Spatial domain methods

The value of a pixel with coordinates (x,y) in the enhanced image $\hat{F}$ is the result of performing some operation on the pixels in the neighbourhood of (x,y) in the input image, F.
Neighbourhoods can be any shape, but usually they are rectangular.

Grey scale manipulation

The simplest form of operation is when the operator T only acts on a $1 \times 1$ pixel neighbourhood in the input image, that is $\hat{F}(x,y)$ only depends on the value of F at (x,y). This is a grey scale transformation or mapping.
The simplest case is thresholding where the intensity profile is replaced by a step function, active at a chosen threshold value. In this case any pixel with a grey level below the threshold in the input image gets mapped to 0 in the output image. Other pixels are mapped to 255.
Other grey scale transformations are outlined in figure 1 below.

**Figure 1:** Tone-scale adjustments.
$\begin{figure} \par \centerline{ \psfig {figure=figure51.ps,width=12cm} } \par\end{figure}$

Histogram Equalization

Histogram equalization is a common technique for enhancing the appearance of images. Suppose we have an image which is predominantly dark. Then its histogram would be skewed towards the lower end of the grey scale and all the image detail is compressed into the dark end of the histogram. If we could `stretch out' the grey levels at the dark end to produce a more uniformly distributed histogram then the image would become much clearer.

**Figure 2:** The original image and its histogram, and the equalized versions. Both images are quantized to 64 grey levels.
$\begin{figure} \par \centerline{ \psfig {figure=figure52.ps,width=12cm} } \par\end{figure}$

Histogram equalization involves finding a grey scale transformation function that creates an output image with a uniform histogram (or nearly so). How do we determine this grey scale transformation function? Assume our grey levels are continuous and have been normalized to lie between 0 and 1.
We must find a transformation T that maps grey values r in the input image F to grey values s = T(r) in the transformed image $\hat{F}$ .
It is assumed that

T is single valued and monotonically increasing, and
$0 \leq T(r) \leq 1$ for $0 \leq r \leq 1$ .

The inverse transformation from s to r is given by

r = T^-1(s).

If one takes the histogram for the input image and normalizes it so that the area under the histogram is 1, we have a probability distribution for grey levels in the input image P_r(r).
If we transform the input image to get s = T(r) what is the probability distribution P_s(s) ?
From probability theory it turns out that

$\begin{displaymath} P_{s}(s) = P_{r}(r)\frac{dr}{ds}, \end{displaymath}$

where r = T^-1(s). Consider the transformation

$\begin{displaymath} s = T(r) = \int_{0}^{r}P_{r}(w)dw. \end{displaymath}$

This is the cumulative distribution function of r. Using this definition of T we see that the derivative of s with respect to r is

$\begin{displaymath} \frac{ds}{dr} = P_{r}(r). \end{displaymath}$

Substituting this back into the expression for P_s, we get

$\begin{displaymath} P_{s}(s) = P_{r}(r) \frac{1}{P_{r}(r)} = 1 \end{displaymath}$

for all $s, where 0 \leq s \leq 1$ .Thus, P_s(s) is now a uniform distribution function, which is what we want.

Discrete Formulation

We first need to determine the probability distribution of grey levels in the input image. Now

$\begin{displaymath} P_{r}(r) = \frac{n_{k}}{N} \end{displaymath}$

where n_k is the number of pixels having grey level k, and N is the total number of pixels in the image. The transformation now becomes

Note that $0 \leq r_{k} \leq 1$ , the index $k = 0,1,2,\ldots ,255$ ,and $0 \leq s_{k} \leq 1$ .
The values of s_k will have to be scaled up by 255 and rounded to the nearest integer so that the output values of this transformation will range from 0 to 255. Thus the discretization and rounding of s_k to the nearest integer will mean that the transformed image will not have a perfectly uniform histogram.

Image Smoothing

The aim of image smoothing is to diminish the effects of camera noise, spurious pixel values, missing pixel values etc. There are many different techniques for image smoothing; we will consider neighbourhood averaging and edge-preserving smoothing.

Neighbourhood Averaging

Each point in the smoothed image, $\hat{F}(x,y)$ is obtained from the average pixel value in a neighbourhood of (x,y) in the input image.
For example, if we use a $3 \times 3$ neighbourhood around each pixel we would use the mask

1/9	1/9	1/9
1/9	1/9	1/9
1/9	1/9	1/9

Each pixel value is multiplied by $\frac{1}{9}$ , summed, and then the result placed in the output image. This mask is successively moved across the image until every pixel has been covered. That is, the image is convolved with this smoothing mask (also known as a spatial filter or kernel).
However, one usually expects the value of a pixel to be more closely related to the values of pixels close to it than to those further away. This is because most points in an image are spatially coherent with their neighbours; indeed it is generally only at edge or feature points where this hypothesis is not valid. Accordingly it is usual to weight the pixels near the centre of the mask more strongly than those at the edge.
Some common weighting functions include the rectangular weighting function above (which just takes the average over the window), a triangular weighting function, or a Gaussian.
In practice one doesn't notice much difference between different weighting functions, although Gaussian smoothing is the most commonly used. Gaussian smoothing has the attribute that the frequency components of the image are modified in a smooth manner.
Smoothing reduces or attenuates the higher frequencies in the image. Mask shapes other than the Gaussian can do odd things to the frequency spectrum, but as far as the appearance of the image is concerned we usually don't notice much.

Edge preserving smoothing

Neighbourhood averaging or Gaussian smoothing will tend to blur edges because the high frequencies in the image are attenuated. An alternative approach is to use median filtering. Here we set the grey level to be the median of the pixel values in the neighbourhood of that pixel.
The median m of a set of values is such that half the values in the set are less than m and half are greater. For example, suppose the pixel values in a $3 \times 3$ neighbourhood are (10, 20, 20, 15, 20, 20, 20, 25, 100). If we sort the values we get (10, 15, 20, 20, |20|, 20, 20, 25, 100) and the median here is 20.
The outcome of median filtering is that pixels with outlying values are forced to become more like their neighbours, but at the same time edges are preserved. Of course, median filters are non-linear.
Median filtering is in fact a morphological operation. When we erode an image, pixel values are replaced with the smallest value in the neighbourhood. Dilating an image corresponds to replacing pixel values with the largest value in the neighbourhood. Median filtering replaces pixels with the median value in the neighbourhood. It is the rank of the value of the pixel used in the neighbourhood that determines the type of morphological operation.

**Figure 3:** Image of Genevieve; with salt and pepper noise; the result of averaging; and the result of median filtering.
$\begin{figure} \par \centerline{ \psfig {figure=figure53.ps,width=5in} } \par\end{figure}$

Image sharpening

The main aim in image sharpening is to highlight fine detail in the image, or to enhance detail that has been blurred (perhaps due to noise or other effects, such as motion). With image sharpening, we want to enhance the high-frequency components; this implies a spatial filter shape that has a high positive component at the centre (see figure 4 below).

**Figure 4:** Frequency domain filters (top) and their corresponding spatial domain counterparts (bottom).
$\begin{figure} \par \centerline{ \psfig {figure=figure53a.ps,width=5in} } \par\end{figure}$

A simple spatial filter that achieves image sharpening is given by

-1/9	-1/9	-1/9
-1/9	8/9	-1/9
-1/9	-1/9	-1/9

Since the sum of all the weights is zero, the resulting signal will have a zero DC value (that is, the average signal value, or the coefficient of the zero frequency term in the Fourier expansion). For display purposes, we might want to add an offset to keep the result in the $0 \ldots 255$ range.

High boost filtering

We can think of high pass filtering in terms of subtracting a low pass image from the original image, that is,

High pass = Original - Low pass.

However, in many cases where a high pass image is required, we also want to retain some of the low frequency components to aid in the interpretation of the image. Thus, if we multiply the original image by an amplification factor A before subtracting the low pass image, we will get a high boost or high frequency emphasis filter. Thus,

Now, if A = 1 we have a simple high pass filter. When A > 1 part of the original image is retained in the output.
A simple filter for high boost filtering is given by

-1/9	-1/9	-1/9
-1/9	$\omega$ /9	-1/9
-1/9	-1/9	-1/9

where $\omega = 9A - 1$ .

Tuesday, March 22, 2011

spatial domain method of image registration