Learn to visualize the Fourier transform and understand its properties. This is a rough draft.

1. The Magical Transform
Felipe Hernández
Wednesday, December 15, 2010

2. Definition
• F(w) = integral (f(x)e^(-jwx)dx)
• As a side note, every sum and integral, unless otherwise noted, go
from negative infinity to positive infinity
Wednesday, December 15, 2010

3. What does it do?
• Suppose the function f(x) looks periodic and smooth (there are actual
conditions, but we won’t worry about them)
• You could model f(x) using sin waves of varying amplitude and
frequency
• This is more commonly used in its discrete form for signal processing
and analysis, but here we discuss the continuous form
Wednesday, December 15, 2010

4. Introduction
• Suppose the function is a sum of sin waves: f(x) = sum(A(w)f(wx))
• f(wx) is a wave of frequency w. For this case, e^(jwx) is what is needed.
The properties are incredibly convenient, and they will be discussed.
• This is better represented with the integral form:
f(x) = integral (A(w)e^(jwx) dw)
• This means that we put the function into the frequency domain. w is the
frequency. F(w) is related to the amplitude of the wave of that frequency.
• A(w) is actually equal to the fourier transform divided by 2pi so that:
f(x) = integral(F(w)sin(wx) dw)/2pi
• The reasons for the division by 2pi will be explained once we explore the
transform in more detail.
Wednesday, December 15, 2010

5. The Visualization
• Lets inspect the definition of the transform: F(w) = integral (f(x)e^(-
jwx)dx)
• There are two variables: w (the frequency) and x
• We’ll visualize the transform starting with a specific frequency.
• What we notice is that e^-jwx is simply -isinw + cosw. As you increase
x, the function goes in a “backwards” unit circle around the complex
plane with frequency w.
• When we multiply this by f(x), we see that we are simply plotting the
function on the complex plane in polar form, except that -wθ = x or θ =
-x/w. Therefore, we graph r = f(-x/w).
• Now, we sum up all the “arrows” that are formed when we plot the
function as x goes from negative infinity to infinity
Wednesday, December 15, 2010

6. Example:
On the right is the function:
y = arctan(x)
To find the Fourier transform F(w) for
w = 1, we plot r = arctan(-θ) and sum the
arrows.
To find the Fourier transform F(w) for
w = 0.5, we plot r = arctan(-2θ) and follow the same
procedure
Wednesday, December 15, 2010

7. The case of sin(wx)
• The most important case is the Fourier transform of f(x) = sin(w0x)
• To start, we will use w = w0 and w = -w0. This should give us non-zero
amplitudes (the frequencies exist there).
• Then, we will find that if |w| is not equal to w0, the amplitude is equal
to zero.
Wednesday, December 15, 2010

8. w = w0 imaginary
Here is the graph of r=sin(-w0x/w) when
w = w0
To find F(w), we add up all of those arrows.
The result is an arrow purely in the
real
-imaginary direction.
The magnitude of the arrow is trickier to
find. First, lets find the magnitude if we only
go around once.
Each arrow contributes only in the y direction. dy = d/dθ (rsin(θ)). r = -sin(θ).
Therefore, we can integrate -sin^2(θ) to find the magnitude of the arrow. For one
revolution, (0 to 2pi), the magnitude is pi. (You do the integral!!)
Therefore, F(w0) = -i*pi*δ(0). The δ(0) part takes care of the rest of the
revolutions. δ(x) is defined as an impulse such that the integral from 0- to 0+ of
δ(x) = 1. We will discuss how it got there in more detail later.
Wednesday, December 15, 2010

9. w = -w0
Here is the graph of r=sin(-w0x/w) when
w = w0
To find F(w), we add up all of those arrows.
The result is an arrow purely in the
imaginary direction.
We obtain the magnitude of the arrow using
the same method as last time.
There is one more case to explore before finding F(sin(wx)).
Wednesday, December 15, 2010

10. |w| ≠ w0
What happens when we choose w = 3w0? We would
expect that F(w) = 0 because that frequency isn’t
represented in f(x) = sin(wx). As it turns out, all of Here, each arrow
represents the sum for
the arrows cancel out, leaving an arrow of magnitude the entire lobe. (I’m
lazy, OK?)
0.
Now what about when w is close to w0, by a
factor of 1.004? We expect it to be 0, but its sooo
close!! As it turns out, the arrows cancel
themselves out in this case as well.
Wednesday, December 15, 2010

11. F[sin(wx)]
• When f(x) = sin(w0x), F(w) = -i*pi[δ(w-w0) - δ(w+w0)]. This fits our
results from the graphical interpretation. It is negative infinity in the
imaginary direction when w = w0, like noticed in the graphical view.
• Let’s take the inverse transform and check the answer. First, we’ll need to
know that integral ( δ(x-a)f(x)dx ) = f(a) -- it’s an impulse at x=a of
magnitude f(a).
• When we check the inverse Fourier transform, we find that we get sin(wx).
(You do it, OK?). By the way, now we can see why that pesky 1/2pi is
there. It “normalizes” the value -- you get pi from -w0 and you get pi from
w0.
• Now lets investigate what happens when apply basic shifts to f(x).
Wednesday, December 15, 2010

12. Horizontal shifts
The first transformation that we will investigate is
f(x-k). We will investigate it arctan(x-k) to show
how it works in the general case.
First, we notice that this causes a shift in the graph.
The first is the graph of r=arctan(-x), and the second
is the graph of r=arctan(-(x+pi/4)). The second arrow
is the first arrow rotated by -pi/4 --clockwise. If we
represent the first arrow by ||A||e^(iθ) (A is the
magnitude), then the second arrow is ||A||e^(i(θ-pi/
4)).
This means that all we have to do is multiple the
first arrow by e^(iθpi/4). To generalize:
F(f(x-k)) = F(w)*e^(-jwk). We can generalize
because the same logic works for all functions.
Wednesday, December 15, 2010

13. Scalar multiplication
• When you multiply f(x) by a constant k, each arrow in the polar form
grows by the same factor.
• The result is that the amplitude is multiplied by k.
• This works for complex numbers, too. (With complex numbers, though, it
is more appropriate to see it as multiplication by ||Z||e^(iθ) -- i.e. a shift
and a “growth”)
Wednesday, December 15, 2010

14. cos(wx)
• Now lets find F[cos(w0x)]. First, we realize that cos(w0x) = sin(pi/2-
w0x) = -sin(w0x-pi/2) = -sin(w0(x-pi/2w0))
• F[cos(w0x)] is then equal to F[-sin(w0(x-pi/2w0))] =-F[sin(w0(x-pi/2w0))]
= -F[sin(w0x)]e^(-iw*pi/2w0).
• F[sin(w0x)] only “matters” at w = w0 and w = -w0. e^(-iw0*pi/2w0) = -i.
e^(iw0*pi/2w0) = i.
• Plugging this into F(w) = -i*pi[δ(w-w0) - δ(w+w0)], we get that F[cos
(w0x) = pi[δ(w-w0) + δ(w+w0)].
• Now we have the basis for all functions!
Wednesday, December 15, 2010

15. Conclusions
• The Fourier transform is awesome because it breaks a signal down into
its frequencies.
• You can visualize the Fourier transform by graphing the function on the
complex plane in polar form.
• Honestly, you could go on and on with all of the interesting patterns you
find with the Fourier transform: the symmetry, the phase shifts, the
meaning of complex frequencies. However, we’ll end here at this
introduction. Maybe I’ll make a presentation on the LaPlace transform
(and why it solves equations) or on the DFT, DTFT, and the FFT.
• It is 2:20 am and I’m tired, so I’ll stop now. I know there are loose
ends, and I’ll fix them.
Wednesday, December 15, 2010