In the previous three lessons, we discussed the Fourier Series, which is for periodic signals. This lesson will cover the Fourier Transform which can be used to analyze aperiodic signals. (Later on, we'll see how we can also use it for periodic signals.) The Fourier Transform is another method for representing signals and systems in the frequency domain.

is the continuous time Fourier transform of
*f*(*t*).

It is an extension of the Fourier Series. The Fourier
transformation creates *F*(*ω*)
in the FREQUENCY domain. We
will see that
*F*(*ω*) can be seen as a "continuous coefficient" of a Fourier Series if we let the period of a periodic function go to infinity so that the resulting function becomes
aperiodic in the limit.

Let *f*_{p}(*t*) be a periodic function. Therefore, we can express it with a Fourier series:

Assuming that
*f*_{p}(*t*)
is
a periodic rectangular pulse train, let's plot
its magnitude spectrum *C*_{k} vs.
*ω=kω _{0}*

As shown, each
*C*_{k} is the frequency component of *f*_{p}(*t*) at the frequency *ω = k ω _{0}*.

Now, if we let *T*_{0},
the
period of *f*_{p}(*t*),
go to infinity (meaning that
*f*_{p}(*t*)
becomes aperiodic),
then the fundamental frequency * ω _{0}*
will go to zero.
Then the lines in the plot will get closer and closer together
and merge into a continuous spectrum.

As *ω _{0} →* 0, the distance between lines goes to 0,
so let's write

Next, we can
replace *C*_{k} in our Fourier Series
representation of *f*_{p}(*t*) by
using the formula we derived in Lesson 13
for the Fourier Series coefficients:

Next,

As *Δω* → 0, we write it as *dω*
and the sum becomes an integral (and *kΔω* approaches *ω* as
*T _{0}* → ∞ where

Let | be the Fourier Transform of f(t). |

So *F*(*ω*) replaces the *C*_{k} as
*Δω* → 0 and is a continuous function of *ω*.

__Finally, we have derived our
Fourier transform pair:__

Notice the similarity between the two formulas except for the sign change in the exponent and the multiplicative factor in front of the synthesis formula.

Because the Fourier Transform is an integral over an infinite range, we must consider whether or not the integral converges. Sufficient conditions for the existence of the Fourier Transform are the Dirichlet conditions. That is, the Fourier Transform exists if:

1. On any finite interval

(a) *f*(*t*) is bounded

(b) *f*(*t*) has a finite number of minima and maxima

(c) *f*(*t*) has a finite number of discontinuities

2.
*f*(*t*) is absolutely integrable, that is

This can be seen because we know that |*e ^{-jωt}*| = 1:

Therefore, if *f*(*t*) is absolutely integrable, then its
Fourier Transform exists.

3. Basically, if you can generate a signal in a laboratory, since it has finite energy, it will have a Fourier Transform.

Now, let's compute our first Fourier Transforms:

__Example 1__

__Example 2__

__Example 3__

__Example 4__

__Example 5__

__Example 6__