Digital Image Processing

Digital Image Processing
Image Enhancement 12/5/2018 Duong Anh Duc - Digital Image Processing

Image Enhancement To process an image so that output is “visually better” than the input, for a specific application. Enhancement is therefore, very much dependent on the particular problem/image at hand. Enhancement can be done in either: Spatial domain: operate on the original image g(m,n) = T[f(m,n)] Frequency domain: operate on the DFT of the original image G(u,v) = T[F(u,v)], where F(u,v) = F[f(m,n)], and G(u,v) = F [g(m,n)], 12/5/2018 Duong Anh Duc - Digital Image Processing

Duong Anh Duc - Digital Image Processing
Image Negative Contrast Stretching Compression of dynamic range Graylevel slicing Image Subtraction Image Averaging Histogram operations Smoothing operations Median Filtering Sharpening operations Derivative operations Histogram operations Low pass Filtering Hi pass Filtering Band pass Filtering Homomorphic Filtering Histogram operations False Coloring Full color Processing 12/5/2018 Duong Anh Duc - Digital Image Processing

Point Operations Output pixel value g(m, n) at pixel (m, n) depends only on the input pixel value at f(m, n) at (m, n) (and not on the neighboring pixel values). We normally write s = T(r), where s is the output pixel value and r is the input pixel value. T is any increasing function that maps [0,1] into [0,1]. 12/5/2018 Duong Anh Duc - Digital Image Processing

Image Negative T(r) = s = L-1-r, L: max grayvalue 12/5/2018
Duong Anh Duc - Digital Image Processing

Negative Image 12/5/2018 Duong Anh Duc - Digital Image Processing

Contrast Stretching Increase the dynamic range of grayvalues in the input image. Suppose you are interested in stretching the input intensity values in the interval [r1, r2]: Note that (r1- r2) < (s1- s2). The grayvalues in the range [r1, r2] is stretched into the range [s1, s2]. 12/5/2018 Duong Anh Duc - Digital Image Processing

Contrast Stretching Special cases: Thresholding or binarization r1 = r2 , s1 = 0 and s2 = 1 Useful when we are only interested in the shape of the objects and on on their actual grayvalues. 12/5/2018 Duong Anh Duc - Digital Image Processing

Contrast Stretching 12/5/2018 Duong Anh Duc - Digital Image Processing

Contrast Stretching Special cases (cont.): Gamma correction: S1 = 0, S2 = 1 and 12/5/2018 Duong Anh Duc - Digital Image Processing

Contrast Stretching Gamma correction 12/5/2018 Duong Anh Duc - Digital Image Processing

Compression of Dynamic Range
When the dynamic range of the input grayvalues is large compared to that of the display, we need to “compress” the grayvalue range --- example: Fourier transform magnitude. Typically we use a log scale. s = T(r) = c log(1+r) 12/5/2018 Duong Anh Duc - Digital Image Processing

Saturn Image Mag. Spectrum Mag. Spectrum in log scale 12/5/2018 Duong Anh Duc - Digital Image Processing

Graylevel Slicing: Highlight a specific range of grayvalues. 12/5/2018 Duong Anh Duc - Digital Image Processing

Example: Highlighted Image (no background) Original Image Highlighted Image (with background) 12/5/2018 Duong Anh Duc - Digital Image Processing

Bitplane Slicing: Display the different bits as individual binary images. 12/5/2018 Duong Anh Duc - Digital Image Processing

12/5/2018 Duong Anh Duc - Digital Image Processing

Image Subtraction In this case, the difference between two “similar” images is computed to highlight or enhance the differences between them: g(m,n) = f1(m,n)-f2(m,n) It has applications in image segmentation and enhancement 12/5/2018 Duong Anh Duc - Digital Image Processing

Example: Mask mode radiography
f1(m, n): Image before dye injection f2(m, n): Image after dye injection g(m, n): Image after dye injection, followed by subtraction 12/5/2018 Duong Anh Duc - Digital Image Processing

Image Averaging for noise reduction
Noise is any random (unpredictable) phenomenon that contaminates an image. Noise is inherent in most practical systems: Image acquisition Image transmission Image recording Noise is typically modeled as an additive process: g(m,n) = f(m,n) + (m,n) Noisy Image Noise-free Image Noise 12/5/2018 Duong Anh Duc - Digital Image Processing

The noise h (m, n) at each pixel (m, n) is modeled as a random variable. Usually, h (m, n) has zero-mean and the noise values at different pixels are uncorrelated. Suppose we have M observations {gi(m, n)}, i=1, 2, …, M, we can (partially) mitigate the effect of noise by “averaging” 12/5/2018 Duong Anh Duc - Digital Image Processing

In this case, we can show that: Therefore, as the number of observations increases (M  ), the effect of noise tends to zero. 12/5/2018 Duong Anh Duc - Digital Image Processing

Image Averaging Example
Noisy Image Noise Variance = 0.05 Noise-free Image 12/5/2018 Duong Anh Duc - Digital Image Processing

Image Averaging Example

Some Averaging Filters

Histograms The histogram of a digital image with grayvalues r0, r1, …, rL-1 is the discrete function The function p(rk) represents the fraction of the total number of pixels with grayvalue rk. Histogram provides a global description of the appearance of the image. 12/5/2018 Duong Anh Duc - Digital Image Processing

Histograms If we consider the grayvalues in the image as realizations of a random variable R, with some probability density, histogram provides an approximation to this probability density. In other words, Pr[R=rk]  p(rk) 12/5/2018 Duong Anh Duc - Digital Image Processing

Some Typical Histograms
The shape of a histogram provides useful information for contrast enhancement. 12/5/2018 Duong Anh Duc - Digital Image Processing

Some Typical Histograms (cont.)
The shape of a histogram provides useful information for contrast enhancement. 12/5/2018 Duong Anh Duc - Digital Image Processing

Example: Histogram Stretching

Histogram Equalization
Idea: find a non-linear transformation g = T(f ) to be applied to each pixel of the input image f(x,y), such that a uniform distribution of gray levels in the entire range results for the output image g(x,y). 12/5/2018 Duong Anh Duc - Digital Image Processing

Let us assume for the moment that the input image to be enhanced has continuous grayvalues, with r = 0 representing black and r = 1 representing white. We need to design a grayvalue transformation s = T(r), based on the histogram of the input image, which will enhance the image. As before, we assume that: T(r) is a monotonically increasing function for 0r1 (preserves order from black to white) T(r) maps [0,1] into [0,1] (preserves the range of allowed grayvalues). 12/5/2018 Duong Anh Duc - Digital Image Processing

Let us denote the inverse transformation by r = T-1(s). We assume that the inverse transformation also satisfies the above two conditions. We consider the grayvalues in the input image and output image as random variables in the interval [0, 1]. Let pin(r) and pout(s) denote the probability density of the grayvalues in the input and output images. 12/5/2018 Duong Anh Duc - Digital Image Processing

If pin(r) and T(r) are known, and T-1(s) satisfies condition 1, we can write (result from probability theory): One way to enhance the image is to design a transformation T(.) such that the grayvalues in the output is uniformly distributed in [0, 1], i.e. pout(s)=1, 0 s1 12/5/2018 Duong Anh Duc - Digital Image Processing

In terms of histograms, the output image will have all grayvalues in “equal proportion.” This technique is called histogram equalization. Consider the transformation 12/5/2018 Duong Anh Duc - Digital Image Processing

Note that this is the cumulative distribution function (CDF) of pin(r) and satisfies the previous two conditions. From the previous equation and using the fundamental theorem of calculus, 12/5/2018 Duong Anh Duc - Digital Image Processing

Therefore, the output histogram is given by The output probability density function is uniform, regardless of the input. 12/5/2018 Duong Anh Duc - Digital Image Processing

Thus, using a transformation function equal to the CDF of input grayvalues r, we can obtain an image with uniform grayvalues. This usually results in an enhanced image, with an increase in the dynamic range of pixel values. 12/5/2018 Duong Anh Duc - Digital Image Processing

Example: Histogram Equalization
Original image Pout after histogram equalization 12/5/2018 Duong Anh Duc - Digital Image Processing

Example: Histogram Equalization

For images with discrete grayvalues, we have L: Total number of graylevels nk: Number of pixels with grayvalue rk n: Total number of pixels in the image 12/5/2018 Duong Anh Duc - Digital Image Processing

The discrete version of the previous transformation based on CDF is given by: 12/5/2018 Duong Anh Duc - Digital Image Processing

Example Consider an 8-level 64 x 64 image with grayvalues (0, 1, …, 7). The normalized grayvalues are (0, 1/7, 2/7, …, 1). The normalized histogram is given in the table: k rk nk p(rk)=nk/n 790 0.19 1 1/7 1023 0.25 2 2/7 850 0.21 3 3/7 656 0.16 4 4/7 329 0.08 5 5/7 245 0.06 6 6/7 122 0.03 7 81 0.02 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) Applying the previous transformation, we have (after rounding off to nearest graylevel): Notice that there are only five distinct graylevels --- (1/7, 3/7, 5/7, 6/7, 1) in the output image. We will relabel them as (s0, s1, …, s4). 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) With this transformation, the output image will have histogram k sk nk p(sk)=nk/n 1/7 790 0.19 1 3/7 1023 0.25 2 5/7 850 0.21 3 6/7 985 0.24 4 448 0.11 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) Note that the histogram of output image is only approximately, and not exactly, uniform. This should not be surprising, since there is no result that claims uniformity in the discrete case. 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) Original image Equalized image 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.) Histogram of original image
Histogram of equalized image 12/5/2018 Duong Anh Duc - Digital Image Processing

Example(cont.)-Histograms
Histogram of original image Histogram of equalized image 12/5/2018 Duong Anh Duc - Digital Image Processing

Histogram equalization may not always produce desirable results, particularly if the given histogram is very narrow. It can produce false edges and regions. It can also increase image “graininess” and “patchiness.” 12/5/2018 Duong Anh Duc - Digital Image Processing

Histogram Specification
Histogram equalization yields an image whose pixels are (in theory) uniformly distributed among all graylevels. Sometimes, this may not be desirable. Instead, we may want a transformation that yields an output image with a prespecified histogram. This technique is called histogram specification. Again, we will assume, for the moment, continuous grayvalues. 12/5/2018 Duong Anh Duc - Digital Image Processing

Suppose, the input image has probability density pin(r). We want to find a transformation z = H(r), such that the probability density of the new image obtained by this transformation is pout(z), which is not necessarily uniform. First apply the transformation This gives an image with a uniform probability density. 12/5/2018 Duong Anh Duc - Digital Image Processing

If the desired output image were available, then the following transformation would generate an image with uniform density: From the grayvalues  we can obtain the grayvalues z by using the inverse transformation, z = G-1(). 12/5/2018 Duong Anh Duc - Digital Image Processing

If instead of using the grayvalues n obtained from (**), we use the grayvalues s obtained from (*) above (both are uniformly distributed!), then the point transformation z = H(r) = G-1[T(r)] will generate an image with the specified density pout(z), from an input image with density pin(r) For discrete graylevels, we have 12/5/2018 Duong Anh Duc - Digital Image Processing

If the transformation zk  G(zk) is one-to-one, the inverse transformation sk  G-1(sk), can be easily determined, since we are dealing with a small set of discrete grayvalues. In practice, this is not usually the case (i.e., zk  G(zk) is not one-to-one) and we assign grayvalues to match the given histogram, as closely as possible. 12/5/2018 Duong Anh Duc - Digital Image Processing

Ex.: Histogram Specification
Consider the previous 8-graylevel 64 x 64 image histogram: k rk nk p(rk)=nk/n 790 0.19 1 1/7 1023 0.25 2 2/7 850 0.21 3 3/7 656 0.16 4 4/7 329 0.08 5 5/7 245 0.06 6 6/7 122 0.03 7 81 0.02 12/5/2018 Duong Anh Duc - Digital Image Processing

It is desired to transform this image into a new image, using a transformation z=H(r)= G-1[T(r)], with histogram as specified below: k zk pout(zk) 0.00 1 1/7 2 2/7 3 3/7 0.15 4 4/7 0.20 5 5/7 0.30 6 6/7 7 12/5/2018 Duong Anh Duc - Digital Image Processing

The transformation T(r) was obtained earlier (reproduced below): risk nk p(sk) r0  s0 = 1/7 790 0.19 r1  s1 = 3/7 1023 0.25 r2  s2 = 5/7 850 0.21 r3,r4  s3 = 6/7 985 0.24 r5, r6,r7  s4=1 448 0.11 12/5/2018 Duong Anh Duc - Digital Image Processing

Next we compute the transformation G as before 12/5/2018 Duong Anh Duc - Digital Image Processing

Notice that G is not invertible. But we will do the best possible by setting G-1(0) = ? (This does not matter since s0) G-1(1/7) = 3/7 G-1(2/7) = 4/7 (This does not matter since s2/7) G-1(3/7) = 4/7 (This is not defined, but we use a close match) G-1(4/7) = ? (This does not matter since s4/7) G-1(5/7) = 5/7 G-1(6/7) = 6/7 G-1(1) = 1 12/5/2018 Duong Anh Duc - Digital Image Processing

Combining the two transformation T and G-1, we get our required transformation H rT(r)=s sG-1(s)=z rG-1 [T(r)]=H(r)=z r0 = 0  1/7 0  ? r0 = 0  z3= 3/7 r1 = 1/7  3/7 1/7  3/7 r1 = 1/7  z4= 4/7 r2 = 2/7  5/7 2/7  4/7 r2 = 2/7  z5= 5/7 r3 = 3/7  6/7 3/7  4/7 r3 = 3/7  z6= 6/7 r4 = 4/7  6/7 4/7  ? r4 = 4/7  z6= 6/7 r5 = 5/7  1 5/7  5/7 r5 = 5/7  z7= 1 r6 = 6/7  1 6/7  6/7 r6 = 6/7  z7= 1 r7 = 1  1 1  1 r7 = 1  z7= 1 12/5/2018 Duong Anh Duc - Digital Image Processing

Applying the transformation H to the original image yields an image with histogram as below: k zk nk nk/n (actual hist.) pout(zk) (spec. hist.) 0.00 1 1/7 2 2/7 3 3/7 790 0.19 0.15 4 4/7 1023 0.25 0.20 5 5/7 850 0.21 0.30 6 6/7 985 0.24 7 448 0.11 12/5/2018 Duong Anh Duc - Digital Image Processing

Again, the actual histogram of the output image does not exactly but only approximately matches with the specified histogram. This is because we are dealing with discrete histograms. 12/5/2018 Duong Anh Duc - Digital Image Processing

Example Original image and its histogram 12/5/2018 Duong Anh Duc - Digital Image Processing

Example Histogram equalized image Actual histogram of output 12/5/2018 Duong Anh Duc - Digital Image Processing

Example Histogram specified image, Actual Histogram, and Specified Histogram 12/5/2018 Duong Anh Duc - Digital Image Processing

Enhancement Using Local Histogram
Used to enhance details over small portions of the image. Define a square or rectangular neighborhood, whose center moves from pixel to pixel. Compute local histogram based on the chosen neighborhood for each point and apply a histogram equalization or histogram specification transformation to the center pixel. Non-overlapping neighborhoods can also be used to reduce computations. But this usually results in some artifacts (checkerboard like pattern). 12/5/2018 Duong Anh Duc - Digital Image Processing

Enhancement Using Local Histogram
Another use of histogram information in image enhancement is the statistical moments associated with the histogram (recall that the histogram can be thought of as a probability density function). For example, we can use the local mean and variance to determine the local brightness/contrast of a pixel. This information can then be used to determine what, if any transformation to apply to that pixel. Note that local histogram based operations are non-uniform in the sense that a different transformation is applied to each pixel. 12/5/2018 Duong Anh Duc - Digital Image Processing

Digital Image Processing

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