# A comparison of fourier series and fourier transform

I have been studying fourier series and fourier transforms and I'm confused about a lot of things mentioned in various questions etc. As an example, I'm a quite puzzled on why fourier transform is used in various applications (signal processing) where we are working with finite functions, and so we might as well assume the signal is a periodic function.

First I repeatedly see people saying "fourier transform of a periodic function" - I don't know what this means. In this question Fourier transform vs Fourier series Lubos Motl says "The Fourier transform of a periodic function is a very special kind of a function, a combination of delta-function". My definition of fourier transform is defined on functions where the integral across $$\mathbb{R}$$ is well-defined, and I don't see how that is the case for a function that is periodic on $$\mathbb{R}$$.

Perhaps it means taking a function defined on $$(-a,a)$$ and extending it by zero on the rest of the domain? In a different answer in the same question this is described.

Second in this question Fourier Transform of a Periodic function

mathreadler answers and shows a picture of a fourier transform of $$f(x) = \cos(kx+1)^{24}$$, again, I don't understand how you fourier transform that function to get the result in the picture.

Third I can see that a fourier transform won't lose any information for a periodic function. As an example, if we extend a function $$f$$ defined on $$(-\pi, \pi)$$ to zero on the rest of the domain, we get that $$\frac{1}{2\pi} \hat{f} (\frac{n}{2 \pi})$$ gives us the sought fourier coefficient $$c_n$$ of this function.

However, it seems that we can can explicitly describe the fourier transform simply from the fourier series. It can be shown that $$\xi \mapsto \int_{-\pi}^{\pi} e^{i \xi x}$$ is a scaled sinc function. Thus, if $$f= \sum a_n e^{inx}$$, we get that $$\hat{f}(\xi) = \sum a_n \int e^{inx - 2\pi i \xi x}$$ which is essentially a sum of scaled sinc functions with variously moved origins. So why even work with a fourier transform when we're working in the context of say signal processing, and our signals are defined on an interval of finite measure?

• The Fourier transform of $h(x) = 1_{[-1/2,1/2]}$ is $\frac{\sin(\pi y)}{\pi y}$ while the Fourier transform of $H(x) = 1$ is $\delta(y)$ the Dirac delta distribution. In one case you get $h(x) = \int_{-\infty}^\infty \frac{\sin(\pi y)}{\pi y}e^{2i \pi xy}dy$ in the other you get $H(x ) = \int_{-\infty}^\infty \delta(y) e^{2i \pi xy}dy=e^{2i\pi x 0}$. Both agree on $[-1/2,1/2]$ and the Fourier series says to recover $h(x)$ on that interval it suffices to look at $\frac{\sin(\pi n)}{\pi n},n \in \mathbb{Z}$. – reuns Jan 16 at 11:29
• In what sense is the fourier transform of H(x) equal to the dirac delta function? Could you formalize this a bit more please? – John P Jan 16 at 11:59
• In the sense of distribution, that is the Fourier transform of $H(x) e^{-\pi x^2/n^2}$ is $n e^{-\pi n^2 y^2} = \delta \ast n e^{-\pi n^2 y^2}$, letting $n \to \infty$ you get something well-defined as an operator acting by convolution on Schwartz functions : a distribution. The point is that the Fourier inversion theorem holds for the distributions. – reuns Jan 16 at 12:07