a companion to fourier analysis for physics...
TRANSCRIPT
A Companion to Fourier Analysis
for Physics Students
(Part Two: Fourier Transform)
Gol Mohammad Nafisi∗
University of Tehran
Spring 2019
Contents
1 Prelude 1
2 Fourier Transform 2
3 Intermezzo 12
4 Variation on the Theme of Dirac Delta Function 13
5 Properties of the Fourier Transform 15
6 Applications of Fourier Transform 17
7 Coda 23
i
1 Prelude
ر دی قمار هوس اال هیچش بنماند بودش چه آن بباخت که قماربازی آن خنک
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Good News, Everyone! The saga of Fourier Analysis continues and in this note, as I promised in the previ-
ous one on Fourier Series, we are going to continue our journey into the realm of Fourier analysis, focusing now
on the Fourier transform. Although the power of this tool is more than its partner in crime, the Fourier series,
I’ve observed that most students do struggle grasping it fully in their first or second encounter1. I’m not the only
one in this observation, as J. F. James once wrote2: “Showing a Fourier transform to a physics student generally
produces the same reaction as showing a crucifix to Count Dracula”. Well, changing a tradition is hard but maybe
it’s worth trying. Who knows?!
Now surely you’re asking “What is a Fourier transform? What does it do? What am I doing with my life?”.
To answer these ontological questions, let’s imagine this analogy3: “Consider an experimental jam in which your
cellist friend plays a steady note (G for example) on his/her Cello (please be it Bach’s Suite No.1 in G major!),
and a microphone connected to an oscilloscope there, produces a voltage and the oscilloscope will display a graph
of the voltages against time, F (t), which is periodic. The waveform contains ‘harmonics’ or ‘overtones’ which are
the multiples of the fundamental frequency, with various amplitudes and in various phases. This waveform can
be analyzed to find the amplitudes of the overtones and the phases of sinusoids which it comprises. So we can
obtain a graph, A(ν), called the ‘sound spectrum’ of the amplitudes against frequency. In fact, A(ν) is the Fourier
transform of F (t)!”. You got it, right?!
I do not claim on originality of the material whatsoever. To quote David Tong, “My primary contribution has
been to borrow, steal and assimilate the best discussions and explanations I could find from the vast literature on
the subject”. Now let’s get back to the business.
1Well, third time’s a charm!2A Student’s Guide to Fourier Transforms, J. F. James, Cambridge University Press3It’s a remixed quote of Mr. James
1
2 Fourier Transform
“Listen to me, Morty. I know new situations can be intimidating. You lookin’ around and it’s all scary and different, but
y’know, meeting them head-on, charging right into ’em like a bull. That’s how we grow as people.”
- Rick Sanchez
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Definition 1 (Fourier Transform). Consider a piecewise continuous function f : R → C , such that f (t) ∈ C , t ∈ R ,
which is absolutely integrable, i.e.∫ +∞
−∞|f (t) |dt < ∞ . Then the Fourier transform of f (t) is defined as:
f (ω) ≡ F (ω) ≡ F f (t) def=
1√2π
∫ +∞
−∞f (t) e−iωt dt (1)
and its inverse transform is given by:
f(t) ≡ F−1(ω) ≡ F−1f (ω)def=
1√2π
∫ +∞
−∞F (ω) e+iωt dω (2)
Some remarks in order.
Remark 1. First, you should know that the Fourier transform is also called Forward Transformation, while its inverse
is called Inverse Transformation. Second, I noticed that most mathematicians prefer to write a factor 1
2πfor inverse
transformation, while dropping the factor from the forward transformation. You should know that there are some debates
on this issue. Third, some authors prefer to use frequency ν , instead of angular frequency ω . Hence by this convention,
they use ω = 2πν , and drop the 1
2πfactor from inverse transformation. However, here we stick with our convention and go
by ω , but keeping the factor 1√2π
for both transformations (with respect to Mr. Arfken’s convention).
Remark 2. In the eyes of a “Shut up and calculate” guy, the Fourier (or in general, Integral) transform is an economic
way to attack the problems which are hard to solve in the given space of configuration. The most important case is solving
differential equations, which we’ll encounter later in this note. I recommend you checking the Figure 20.1 of Arfken’s (7th
edition), to get a grasp on what one tries to do by introducing the concept of Integral Transforms. The Fourier transform
is one of the important classes of such transforms.
Sometimes it’s useful to use the Fourier transformation, for functions that we know they are even or odd, in terms of Sine
and Cosine integrals:
Definition 2 (Fourier Cosine Transform).
Fc (ω) =
√2
π
∫ ∞
0
f (t) cos(ωt) dt ; f (t) ≡ F−1c (ω) =
√2
π
∫ ∞
0
Fc (ω) cos(ωt) dω (3)
2
Definition 3 (Fourier Sine Transform).
Fs (ω) =
√2
π
∫ ∞
0
f (t) sin(ωt) dt ; f (t) ≡ F−1s (ω) =
√2
π
∫ ∞
0
Fs (ω) sin(ωt) dω (4)
Now the wait is over and it’s time for some examples.
Example 1. Find the Fourier transform of the function
f (t) =
1 + t −1 ≤ t ≤ 0
1− t 0 < t < 1
0 otherwise
(5)
Solution 1. By definition, we have:
F (ω) =1√2π
∫ +∞
−∞f (t) e−iωt dt =
1√2π
∫ 0
−1
(1 + t) e−iωt dt+1√2π
∫ 1
0
(1− t) e−iωt dt
=1√2π
∫ 0
−1
e−iωt dt+
∫ 0
−1
t e−iωt dt+
∫ 1
0
e−iωt dt−∫ 1
0
t e−iωt dt
(6)
The first and third integrals can be computed via standard integration, while the second and fourth can be done via integration
by parts. Hence
F (ω) =2
ω2√2π
[1− cos (ω)] (7)
Example 2. Consider the function
x(t) = e−γι cos (ω0t) θ(t) (8)
which describes the position of an underdamped oscillator, where the unit step function is defined by
θ(t) =
1 t > 0
0 t ≤ 0
(9)
Find the Fourier transform of x(t) .
Solution 2. First we rewrite the function as
x(t) = e−γt cos (ω0t) θ(t) =1
2e−γteiω0tθ(t) +
1
2e−γte−iω0tθ(t) (10)
Now taking the FT of the first term we get
F1(ω) =1√8π
∫ ∞
−∞dt e−γte−i(ω−ω0)tθ(t) =
1√8π
∫ ∞
0
dt e(−γ−iω+iω0)t
=1√8π
1
−γ − i (ω − ω0)e(−γ−iω+iω0)t
∣∣∣∞0
=1√8π
1
γ + i (ω − ω0)
(11)
3
Using the similar procedure we obtain for the second term
F2(ω) =1√8π
1
γ + i (ω + ω0)(12)
Hence
F (ω) = F1 + F2 =1√8π
[1
γ + i (ω − ω0)+
1
γ + i (ω + ω0)
]=
1
i√2π
ω − iγ
(ω − iγ)2 − ω20
(13)
Example 3 (Arfken 20.2.1).
(a) Show that g(−ω) = g∗(ω) is a necessary and sufficient condition for f(x) to be real.
(b) Show that g(−ω) = −g∗(ω) is a necessary and sufficient condition for f(x) to be pure imaginary.
Solution 3. First note that in Arfken’s convention, g(ω) is equivalent to our F (ω) . Now you can think of this problem
as proving an iff (if and only if) theorem, meaning that one proves a statement of the form “P iff Q”, also denoted as
“P ⇐⇒ Q”, by proving “if P, then Q” AND “if Q, then P”.
(a) The 2-step algorithm here is to prove: (1) if g∗ (ω) = g (−ω) then f (x) is real , and (2) if f (x) is real then g∗ (ω) =
g (−ω) . To prove (1) we have:
g (ω) =1√2π
∫ +∞
−∞f (x) e−iωx dx → g∗ (ω) =
1√2π
∫ +∞
−∞f∗ (x) e+iωx dx ; g (−ω) =
1√2π
∫ +∞
−∞f (x) e+iωx dx (14)
As you can see, the condition g∗ (ω) = g (−ω) can be satisfied only when f∗ (x) = f (x) meaning that f is real. To
prove (2) we have:
f (x) =1√2π
∫ +∞
−∞g (ω) e+iωx dω → f∗ (x) =
1√2π
∫ +∞
−∞g∗ (ω) e−iωx dω (15)
As you can see, only when we put g∗ (ω) = g (−ω) we get:
f∗ (x) =1
2π
∫ ∞
−∞
∫ ∞
−∞
[f (x′) e+iωx′
dx′]e−iωx dω
=1
2π
∫ ∞
−∞f (x′) dx′
∫ ∞
−∞eiω(x−x′) dω =
∫ ∞
−∞f (x′) δ(x− x′) dx′ = f (x)
(16)
Hence the statement is true.
(b) Pure imaginary means that f∗ (x) = −f (x) . The procedure is same as part (a).
Remark 3. Note that Mr. Arfken prefers to work with x instead of t, and uses the notation g (ω) for the forward Fourier
transform, instead of our notation F (ω) here. It’s your choice, but I’m happy with my t and F (ω) . So please don’t get
confused.
4
Example 4 (Arfken 20.2.2). The function
f(t) =
1, |t| < 1
0, |t| > 1(17)
is a symmetrical finite step function.
(a) Find gc(ω) , Fourier cosine transform of f(t) .
(b) Taking the inverse cosine transform, show that
f(t) =2
π
∫ ∞
0
sinω cosωtω
dω (18)
(c) From part (b) show that
∫ ∞
0
sinω cosωtω
dω =
0 |t| > 1
π
4|t| = 1
π
2|t| < 1
(19)
Solution 4.
(a) Using the definition we have
Fc(ω) =
√2
π
∫ ∞
0
f(t) cos(ωt)dt ⇒ Fc(ω) =
√2
π
∫ 1
0
cos(ωt)dt =√
2
π
sin(ω)ω
(20)
(b) Again
f(t) ≡ F−1c (ω) =
√2
π
∫ ∞
0
Fc(ω) cos(ωt)dω → f(t) =
√2
π
∫ ∞
0
√2
π
sin(ω)ω
cos(ωt)dω =2
π
∫ ∞
0
sin(ω)ω
cos(ωt)dω
(21)
(c) From the statement and (b) we get
∀|t| < 1 : 1 =
2
π
∫ ∞
0
sin(ω)ω
cos(ωt)dω →∫ ∞
0
sin(ω)ω
cos(ωt)dω =π
2
∀|t| > 1 : 0 =
∫ ∞
0
sin(ω)ω
cos(ωt)dω(22)
What about |t| = 1 ? Since our function has discontinuity at this point, we turn ourselves to Dirichlet’s theorem which
states that if x0 is a discontinuity for f , then the value of Fourier integral of f(x) at x = x0 is equal to
limx→x0f(x) + limx→x+
0f(x)
2
Hence
f(t = 1) =2
π
∫ ∞
0
sin(ω)ω
cos(ω)dω =1
2→
∫ ∞
0
sin(ω)ω
cos(ω)dω =π
4(23)
5
Now let’s put our results together
∫ ∞
0
sin(ω)ω
cos(ωt)dω =
0 |t| > 1
π
4|t| = 1
π
2|t| < 1
(24)
Example 5 (Arfken 20.2.3).
(a) Show that the Fourier sine and cosine transforms of e−at are
gs(ω) =
√2
π
ω
ω2 + a2, gc(ω) =
√2
π
a
ω2 + a2(25)
(b) Show that ∫ ∞
0
ω sinωt
ω2 + a2dω =
π
2e−at t > 0∫ ∞
0
cosωtω2 + a2
dω =π
2ae−at t > 0
(26)
Solution 5. First, it’s worth to note that the Sine and Cosine Fourier transforms of the function f(t) = e−at are actually
the Laplace transforms of the Sine and Cosine functions.
(a)
Fc(ω) =
√2
π
∫ ∞
0
f(t) cos(ωt)dt =√
2
π
∫ ∞
0
e−at cos(ωt)dt (27)
All we have to do is to compute the integral via integration by parts. I assume that you are quite familiar with the
method. The result is
Fc(ω) =
√2
π
a
a2 + ω2(28)
(b) This part can be done by two different methods: using our previous results in the Sine and Cosine Fourier transform of
e−at , or via the method of contour integration. I would like to do it in both ways, because it can serve you as a review
on what you’ve learned.
(i) via Fourier transform:
f(t) =
√2
π
∫ ∞
0
Fs(ω) sin(ωt)dω → e−at =
√2
π
∫ ∞
0
√2
π
ω
a2 + ω2sin(ωt)dω →
∫ ∞
0
ω
a2 + ω2sin(ωt)dω =
π
2e−at
(29)
You can do the same procedure, now Cosine Fourier transform, and obtain∫ ∞
0
1
a2 + ω2cos(ωt)dω =
π
2ae−at .
(ii) via contour integration: As you have learned, we can do cool MATHEMAGICS with Complex Analysis tools.
One of which is computing some tough improper real integrals. Here we’re going to conquer∫ ∞
0
ω
a2 + ω2sin(ωt)dω .
Consider the integral I =
∮C
zeitz
z2 + a2dz around a loop C in the complex plane. Let’s choose our contour to be a
semicircle of radius R, centered at origin. You might wonder why didn’t we choose a circle? Because the less poles
reside in our contour, is better! As you can see, we have two simple poles here: z1 = ia and z2 = −ia . Only z1
resides in our contour therefore we can rewrite our integral as I =
∮F (z)
z − iadz where F (z) =
zeitz
z + ia. By Cauchy
integral formula we get I = iπe−at .
6
Now let’s decompose our integral into two pieces as:
I =
∫ R
−R
xeitx
x2 + a2dx+
∫ z=Reiπ
z=R
zeitz
z2 + a2dz (30)
where the first integral is taken on the real line and the second one is taken on the curve (part of our semicircle
circumference). I argue that the second integral tends to zero, if the radius tends to infinity. Don’t you agree? Fair
enough, let’s hear my argument then: Suppose that R ≤ 1 , then
|z| = R ≤ 1∣∣z2 + a2
∣∣ ≥ ∣∣z2 − 1∣∣ ⇒
∣∣∣∣ zeitz
z2 + a2
∣∣∣∣ ≤ ∣∣∣∣ z
z2 + a2
∣∣∣∣ ≤ ∣∣∣∣ 1
|z|2 − 1
∣∣∣∣ = 1
R2 − 1(31)
Now we can use the “estimation lemma” (a.k.a ML inequality), which states that on an arc C:∣∣∣∣∫ f(z)dz
∣∣∣∣ ≤ ML
in which M is an upper bound of |f(z)| along the arc and L is the arc length. Here we found an upper bound
M =1
R2 − 1and L = πR , hence
∣∣∣∣∫ zeitz
z2 + a2dz
∣∣∣∣ ≤ πR
R2 − 1. What will happen if we extend our radius to infinity?
You’re right! limR→∞
∣∣∣∣∫ zeitz
z2 + a2dz
∣∣∣∣ = 0 therefore
I =
∫ R
−R
xeitx
x2 + a2dx =
∫ 0
−R
xeitx
x2 + a2dx+
∫ R
0
xeitx
x2 + a2dx = −
∫ R
0
xe−itx
x2 + a2dx+
∫ R
0
xeitx
x2 + a2dx
=
∫ R
0
x(eitx − e−itx
)x2 + a2
dx =
∫ R
0
x[2i sin(tx)]x2 + a2
dx
(32)
Since we obtained that I = iπe−at , then
R → ∞ : 2i
∫ ∞
0
x sin(tx)x2 + a2
dx = πie−at →∫ ∞
0
x sin(tx)x2 + a2
dx =π
2e−at (33)
For the other integral,∫ ∞
0
1
a2 + ω2cos(ωt)dω , you can calculate I =
∮eitz
z2 + a2dz instead by using the same
contour method as above and obtain ∫ ∞
0
cos(tx)x2 + a2
dx =π
2ae−at (34)
Example 6 (Arfken 20.2.9). Prove that
h
2πi
∫ ∞
−∞
e−iωtdω
E0 − iΓ/2− hω=
exp
(−Γt
2h
)exp
(−i
E0t
h
)t > 0
0 t < 0
(35)
This Fourier integral appears in a variety of problems in quantum mechanics: barrier penetration, scattering, time-dependent
perturbation theory, and so on.
Solution 6. Fortunately, I explained in the previous example how to use contour integration to compute an improper real
integral. You can follow the procedure here and obtain the desired result. However, I’m going to choose an easier way here.
There’s a theorem in Complex Analysis which states that4:4See Theorem 4.8c of Peter Henrici’s Applied and Computational Complex Analysis, Vol.1
7
Theorem 1. Let r be a rational function with no poles on the real axis and a zero of order ≥ 1 at infinity. Then for ω > 0 :
∫ +∞
−∞r(x)eiωxdx = 2πi
∑Im(z)>0
Reseiωzr(z)
(36)
where the sum involves the residues at all poles of r in the upper half-plane.
It’s clear that here for t > 0 if we let x 7→ −x we should sum in the lower half-plane. We have a simple pole at z0 =E0
h− iΓ
2h
(in the lower half-plane, and obviously not on the real axis), with the residue
Res(z0) = − exp−it
(E0
h− iΓ
2h
)(37)
Hence ∫ +∞
−∞
e−iωtdω
E0 − iΓ2 − hω
= (2πi) exp[−it
(E0
h− iΓ
2h
)]s.t. Γ > 0 (38)
The minus sign of residue vanished because our contour was clockwise in the lower half-plane. What about t < 0 ? You see
that in this case, the exact form of the theorem applies, meaning that we should count the residues on the upper half-plane
and since we have no pole there, therefore our function will be entire and the integral vanishes (Cauchy-Goursat theorem).
Example 7 (Arfken 20.2.10). Verify that the following are Fourier integral transforms of one another:
(a)
√
2
π
1√a2 − x2
|x| < a
0 |x| > a
and J0(ay)
(b)
0 |x| < a
−√
2
π
1√x2 + a2
|x| > a
and Y0(a|y|)
(c)√
π
2
1√x2 + a2
and K0(a|y|)
(d) Can you suggest why I0(ay) is not included in this list?
Solution 7. You should be able to do it by yourself. Just note that it’s better to use the integral representations of the
Bessel functions and take their forward or inverse Fourier transforms. You can find these representations in any Handbooks
covering the special functions. For example, see Abramowitz & Stegun’s Handbook of Mathematical Functions, Alan Jeffrey
& Hui Hui Dai’s Handbook of Mathematical Formulas and Integrals, N. M. Temme’s Special Functions, NIST Handbook of
Mathematical Functions5 or whatever resources you find suitable. Here you can try these special cases (x > 0) :
J0(x) =2
π
∫ ∞
0
sin[x cosh(t)] dt
Y0(x) = − 2
π
∫ ∞
0
cos[x cosh(t)] dt
K0(x) =
∫ ∞
0
cos[x sinh(t)] dt
I0(x) =1
π
∫ π
0
cosh[x cos(θ)] dθ
(39)
5Online version is available at https://dlmf.nist.gov
8
Note that it’s better to make a change of variable in above integrals so that you could obtain results similar to what Mr.
Arfken suggested. For example, let
cosh(t) = q → dq = sinh(t)dt =√q2 − 1dt →
∫ ∞
0
sin[x cosh(t)]dt =∫ ∞
1
sin(xq) dq√q2 − 1
(40)
and so on. For part (d), we should consider the asymptotic behavior of the given Bessel functions, since as I mentioned
at the beginning, f(t) must be absolutely integrable, i.e.∫ ∞
−∞|f |dt < ∞ . Now let’s see the asymptotic behaviors of our
functions for large arguments (x 0) :
Jv(x) ≈√
2
πx
[cos
(x− 1
2vπ − 1
4π
)]Yv(x) ≈
√2
πx
[sin
(x− 1
2vπ − 1
4π
)]Kv(x) ≈
√π
2xe−x
[1 +
4v2 − 1
8x
]Iv(x) ≈
1√2πx
ex[1− 4v2 − 1
8x
](41)
Now as you can see, improper integrals of each functions, i.e. lima→−∞
∫ c
a
f dt+ limb→+∞
∫ b
c
f dt , converges except for the Iv(x)
and neither its absolute integral. Thus it does not have a Fourier transform.
Example 8 (Arfken 20.2.13). (a) Show that f(x) = x− 12 is a self-reciprocal under both Fourier cosine and sine transforms;
that is: √2
π
∫ ∞
0
x−1/2 cosxtdx = t−1/2 ;
√2
π
∫ ∞
0
x−1/2 sinxtds = t−1/2 (42)
(b) Use the preceding results to evaluate the Fresnel integrals
∫ ∞
0
cos(y2) dy and∫ ∞
0
sin(y2) dy (43)
Solution 8.
(a) We use the contour integration method. The suitable contour for this problem can be taken as shown in Fig.1. Now
let’s consider the integral I =
∮eitz√zdz . Obviously the branch point z = 0 does not reside inside or on the contour,
thus∮
eitz√zdz = 0 . By dividing the integral on four paths shown in the photo, we’ll get
(z1 = Reiθ, z2 = reiθ
):
∮eitz√zdz =
∫ R
r
eitx√xdx+
∫ π2
0
eitReiθRieiθ√Reiθ
dθ +
∫ r
R
e−ty
√iy
idy +
∫ 0
π2
eitreiθ
rieiθ√reiθ
dθ = 0 (44)
We can write the second integral as:
∫ π2
0
eitR cos θ−R sin θRi√eiθ√
Rdθ =
∫ π2
0
eitR cos θ√Ri√eiθ
eR sin θdθ (45)
9
Figure 1: A suitable contour
Since∣∣eitR cos θ∣∣ = 1, so if we tend R → ∞ (which we should do) we see that above integral tends to zero. The same
argument holds for the 4th integral as r → 0 , hence
∫ ∞
0
eitx√xdx+
∫ 0
∞
e−ty
√iy
idy = 0 (46)
Let’s make a change of variable
y = q2 → dy = 2qdq →∫ 0
∞
e−ty
√iy
idy =
∫ 0
∞
e−tq2
√iq
2iqdq = −2√i
∫ ∞
0
e−tq2dq = −√i
√π
t(47)
On the other hand
∫ ∞
0
eitx√xdx =
∫ ∞
0
cos(tx) + i sin(tx)√x
dx =
∫ ∞
0
cos(tx)√x
dx+ i
∫ ∞
0
sin(tx)√x
dx =√i
√π
t(48)
Now since√i = e
iπ4 =
1√2(1 + i) then
∫ ∞
0
cos(tx)√x
dx+ i
∫ ∞
0
sin(tx)√x
dx =
√π
2t+ i
√π
2t(49)
Finally if we equate the real and imaginary parts of both sides, we’ll get:
∫ ∞
0
cos(tx)√x
dx =
√π
2t;
∫ ∞
0
sin(tx)√x
dx =
√π
2t(50)
(b) In∫ ∞
0
cos(tx)√x
dx , make the following change of variable
xt = y2 → tdx = 2ydy = 2(√x√t)dy → 1√
xdx =
2√tdy (51)
Now substituting back into integral will yield:
∫ ∞
0
cos(tx)√x
dx =2√t
∫ ∞
0
cos(y2)dy =
√π
2t⇒
∫ ∞
0
cos(y2)dy =
1
2
√π
2(52)
10
For∫ ∞
0
sin(tx)√x
dx , following the same procedure will give you the right answer:
∫ ∞
0
sin(y2)dy =
1
2
√π
2(53)
Example 9 (Arfken 20.2.14). Show that[1
r2
]T(k) =
(π2
)1/2 1
k.
Solution 9. In ancient times, maybe pre-Egyptians,[1
r2
]Tmeant taking the Fourier transform of 1
r2. Now the 3D version
of Fourier transform of our radial function (presented in IMAX) is
F (k) =1
(2π)32
∫eik·r
r2d3r (54)
Let’s go to the spherical coordinates so that we could rewrite the integral as:
F (k) =1
(2π)32
∫eikr cos θ
r2r2drdΩ =
1
(2π)32
∫ ∞
0
dr
∫ π
0
sin(θ)eikr cos θdθ
∫ 2π
0
dφ =1
(2π)12
∫ ∞
0
2 sin(kr)kr
dr (55)
In order to calculate this integral, if you write down the Fourier transform of the function f(t) =
1 |t| < k
0 |t| > k
, you’ll get
F (ω) =2√2π
sin(ωk)ω
inverse====⇒ f(t) =1
π
∫ ∞
−∞
sin(ωk)ω
eiωtdω (56)
Now applying the Dirichlet’s theorem yields
f(0) = 1 =k
π
∫ ∞
−∞
sin(ωk)kω
dωintegrand is even=============⇒
∫ ∞
0
sin(ωk)kω
dω =π
2k(57)
Finally by substituting this into our obtained 3D Fourier, we get:
F (k) =2√2π
π
2k=
1
k
√π
2(58)
11
3 Intermezzo
“IT’S-A-GOODA-SHOW!”
- The Italian Chef
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I propose at this point of our journey that we all should...
12
4 Variation on the Theme of Dirac Delta Function
“I’m gonna make him an offer he can’t refuse.”
- Don Corleone
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Figure 2: δτ (t) as τ → 0
That’s right! I’m gonna offer you another approach, which not many souls dared to offer, for comprehending the infamous
delta function. First, note that the “Dirac delta function”, is not a function at all. So you cannot simply try to integrate
this creature in space using the standard methods of Calculus. Instead, it can be regarded as a distribution or as a measure.
What is a measure? Well, consider a set M , for convenience let M = R . Now let δ : A ⊆ R 7−→ 0, 1 defines a measure
such that, it takes a subset A and returns δ(A) = 1 if 0 ∈ A and δ(A) = 0 if otherwise. I assume that you have seen some
definitions for this δ . The approach I take here is based on a function called “Unit Impulse Function” which has the form
δτ (t− a) =
1
2τa− τ < t < τ + a
0 otherwise(59)
You can easily check that
I(τ) =
∫ ∞
−∞δτ (t− a)dt = 1 (60)
The interesting thing here is that we let τ → 0 . In Fig.2, you can see what will happen. δτ (t) gets taller and narrower.
Hence for t 6= 0 : limτ→0
δτ (t) = 0, limτ→0
I(τ) = 1 . Now we can define our Dirac delta function about point a as:
δ(t− a) = limτ→0
δτ (t− a) (61)
In this spirit, we can take the Fourier transform of Delta function precisely as:
13
Fδ(t− a) = limτ→0
F δτ (t− a) =1√2π
limτ→0
∫ +∞
−∞δτ (t− a)e−iωtdt
=1√2π
limτ→0
1
2τ
∫ a+τ
a−τ
e−iωtdt
=1√2π
limτ→0
1
2τ
(1
−iω
)(e−iω(a+τ) − e−iω(a−τ)
)=
1√2π
limτ→0
1
τe−iωa sin(ωτ)
=1√2π
e−iωa
(62)
Now taking the inverse Fourier transform yields:
δ(t− a) = limτ→0
δτ (t− a) =1
2π
∫ ∞
−∞eiω(t−a)dω (63)
Nice! You’re going to see or use this relation frequently in your Physics courses.
Example 10 (Arfken 20.2.7). Using the formula
δ(t− x) =1
2π
∫ ∞
−∞eiω(t−x)dω =
1
2π
∫ ∞
−∞eiωte−iωxdω (64)
Derive Sine and Cosine representations of δ(t− x) .
Solution 10. We have
δs(ω) =
√2
π
1
2π
∫ ∞
0
∫ ∞
−∞sin(ωt)e−iωxeiωt dω dt (65)
and using sin(ωt) = eiωt − e−iωt
2i, together with integral property of delta function to obtain Sine form of delta. Finally by
taking the inverse transform, we’ll have a Sine representation of Delta function
δs(ω) =2
π
∫ ∞
0
sin(ωt) sin(ωx) dω (66)
Same procedure goes for Cosine representation and
δc(ω) =2
π
∫ ∞
0
cos(ωt) cos(ωx) dω (67)
14
5 Properties of the Fourier Transform
“When things get too heavy, just call me helium, the lightest known gas to man.”
- Jimi Hendrix
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I’ve gathered for you, and only for you, some of the Fourier transform properties because they can help us avoiding some
tiresome integrations. They might look heavy though, but the reward is that they will be very helpful when we want to solve
some (partial) differential equations. We have the following properties and theorems (I work with time variable t, however
you can replace it by x, and for the 3D case, consult Arfken):
Property 1 (Linearity).
Faf(t) + bg(t) = aF (ω) + bG(ω) (68)
Property 2 (Time Shift (Translation)).
F f (t± t0) = e±iωt0F (ω) (69)
Property 3 (Frequency Shift).
F−1(ω ± a) = f(t)e∓iat (70)
Property 4 (Change of Scale).
Ff(ax) =1
aF(ωa
)(71)
Property 5 (Complex Conjugation).
F f∗(x) = F ∗(−ω) (72)
Property 6 (Symmetry (Duality)). If Ff(t) = F (ω) then
FF (t) = 2πf(−ω) (73)
Theorem 2 (Multiplication Theorem). If Ff(t) = F (ω) and Fg(t) = G(ω) then
∫ ∞
−∞f(t) g(t) dt =
1
2π
∫ ∞
−∞F (ω)G(−ω) dω∫ ∞
−∞f(t) g(−t) dt =
1
2π
∫ ∞
−∞F (ω)G(ω) dω∫ ∞
−∞f(t) g(−t) dt =
1
2π
∫ ∞
−∞F (ω)G(ω) dω∫ ∞
−∞f(t)G(t) dt =
1
2π
∫ ∞
−∞F (t) g(t) dt∫ ∞
−∞f(t) g(t) dt =
1
2π
∫ ∞
−∞F (ω)G∗(ω) dω
(74)
15
Property 7 (Parseval’s). ∫ ∞
−∞[f(t)]2 dt =
1
2π
∫ ∞
−∞|F (ω)|2 dω (75)
Property 8 (Parseval’s for Sine and Cosine ).
∫ ∞
0
[f(t)]2dt =2
π
∫ ∞
0
|Fs(ω)|2 dω ;
∫ ∞
0
[f(t)]2 dt =2
π
∫ ∞
0
|Fc(ω)|2 dω (76)
Theorem 3 (Cross-Correlation Theorem). Cross-correlation of two functions f(t) and g(t) is defined by:
(f ⋆ g)(τ)def=
∫ ∞
−∞f∗(t) g(t+ τ) dt (77)
Then
Ff ⋆ g = F ∗(ω)G(ω) (78)
Theorem 4 (Convolution Theorem). Convolution of two functions is defined by
(f ∗ g)(t) dete=
∫ ∞
−∞f(τ)g(t− τ)dτ (79)
Then
Ff ∗ g = F (ω)G(ω) ; Ff(t)g(t) = F (ω) ∗G(ω) (80)
Remark 4. Convolution is in fact an operation on two signals, yielding a third one which is a modified version of the one
of the original signals. The important thing is, in the Fourier transform of the input and output signals of this operation,
no new frequency components are created. The existing ones are only modified (in amplitude and/or phase) and in general,
the multiplication of two functions in the time domain produces a convolution in the Fourier (frequency) domain, and
multiplication of two functions in the Fourier (frequency) domain will give us the convolution in time domain.
Property 9 (Time and Frequency Derivative). If we denote dn
dtnf(t) = f (n)(t) and dn
dωnF (ω) = F (n)(ω) , then
Ff (n)(t)
= (iω)nF (ω) ; F tnf(t) = (i)nF (n)(ω) (81)
Property 10 (Time Integration).
F
∫ t
−∞f(τ)dτ
=
1
iωF (ω) + πF (0)δ(ω) (82)
Now get ready to see some action!
16
6 Applications of Fourier Transform
“Life is a combination of magic and pasta.”
- Federico Fellini
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
One of the best things one can do with Fourier transform, is to solve differential equations by them. You should note
that often in the physical situations, we have the equations dependent on spatial x and temporal t variables. To avoid any
confusion, from now on we let our Fourier transform operates on the spatial variable instead of our previous convention6.
Therefore we will use Ff(x) = F (ω) from now on. The reason is that due to the conventions, it is said (in the case
of solving PDEs) the Fourier transform operates on the domain of space while leaving time unaffected, and the Laplace
transform operates on the domain of time, while leaving space unaffected.
The method of solving is crystal clear: Apply the forward Fourier transform to the given differential equation. In doing so,
you will obtain an equation for F (ω) and by solving it algebraically (if it’s necessary), obtain its explicit form. Then by
applying the inverse Fourier transform, you will obtain the solution for given differential equation. You will also need to
apply FT to the initial or boundary conditions during your calculations to get rid of the constants and coefficients appeared
in the process. Also, it is very useful to have a table of Fourier pairs7 to obtain quickly the forward or inverse transform
of the desired functions. However, in this course, you have to do the transforms in detail completely by yourselves, to get
your hands synchronized with your brain. Enough is enough, let’s have some fun!
Example 11 (Arfken 20.3.6). The 1-D neutron diffusion equation with a (plane) source is
−Dd2φ(x)
dx2+K2Dφ(x) = Qδ(x) (83)
where ϕ(x) is the neutron flux, Qδ(x) is the (plane) source at x = 0 and D and K2 are constants. Apply a Fourier transform.
Solve the equation in transform space. Transform your solution back into x-space.
Solution 11. Applying the forward Fourier transform to the given DE yields:
(−D)F
d2φ(x)
dx2
+(K2D
)Fφ(x) = QFδ(x) (84)
We know from the properties of the transform,
F f ′′(x) = −ω2Ff(x) ; Fδ(x) =1√2π
(85)
6I know, what a cheap twist and wasted turn. But it’s not my fault, it’s Physicist’s!7You can find a nice table of Laplace transform and Fourier pairs here: https://onlinelibrary.wiley.com/doi/pdf/10.1002/
9780470117866.app2
17
Hence letting Ff(x) = F (ω) we have
Dω2 F (ω) +K2 DF (ω) =Q√2π
→ F (ω) =Q
D√2π
(1
ω2 +K2
)(86)
This is the solution in transform space. In order to find the solution in x-space, we need to apply the inverse transform. We
know from the table (while you should know it from your muscle memory) that:
∀K > 0 : Fe−K|x|
=
√2
π
K
K2 + ω2(87)
Therefore
ϕ(x) = F−1
Q
D√2π
(1
ω2 +K2
)⇒ ϕ(x) =
Q
2KDe−K|x| (88)
Example 12. Solve the following Heat equation via Fourier transform:
c∂2u(x, t)
∂x2=
∂u(x, t)
∂tx ∈ R
u(x, t = 0) = f(x) s.t. f(x) =
u0 |x| < 1
0 |x| > 1
(89)
Solution 12. Let Fu(x, t) = U(ω, t) . By taking the Fourier transform of the PDE, we’ll get:
∂U(ω, t)
∂t+ c ω2U(ω, t) = 0 (90)
This equation has a solution of the form U(ω, t) = Ae−cω2t . Next step is taking the FT of initial condition:
U(ω, 0) =
∫ ∞
−∞f(x)e−iωxdx =
∫ 1
−1
(u0) e−iωxdx → U(ω, 0) = u0
(2 sinω
ω
)(91)
By using this in our general solution we’ll get:
Ae−cω2×0 = u0
(2 sinω
ω
)→ A =
2u0 sinω
ω(92)
Hence
U(ω, t) =
(2u0 sinω
ω
)e−cω2t (93)
This is the solution in transform space. To get the solution in x-space we need to take the inverse transform of eq.93:
u(x, t) =1
2π
∫ ∞
−∞
(2u0 sinω
ωe−cω2t
)eiωxdω (94)
18
Now using Euler identity eiωx = cos(ωx) + i sin(ωx) :
u(x, t) =u0
π
∫ ∞
−∞
sinω
ωe−cω2t cos(ωx)dω + i
∫ ∞
−∞
sinω
ωe−cω2t sin(ωx)dω
(95)
The second integrand is odd, therefore its integral over R vanishes. Hence the final solution will be
u(x, t) =u0
π
∫ ∞
−∞
(sin(ω) cos(ωx)
ω
)e−cω2tdω (96)
This integral solution is indeed acceptable and we’re done, but if you’re eager to know the explicit solution (i.e. by performing
the integration which might invoke some demons), it’s given by8:
u(x, t) =u0
2
(erf
(x+ 1
2√ct
)− erf
(x− 1
2√ct
))(97)
where
erf(z) ≡ 2√π
∫ z
0
e−t2dt (98)
is called error function which can also be defined as a Maclaurin series:
erf(z) = 2√π
∞∑n=0
(−1)nz2n+1
n!(2n+ 1)(99)
Hence
u(x, t) ≈ u0 erf(
1
2√ct
)−
x2(u0 e
− 14ct
√ct)
4 (√πc2t2)
+O(x3
)(100)
Remark 5. Sometimes it’s useful to use the Sine and Cosine Fourier transforms when we are dealing with a PDE defined
on R+ . Just note that if the conditions are given on the function itself, it’s better to use the Sine transform, and if those
conditions are given on the derivative of the function, it’s better to use the Cosine transform. Following formulas might be
useful:Fsu(x, t) = u(ω, t) =
∫ ∞
0
u(x, t) sin(ωx)dx
Fcu(x, t) = u(ω, t) =
∫ ∞
0
u(x, t) cos(ωx)dx;
F−1S u(ω, t) = u(x, t) =
2
π
∫ ∞
0
u(ω, t) sin(ωx)dω
F−1c u(ω, t) = u(x, t) =
2
π
∫ ∞
0
u(ω, t) cos(ωx)dω(101)
Fs uxx(x, t) = ω u(0, t)− ω2Fsu(x, t)
Fc uxx(x, t) = −ux(0, t)− ω2Fcu(x, t)(102)
8I used Wolfram Mathematica 10
19
Example 13. Solve the following Wave equation:
uxx = utt ∀x > 0
ux(0, t) = 0 ; ut(x, 0) = 0
|u(x, t)| < M
u(x, 0) =
1− x2 |x| < 1
0 |x| > 1
(103)
Solution 13. Let’s take the Cosine transform of PDE and for convenience denote it by u(ω, t) . Hence
utt(ω, t) + ω2 u(ω, t) = 0 ⇒ u(ω, t) = A sin(ωt) +B cos(ωt) (104)
Taking the same transform of the conditions yields
Fcu(x, 0) = u(ω, 0) =
∫ 1
−1
(1− x2
)cos(ωx)dx = 2
(sinω − ω cosω
ω3
)Fc ut(x, 0) = Fc0 = ut(ω, 0) = 0
(105)
By comparing these with the general solution we get
A = 0 , B = 2
(sinω − ω cosω
ω3
)⇒ u(ω, t) = 2
(sinω − ω cosω
ω3
)cos(ωt) (106)
Finally by taking the inverse transform of eq.106 we get
u(x, t) = F−1c u(ω, t) =
4
π
∫ ∞
0
[(sinω − ω cosω
ω3
)cos(ωt)
]cos(ωx)dω (107)
You don’t need to worry about the explicit solution, this integral form suffices. Though I’ve plotted the explicit solution for
you here just for fun!
20
Remark 6. You should note that all we did was for the homogeneous equations. For the case of inhomogeneous equations
with inhomogeneous boundary and initial conditions, one should consider some extra help e.g. change of variables or
Duhamel’s principle. I’m not going to elaborate on these issues.
One last interesting thing. You can decompose the Fourier transform of a function into Sine and Cosine transforms as
F (ω) = Fc(ω) + i Fs(ω) (108)
From Complex Analysis we know that one can use the polar form to represent a complex function. So
F (ω) = A(ω) eiϕ(ω) (109)
where A(ω) = |F (ω)| is the magnitude and ϕ(ω) is the phase. Now what if I told you that we could take the Fourier transform
of an image? Hear me out9:
The building blocks of a photo are pixels. Let’s say we have a gray-scale (to avoid complexities due to color) photo which
has w× h pixels. Now in computer language, this photo can be described by a function f(x, y) whose range takes the values
between 0 which is the black color to 255 which is the white color, where 0 ≤ x ≤ w and 0 ≤ y ≤ h . Taking the Fourier
transform of this image, which is called 2D Fourier transform, yields
F (ωx, ωy) =1√2π
∫ ∞
−∞dx
∫ ∞
−∞dy f(x, y) e−iωxxe−ωyy (110)
Now I’ve taken the FT of an owl and a husky photos with Mathematica, which you can see below:
(a) Original photo: fowl (b) Its magnitude: Aowl (c) Its phase: ϕowl
(a) Original photo: fhusky (b) Its magnitude: Ahusky (c) Its phase: ϕhusky
9I’ve inspired by an example in Matthew Schwartz’s The Physics of Waves course notes available at http://users.physics.
harvard.edu/~schwartz/teaching
21
Well, what will happen if we combine the magnitude of one photo with the phase of the other one and then take the inverse
Fourier transform? Cool stuff happens!
(a) Aowl and ϕhusky (b) Ahusky and ϕowl
As you can see, the phase is more dominant than the magnitude.
22
7 Coda
We have reached the end of our adventures in the realm of Fourier Analysis. I hope that these notes would have helped
you even in an infinitesimal way possible. Now following the routine, I would like to end my note with a quote by the great
Fyodor Dostoyevsky, which I enjoyed while reading it in his masterpiece “The Brothers Karamazov”. Have fun!
“The centripetal force on our planet is still fearfully strong, Alyosha. I have a longing for life, and I go on living in
spite of logic. Though I may not believe in the order of the universe, yet I love the sticky little leaves as they open in spring. I
love the blue sky, I love some people, whom one loves you know sometimes without knowing why. I love some great deeds done
by men, though I’ve long ceased perhaps to have faith in them, yet from old habit one’s heart prizes them.. I want to travel
in Europe, Alyosha, I shall set off from here. And yet I know that I am only going to a graveyard, but it’s a most precious
graveyard, that’s what it is! Precious are the dead that lie there, every stone over them speaks of such burning life in the
past, of such passionate faith in their work, their truth, their struggle and their science, that I know I shall fall on the ground
and kiss those stones and weep over them; though I’m convinced in my heart that it’s long been nothing but a graveyard.
And I shall not weep from despair, but simply because I shall be happy in my tears, I shall steep my soul in emotion. I love
the sticky leaves in spring, the blue sky - that’s all it is. It’s not a matter of intellect or logic, it’s loving with one’s inside,
with one’s stomach.. I believe like a child that suffering will be healed and made up for, that all the humiliating absurdity
of human contradictions will vanish like a pitiful mirage, like the despicable fabrication of the impotent and infinitely small
Euclidean mind of man, that in the world’s finale, at the moment of eternal harmony, something so precious will come to
pass that it will suffice for all hearts, for the comforting of all resentments, for the atonement of all the crimes of humanity,
for all the blood that they’ve shed; that it will make it not only possible to forgive but to justify all that has happened.”
23