# Fourier transform of periodic functions

The Fourier transform is derived from the Fourier series by considering a non-periodic signal, thinking of it as a infinitely long periodic signal, putting it into the Fourier series and making this series as a Riemann sum (the definition of integral) and thus, deriving the Fourier transform integral.

My question is, since we derived the Fourier transform integral from the Fourier series for the functions that are not periodic (infinitely long periodic), how can we use this transform for regular periodic signals such as $$\cos(x)$$?

• There is no need to derive the Fourier Transform from the Fourier Series. The FT is simply the representation of a signal as a set of orthogonal basis functions. The basis functions are complex exponentials. Sep 20 at 11:25

The Fourier transform of a periodic signal is zero everywhere except at integer multiples of the fundamental frequency. Using the Dirac delta impulse, it is straightforward to derive it from the Fourier series. Let $$x(t)$$ be a $$T$$-periodic function with Fourier coefficients $$c_k$$:

$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{jk\omega_0 t},\quad \omega_0=\frac{2\pi}{T}\tag{1}$$

With

$$\mathcal{F}\big\{e^{j\omega_0t}\big\}=2\pi\delta(\omega-\omega_0)\tag{2}$$

we get from $$(1)$$ and $$(2)$$

$$X(j\omega)=\mathcal{F}\big\{x(t)\big\}=2\pi\sum_{k=-\infty}^{\infty}c_k\delta(\omega-k\omega_0)\tag{3}$$

Reacting to a comment about the Laplace transform, it's important to realize that the Laplace transform of a truly periodic function doesn't exist. E.g., there is no Laplace transform of $$\sin(\omega_0t)$$ or $$\cos(\omega_0t)$$. If a Laplace transform table says otherwise, then what they really mean is the Laplace transform of $$\sin(\omega_0t)u(t)$$ or $$\cos(\omega_0t)u(t)$$, where $$u(t)$$ is the unit step function. These functions are not periodic because they are zero for $$t<0$$.

The Laplace transform of such a pseudo-periodic function satisfying $$f(t+T)=f(t)$$ for $$T>0$$ and $$t>0$$ is given by

$$F(s)=\frac{F_0(s)}{1-e^{-sT}}\tag{4}$$

where $$F_0(s)$$ is the Laplace transform of one period of $$f(t)$$ in the interval $$[0,T]$$:

$$F_0(s)=\int_0^Tf(t)e^{-st}dt\tag{5}$$

• what I dont understand is, derivation of fourier transform integral is made from fourier series but this derivation is made via the signals that are non-periodic. But I have seen many times that, people keep taking fourier transform of periodic functions which should be wrong according to the derivation Sep 20 at 10:55
• @ozguncan: That's why you need the distribution $\delta(\omega)$. The corresponding integrals don't converge in the conventional sense. Sep 20 at 11:05
• my real question is, we use laplace transform and then convert it back into the time domain to find circuit's response easily. and the input(the source) can simply be sinusoidal as well But the problem is, laplace transform is derived from fourier transform by adding the damping term. But how can we able to derive some finite and meaningful resullts for periodic-input signals by using laplace transform which is derived from fourier transform that is applicable for non-periodic functions. Sep 20 at 11:16
• @ozguncan: You should reformulate your question if it's really about the Laplace transform instead of the Fourier transform. You don't need to derive the Laplace transform from the Fourier transform. Furthermore, the Laplace transform of periodic signals doesn't even exist, unlike the Fourier transform. Sep 20 at 11:50
• I think part of your confusion is your "is derived from". Maybe historically, yes. Perhaps in your coursework, yes. But they don't have to be, and they are their own mathematical entities that can stand on their own. The Laplace transform can be derived as a stand-alone entity (that, oh gosh, happens to share a lot of properties with the Fourier transform). Similarly the Fourier Transform can be derived from the Fourier Series (with either some questionable hand-waving or a whole lot of very very careful mathematical justification). But they don't have to be. Sep 20 at 14:38

Laplace transform of periodic signals is performed on a single cycle T of the signal, provided on knows the cycle, and then apply 1/(1+exp(s*T))