Hello, welcome to the video lecture series on digital image processing. For last few
classes, we were discussing about image enhancement techniques and we have completed our discussion
on point processing techniques for image enhancement.
So, what we have done till now is the point processing techniques for image enhancement
and under this, the first operation that we have done is image negatives and there we
have seen that image negative operation is useful incase the image contains information
which are either contained in the gray level or in the white pixels that are embedded in
dark image regions.
So, if you take the negative of such images, then the information content becomes dark
where as background becomes white and visualization of that information in those negative images
is much more easier.
The second operation that we have done is the logarithmic transformation for dynamic
range compression. We have done this logarithmic transformation because we have seen that in
some cases, the dynamic range that is the difference between the minimum intensity value
and the maximum intensity value of an image is so high that a display device is normally
not capable to deal with such a high dynamic range.
So, for such cases, we have to reduce the dynamic range of the image so that the image
can be displayed properly on the given display device and this logarithmic transformation
operation gives such a dynamic range compression operation.
Then the next technique we have talked about is the power-law transformation. In case of
power-law transformation, we have seen that many of the devices like whether the image
printing device or the image display device or the image acquisition device; these different
devices, they themselves introduce some sort of power-law operation on the image that is
to be displayed.
As a result, the image that we want to display or the image that we want to print, they become
distorted. The appearance of the output image is not same as the image that is to be outputted
that is intended to be outputted. So, this power-law which is introduced by those devices
has to be corrected by some power-law operation.
So, we have seen that in case of this power-law compensation or power-law operation, we introduced
a pre processed image having a power-law which is inverse of the power-law that is introduced
by the device and as a result; the pre process image, when it goes to the device, then the
output of the device will be almost similar to the image that is intended.
So, for this kind of operation, to nullify the effect of the device, we go for power-law
compensation technique. Then the next operation that we have done is contrast stretching.
So, in case of contrast stretching, we have seen that in many cases, we can get a very
very dark image because of the fact that may be the scene when the image was taken was
not properly illuminated or the scene was very very poorly illuminated.
The other reason why we can get such a dark image is that while taking the photograph,
the camera lens was not properly set that is the aperture of the lens was not properly
set or we can also get dark image because of the limitation of the sensor itself. The
image sensor, if the dynamic range of the image sensor is a very narrow; in that case,
such a kind of sensor also leads to an image which is a dark image.
So, to enhance such dark images so that the image can be visualized properly, we go for
the contrast enhancement technique. The other kind of image enhancement we have talked about
is the grey level slicing operation and this kind of grey level operation is useful in
cases if the application needs to highlight certain range of grey levels in the image.
So, in such cases, in grey level slicing, we have seen 2 kind of techniques ((05:40))…
region is highlighted whereas the grey levels outside that particular specified range is
suppressed and the second kind of grey level slicing operation that we have seen is the
grey levels within the specified range is highlighted whereas the grey levels outside
the range remains as it is.
So, these are the 2 different types of grey level slicing operations we have talked about
and as I said that if the application demands that the application needs enhancement of
certain range of grey levels, the application is not interested in other grey level values
or other intensity values; in that case, what we go for is the grey level slicing kind of
operation. Then we have talked about other enhancement
techniques where these enhancement techniques are based on histogram based processing operations.
In other point processing techniques, we define a transformation function where the transformation
function simply works on a particular pixel of the input image to generate a processed
pixel of the output image and those transformation functions, they do not take care of or they
do not consider the overall appearance of the image and we have seen that the overall
appearance of the image is actually reflected in what is called the histogram of the image.
So, this histogram based processing techniques, they try to modify or they try to highlight
the overall appearance of the image by modifying the histogram of that particular image and
under this category, we have talked about 2 kinds of histogram based processing techniques;
one was one of them was histogram equalization technique and the second one was histogram
Then we have talked about 2 other kinds of image enhancement operations where these enhancement
operations does not perform on a single image but it performs on multiple number of images.
So, one of them we have talked about was image differencing operation. So, this image differencing
operation, it highlights those regions in the image where there is a difference between
the given 2 images. So, only those regions where the given 2 images are different, those
regions will be highlighted and where the 2 images are similar, those regions will be
The other kind of operation that we have done was image averaging operation and we have
said that this kind of image averaging operation is very very useful where the object which
is imaged that is of very very low intensity. So, for such kind of objects or while imaging
such kind of objects, the image that you get is likely to be dominated by noise.
So, if I get multiple number of frames of such noisy images and if the noise that is
added to the image is a 0 mean noise, then taking average of multiple number of frames
of such noisy images is likely to cancel the noise part and ultimately what comes out after
the averaging operation is the actual image that is desired. So, these are the different
kind of operations, point processing techniques for image enhancement that we have done till
our last class.
Now, in today’s class we will talk about another special domain technique which is
called mask processing technique. The previous lectures also we were dealing with the special
domain techniques and we have said that image enhancement techniques can broadly be categories
into special domain techniques and frequency domain techniques. The frequency domain techniques
we will talk about later on.
So, in today’s class, we will talk about another special domain techniques which are
known as mask processing techniques and other under this, we will discuss 3 different types
of operations. The first one is the linear smoothing operation; the second one is a nonlinear
operation which is based on the statistical features of the image which is known as the
median filtering operation and the third kind of mask processing technique that we will
talk about is the sharpening filter. Now, let us see what this mask processing
Now, in our earlier discussions we have mentioned that while going for this contrast enhancement,
what we basically do is given an input image say f (x, y), we transform this input image
by a transformation operator say T which gives us an output image g (x, y) and what will
be the nature of this output image g (x, y) that depends upon what is this transformation
In point processing technique, we have said that this transformation operation T that
operates on a single pixel in the image. That is it operates on a single pixel intensity
value. But as we earlier said that T is an operator which operates on a neighborhood
of the pixels at location (x, y); so for point processing operation, the neighborhood size
was 1 by 1. So, if we consider a neighborhood of size more than 1 that is we can consider
a neighborhood of size say 3 by 3, we may consider a neighborhood of size say 5 by 5,
we may consider a neighborhood of size 7 by 7 and so on.
So, if we consider a neighborhood of size more than 1, then the kind of operation that
we are going to have that is known as mask processing operation. So, let us see what
does this mask processing operation actually mean.
Here, we have shown a 3 by 3 neighborhood around a pixel location (x, y). So, this outer
rectangle represents a particular image and in the middle of this, we have shown a 3 by
3 neighborhood and this 3 by 3 neighborhood is taken around a pixel at location (x, y).
By mask processing what we mean is; so if I consider a neighborhood of size 3 by 3,
I also consider a mask of size 3 by 3. So, we find that here on the right hand side,
we have shown a mask. So, this is a given mask of size 3 by 3 and these different elements
in the mask that is W minus 1, minus 1 W minus 1, 0 W minus 1, 1 W 0, minus 1 and so on upto
W 1, 1; these elements represent the coefficients of this mask.
So, for all these mask processing techniques what we do is we place this mask on this image
where the mask center coincides with the pixel location (x, y). Once you place this mask
on this particular image, then you multiply every coefficient of this mask by the corresponding
pixel on the image and then you take the sum of all these products.
So, the sum of all these products is given by this particular expression and whatever
sum you get that is placed at location (x, y) in the image g(x, y). So, by mask processing
operation, this is the mathematical expression we get that g (x, y) equal to W ij into f
(x plus I, y plus j). You have to take the summation of this product over j varying from
minus 1 to 1 and i varying from minus 1 to 1.
So, this is the operation that has to be done for a 3 by 3 neighborhood in which case we
get a mask again of size 3 by 3. Of course, as we said that we can have masks of higher
dimension; we can have a mask of 5 by 5, if I consider a 5 by 5 neighborhood. I have to
consider a mask of size 7 by 7, if I consider a 7 by 7 neighborhood and so on.
So, if this particular operation is done at every pixel location (x, y) in the image,
then the output image g (x, y) for various values of x and y that we get is the processed
image g. So, this is what we mean by mask processing operation.
Now, the first of the mask processing operation that we will consider is the image averaging
or image smoothing operation. So, image smoothing is a special filtering operation where the
value at a particular location (x, y) in the processed image is the average of all the
pixel values in the neighborhood of x and y. So, because it is the average, this is
also known as averaging filter and later on we will see that this averaging filter is
nothing but a low pass filter. So, when we have such averaging filter, the corresponding
mask can be represented in this form.
So, again here we are showing a mask, a 3 by 3 mask and here we find that all the coefficients
in this 3 by 3 mask are equal to 1 and by going back to our mathematical expression,
I get an expression of this form that is g (x, y) equal to 1 upon 9 into f (x plus I,
y plus j). Take the summation over j equal to minus 1 to 1 and i equal to minus 1 to
So naturally, as this expression says you find that what we are doing; we are taking
the summation of all the pixels in the 3 by 3 neighborhood of the pixel location (x, y)
and then dividing these summation by 9. So, which is nothing but average of all the pixel
values in the neighborhood of (x, y) in the 3 by 3 neighborhood of (x, y) including the
pixel at location (x, y) and this average is placed at location (x, y) in the processed
So, this is what is known as averaging filter and also this is called a smoothing filter
and the filter could and the particular mask for which all the filter coefficients or mask
coefficients are same or equal to 1 in this particular case, this is known as a box filter.
So, this particular filtering operation that we are getting, this particular mask is known
as a box filter.
Now, when we perform this kind of operation, then naturally because we are going for averaging
of all the pixels in the neighborhood; so the output image is likely to be a smoothed
image that means it will have a blurring effect, all the sharp transitions in the images will
be removed and they will be replaced by a blurred image.
As a result, if there is any sharp edge in the image; the sharp images, the sharp edges
will also be blurred. So, to avoid the effect of blurring, there is another kind of mask;
averaging mask or smoothing mask which performs weighted average.
So, such a kind of mask is given by this particular mask. So, here you find that in this mask,
the center coefficient is equal to 4. The coefficients vertically up, vertically down
or horizontally left, horizontally right; they are equal to 2 and all the diagonal elements
of the center elements in this mask are equal to 1. So effectively, what we are doing is
when we are taking the average, we are weighting every pixel in the neighborhood by the corresponding
coefficients and what we get is a weighted average.
So, the center pixel that is the pixel at location (x, y) gets the maximum weightage
and as you move away from the pixel locations, from the center location, the weightage of
the pixels are reduced. So, when we apply this kind of mask, then our general expression
of this mask operation that is valid which becomes W ij f (x minus i y minus f (x plus
I, y plus j). Take the summation from j equal to minus 1 to 1 and i equal to minus 1 to
1 and take 1 upon 16 of this and that will give the value which is to be placed at location
(x, y) in the processed image g. So, this becomes the expression of g(x, y).
Now, the purpose of going for this kind of weighed averaging is that because here we
are weighting the different pixels in the image for taking the average, the blurring
effect will be reduced in this particular case. So in case of box filter, the image
will be very very blurred and of course the blurring will be more if I go for bigger and
bigger neighborhood size or bigger and bigger mask size. When we go for averaging, weighted
averaging; in such cases, the blurring effect will be reduced. Now, let us say what kind
of result we get.
So, this gives the general expression that when we will consider W ij, we have to have
a normalization factor that is this summation has to be divided by sum of the coefficients
and as we said that 3 by 3 neighborhood is only a special case, I can have neighborhoods
of other sizes; so here it shows that we can have a neighborhood of size M by N where M
equal to 2a plus 1 and N equal to 2b plus 1 where a and b are some positive integers
and obviously here, you show the here it is shown that the mask is usually of odd dimension,
it is not even dimension and that is normally the mask of odd dimension which is normally
used in case of image processing.
Now, using this kind of mask operation, here we have shown some results. You find that
the top left image is noise image. When you do the masking operation or averaging operation
on this noisy image, the right top image shows the averaging with a mask of size 3 by 3,
the left bottom image is obtained using a mask of size 5 by 5 and the right bottom image
is obtained using a mask of size 7 by 7.
So, from these images, it is quite obvious that as I increase the size of the mask, the
blurring effect becomes more and more. So, we find that the right bottom image which
is obtained by a mask of size 7 by 7 is much more blurred compared to the other 2 images
and this effect is more prominent if you look at the edge regions of this images.
Say, if I compare this particular region with the similar region in the upper image or the
similar regions in the original image. You find that here, in the original image that
is very very sharp whereas when I do the smoothing using a 7 by 7 mask, it becomes very blurred
whereas the blurring effect when I use the 3 by 3 mask is much less. Similar such result
is obtained with other images also.
So, here is another image. Again, we do the masking operation or the smoothing operation
with different mask sizes. On the top left, we have an original noisy image and the other
images are the smoothed images using various mask sizes. So, on the right top, this is
obtained using a mask of size 3 by 3, the left bottom is an image obtained using a mask
of size 5 by 5 and the right bottom is an image obtained using a mask of size 7 by 7.
So, we find that as we increase the mask size, the reduction is in noise or the noise is
reduced to a greater extend but at the cost of addition of blurring effect. So, though
the noise is reduced but image becomes very blurred. So, that is the effect of using the
box filters or the smoothing filters that though the noise will be removed but the images
will be blurred or the sharp contrast in that image will be reduced.
So, there is a second kind of image, second kind of masking operations which are based
on order statistics which will reduce this kind of blurring effects. So, let us consider
one such filter based on order statistics.
So, these kind of filters unlike in case of the earlier filters; these filters are nonlinear
filters. So, here in case of this order statistic filters; the response is based on the ordering
of the intensities, ordering of the pixel values in the neighborhood of the point under
consideration. So, what we do is we take the set of intensity values which is in the neighborhood
which are in the neighborhood of the point (x, y), then order all those intensity values
in a particular order and based on this ordering, you select a value which will be put at location
(x, y) in the processed image g and that is how the output image that you get is a processed
But here the processing is done using the order statistics filter. A widely used filter
under this order statistics is what is known as a median filter. So in case of a median
filter, what we have to do is I have an image and what I do is around point (x, y), I take
a 3 by 3 neighborhood and consider all the 9 pixels, intensity values of all the 9 pixels
in this 3 by 3 neighborhood.
Then, I arrange this pixel values, the pixel intensity values in a certain order and take
the median of this pixel intensity values. Now, how do you define the median? We define
the median say zeta of a set of values such that half of the values in the set will be
less than or equal to zeta and the remaining half of the values of the set will be greater
than or equal to zeta.
So, let us take a particular example. Suppose, I take a 3 by 3 neighborhood around a pixel
location (x, y) and the intensity values in this 3 by 3 neighborhood, let us assume that
this is 100, this is a 85, this is a 98, this may have a value 99, this may have a value
say 105, this may have a value say 102, this may have a value say 90, this may have a value
say 101, this may have a value say 108 and suppose this represents a part of my image
say f (x, y).
Now, what I do is I take all these pixel values, all this intensity values and put them in
ascending order of magnitude. So, if I put them in ascending order of magnitude, you
find that the minimum of these values is 85, the next value say 90, and next one is 98,
the next one is 99, the next one is 101, the next one is 102, the next one is 105 and the
next one is 108. So, all these 9 intensity values, I have put in ascending order of magnitude.
Here there will be one more, so there is one more value - 100. So, these are the 9 intensity
values which are put in the ascending order of magnitude. So once I put them into ascending
order of magnitude, from this I take the fifth maximum value which is equal to 100.
So, if I take the fifth maximum value, you find that there will be equal number of values
which is greater than this fifth value; greater than or equal to this fifth number and there
will be same number of values which will be less than or equal to this fifth number. So,
I consider this particular pixel value 100 and when I generate the image g (x, y), in
g (x, y) at location (x, y), I put this value 100 which is the median of the pixel values
within this neighborhood. So, this gives my processed image g (x, y).
Of course, the intensities in other locations in other pixel regions will be decided by
the median value of the neighborhood of the corresponding pixels. That is if I want to
find out what will be the pixel value at this location, then the neighborhood that I have
to consider will be this particular neighborhood.
So, this is how I can get the median filtered output and as you can see that this kind for
filtering operation is based on statistics. Now, let us see that what kind of result that
we can have using this median filter.
So here, you find that it is again on the same building image. The left top is our original
noised image; on the right hand side, it is the smoothed image using box filter and on
the bottom, we have the image using this median filter.
So here again, as you see that the image obtained using the processed image obtained using median
filter operation, maintains the sharpness of the image to a greater extend then that
obtained using the smoothing images.
Coming to the second one, again this is one of the images that we have shown earlier a
noise image having 4 coins. Here again, you find that after doing the smoothing operation,
the edges becomes blurred and at the same time, the noises are not reduced to a great
extends. Still this particular image is noisy.
So if I want to remove all these noise, what I have to do is I have to smooth this images
using higher neighborhood size and the moment I will go for the larger neighborhood size,
the blurring effect will be more and more. On the right hand side, the image that we
have; so this particular image is also the processed image but here the filtering operation
which is done is median filter.
So here, we find that because of the median filtering operation; the output image that
we get, the noise in this output image is almost vanished but at the same time the contrast
of the image or the sharpness of the image remains more or less interact. So, this is
an advantage that you get if we go for median filtering rather than smooth smoothing filtering
or averaging filtering. To show the advantage of this median filtering, we will take another
So, this is the image of the butterfly, a noisy image of a butterfly. On bottom left,
the image that is shown, this is an averaged image where the averaging is done over a neighborhood
of size 5 by 5. On the bottom right is the image which is filtered by using median filter.
So, this particular image clearly shows, this result clearly shows the superiority of the
median filtering over the smoothing operation or averaging operation and such median filtering
is very very useful for a particular kind of noise where the noise is a random noise
which are known as salt and pepper noise because of the appearance of the noise in the image.
So, these are the different filtering operations which reduces the noise in the particular
image or the filtering operations which introduce blurring or smoothing over the image. We will
now consider another kind of spatial filters which increases the sharpness of the image.
So, the spatial filter that we will consider now is called sharpening spatial filter.
So, we will consider sharpening spatial filter. So, the objective of this sharpening spatial
filter is to highlight the details, the intensity details or variation details in an image.
Now, through our earlier discussion, we have seen that if I do averaging over an image
or smoothing over an image, then the image becomes blurred or the details in the image
are removed. Now, this averaging operation is equivalent to integration operation.
So, if I integrate the image, then what I do is what I am going to get is a blurring
effect or a smoothing effect of the image. So, if integration gives a smoothing effect,
so it is quite logical to think that if I do the opposite operation that is instead
of integration, if I do differentiation operation, then the sharpness of the image is likely
to be increased. So, it is the derivative operations or the differentiations which are
used to increase the sharpness of an image.
Now, when I go for the derivative operations, I can use 2 types to derivatives. I can use
the first order derivative or I can also use the second order derivative. So, I can either
use the first order derivative operation or I can use the second order derivative operation
to obtain or to enhance the sharpness of the image. Now, let us see what are the desirable
effects that these derivative operations are going to give.
If I use a first order derivative operation or a first order derivative filter, then the
desirable effect of this first order derivative filter is it must be 0, the response must
be 0 in areas of constant grey level in the image and the response must be non zero at
the onset of the grey level step or or at the onset of a grey level ramp and it should
be non zero along ramps. Whereas, if I use a second order derivative filter; then the
second order derivative filter response should be 0 in the flat areas, it should be non zero
at the onset and end of a grey level step or grey level ramp and it should be 0 along
ramps of constant slope. So, these are the desirable features or the desirable responses
of a first order derivative filter and the desirable response of a second order derivative
Now, whichever derivative filter I use; whether it is a first order derivative filter or a
second order derivative filter, I have to look for discrete domain formulation of those
derivative operations. So, let us see how we can formulate the derivative operations;
the first order derivative or the second order derivative in discrete domain.
Now, we know that in continuous domain the derivative is given by let us consider a 1
dimensional case that is if I have a function f (x) which is a function of variable x, then
I can have the derivative of this which is given by df (x)/ dx which is given by limit
delta x tends to zero f (x plus delta x) minus f (x) upon delta x. So, this is the definition
of derivative in continuous domain.
Now, when I come to discrete domain; in case of our digital images, the digital images
are represented by a discrete set of points or pixels which are represented at different
grid locations and the minimum distance between 2 pixels is equal to 1.
So, in our case, we will consider the value of delta x equal to 1 and this derivative
operation in case of 1 dimension, now reduces to del f del x is equal to f of x plus 1 minus
f of x. Now, here I use the partial derivative notation because our image is a 2 dimensional
image. So, when I take the derivative in 2 dimensions, we will have partial derivatives
along x and we will have partial derivatives along y. So, the first derivative, the first
order derivative for 1 dimensional discrete signal is given by this particular expression.
Similarly, the second order derivative of a discrete signal in1dimension can be approximated
by del 2 f upon del x 2 which is given by f (x plus 1) plus f (x minus1) minus 2 f (x).
So, this is the first order derivative and this is the second order derivative and you
find that these 2 derivations, these 2 definitions of the derivative operations, they satisfy
the desirable properties that we have discussed earlier. Now, to illustrate the response of
these derivative operations, let us take an example.
So, this is a 1 dimensional signal where the values of the 1 dimensional signals for various
values of x are given in the form of an array like this and the plot of these functional
values, these discrete values are given on the top. Now, if you take the first order
derivative of this as we have just defined, the first order derivative is given in the
second array and the second order derivative is given in the third array.
So, if you look at this functional value, the plot of this functional value, this represents
various regions, Say for example; here, this part is a flat region, this particular portion
is a flat region, this is also a flat region, this is also a flat region. This is a ramp
region, this represents an isolated point, this area represents a very thin line and
here we have a step kind of discontinuity.
So now, if you compare the response of the first order derivative and the second order
derivative of this particular discrete function; you find that the first order derivative is
non zero during ramp, whereas the first order derivative is 0 along a ramp. The second order
derivative is 0 along a ramp; the second order derivative is non zero at the onset and end
of the ramp.
Similarly coming to this isolated point, if I compare the response of the first order
derivative and the response of the second order derivative, you find that the response
of the second order derivative for an isolated point is much stronger than the response of
the first order derivative.
Similar is the case for a thin line. The response of the second order derivative is greater
than the response of the first order derivative. Coming to this step edge, the response of
the first order derivative and response of the second order derivative is almost same
but the difference is in case of second order derivative, I have a transition from a positive
polarity to a negative polarity.
Now, because of this transition from positive polarity to negative polarity, the second
order derivatives normally leads to double lines the moment in case of a step discontinuity
in a image whereas, the first order derivative that leads to a single line. Of course, this
double line, getting this double line; usefulness of this, we will discuss later.
Now, but as we have seen, the first order the second order derivative gives a stronger
response to isolated points or to thin lines and because the details in an image normally
has the property that it will be either isolated points or thin lines to which the second order
gives second order derivative gives a stronger response; so, it is quite natural to think
that the second order derivative based operator will be most suitable for image enhancement
So, our observation is as we have discussed previously that first order derivative generally
produce a thicker edge because we have seen that during a ramp or along a ramp, the first
order derivative is non zero whereas, the second order derivative along a ramp is 0
but it gives non zero values at the starting of the ramp and the end of the ramp.
So, that is why the first order derivatives generally produce a thicker edge in an image.
The second order derivatives gives stronger response to find it is such as thin lines
and isolated points. The first order derivative have stronger response to gray level step
and the second order derivative produce a double response at step edges and as we have
already said that as the details in the image are either in the form of isolated points
or thin lines; so the second order derivatives are better suited for image enhancement operations.
So, we will mainly discuss about the second order derivatives for image enhancement. But
to use this for image enhancement operation; obviously, because our images are digital
and as we have said in many times that we have to have a discrete formulation of this
second order derivative operations and the filter that will design that should be isotropic.
That means the response of the second order derivative filter should be independent of
the orientation of the discontinuity in the image and the most widely used or the popularly
known second order derivative operator of isotropic nature is what is known as a Laplacian
So, we will discuss about the Laplacian operator and as we know that the Laplacian of a function
is given by del square f equal to del 2 f del x 2 plus del 2 f del y 2. So, this is
the laplacian operator in continuous domain but what we have to have is the laplacian
operator in the discrete domain. And, as we have already seen that del 2 f
del x 2 in case of discrete domain is approximated as f (x minus1) or f (x plus1) plus f (x minus1)
minus 2 f (x).
So, this is in case of a 1 dimensional signal. In our case, our function is a 2 dimensional
function that is the function of variables x and y. So for this, we can write for 2 dimensional
signal del 2 f del x 2 which will be a simply f of (x plus1, y) plus f of (x minus1, y)
minus 2 f (x, y). Similarly, del 2 f del y 2 will be given by f of (x, y plus1) plus
f of (x, y minus 1) minus 2 f (x, y).
And, if I add these 2, I get the Laplacian operator in discrete domain which is given
by del 2 f is equal to del 2 f del x 2 plus del 2 f del y 2 and we will find that this
will be given as f (x plus 1, y) plus f (x minus 1, y) plus f (x, y plus 1) plus f (x,
y minus 1) minus 4 f (x, y) and this particular operation can be represented again in the
form of a 2 dimensional mask. That is for this Laplacian operator, we can have a 2 dimensional
mask and the 2 dimensional mask in this particular case will be given like this.
So, on the left hand side, the mask that is shown, this mask considers the Laplacian operation
only in the vertical direction and in the horizontal direction and if we also include
the diagonal directions, then the Laplacian mask is given on the right hand side. So,
we find that using this particular mask which is shown on the left hand side, I can always
derive the expression that we have just shown.
Now, here I can have 2 different types of mask. Depending upon polarity of the coefficient
at the center pixel, I can have the center pixel to have a polarity either negative or
positive. So, if the polarity is positive, same of the center coefficient; then I can
have a mask of this form where the center pixel will have a positive polarity but otherwise
the nature of the mask remains the same.
Now, if I have these kinds of operation, then you find that the image that you get that
will have that will just highlight the discontinuous regions in the image whereas, all the smooth
regions in the image will be suppressed. So, this shows an original image.
On the right hand side, we have the output of the laplacian and if you closely look at
this particular image, you will find that all the discontinuous regions will have some
value. However this particular image cannot be displayed properly.
So, we have to have some scaling operation because I will have say gamma H greater than
1 and gamma L less than 1. This will amplify all the high frequency components that is
the contribution of the reflectance and it will attenuate the low frequency components
that is contribution due to the illumination. Now, using this time of type of filtering,
the kind of result that we get is something like this.
Here, on the left hand side is the original image and on the right hand side is the enhanced
image and if you look in the boxes, you find that
many of the details in the boxes which are not available in the original image is now
available in the enhanced image. So, using such homomorphic filtering, we can even go
for this kind of enhancement or the illumination, the contribution due to illumination will
be reduced; so even in the dark areas, we can take out the details.
So with this, we come to an end to our discussion on image enhancements. Now, let us go to some
questions of our today’s lecture.
The first question is a digital image contains an unwanted region of size 7 pixels. What
should be the smoothing mask size to remove this region? Why Laplacian operator is normally
used for image sharpening operation? Third question - what is unsharp masking? Fourth
question - give a 3 by 3 mask for performing unsharp masking in a single pass through an
image. Fifth, state some applications of first derivative in image processing.
Then, what is ringing? Why ideal low pass and high pass filters lead to ringing effects?
How does blurring vary with cut off frequency? Does Gaussian filter lead to ringing effect?
Give the transfer function filter and what is the principle of homomorphic filter?