# Triangle wave Fourier transform

1. Mar 11, 2015

### Raz91

Hello,
Im not sure if it is the right place to ask it but anyway ...

i got this function:

M(t)=\sum\limits_{q=1}^N \frac{v^2}{N+ \frac{1}{2}} \cot^2 \left(\frac{\alpha_q}{2}\right) {\sin^2\left(\sin\left(\frac{\alpha_q}{2}\right)t\right)}

where:

\alpha_q =\frac{\pi(q-\frac{1}{2})}{N+\frac{1}{2}}

when I ploted it numerically I found out that it looks like a triangle for large N.Now , I try to figure out why :
I tried this :
For N>>1 we get $\alpha_q\ll1$, so we can approximate $\sin\left(\frac{\alpha_q}{2}\right)\approx \frac{\alpha_q}{2}$ and $\cot\left(\frac{\alpha_q}{2}\right) \approx \frac{2}{\alpha_q}$ and get :

\begin{eqnarray}
M(t)&=&\sum\limits_{q=1}^\infty\frac{v^2}{N+ \frac{1}{2}} \left(\frac{2}{\alpha_q}\right)^2 {\sin^2\left(\frac{\alpha_q}{2}t\right)}\\
&=& \sum\limits_{q=1}^\infty\frac{v^2}{N+ \frac{1}{2}} \left(\frac{2}{\alpha_q^2}\right)\left( 1-\cos (\alpha_q t) \right) \\
&=& \sum\limits_{q=1}^\infty2v^2 \frac{N+\frac{1}{2}}{\left(\pi(q-\frac{1}{2}) \right)^2} \left( 1-\cos\left(\frac{\pi(q-\frac{1}{2})}{N+\frac{1}{2}} t\right) \right)
\end{eqnarray}
Now we define :

M^*(t)=\frac{M(t)}{N+\frac{1}{2}}

t^*=\frac{t}{N+\frac{1}{2}}

and we get
\begin{eqnarray}
M^*(t^*)
=\frac{2v^2}{\pi^2} \sum\limits_{q=1}^\infty\frac{1}{\left(q-\frac{1}{2}\right)^2} - \frac{2v^2}{\pi^2} \sum\limits_{q=1}^\infty\frac{4}{\left(2q-1\right)^2} \cos\left(\frac{2\pi(2q-1)}{4} t^*\right)
\end{eqnarray}
now we use the identity:

\sum\limits_{q=1}^\infty \frac{1}{\left(q-\frac{1}{2}\right)^2}=\frac{\pi^2}{2}

and get

M^*(t^*)=v^2 - \frac{2v^2}{\pi^2} \sum\limits_{q=1}^\infty \frac{4}{\left(2q-1\right)^2} \cos \left(\frac{2\pi(2q-1)}{4} t^*\right)

and this is exactly the Fourier series of a triangle wave with an amplitude of C^*=2v^2 and period time of T^*=4
************
The Fourier series of a triangle wave with an amplitude $A$ and period time $T$ is given by: \\

f(t)=\frac{A}{2} - \frac{A}{\pi^2} \sum\limits_{q=1}^\infty \frac{4}{\left(2q-1\right)^2} \cos \left(\frac{2\pi(2q-1)}{T} t\right)

*************
We denote with $\Lambda_{T}(t)$ a triangle wave with an amplitude $C=1$ and period time $T$, thus we can write:

M^*(t^*)=2v^2\Lambda_{4}(t^*)

so finally M(t) for N>>1:

\boxed{
M(t)=v^2(2N+1)\Lambda_{4N+2}(t)}

BUT , I know my approximation doesnt hold for any q , so I tought maybe the large $\alpha_q$ for which $q~N$ are not important because of the $\cot^2 \left(\frac{\alpha_q}{2}\right)$ which amplify the terms for which q is small... but still i dunno how to show mathematically properly...

THANK YOU!

2. Mar 16, 2015