Fourier transform II - The Fourier transform and inverse transform

Inspired by the previous example and ignoring any technical issues (such as convergence), we have been led to define the following:

Definition of the Fourier transform and inverse transform

The Fourier transform of a function f(t) is the function F(s) defined by the integral

F(s)=βˆ«βˆ’βˆžβˆžf(t)eβˆ’2Ο€itsdt,

if the integral exists. The function F is sometimes denoted Ff or f^.

The inverse Fourier transform of a function F(s) is the function f(t) defined by the integral

f(t)=βˆ«βˆ’βˆžβˆžF(s)e2Ο€itsds,

if the integral exists. The function f is sometimes denoted Fβˆ’1F or FΛ‡.

Convention warning

Although our definition of the Fourier transform seems like the most natural one, as an evolution of the Fourier series, in practice there are many competing definitions. Some sources switch the roles of what we have called the Fourier transform and the inverse Fourier transform. Some use the integral

F(s)=12Ο€βˆ«βˆ’βˆžβˆžf(t)eβˆ’itsdt,

or even simply the integral

F(s)=βˆ«βˆ’βˆžβˆžeβˆ’itsdt.

In these notes will stick with the definitions we made above, but be wary when reading other resources.

Examples


The rectangle function

Based on our work in the previous example, the Fourier transform of the rectangle function Ξ (t) is the function F(s)=sin⁑(Ο€s)Ο€s:

(FΞ )(s)=βˆ«βˆ’βˆžβˆžΞ (t)eβˆ’2Ο€itsdt=sin⁑(Ο€s)Ο€s.

If we believe the rectangle function Ξ (t) is a "fundamental" function, then its Fourier transform probably deserves to be considered fundamental, as well. Indeed, it is often given a name:

Definition of the sinc function

Define the function sinc (pronounced "sink") by

sinc(s)={sin⁑(Ο€s)Ο€s,Β forΒ sβ‰ 01,Β forΒ s=0.
Convention warning

As with the definition of the Fourier transform, not all sources agree on this definition of the sinc function. Some instead use the function sin⁑(s)s.

So, we can now say FΞ =sinc. Also, for reference, here's a bit of the graph of the sinc function:

sinc.png|900

We should note that in our build-up to the discovery of the Fourier transform, we also "proved" that the inverse Fourier transform of the sinc function is the rectangle function Ξ , i.e.,

Fβˆ’1sinc=Ξ 

This would be fairly difficult to verify directly, since it would amount to calculating the integral

(Fβˆ’1sinc)(t)=βˆ«βˆ’βˆžβˆžsinc(s)e2Ο€itsds=βˆ«βˆ’βˆžβˆžsin⁑(Ο€s)Ο€se2Ο€itsds.

In fact, the above integral doesn't quite exactly agree with our rectangle function. It turns out that the above integral takes the value 0 outside (βˆ’12,12), takes the value 1 inside (βˆ’12,12), and takes the value 12 at t=Β±12. In other words, it agrees with Ξ (t) except at the two jump discontinuities (where the inverse transform takes the average of the values across the jump).

We won't worry about this for now.

A triangle function

Consider the "triangle function" defined piecewise by

Ξ›(t)={1+t,Β forΒ βˆ’1≀t≀01βˆ’t,Β forΒ 0<t≀10,Β otherwise

The graph of this function looks like a isosceles right triangle of height 1 and base 2 centered at the origin:

triangleFunction.png|900

We have chosen to use the capital lambda symbol for this function's name because that Greek character also happens to look like a triangle.

In any case, we can compute the Fourier transform of this function:

(FΞ›)(s)=βˆ«βˆ’βˆžβˆžΞ›(t)eβˆ’2Ο€itsdt=βˆ«βˆ’10(1+t)eβˆ’2Ο€itsdt+∫01(1βˆ’t)eβˆ’2Ο€itsdt=βˆ«βˆ’11eβˆ’2Ο€itsdt+βˆ«βˆ’10teβˆ’2Ο€itsdtβˆ’βˆ«01teβˆ’2Ο€itsdt=1βˆ’2Ο€is[eβˆ’2Ο€its]t=βˆ’1t=1+1(βˆ’2Ο€is)2∫2Ο€is0ueuduβˆ’1(βˆ’2Ο€is)2∫0βˆ’2Ο€isueudu=1βˆ’2Ο€is(eβˆ’2Ο€isβˆ’e2Ο€is)+1(βˆ’2Ο€is)2[ueuβˆ’eu]u=2Ο€isu=0βˆ’1(βˆ’2Ο€is)2[ueuβˆ’eu]u=0u=βˆ’2Ο€is=12Ο€is(e2Ο€isβˆ’eβˆ’2Ο€is)+1(βˆ’2Ο€is)2(βˆ’1βˆ’2Ο€ise2Ο€is+e2Ο€is)βˆ’1(βˆ’2Ο€is)2(βˆ’2Ο€iseβˆ’2Ο€isβˆ’eβˆ’2Ο€is+1)=12Ο€ise2Ο€isβˆ’12Ο€iseβˆ’2Ο€isβˆ’1(βˆ’2Ο€is)2βˆ’12Ο€ise2Ο€is+1(βˆ’2Ο€is)2e2Ο€is+12Ο€iseβˆ’2Ο€is+1(2Ο€is)2eβˆ’2Ο€isβˆ’1(βˆ’2Ο€is)2=1(βˆ’2Ο€is)2(e2Ο€isβˆ’2+eβˆ’2Ο€is),

We could leave the answer like this, but this expression can be simplified even further, as follows:

(FΞ›)(s)=1(βˆ’2Ο€is)2(e2Ο€isβˆ’2+eβˆ’2Ο€is)=1(βˆ’2Ο€is)2(eΟ€isβˆ’eβˆ’Ο€is)2=1βˆ’4Ο€2s2(2isin⁑(Ο€s))2=sin2⁑(Ο€s)(Ο€s)2=sinc2(s)

So the Fourier transform of the triangle function Ξ› is exactly the square of the Fourier transform of the rectangle function Ξ . We'll soon see that there is a secret reason for this!

One-sided exponential decay

Consider the one-sided exponential decay function

f(t)={eβˆ’t,Β forΒ tβ‰₯00,Β otherwise.

The graph of f looks like

oneSidedExponentialDecay.png|900

(We have not chosen a fun character to name this function, since there aren't any characters that look like the above graph.)

We compute the Fourier transform of this function:

(Ff)(s)=βˆ«βˆ’βˆžβˆžf(t)eβˆ’2Ο€itsdt=∫0∞eβˆ’teβˆ’2Ο€itsdt=∫0∞e(βˆ’1βˆ’2Ο€is)tdt=1βˆ’1βˆ’2Ο€islimbβ†’βˆž(e(βˆ’1βˆ’2Ο€is)bβˆ’1)=1βˆ’1βˆ’2Ο€islimbβ†’βˆž(eβˆ’beβˆ’2Ο€ibsβˆ’1)=1βˆ’1βˆ’2Ο€is(0βˆ’1)=11+2Ο€is.

(Note that here we are assuming s only takes real values, in which case the exponential function eβˆ’2Ο€ibs=cos⁑(2Ο€bs)βˆ’isin⁑(2Ο€bs) is always a complex number on the unit circle. This is how we concluded that the term eβˆ’beβˆ’2Ο€ibs went to 0 as bβ†’βˆž.)

A Gaussian function

Consider the classic Gaussian function f(t)=eβˆ’Ο€t2, whose graph is the usual bell curve:

gaussian.png|900

Note that the constant Ο€ was chosen so that the area under the bell curve is exactly 1, i.e.,

βˆ«βˆ’βˆžβˆžeβˆ’Ο€t2dt=1.

The Fourier transform of this function is difficult to compute directly:

(Ff)(s)=βˆ«βˆ’βˆžβˆžeβˆ’Ο€t2eβˆ’2Ο€itsdt

We can determine Ff without computing the above integral in a very sneaky way. Let's temporarily denote the above integral by F(s). First we consider differentiating the mystery function F, finding

Fβ€²(s)=dds(βˆ«βˆ’βˆžβˆžeβˆ’Ο€t2eβˆ’2Ο€itsdt)

If we live life reckless and pass the differentiation inside the integral, we find that

Fβ€²(s)=βˆ«βˆ’βˆžβˆžeβˆ’Ο€t2β‹…(βˆ’2Ο€it)eβˆ’2Ο€itsdt.

This new integral can be computed using integration-by-parts. Specifically, if we let u=eβˆ’2Ο€its and dv=βˆ’2Ο€iteβˆ’Ο€t2, then du=βˆ’2Ο€iseβˆ’2Ο€itsdt and v=ieβˆ’Ο€t2, and so we have

∫eβˆ’Ο€t2β‹…(βˆ’2Ο€it)eβˆ’2Ο€itsdt=eβˆ’2Ο€itsβ‹…ieβˆ’Ο€t2βˆ’βˆ«ieβˆ’Ο€t2β‹…(βˆ’2Ο€is)eβˆ’2Ο€itsdt=ieβˆ’Ο€t2eβˆ’2Ο€itsβˆ’2Ο€s∫eβˆ’Ο€t2eβˆ’2Ο€itsdt.

This is the antiderivative formula, so what happens to the definite (improper) integral? We'll be a bit fast and loose here, but the main point is that as tβ†’Β±βˆž, the boundary term ieβˆ’Ο€t2eβˆ’2Ο€ts rapidly tends to 0. Indeed, if you converted it the usual real-and-imaginary parts, you would see that

ieβˆ’Ο€t2eβˆ’2Ο€its=ieβˆ’Ο€t2(cos⁑(2Ο€ts)βˆ’isin⁑(2Ο€ts))=eβˆ’Ο€t2sin⁑(2Ο€ts)+ieβˆ’Ο€t2cos⁑(2Ο€ts).

The net effect is that

Fβ€²(s)=βˆ’2Ο€sβˆ«βˆ’βˆžβˆžeβˆ’Ο€t2eβˆ’2Ο€itsdt.

But this new integral is just the integral we started with, which defined F(s). So we've discovered that

Fβ€²(s)=βˆ’2Ο€sβ‹…F(s).

The general solution to this ordinary differential equation is

F(s)=Ceβˆ’Ο€s2.

We can even determine the mystery constant C, since F(0)=C and

F(0)=βˆ«βˆ’βˆžβˆžeβˆ’Ο€t2dt=1.

So, we can finally conclude that F(s)=eβˆ’Ο€s2. In other words, the Fourier transform of the Gaussian f(t)=eβˆ’Ο€t2 is the identical Gaussian F(s)=eβˆ’Ο€s2.

If you think of the Fourier transform as some kind of "operator" that transforms functions into other functions, then you can also start to think of this particular Gaussian function as something like an "eigenfunction" with eigenvalue 1.

Suggested next notes


Fourier transform III - Properties of the Fourier transform