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Chapter 2 ■ Edge-Detection Techniques 43

Canny assumed a step edge subject to white Gaussian noise. The edge
detector was assumed to be a convolution filter f which would smooth the
noise and locate the edge. The problem is to identify the one filter that
optimizes the three edge-detection criteria.
In one dimension, the response of the filter f to an edge G is given by a
convolution integral:
W

H = G(−x)f(x)dx (EQ 2.17)
−W
The filter is assumed to be zero outside of the region [−W, W]. Mathemati-
cally, the three criteria are expressed as:
0

A f(x)dx
−W
SNR = w (EQ 2.18)
2
n 0 f (x)dx
−W
A|f(0)|
Localization = W (EQ 2.19)
2
n 0 f dx
−W

 ∞ f (x)dx  1 2

2
x zc = π  −∞  (EQ 2.20)

f (x)dx
 ∞ 
2
−∞
The value of SNR is the output signal-to-noise ratio (error rate), and should
be as large as possible: we need a lot of signal and little noise. The localization
value represents the reciprocal of the distance of the located edge from the true
edge, and should also be as large as possible, which means that the distance
wouldbeassmall as possible. The value x zc is a constraint; it represents the
mean distance between zero crossings of f, and is essentially a statement that
the edge detector f will not have too many responses to the same edge in a
small region.
Canny attempts to find the filter f that maximizes the product SNR ∗ local-
ization subject to the multiple response constraint. Although the result is too
complex to be solved analytically, an efficient approximation turns out to be
the first derivative of a Gaussian function. Recall that a Gaussian has the form:

x 2
−
G(x) = e 2σ 2 (EQ 2.21)