Description
1 Complex Conjugate Property of the Fourier Transform
Let g(x) be a complex-valued function, that is g(x) maps the reals x (i.e. x ∈ R) to complex values;
mathematically, we write:
g : R 7−→ C.
If g(x) ←→ G(ω), i.e. the Fourier transform (FT) of g(x) is G(ω), then from the definition of the FT,
demonstrate that,
g
∗
(x) ←→ G
∗
(−ω),
where g
∗
(x) is the complex conjugate of g(x). (Remember that for a complex number x = xR + jxI
, whose
real part is xR and imaginary part is xI
, the complex conjugate of x denoted by x
∗ = xR − jxI
. The
imaginary unit j = √
−1.)
2 Fourier Transform of Real Functions
Let
u(x) = (
1, for x ≥ 0;
0, otherwise,
represent the unit step function. Using the definition of the Fourier transform, compute the FT of the
following functions:
(a) f(x) = e
−axu(x)
(b) f(x) = sinc(x) = sin(x)
x
(c) f(x) = rect(x) = (
1, for − 1/2 ≤ x ≤ 1/2;
0, otherwise.
(d) f(x) = Λ(x) =
1 + x, for − 1 ≤ x < 0;
1 − x, for 0 ≤ x ≤ 1;
0, otherwise.
The function Λ(x) is also known as the triangle (or ‘tent’) function, sketch it to see why.
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3 Inverse Fourier Transforms
Using the definition of the inverse Fourier transform (IFT), compute the IFT f(x) for the following
functions:
(a) F(ω) = rect(ω) = (
1, for − 1/2 ≤ ω ≤ 1/2;
0, otherwise.
(b) F(ω) = e
−a|ω|
, where is a > 0. Recall that |ω| denotes the absolute value of ω, i.e.
|ω| =
(
ω, when ω ≥ 0;
−ω, when ω < 0.
4 2D Fourier Transforms
In this question, we will compute the 2D FT for some important 2D functions f(x, y). Two dimensional
functions accept two arguments and return a scalar, mathematically f : R
2
7−→ R.
From the definition of the 2D FT, compute the FT of the following 2D functions:
(a) f(x, y) = rect(ax)rect(by), where the function
rect(x) = (
1, for − 1/2 ≤ x ≤ 1/2;
0, otherwise.
(b) f(x, y) = Λ(x)Λ(y).
5 Fourier Transform of some Non-Square-Integrable Functions
In this question, we aim to compute the Fourier transform of functions g(x) that are not in the space
L
2
(R), i.e. non-square-integrable functions. Examples include the signum1
function, defined as
sign(x) =
1, for x > 0;
0, for x = 0;
−1, for x < 0;
,
and the unit step function
u(x) = (
1, for x ≥ 0;
0, otherwise.
For such functions, the Fourier transform integral does not exist because it is not finite. However, we can
sometimes derive, so called, generalized Fourier transforms G(ω), that are still very useful.
To begin, consider the function
fa(x) =
e
−ax
, for x > 0;
0, for x = 0;
−e
ax
, for x < 0,
where a is strictly positive.
(a) Sketch the function fa(x), for some a > 0.
1The signum function simply returns the sign of its argument/input.
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(b) Let g(x) = lima→0
fa(x). Sketch g(x) and argue that it is equal to the signum function. That is, argue
that as a goes to zero, the function fa(x) goes tends to sign(x).
(c) Using the result in part (b), above, prove that the Fourier transform of the signum function,
F {sign(x)} =
2
jω
.
(d) By noting that the unit step function u(x) may be written as,
u(x) = 1/2 (1 + sign(x)),
show that the Fourier transform of u(x) is
U(ω) = πδ(ω) + 1
jω
.
[Hint: you may use the following Fourier transform pair: 1 ←→ 2πδ(ω). To be explicit, this means
that the FT of 1 is 2πδ(ω).]
6 Fast Fourier Transform (Computational Question)
(a) Consider the time-varying signal g(t) = 3 sin(200πt) + sin(40πt).
(i) Using Matlab, obtain samples of the signal g(t) at a frequency fs = 1000 Hz, for 1 s. How many
samples should you have?
(ii) Using Matlab, compute the FFT of the sampled signal, and plot the magnitude spectrum in Hz.
(iii) Comment on the position and amplitude/size of the peaks, in the plot.
7 2D Fourier Transform of the Circle Function (Optional, for extra
credit)
In general, imaging systems have an aperture through which light passes in order to enter the imaging
system. The circle function is a good model for such apertures and it’s Fourier transform can be important
when attempting to derive a mathematical model for the imaging systems.
Given that the circle function
can be written mathematically as
circ p
x
2 + y
2
=
1, when x
2 + y
2 < 1;
1/2, when x
2 + y
2 = 1;
0, otherwise
Show that the Fourier transform of the circle function is
F
n
circ p
x
2 + y
2
o =
2π
ρ
J1(ρ)
2π q
ω2
x + ω2
y
J1
q
ω2
x + ω2
y
,
where J1(x) is the first order Bessel function of the first kind, given by
Z x
0
ξJ0(ξ) dξ = xJ1(x),
and J0(a) = 1
2π
R 2π
0
e
−ja cos(θ−φ) dθ is the zero order Bessel function of the first kind. [Hint: Use the
substitutions x = r cos(θ), y = r sin(θ), ωx = ρ cos(φ), and ωy = ρ sin(φ).]
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