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Triangle wave Fourier transform

  1. Mar 11, 2015 #1
    Hello,
    Im not sure if it is the right place to ask it but anyway ...

    i got this function:
    \begin{equation}
    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)}
    \end{equation}

    where:
    \begin{equation}
    \alpha_q =\frac{\pi(q-\frac{1}{2})}{N+\frac{1}{2}}
    \end{equation}

    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 :
    \begin{equation}
    M^*(t)=\frac{M(t)}{N+\frac{1}{2}}
    \end{equation}
    \begin{equation}
    t^*=\frac{t}{N+\frac{1}{2}}
    \end{equation}
    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:
    \begin{equation}
    \sum\limits_{q=1}^\infty \frac{1}{\left(q-\frac{1}{2}\right)^2}=\frac{\pi^2}{2}
    \end{equation}
    and get
    \begin{equation}
    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)
    \end{equation}
    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: \\
    \begin{equation}
    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)
    \end{equation}
    *************
    We denote with $\Lambda_{T}(t)$ a triangle wave with an amplitude $C=1$ and period time $T$, thus we can write:
    \begin{equation}
    M^*(t^*)=2v^2\Lambda_{4}(t^*)
    \end{equation}
    so finally M(t) for N>>1:
    \begin{equation}
    \boxed{
    M(t)=v^2(2N+1)\Lambda_{4N+2}(t)}
    \end{equation}

    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. jcsd
  3. Mar 16, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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