**WHY Fourier Transform?**

If a function f (t) is not a periodic and is defined on an infinite interval, we cannot represent it by Fourier series. It may be possible, however, to consider the function to be periodic with an infinite period. In this section we shall consider this case in a non-rigorous way, but the results may be obtained rigorously if f (t) satisfies the following conditions:

- $\int\limits_{-\infty }^{\infty }{\left| f(t) \right|}dt$ is finite means $\int\limits_{-\infty }^{\infty }{\left| f(t) \right|}dt<\infty $
- In any finite interval, f(t) has at most a finite number of finite discontinuities
- In any finite interval, f(t) has at most a finite number of maxima and minima

Let us begin with the exponential series for a function f_{T }(t) defined to be f (t) for

$-T/2<t<T/2$

The result is

${{f}_{T}}(t)=\sum\limits_{-\infty }^{\infty }{{{c}_{n}}{{e}^{{}^{j2\pi nt}/{}_{T}}}}\text{ }\cdots \text{ (1)}$

Where

\[{{c}_{n}}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{{{f}_{T}}(x)}{{e}^{{}^{-j2\pi nx}/{}_{T}}}dx\text{ }\cdots \text{ }(2)\]

We have replaced ω_{o} by 2π/T and are using the dummy variable x instead of t in the coefficient expression. Our intention is to let T→∞, in which case f_{T }(t) →f (t).

Since the limiting process requires that ωo=2π/T→∞, for emphasis we replace 2π/T by ∆ω. Therefore substituting (2) into (1), we have

$\begin{align} & {{f}_{T}}(t)=\sum\limits_{-\infty }^{\infty }{\left[ \frac{\Delta \omega }{2\pi }\int\limits_{-T/2}^{T/2}{{{f}_{T}}(x)}{{e}^{-jxn\Delta \omega }}dx \right]{{e}^{jtn\Delta \omega }}} \\ & \text{=}\sum\limits_{-\infty }^{\infty }{\left[ \frac{1}{2\pi }\int\limits_{-T/2}^{T/2}{{{f}_{T}}(x)}{{e}^{-j(x-t)n\Delta \omega }}dx \right]}\Delta \omega \text{ }\cdots \text{ (3)} \\\end{align}$

If we define the function

\[g(\omega ,t)=\frac{1}{2\pi }\int\limits_{-T/2}^{T/2}{{{f}_{T}}(x)}{{e}^{-j\omega (x-t)}}dx\text{ }\cdots \text{ }(4)\]

Then clearly the limit of (3) is given by

$f(t)=\underset{T\to \infty }{\mathop{\lim }}\,\sum\limits_{n=-\infty }^{\infty }{g(n\Delta \omega ,t)\Delta \omega \text{ }\cdots \text{ (5)}}$

By the fundamental theorem of integral calculus the last result appears to be

$f(t)=\int\limits_{-\infty }^{\infty }{g(\omega ,t)d}\omega \text{ }\cdots \text{ (6)}$

But in the limit, f_{T}→ f and T→∞ in (4) so that what appears to be g (ω, t) in (6) is really its limit, which by (4) is

\[\underset{T\to \infty }{\mathop{\lim }}\,g(\omega ,t)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{f(x)}{{e}^{-j\omega (x-t)}}dx\]

Therefore (6) is actually

$f(t)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{\left[ \int\limits_{-\infty }^{\infty }{f(x)}{{e}^{-j\omega (x-t)}}dx \right]d}\omega \text{ }\cdots \text{ (7)}$

As we said, this is non-rigorous development, but the results may be obtained rigorously.

Let us rewrite (7) in the form

$f(t)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{\left[ \int\limits_{-\infty }^{\infty }{f(x)}{{e}^{-j\omega x}}dx \right]{{e}^{j\omega t}}d}\omega \text{ }\cdots \text{ (8)}$

Now, let us define the expression in brackets to be the function

Where we have changed the dummy variable from x to t. then (8) becomes

The function F (jω) is called the Fourier Transform of f (t), and f (t) is called the inverse Fourier Transform of F (jω). These facts are often stated symbolically as

$\begin{matrix} \begin{align} & F(j\omega )=\Im [f(t)] \\ & f(t)={{\Im }^{-1}}[F(j\omega )] \\\end{align} & \cdots & (11) \\\end{matrix}$

Also, (9) and (10) are collectively called the Fourier Transform Pair, the symbolism for which is

$f(t)\leftrightarrow F(j\omega )\text{ }\cdots \text{ (12)}$

The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). Equation (10) is, of course, another form of (7). Another description for these analogies is to say that the Fourier Transform is a continuous representation (ω being a continuous variable), whereas the Fourier series is a discrete representation (nω_{o}, for n an integer, being a discrete variable).

**Fourier Transform Example**

As an example, let us find the transform of

$f(t)={{e}^{-at}}u(t)$

Whereas a>0. By definition we have

$\begin{align} & \Im [{{e}^{-at}}u(t)]=\int\limits_{-\infty }^{\infty }{{{e}^{-at}}u(t){{e}^{-j\omega t}}dt} \\ & =\int\limits_{0}^{\infty }{{{e}^{-(a+j\omega )t}}dt} \\\end{align}$

Or

$\Im [{{e}^{-at}}u(t)]=\left. \frac{1}{-(a+j\omega )}{{e}^{-(a+j\omega )t}} \right|_{0}^{\infty }$

The upper limit is given by

$\underset{t\to \infty }{\mathop{\lim }}\,{{e}^{-at}}(\cos \omega t-j\sin \omega t)=0$

Since the expression in parentheses is bounded while the exponential goes to zero. Thus we have

$\Im [{{e}^{-at}}u(t)]=\frac{1}{(a+j\omega )}$

Or

${{e}^{-at}}u(t)\leftrightarrow \frac{1}{(a+j\omega )}$