0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 at 7:56
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.