Fourier Transform

The Fourier Transform is a method for representing aperiodic signals as a sum of exponentials of the form $e^{j\omega t}$.

Formulation

Given an aperiodic signal $x(t)$, we can think of it as a periodic signal with a period of infinity:

$$x(t)=\underset{T_0\to \infty }{\lim }{x}_{T_0}(t)$$

The exponential Fourier series for $x_{T_0}(t)$ is given by

$$x_{T_0}(t)=\sum _{n=-\infty }^{\infty }{D}_n{e}^{jn{\omega }_{0}t}$$

where ${\omega }_0=\frac{2\pi }{T_0}$ and

$$D_n=\frac{1}{T_0}{\int }_{-\frac{T_0}{2}}^{\frac{T_0}{2}}{x}_{T_0}(t)e^{-jn{\omega }_{0}t}dt$$

Integrating $x_{T_0}$ over $(-T_0/2,T_0/2)$ is equivalent to integrating $x(t)$ over $(-\infty ,\infty )$. Therefore, the above equation can also be written as:

$$D_n=\frac{1}{T_0}{\int }_{-\infty }^{\infty }x(t)e^{-jn{\omega }_{0}t}dt$$

It is interesting to see how the nature of the spectrum changes as $T_0$ increases. To understand this fascinating behavior, let us define $X(\omega )$, a continuous function of $\omega$, as:

$$\boxed{X(\omega )={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt}$$

This is the Fourier Transform.

Inverse Fourier Transform

Using the Fourier Transform $X(\omega )$, we can rewrite $D_n$ as:

$$D_n=\frac{1}{T_0}X(n{\omega }_0)$$

And we can rewrite $x_{T_0}(t)$ as:

$$x_{T_0}(t)=\sum _{n=-\infty }^{\infty }\frac{1}{T_0}X(n{\omega }_0)e^{jn{\omega }_{0}t}$$

Given that we have $\frac{2\pi }{T_0}={\omega }_0$ or $T_0=\frac{2\pi }{{\omega }_0}$, we then have:

$$x_{T_0}(t)=\frac{1}{2\pi }\sum _{n=-\infty }^{\infty }X(n{\omega }_0)e^{jn{\omega }_{0}t}{\omega }_0$$

Since our signal is aperiodic and $T_0$ approaches infinity, this means that ${\omega }_0$ must approach zero. If we say ${\omega }_0=\Delta \omega$ (incrementing notation basically?), we then have:

$$x_{T_0}(t)=\frac{1}{2\pi }\sum _{n=-\infty }^{\infty }X(n\Delta \omega )e^{jn\Delta \omega t}\Delta \omega$$

As $T_0\to \infty$, this sum becomes an integral:

$$\boxed{x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(\omega )e^{j\omega t}d\omega }$$

This is the Fourier Integral, which provides the Inverse Fourier Transform.

Notation

We can say that:

$$\mathcal{F}[x(t)]=X(\omega )$$

This is similar to the Laplace Transform, where we have $\mathcal{L}[x(t)]={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$. Fourier is a special case of the Laplace Transform for $s=j\omega$, only when ROC of $X(s)$ includes the imaginary axis such that $X(j\omega )=X(\omega )$.