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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:

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

The capital letter denotes the Fourier transform.

Note:

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

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

Attempt:

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:

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

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?

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  • $\begingroup$ This is messy. $f(t)$ is a sawtooth, $F(\nu)$ its Fourier transform, $f_p(s)$ is the periodized sawtooth, $c_k$ are its Fourier series coefficients ? But with $\sum^\infty_{k=-\infty} F \left( \frac{k}{T} \right) \delta(\nu - \frac{k}{T})$ you are periodizing $F$ : in the Fourier domain... So maybe you meant $f_p$ is the sampled sawtooth ? $\endgroup$
    – reuns
    Apr 10, 2017 at 7:38
  • $\begingroup$ $f_p$ is just a periodic sawtooth function, so it can be regarded as the convolution of $f$ (a single period of $f_p$) with an infinite train of delta functions: $$f_p(t) = \sum f(kT)*\delta(t-kT)= \sum f(t-kT).$$ $F_p(\nu)$ is the FT of $f_p$ whose spectrum consists of discrete components at multiples of $1/T.$ Equation (1) above gives the series, and (2) gives the corresponding coefficients $c_k.$ $\endgroup$
    – Merin
    Apr 10, 2017 at 7:56

1 Answer 1

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There are some mistakes in what you wrote.


(Assuming everything converges)

If $$f_p(t) = \sum_{k=-\infty}^\infty f(t-kT)$$ Then we have the Fourier series $$f_p(t) = \sum_{n=-\infty}^\infty c_n e^{2i \pi n t/T}$$ where $$c_n = \frac{1}{T}\int_0^T f_p(t) e^{-2i \pi n t / T}dt = \frac{1}{T}\int_0^T \sum_{k=-\infty}^\infty f(t-kT) e^{-2i \pi n t / T}dt \\= \frac{1}{T}\sum_{k=-\infty}^\infty \int_0^T f(t-kT) e^{-2i \pi n t / T}dt =\frac{1}{T}\sum_{k=-\infty}^\infty \int_{-kT}^{-kT+T} f(t) e^{-2i \pi n t / T}dt \\=\frac{1}{T}\int_{-\infty}^\infty f(t) e^{-2 i\pi n t/T}dt =\frac{1}{T} F(n/T)$$ In other words : the Fourier transform (in the sense of dsitributions) of $f_p(t)$ is $F_p(\nu) = \sum_{n=-\infty}^\infty c_n \delta(\nu-n/T)= F(\nu)\frac{1}{T}\sum_{n=-\infty}^\infty \delta(\nu-n/T)$

All this means that the Fourier series theorem is equivalent to the statement that the Dirac comb $\displaystyle\sum_{k=-\infty}^\infty \delta(t-k)$ is its own Fourier tranform, and that periodizing in time domain amounts to sampling in the Fourier domain

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