A Fourier transform hit parade

With the definition in place it’s time to reap the benefits and find some Fourier transforms explicitly. We note one general property

• F is linear on tempered distributions.

This means that

F (T1+ T2) = F T1+ F T2 and F (αT ) = αF T , α a number. These follow directly from the definition. To wit:

hF (T₁+ T₂), ϕi = hT₁+ T₂, F ϕi = hT₁, F ϕi + hT₂, F ϕi = hF T₁, ϕi + hF T₂, ϕi = hF T₁+ F T₂, ϕi hF (αT ), ϕi = hαT , F ϕi = αhT , F ϕi = αhF T , ϕi = hαF T , ϕi

The Fourier transform of δ As a first illustration of computing with the generalized Fourier transform we’ll find F δ. The result is:

• The Fourier transform of δ is

F δ = 1 .

This must be understood as an equality between distributions, i.e., as saying that F δ and 1 produce the same values when paired with any Schwartz function ϕ. Realize that “1” is the constant function, and this defines a tempered distribution via integration:

h1, ϕi = Z ∞

−∞

1 · ϕ(x) dx

That integral converges because ϕ(x) is integrable (it’s much more than integrable, but it’s certainly integrable).

We derive the formula by appealing to the definition of the Fourier transform and the definition of δ. On the one hand,

hF δ, ϕi = hδ, F ϕi = F ϕ(0) = Z ∞

−∞

ϕ(x) dx . On the other hand, as we’ve just noted,

h1, ϕi = Z ∞

−∞

1 · ϕ(x) dx = Z ∞

−∞

ϕ(x) dx .

The results are the same, and we conclude that F δ = 1 as distributions. According to the inversion theorem we can also say that F⁻¹1 = δ.

We can also show that

F 1 = δ . Here’s how. By definition,

hF 1, ϕi = h1, F ϕi = Z ∞

−∞

F ϕ(s) ds .

But we recognize the integral as giving the inverse Fourier transform of F ϕ at 0:

F⁻¹F ϕ(t) = Z ∞

−∞

e^2πistF ϕ(s) ds and at t = 0 F⁻¹F ϕ(0) = Z ∞

−∞

F ϕ(s) ds . And now by Fourier inversion on S,

F⁻¹F ϕ(0) = ϕ(0) . Thus

hF 1, ϕi = ϕ(0) = hδ, ϕi

and we conclude that F 1 = δ. (We’ll also get this by duality and the evenness of δ once we introduce the reverse of a distribution.)

The equations F δ = 1 and F 1 = δ are the extreme cases of the trade-off between timelimited and bandlimited signals. δ is the idealization of the most concentrated function possible — it’s the ultimate timelimited signal. The function 1, on the other hand, is uniformly spread out over its domain.

It’s rather satisfying that the simplest tempered distribution, δ, has the simplest Fourier transform, 1.

(Simplest other than the function that is identically zero.) Before there were tempered distributions, however, there was δ, and before there was the Fourier transform of tempered distributions there was F δ = 1. In the vacuum tube days this had to be established by limiting arguments, accompanied by an uneasiness (among some) over the nature of the limit and what exactly it produced. Our computation of F δ = 1 is simple and direct and leaves nothing in question about the meaning of all the quantities involved. Whether it is conceptually simpler than the older approach is something you will have to decide for yourself.

The Fourier transform of δa Recall the distribution δa is defined by hδ_a, ϕi = ϕ(a) .

What is the Fourier transform of δa? One way to obtain F δa is via a generalization of the shift theorem, which we’ll develop later. Even without that we can find F δ_a directly from the definition, as follows.

The calculation is along the same lines as the one for δ. We have hF δ_a, ϕi = hδ_a, F ϕi = F ϕ(a) =

Z ∞

−∞

e^−2πiaxϕ(x) dx .

This last integral, which is nothing but the definition of the Fourier transform of ϕ, can also be interpreted as the pairing of the function e^−2πiax with the Schwartz function ϕ(x). That is,

hF δa, ϕi = he^−2πiax, ϕi hence

F δa= e^−2πisa.

To emphasize once again what all is going on here, e^−2πiax is not integrable, but it defines a tempered

distribution through Z ∞

−∞

e^−2πiaxϕ(x) dx

which exists because ϕ(x) is integrable. So, again, the equality of F δ_a and e^−2πisa means they have the same effect when paired with a function in S.

To complete the picture, we can also show that

F e^2πixa= δa.

(There’s the usual notational problem here with variables, writing the variable x on the left hand side. The

“variable problem” doesn’t go away in this more general setting.) This argument should look familiar: if ϕ is in S then

hF e^2πixa, ϕi = he^2πixa, F ϕi

= Z ∞

−∞

e^2πixaF ϕ(x) dx (the pairing here is with respect to x)

But this last integral is the inverse Fourier transform of F ϕ at a, and so we get back ϕ(a). Hence hF e^2πixa, ϕi = ϕ(a) = hδa, ϕi

whence

F e^2πixa= δ_a.

Remark on notation You might be happier using the more traditional notation δ(x) for δ and δ(x − a) for δ_a (and δ(x + a) for δ_−a). I don’t have any objection to this — it is a useful notation for many problems — but try to remember that the δ-function is not a function and, really, it is not to be evaluated

“at points”; the notation δ(x) or δ(x − a) doesn’t really make sense from the distributional point of view.

In this notation the results so far appear as:

F δ(x ± a) = e^±2πisa, F e^±2πixa= δ(s ∓ a)

Careful how the + and − enter.

You may also be happier writing Z ∞

−∞

δ(x)ϕ(x) dx = ϕ(0) and Z ∞

−∞

δ(a − x)ϕ(x) dx = ϕ(a) .

I want you to be happy.

The Fourier transform of sine and cosine We can combine the results above to find the Fourier transform pairs for the sine and cosine.

F ¹

2(δa+ δ−a)

= ¹₂(e^−2πisa+ e^2πisa) = cos 2πsa . I’ll even write the results “at points”:

F ¹

2(δ(x − a) + δ(x + a))

= cos 2πsa . Going the other way,

F cos 2πax = F ¹₂(e^2πixa+ e^−2πixa)

= ¹₂(δ_a+ δ_−a) . Also written as

F cos 2πax = ¹₂(δ(s − a) + δ(s + a))) . The Fourier transform of the cosine is often represented graphically as:

I tagged the spikes with 1/2 to indicate that they have been scaled.¹⁷ For the sine function we have, in a similar way,

F₁

2i(δ(x + a) − δ(x − a))

= 1

2i(e^2πisa− e^−2πisa) = sin 2πsa , and

F sin 2πax = F₁

2i(e^2πixa− e^−2πixa)

= 1

2i(δ(s − a) − δ(s + a)) . The picture of F sin 2πx is

17Of course, the height of a δais infinite, if height means anything at all, so scaling the height doesn’t mean much. Sometimes people speak of αδ, for example, as a δ-function “of strength α”, meaning just hαδ, ϕi = αϕ(0).

Remember that 1/i = −i. I’ve tagged the spike δa with −i/2 and the spike δ−a with i/2.

We’ll discuss symmetries of the generalized Fourier transform later, but you can think of F cos 2πax as real and even and F sin 2πax as purely imaginary and odd.

We should reflect a little on what we’ve done here and not be too quick to move on. The sine and cosine do not have Fourier transforms in the original, classical sense. It is impossible to do anything with the

integrals Z ∞

−∞

e^−2πisxcos 2πx dx or Z ∞

−∞

e^−2πisxsin 2πx dx .

To find the Fourier transform of such basic, important functions we must abandon the familiar, classical terrain and plant some spikes in new territory. It’s worth the effort.