd}^{\infty} [\frac{1}{n}sin(n\omega_0t)]$$
Finally, we get the right answer:
$$V(t)=\frac{4V_0}{\pi}\sum_{n=1,n=odd}^{\infty} \frac{sin(n\omega_0t)}{n}$$
Note: there is an error in the book's answer, the sum should not start from $-\infty$ but rather from $1$.