# Fourier Transform

Thread Starter

#### RdAdr

Joined May 19, 2013
214
In a book the Fourier transform is defined like this. Let g(t) be a nonperiodic deterministic signal... and then the integrals are presented.

So, I understand that the signal must be deterministic and not random. But why it has to be nonperiodic (aperiodic).
The sin function is periodic and we can calculate its Fourier transform.

Is it because a nonperiodic signal is absolutely integrable?

And with the sin function. Yes, I can calculate. But deltas appear.

This is the answer?

#### WBahn

Joined Mar 31, 2012
26,915
A non-periodic signal doesn't have to be absolutely integrable -- consider f(t) = At, for instance.

If g(t) IS absolutely integrable, then it's Fourier transform exists because this is a sufficient condition. However, it is not a necessary condition.

Periodic signals, except the trivial case of f(t) = 0, are not absolutely integrable and, hence, do not satisfy the sufficient condition and, hence, the Fourier transform is not guaranteed to exist -- but it may. By introducing special functions, such as the Dirac delta function, we can express the Fourier transforms of many functions that are not absolutely integrable in a useful and meaningful way.

Thread Starter

#### RdAdr

Joined May 19, 2013
214
Ok. I understand. Thanks for the answer.

Also. When deriving the Fourier transform from the Fourier series, we have a finite-length signal and repeat it multiple times over the time axis. And then expand it into a Fourier series. And then calculations. And then we get the Fourier transform.

So the Fourier transform is for finite-length signals.

The fact that we can calculate Fourier transforms for periodic signals or signals like the unit step is because we involve deltas functions there. Like you said.

But what about the decaying exponential? Its Fourier transform does not involve deltas and it is not of finite length.
How can this decaying exponential be viewed as a finite-length signal that gets repeated multiple times over the time axis. Its period is infinite.

So my question. How does one attach a Fourier transform to such decaying exponential? It could be the fact that for such signals we actually have other derivation of the Fourier transform but we haven't found it yet?
We derived the Fourier transform for finite-length signals and with it we just calculated the Fourier transform for decaying exponential?
Or one can think of having multiple infinities into one infinite. So we have that notion that some infinities are bigger than others?

#### WBahn

Joined Mar 31, 2012
26,915
Ok. I understand. Thanks for the answer.

Also. When deriving the Fourier transform from the Fourier series, we have a finite-length signal and repeat it multiple times over the time axis. And then expand it into a Fourier series. And then calculations. And then we get the Fourier transform.

So the Fourier transform is for finite-length signals.

The fact that we can calculate Fourier transforms for periodic signals or signals like the unit step is because we involve deltas functions there. Like you said.

But what about the decaying exponential? Its Fourier transform does not involve deltas and it is not of finite length.
How can this decaying exponential be viewed as a finite-length signal that gets repeated multiple times over the time axis. Its period is infinite.

So my question. How does one attach a Fourier transform to such decaying exponential? It could be the fact that for such signals we actually have other derivation of the Fourier transform but we haven't found it yet?
We derived the Fourier transform for finite-length signals and with it we just calculated the Fourier transform for decaying exponential?
Or one can think of having multiple infinities into one infinite. So we have that notion that some infinities are bigger than others?
Go back to your first post where you talked about g(t) being a deterministic nonperiodic signal.

Is a decaying exponential absolutely integrable?

If so, then the Fourier transform exists because that is a sufficient condition for existence.

Thread Starter

#### RdAdr

Joined May 19, 2013
214
I understand. Once we find the Fourier transform formula, we can see that a sufficient condition for it to exist is that the signal is absolutely integrable.

But to find the Fourier transform, first, we must go through the derivation.

And the derivation of the Fourier transform is based on repeating this decaying exponential (which has infinite duration) on the time axis, so that we obtain a periodic signal that we can expand into a Fourier series. But this periodic signal has an infinite period.

How does one obtains a periodic signal from a superposition of infinite-length signals? One can obtain such periodic signal from a superposition of finite-length signals, but infinite-length?

#### WBahn

Joined Mar 31, 2012
26,915
The derivation of the Fourier transform equations involves math that you haven't seen yet. What you are seeing at the level you are at is basically a layman's rationale to justify going from a periodic to an aperiodic signal. A similar thing happens with the Laplace transform, except it is even more obscure. Once you get past Diffy-Q and take a course in Math Physics or Complex Variables, you will get to see the actual derivation of both.

Thread Starter

#### RdAdr

Joined May 19, 2013
214
Can I find somewhere the actual derivation?

I took a course in complex analysis.

The derivation involves the complex exponential Fourier series of the signal. And then they take the limit. They make the frequency go to zero. And the Fourier transform pops up.

Last edited:

#### WBahn

Joined Mar 31, 2012
26,915
It's been a couple decades since I went through it, so I might be getting some things mixed up. But IIRC, the derivation of the Laplace transform (of which the Fourier transform is a special case) involves contour integration.

You can probably find plenty of references if you Google "derivation of Laplace transform".

Thread Starter

#### RdAdr

Joined May 19, 2013
214
I can't find any on Google.

Even in the math textbook that I had when taking the complex analysis course, the Laplace transform was only defined and then properties were given. But no derivation.

Thanks anyway.

#### WBahn

Joined Mar 31, 2012
26,915
You might look in the Math Physics text by Butkov (sp?). I think that was the text we used when I went through it.

Thread Starter

#### RdAdr

Joined May 19, 2013
214
Ok. Thanks.

Thread Starter

#### RdAdr

Joined May 19, 2013
214
These two videos are interesting: