2024-master dityv5y65v56u4b6u64u46p 0318-25.pdf

Digital image processing
GJ1644004A
Zhina Song，
504252932@qq.com

03 Image Intensity
Transformations

3.1 Review
Geometric transformation vs Intensity transformation
Spatial domain
The value at the
corresponding position
of the image does not
change, but the pixel
position changes
The pixel position in
the image does not
change, but the value
changes

3.2 The key points and difficulties of this class
 Be familiar with the principal techniques used for intensity transformations
 Learn basic log transformations and power-law transformations
 Understand the realization process of the two transformations

3.3 Intensity Transformation
Graphical display of basic intensity transformation
0 L-1
L-1
Identical
transformation
0 L-1
L-1
0 L-1
L-1

Log transformations 𝑔𝑔 𝑥𝑥, 𝑦𝑦 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(1 + 𝑓𝑓(𝑥𝑥, 𝑦𝑦))

Power-law (gamma) transformations 𝑔𝑔 𝑥𝑥, 𝑦𝑦 = 𝑐𝑐𝑓𝑓(𝑥𝑥,𝑦𝑦)𝛾𝛾

𝛾𝛾 = 3.0 , 4.0 and
5.0 respectively.

𝛾𝛾 = 0.6

Piecewise linear transformation functions

Intensity-Level Slicing

Discussion
What are the advantages and disadvantages
of these transformations?
Trial and error
A certain basis
intensity distribution
peaks and valleys
Discuss the pros and cons of these methods:
 Reasonable or not
 Automatic degree
 Robustness

Trial and error
A certain basis
intensity distribution
Discussion
 You may ask, to achieve this result, I can directly use PS, what is the meaning of
learning these transformations?
 We know how to use PS to achieve effects more
quickly and accurately without a lot of trying
 We can perform different transformations in
different regions
 We can perform different transformations on
different grayscale ranges

Image
intensity
distribution
Histogram
Intensity histogram
Discrete values
3.4 Histogram equalization

Enhance information
Restore information
Understand information
Image
processing
How to effectively use histogram information to enhance the image and make it "clear"?

3.2 Histogram Processing
Image gray histogram
 No spatial information involved
 The same histogram distribution may
correspond to different images
 Information additivity
 Related to the amount of information

Describe image with gray histogram
The grayscale of the image is concentrated
in the brighter area, and a considerable part
of them are concentrated in the part close to
1, resulting in overexposure of the image
The pixel distribution in the image
is “polarized”, resulting in the loss
of image details
The distribution of image histogram is related to the quality of image to some extent

A “clear” image
The histogram reflects the clarity of the image, when it is evenly distributed, the image is “clearer”
Histogram
equalization
each gray level should have a
certain number of gray values
Different objects should have
distinguishable grayscale variations

 Transform discrete distributions into continuous probability distributions for analysis
𝑝𝑝 𝑟𝑟 ∶ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑃𝑃 𝑟𝑟2 − 𝑃𝑃 𝑟𝑟1 = ∫
𝑟𝑟1
𝑟𝑟2
𝑝𝑝 𝑟𝑟 𝑑𝑑𝑑𝑑
Relationship before and after transformation:
𝑃𝑃 𝐷𝐷𝐴𝐴1 < 𝐷𝐷𝐴𝐴 < 𝐷𝐷𝐴𝐴𝐴 = 𝑃𝑃 𝐷𝐷𝐴𝐴𝐴 − 𝑃𝑃 𝐷𝐷𝐴𝐴1 = ∫
𝐷𝐷𝐴𝐴𝐴
𝐷𝐷𝐴𝐴𝐴
𝑝𝑝 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑𝑑𝑑 = ∫
𝐷𝐷𝐵𝐵1
𝐷𝐷𝐵𝐵2
𝑝𝑝 𝐷𝐷𝐵𝐵 𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑃𝑃 𝐷𝐷𝐵𝐵2 − 𝑃𝑃 𝐷𝐷𝐵𝐵1
𝐻𝐻𝐻𝐻𝐻𝐻 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
H𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎,

original image target image
For a random
distribution transform
to uniform distribution
original histogram
target histogram
L r s
S=T(r)
𝑝𝑝(𝑟𝑟𝑖𝑖) ≠ 𝑝𝑝(𝑟𝑟𝑗𝑗) 𝑝𝑝(𝑠𝑠𝑖𝑖) = 𝑝𝑝(𝑠𝑠𝑗𝑗)
𝑃𝑃 𝑇𝑇 𝑟𝑟𝑖𝑖 < 𝑠𝑠 < 𝑇𝑇 𝑟𝑟𝑗𝑗 =
∫𝑟𝑟𝒊𝒊
𝑟𝑟𝑗𝑗
𝑝𝑝 𝑟𝑟 𝑑𝑑𝑑𝑑 =
1
𝐿𝐿−1
× (𝑇𝑇 𝑟𝑟𝒋𝒋 − 𝑇𝑇 𝑟𝑟𝑖𝑖 )
𝑖𝑖𝑖𝑖 𝑟𝑟𝑗𝑗 > 𝑟𝑟𝑖𝑖, 𝑠𝑠𝑗𝑗> 𝑠𝑠𝑖𝑖
a b
𝑃𝑃(𝑎𝑎 ≤ 𝑠𝑠 ≤ 𝑏𝑏)
𝑠𝑠 = 𝑇𝑇 𝑟𝑟 = (𝐿𝐿 − 1) �
0
𝑟𝑟
𝑝𝑝(𝑟𝑟) 𝑑𝑑𝑑𝑑

k refers to the different gray values in the original image
P(k) corresponds to the frequency of the value in all pixels of the original image
Histogram
（ normalized ）
Unique Pixel of
𝒇𝒇(𝒙𝒙,𝒚𝒚)
𝒓𝒓𝟏𝟏 𝒓𝒓𝟐𝟐 … 𝒓𝒓𝒋𝒋 … 𝒓𝒓𝒌𝒌
frequency of
𝒑𝒑𝟏𝟏 𝒑𝒑𝟐𝟐 … 𝒑𝒑𝒋𝒋 … 𝒑𝒑𝒌𝒌
Pixel of
𝒈𝒈(𝒙𝒙, 𝒚𝒚)/(𝑳𝑳 − 𝟏𝟏)
𝒑𝒑𝟏𝟏 𝒑𝒑𝟏𝟏+𝒑𝒑𝟐𝟐 … +…+𝒑𝒑𝒋𝒋 … +…+𝒑𝒑𝒌𝒌
Discrete situation: Gray value quantification The quantization value closest to its
value is taken as the final gray value
𝑠𝑠 = 𝑇𝑇 𝑟𝑟 = (𝐿𝐿 − 1) �
0
𝑟𝑟
𝑝𝑝(𝑟𝑟) 𝑑𝑑𝑑𝑑

Example
Gray value range 0~7

Quantification: arranging the closest values
Example

summary
The
transformed
frequencies are
not equal. Why?
Example

1. After histogram equalization, how does the gray level of
the new imagechange?
2. What are the advantages and disadvantages of gray
histogram equalization? (Human intervention required;
reversible; Is it valid in all cases?)
 Purpose of Histogram Equalization
 Principle of Histogram Equalization
 Specific operation of histogram equalization
Summary and Discussion

Other image enhancement methods

Linear stretch Histogram equalization
Transformation
function
Comparisonof
image
enhancement
 Simple transformation
 Can be transformed
back to the original
image
 Need to manually set
parameters
 Poor generality
 Less information loss
 Automated, no
parameters required
 Unable to restore
 Poor generality

original image HE
Some failed examples using HE

Some failed examples using HE

Some improvement methods
LOCAL HISTOGRAM PROCESSING
Some differences and consistency in the
local area need to be preserved, but they
are often destroyed because the global
calculated value is obviously different from
the local calculated value.
p.150-153

Some improvement methods
LOCAL HISTOGRAM PROCESSING

Histogram matching (specification)
When doing change
detection on two
geometrically
aligned images
|𝑰𝑰𝟏𝟏 − 𝑰𝑰𝟐𝟐|

Histogram equalization where we take any histogram, any pixel distribution, and
we match it to something which is as uniform as possible.
Input
image
𝒈𝒈(𝒙𝒙, 𝒚𝒚)
Target
image
𝑠𝑠 = 𝑇𝑇 𝑟𝑟 = �
0
𝑟𝑟
𝑝𝑝𝑟𝑟(𝑢𝑢) 𝑑𝑑𝑑𝑑
𝑠𝑠 = 𝑇𝑇 z = �
0
𝑟𝑟
𝑝𝑝𝑧𝑧(𝑣𝑣) 𝑑𝑑𝑑𝑑
Histogram matching (specification)

Histogram matching (specification)
Just by doing the histogram equalizations, we can match any two desired distributions
Step1: Compute the histogram of the input image r, and do histogram equalization to
get the histogram-equalized image s1.
Step2: Compute the histogram of the target image z, and do histogram equalization
to get the histogram-equalized image s2.
Step3: For every value of s1, use the stored values of s2 from Step 2 to find the
corresponding value closest to s1 . Store these mappings from s1 to z
Step4: For every value of the image r, let the {𝑟𝑟𝑘𝑘} to be {𝑧𝑧𝑘𝑘
′
} ,by using the mappings
found in Step 3 to get the histogram-specified image.

The expressions of spatial domain processing
neighborhoodis of size 1 × 1
The function T can be linear or non-linear,
new gray value can be obtained by transforming
the original pixel value, or it can be obtained by
transforming the neighborhoodpixels.

3.6 Fundamentals of Spatial Filtering
If a pixel value in an
image is lost (or affected
by noise), can we use the
information in other
place to estimate its
value?
It can be approximately equal to the
average of all values of the entire image
It can be approximately given by the
average of several nearby pixel values
The value of each pixel changes by
globally
or
locally Related to its location

Spatial filtering modifies an image by
replacing the value of each pixel by a
function of the values of the pixel and its
neighbors.
linear spatial filter
nonlinear spatial filter
𝑌𝑌 = 𝑊𝑊𝑊𝑊 + 𝑏𝑏

The mechanics of linear spatial
filtering
A linear spatial filter performs a sum-of-
products operation between an image f
and a filter kernel w
kernel :
an array ;
size defines the neighborhood of operation;
coefficients determine the nature of the filter;
also can be called mask, template, window
一种特征提取器

The mechanics of linear spatial filtering
The size of the kernel is odd, because we
must ensure that the current point we are
dealing with is in the exact center
m=2a+1; n=2b+1

The mechanics of linear spatial filtering
with box kernels
of sizes 3 × 3,
11 × 11,
and 21 × 21
the larger the neighborhood, the
more pixels we are averaging

Spatial correlation and convolution
Correlationconsists of moving the
center of a kernel over an image, and
computing the sum of products at each
location.
VS
spatial convolution are the same,
except that the correlation kernel is
rotated by 180°
when the values of a kernel are symmetricaboutits center, correlationand convolutionyield sameresult

We can define correlation and convolution
so that every element of w(instead of just
its center) visits every pixel in f. This
requires that the starting configuration be
such that the right, lower corner of the kernel
coincides with the origin of the image.
the size of the resulting full correlation or
convolution array will be of size（by padding）
Sv ×S h ：

Spatial correlation and convolution
“convolving a kernel with an image” often is used to denote the sliding, sum-of-products process
Sometimes an image is filtered (i.e., convolved) sequentially, multistage filtering can be done in a
single filtering operation,
These convolution kernels can be combined and of course can be separated

3.7 Smoothing (Lowpass) Spatial Filters
Average kernel
Becauserandomnoisetypicallyconsistsof sharp transitionsin intensity, an obviousapplicationof
smoothingis noisereduction.
The differencebetween each pixeland its surroundingpixels will be smaller than theoriginalones,
so smoothingfiltercan be used to smooththe imageand remove somefalsecontours.
BOX FILTER KERNELS
Smoothing is used to reduce irrelevantdetailin an image
The kernel should be normalized
with box kernels
of sizes 3 × 3,
11 × 11,
and 21 × 21

Average kernel
BOX FILTER KERNELS -Set manually
image kernel no padding padding
?

Average kernel
LOWPASS GAUSSIAN FILTER KERNELS
circularlysymmetric(also
called isotropic)kernel
Distances from
the center for
various sizes of
square kernels.
圆形对称（也称为各向同性）核

Average kernel
LOWPASS GAUSSIAN FILTER KERNELS
K = 1
𝜎𝜎 = 1
如果所有内核都是高斯，我们可以
在表中使用结果来计算复合内核的
标准偏差（并定义它），而无需实
际执行所有内核的卷积。
If all kernels are Gaussian, we can use the
composite kernel (and define it), without actually
performing the convolution of all kernels.

Average kernel
kernel of size 21 × 21,
standard deviations 3.5
kernel of size 43 × 43,
standard deviations 3.5
box kernels
of sizes 11 × 11,
21 × 21
Comparison

Average kernel Comparison
with a box kernel of size 71 × 71
Gaussian kernel of size 151
× 151, with K = 1 and 𝜎𝜎= 25
• box filter producedlinearsmoothing, with the transitionfrom blackto whitehavingthe shapeof a ramp
• the Gaussianfilter yieldedsignificantlysmoother results aroundthe edge transitions

Applications
Using lowpass filtering and thresholding for region extraction
2566 × 2758 Hubble Telescope
image
Result of lowpass filtering
with a Gaussian kernel
size 151 × 151, 𝜎𝜎 = 25
Result of thresholding the filtered
image
Average kernel

Applications
Shading correction using lowpass filtering
Lowpass filtering is a rugged, simple
method for estimating shading patterns
512 × 512 Gaussian kernel (four
times the size of squares), K = 1, and
𝜎𝜎= 128 (equal to the size of squares)
Average kernel

Order-statistic (nonlinear) filters
 response is based on ordering (ranking) the pixels contained in the region
encompassed by the filter
 Smoothing is achieved by replacing the value of the center pixel with the value
determined by the ranking result.
 median filter: replaces the value of the center pixel by the median of the intensity
values in the neighborhood of that pixel
FORCE POINTS TO BE MORE LIKE THEIR NEIGHBORS
median filter

Order-statistic (nonlinear) filters median filter
image corrupted by salt-
and-pepper noise
result using 19 × 19
Gaussian lowpass filter
kernel with 𝜎𝜎= 3
result using 7 × 7
median filter

3.8 Sharpening (Highpass) Spatial Filters
Distribution of grayscale changes in the image
Scan line
The gray distribution of the image
in the direction of the scan line
First derivative
Second derivative

Step pulse slope
First
derivative
Second
derivative

The gradientof an image f at coordinates(x, y) is defined as the two dimensional column vector
Image gradient
The magnitude (length) of vector f , denotedas M(x, y)
First derivative

Image gradient： derivative operation --> differential operation
For discrete images, differentiation can be approximated by difference
||𝛻𝛻𝑓𝑓|| = (𝑓𝑓 𝑥𝑥,𝑦𝑦 − 𝑓𝑓 𝑥𝑥 + 1, 𝑦𝑦 )2+(𝑓𝑓 𝑥𝑥, 𝑦𝑦 − 𝑓𝑓 𝑥𝑥, 𝑦𝑦 + 1 )2
computationally to approximate the squares and square root operations by absolute values
𝛻𝛻𝑓𝑓 ≈ 𝑓𝑓 𝑥𝑥, 𝑦𝑦 − 𝑓𝑓 𝑥𝑥 + 1,𝑦𝑦 + |𝑓𝑓 𝑥𝑥, 𝑦𝑦 − 𝑓𝑓 𝑥𝑥, 𝑦𝑦 + 1 |
The magnitude of the gradient is approximated as the (absolute) sum
of the adjacent pixel differencesalong the horizontal and vertical axes

① The pixel value of the new image is directly replaced by the gradient of the original image
② The output image is according to the gradient threshold
Image Sharpening 图像锐化

Image Sharpening using gradient
The edges of the image are enhanced, and some noise is also amplified

Robert Operator
The differential sum of the two
directions after rotating ±45°
The area involved in the calculation is too
small, and the obtained edge is weak

Image Sharpening using gradient 3*3 kernel
image
x
y
Maintaining directional consistency in the calculation, 3*3 can be viewed as a
superposition of multiple 2*2 regions with respect to the current pixel position.

image
x
y Sobel operators

Sobel 算子垂直方向
Sobel 算子水平方向 Sobel 算子综合叠加

Results obtained by extracting the
edges using Sobel operator
Enhancedimage

second-order derivative of f (x) 二阶微分，寻找突变区域

3.8Sharpening (Highpass) Spatial Filters
second-order derivative of f (x)
Laplacian operators

second-order derivative of f (x) Laplacian operators

second-order derivative of f (x)
Flexible extensions of the Laplace operator
1 -2 1
-2 4 -2
1 -2 1
Background features can be “recovered” while still preserving the sharpening effect of the
Laplacian by adding the Laplacian image to the original.
Let c = −1

The latter has a more
pronounced sharpening effect

一阶微分和二阶微分算子的锐化区别

3.7 HIGHPASS, BANDREJECT, AND BANDPASS FILTERS

3.7 Unsharp Masking And Highboost Filtering
Add a weighted portion of the mask back to the original image:
The mask

3.7 Unsharp Masking And Highboost Filtering

3.8 Combining Spatial Enhancement Methods
a nuclear
whole body
bone scan
image
Objective: show more of the skeletal detail
method: enhance the edges
Laplacian of image Sharpened image

Objective: show more of the skeletal detail
method: enhance the edges and suppress noise
Sobel gradient of image Sobel image smoothed with a 5 × 5 box filter
Mask image formed by
the product of (b) and (e).

Objective: show more of the skeletal detail method: enhance the edges and suppress noise
Sharpened image obtained
by the adding images (a) and (f).

summry
Image basic geometric transformation
Image Intensity Transformations
Spatial Filtering

Homework Deadline： before 9 April
1. Consider that the maximum value of an image 𝑰𝑰𝟏𝟏is M and its minimum is m
(m≠M). An intensity transform that maps the image 𝑰𝑰𝟏𝟏 onto 𝑰𝑰𝟐𝟐 such that the
maximal value of 𝑰𝑰𝟐𝟐 is L and the minimal value is：
2. Why global discrete histogram equalization does not, in general, yield a flat
(uniform) histogram?
A Because images are in color.
B Becausethe histogramequalizationmathematicalderivationdoesn’texist for discretesignals.
C In global histogramequalization, all pixels with the same value are mapped to same value.
D Actually, global discretehistogramequalizationalways yields flat histograms by definition.

Homework
3. Discrete histogram equalization is an invertible operation, meaning we can
recover the original image from the equalized one by inverting the operation,
since?
A Actually, histogram equalization is in general non-invertible.
B There is a unique histogram equalization formula per image.
C Pixels with different values are mapped to pixels with different values.
D Images have unique histograms.
4. Given an image with only 3 pixels and 4 possible values for each one. Determine
the number of possible different images and the number of possible different
histograms. How many images and histograms?

Homework
5. This image is a 6*6 grayscale image I(x, y) , with 4 gray levels
(x = 0, 1, 2, ... 5; y = 0, 1, 2, ..., 5) , the value of each point in the
figure represents the gray value of the image pixels.
1) Calculate the histogram of the image
2) Using histogram equalization to process this image (write the
process details )
3) Write the new histogram after histogram equalization.

Homework
6. Which integer number minimizes
7. Which integer number minimizes
8. Applying a 3×3 averaging filter to an image a large (infinity) number of times is:
A Equivalent to replacing all the pixel values by 0..
B Equivalent to replacing all the pixel values by the average of the values in the
original image.
C The same as applying it a single time.
D The same as applying a median filter.

9. In the original image used to generate the three blurred images shown, the vertical
bars are 5 pixels wide, 100 pixels high, and their separation is 20 pixels. The image was
blurred using square box kernels of sizes 23, 25, and 45 elements on the side,
respectively. The vertical bars on the left, lower part of (a) and (c) are blurred, but a
clear separation exists between them. However, the bars have merged in image (b),
despite the fact that the kernel used to generate this image is much smaller than the
kernel that produced image (c). Explain the reason for this.
Homework

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