# 0.16 Lab 10b - image processing (part 2)  (Page 2/5)

 Page 2 / 5

In $Y{C}_{b}{C}_{r}$ , the luminance parameter is related to an overall intensity of the image.The chrominance components are a measure of the relative intensities of the blue and red components. The inverse of the transformation in [link] can easily be shown to be the following.

$\begin{array}{ccc}R\hfill & =\hfill & Y+1.4025\left({C}_{r}-128\right)\hfill \\ G\hfill & =\hfill & Y-0.3443\left({C}_{b}-128\right)-0.7144*\left({C}_{r}-128\right)\hfill \\ B\hfill & =\hfill & Y+1.7730\left({C}_{b}-128\right)\hfill \end{array}$

## Color exercise

Download the files girl.tif and ycbcr.mat . For help on image command select the link.

You will be displaying both color and monochrome images in the following exercises.Matlab's image command can be used for both image types, but care must be taken for the command to work properly.Please see the help on the image command for details.

Download the $RGB$ color image file girl.tif , and load it into Matlab using the imread command. Check the size of the Matlab array for this image by typing whos . Notice that this is a three dimensional array of type uint8 . It contains three gray scale image planes corresponding to the red, green, andblue components for each pixel. Since each color pixel is represented bythree bytes, this is commonly known as a 24-bit image. Display the color image using

image(A);

axis('image');

where $A$ is the 3-D $RGB$ array.

You can extract each of the color components using the following commands.

RGB = imread('girl.tif'); % color image is loaded into matrix RGB

R = RGB(:,:,1); % extract red component from RGB

G = RGB(:,:,2); % extract green component from RGB

B = RGB(:,:,3); % extract blue component from RGB

Use the subplot and image commands to plot the original image, along with each of the three color components.Note that while the original is a color image, each color component separately is a monochrome image.Use the syntax subplot(2,2,n) , where $n=1,2,3,4$ , to place the four images in the same figure.Place a title on each of the images, and print the figure(use a color printer).

We will now examine the $Y{C}_{b}{C}_{r}$ color space representation. Download the file ycbcr.mat , and load it intoMatlab using load ycbcr . This file contains a Matlab array for a color image in $Y{C}_{b}{C}_{r}$ format. The array containsthree gray scale image planes that correspond to the luminance ( $Y$ ) and two chrominance ( ${C}_{b}{C}_{r}$ ) components. Use subplot(3,1,n) and image to display each of the components in the same figure.Place a title on each of the three monochrome images, and print the figure.

In order to properly display this color image, we need to convert it to $RGB$ format. Write a Matlab function that will perform the transformation of [link] . It should accept a 3-D $Y{C}_{b}{C}_{r}$ image array as input, and return a 3-D $RGB$ image array.

Now, convert the ycbcr array to an $RGB$ representation and display the color image.Remember to convert the result to type uint8 before using the image command.

An interesting property of the human visual system, with respect to the $Y{C}_{b}{C}_{r}$ color space, is that we are much more sensitive to distortion in the luminance component than in the chrominancecomponents. To illustrate this, we will smooth each of these components with a Gaussian filter and view the results.

#### Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
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J, combine like terms 7x-4y
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
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Azam
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Purdue digital signal processing labs (ece 438). OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10593/1.4
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