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254 Chapter 7 ■ Image Restoration

it would be represented in the frequency domain as a sine or cosine function
having a low frequency. Something that changes quickly, such as an edge, will
have high-frequency components.
It is therefore possible to build filters that will remove or enhance certain
frequencies in the image, and this will sometimes have a restorative effect.
Indeed, noise consists of mainly high-frequency information, and so filtering
out of the very high frequencies should have a noise reduction effect. It
unfortunately also has an edge-reduction effect.
There are many other reasons to use a Fourier transform. A convolution can
be carried out directly on an image by moving the convolution matrix (image)
so that it is centered on each pixel of the image in turn, and then multiplying
the corresponding elements and summing the products. This was described
in Equation 2.13, for example. It is a time-consuming process for large images,
and one that can be speeded up greatly by using the Fourier transform.

7.2.1 The Fourier Transform
The Fourier transform breaks up an image (or, in one dimension, a signal) into
a set of sine and cosine components. It is important to keep these components
separate, and so a vector of the form (cosine, sine)isusedateachpoint in the
frequency domain image; that is, the values of the pixels in the frequency
domain image are two component vectors. A convenient way to represent
these is as complex numbers.
Each complex number consists of a real part and an imaginary part, which
can be thought of as a vector. A typical complex number could be written as:
z = (x, jy) = x + jy (EQ 7.1)

where j is the imaginary number −1. The exponential of an imaginary number
can be shown to be the sum of a sine and cosine, which is exactly what we want:

jθ
e = cos θ + jsin θ (EQ 7.2)
This polar form is what will be used from here on, but it is important to
remember that it is really a shorthand for the sum of the sine and cosine parts.
In one dimension, the Fourier transform of a continuous function f(x)is:
∞

F(w) = f(t) e −jwt dt (EQ 7.3)
t = 0
If the function has been sampled so that it is now discrete, the integral
becomes a sum over all the sampled points:
N − 1
2πjwk
F(w) = f(k) e N (EQ 7.4)
k = 0