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Fourier Series of Sawtooth Wave from Inverse FT

  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    I want to find the Fourier series of the sawtooth function in terms of real sine and cosine functions by using the formula:

    $$f_p (t)=\sum^\infty_{k=-\infty} c_k \exp \left(j2\pi \frac{k}{T}t \right) \tag{1}$$

    This gives the Fourier series of a periodic function, with the coefficients:

    $$c_k = \frac{1}{T} F \left( \frac{k}{T} \right) \tag{2}$$

    where the capital letter denotes the Fourier transform.

    2. Relevant equations

    Equation (1) and (2) above are found by considering the IFT which recovers ##f_p## from ##F_p##:

    $$\intop^\infty_{-\infty} F_p (\nu) e^{j2\pi \nu t} \ d\nu = \intop^\infty_{-\infty} \frac{1}{T} \sum^\infty_{-\infty} F \left( \frac{k}{T} \right) \delta(\nu - \frac{k}{T}) e^{j2\pi \nu t} \ d\nu=\sum^\infty_{-\infty} \frac{1}{T} F \left( \frac{k}{T} \right) e^{j2\pi \frac{k}{T} t}$$

    3. The attempt at a solution

    By using a low pass rectangular filter, a single period of the sawtooth function is given by

    $$f(t)=t\Pi\left(\frac{t}{T}\right)$$

    Since we have the following Fourier transform pair:

    $$t \leftrightarrow \frac{j \delta'(\nu)}{2 \pi}$$

    We can write the FT of a single period of the sawtooth wave as:

    $$F(\nu)=\frac{j\delta'\left(\nu\right)}{2\pi}*T\ sinc\left(\nu T\right)=\frac{jT}{2\pi}\ sinc^{\prime}(\nu T)=\left(\frac{jT}{2\pi}\right)\left(\frac{\cos(\pi\nu T)}{\nu T}-\frac{\sin(\pi\nu T)}{\pi\nu^{2}T^{2}}\right)$$

    Using equation (2), we get the coefficients:

    $$c_{k}=\frac{1}{T}\left(\frac{jT}{2\pi}\right)\left(\frac{\cos(\pi\frac{k}{T}T)}{\frac{k}{T}T}-\frac{\sin(\pi\frac{k}{T}T)}{\pi\left(\frac{k}{T}\right)^{AC2}T^{2}}\right)=\frac{j}{2\pi}\left(\frac{\cos(\pi k)}{k}-\frac{\sin(\pi k)}{\pi k^{2}}\right).$$

    And therefore, the Fourier series becomes:

    $$f_{p}(t)=c_{k}\ e^{j2\pi\frac{k}{T}t}=\frac{j}{2\pi}\left(\frac{\cos(\pi k)}{k}-\frac{\sin(\pi k)}{\pi k^{2}}\right)e^{j2\pi\frac{k}{T}t}$$

    But this does not look correct (it is very different than the Fourier series of the sawtooth given here). Since the sawtooth function is odd, I think we must only have the sine terms present. What is wrong here?

    Also, because I need to plot this function, how can I get rid of the ##j## terms?

    Any help would be appreciated.
     
  2. jcsd
  3. Apr 10, 2017 #2

    BvU

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    Can you explain this ? With some pictures preferably ? It doesn't look to me like ##s(x) ={x\over \pi}## at all ?
     
    Last edited: Apr 10, 2017
  4. Apr 10, 2017 #3

    Dr Transport

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    Where'd the delta function come from?? it isn't part of the expression you derived for the coefficients. Notation, notation, notation, be consistent and explicit. Do the integrals and the answer will pop out quickly, it isn't a hard problem to solve.
     
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