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− | + | [[Category:ECE]] | |

+ | [[Category:QE]] | ||

+ | [[Category:CNSIP]] | ||

+ | [[Category:problem solving]] | ||

+ | [[Category:image processing]] | ||

+ | |||

+ | <center> | ||

+ | <font size= 4> | ||

+ | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||

+ | </font size> | ||

+ | |||

+ | <font size= 4> | ||

+ | Communication, Networking, Signal and Image Processing (CS) | ||

+ | |||

+ | Question 5: Image Processing | ||

+ | </font size> | ||

+ | |||

+ | August 2016 (Published in Jul 2019) | ||

+ | </center> | ||

+ | ==Problem 1== | ||

+ | |||

+ | #Calcualte an expression for <math>\lambda_n^c</math>, the X-ray energy corrected for the dark current | ||

+ | |||

+ | <center> | ||

+ | <math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math> | ||

+ | </center> | ||

+ | |||

+ | #Calculate an expression for <math>G_n</math>, the X-ray attenuation due to the object's presence | ||

+ | |||

+ | <center> | ||

+ | <math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n * \Delta d)\lambda_n^c</math> | ||

+ | </center> | ||

+ | |||

+ | #Calculate an expression for <math>\hat{P}_n</math>, an estimate of the integral intensity in terms of <math>\lambda_n</math>, <math>\lambda_n^b</math>, and <math>\lambda_n^d</math> | ||

+ | |||

+ | |||

+ | <center> | ||

+ | <math>\lambda_n = (\lambda_n^b-\lambda_n^d) e^{-\int_{0}^{x}\mu(t)dt}d)\lambda_n^c</math> | ||

+ | |||

+ | <math>\hat{P}_n = \int_{0}^{x}\mu(t)dt= -log(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d})</math> | ||

+ | </center> | ||

+ | |||

+ | #For this part, assume that the object is of constant density with <math>\mu(x,y) = \mu_0</math>. Then sketch a plot of <math>\hat{P}_n</math> versus the object thickness, <math>T_n</math>, in mm, for the <math>n^{th}</math> detector. Label key features of the curve such as its slope and intersection. | ||

+ | |||

+ | ==Problem 2== | ||

+ | |||

+ | #Specify the size of <math>YY^t</math> and <math>Y^tY</math>. Which matrix is smaller | ||

+ | |||

+ | <center> | ||

+ | Y is of size <math>p \times N</math>, so the size of <math>YY^t</math> is <math>p \times p</math> | ||

+ | |||

+ | Y is of size <math>p \times N</math>, so the size of <math>Y^tY</math> is <math>N \times N</math> | ||

+ | |||

+ | Obviously, the size of <math>Y^tY</math> is much smaller, since N << p. | ||

+ | </center> | ||

+ | |||

+ | #Prove that both <math>YY^t</math> and <math>Y^tY</math> are both symmetric and positive semi-definite matrices. | ||

+ | |||

+ | <center> | ||

+ | To prove it is symmetric: | ||

+ | |||

+ | <math>(YY^t)^t = YY^t</math> | ||

+ | |||

+ | To prove it is positive semi-definite: | ||

+ | |||

+ | Let x be an arbitrary vector | ||

+ | |||

+ | <math>x^tYY^tx = (Y^tx)^T(Y^tx) \geq 0</math> so the matrix of <math>YY^t</math> is positive semi-definite. | ||

+ | |||

+ | The proving procedures for <math>Y^tY</math> are the same | ||

+ | </center> | ||

+ | |||

+ | #Derive expressions for <math>V</math> and <math>\Sigma</math> in terms of <math>T</math>, and <math>D</math>. | ||

+ | |||

+ | <center> | ||

+ | <math>Y^tY = (U \Sigma V^t)^T U \Sigma V^t = V\Sigma^2 V^t = TDT^t</math> therefore <math>V =T</math> and <math>\Sigma = D^{\frac{1}{2}}</math> | ||

+ | </center> | ||

+ | |||

+ | #Derive an expression for <math>U</math> in terms of <math>Y</math>, <math>T</math>, <math>D</math>. | ||

+ | <center> | ||

+ | |||

+ | <math>Y = U\Sigma V^t= UD^{\frac{1}{2}}T^t</math> | ||

+ | |||

+ | <math>\therefore U = Y(D^{\frac{1}{2}}T^t)^{-1}</math> | ||

+ | </center> | ||

+ | |||

+ | #Derive expressions for E in terms of Y, T, and D. | ||

+ | <center> | ||

+ | <math>YY^t = U\Sigma V^t(U\Sigma V^t)^t= U\Sigma^2 U^t = E\Gamma E^t</math> | ||

+ | |||

+ | therefore | ||

+ | |||

+ | <math>E = U = Y(D^{\frac{1}{2}}T^t)^{-1}</math> | ||

+ | </center> | ||

+ | |||

+ | #If the columns of Y are images from a training database, then what name do we give to the columns of U? | ||

+ | <center> | ||

+ | They are called '''eigenimages''' | ||

+ | </center> |

## Latest revision as of 20:50, 9 July 2019

Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published in Jul 2019)

## Problem 1

- Calcualte an expression for $ \lambda_n^c $, the X-ray energy corrected for the dark current

$ \lambda_n^c=\lambda_n^b-\lambda_n^d $

- Calculate an expression for $ G_n $, the X-ray attenuation due to the object's presence

$ G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n * \Delta d)\lambda_n^c $

- Calculate an expression for $ \hat{P}_n $, an estimate of the integral intensity in terms of $ \lambda_n $, $ \lambda_n^b $, and $ \lambda_n^d $

$ \lambda_n = (\lambda_n^b-\lambda_n^d) e^{-\int_{0}^{x}\mu(t)dt}d)\lambda_n^c $

$ \hat{P}_n = \int_{0}^{x}\mu(t)dt= -log(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d}) $

- For this part, assume that the object is of constant density with $ \mu(x,y) = \mu_0 $. Then sketch a plot of $ \hat{P}_n $ versus the object thickness, $ T_n $, in mm, for the $ n^{th} $ detector. Label key features of the curve such as its slope and intersection.

## Problem 2

- Specify the size of $ YY^t $ and $ Y^tY $. Which matrix is smaller

Y is of size $ p \times N $, so the size of $ YY^t $ is $ p \times p $

Y is of size $ p \times N $, so the size of $ Y^tY $ is $ N \times N $

Obviously, the size of $ Y^tY $ is much smaller, since N << p.

- Prove that both $ YY^t $ and $ Y^tY $ are both symmetric and positive semi-definite matrices.

To prove it is symmetric:

$ (YY^t)^t = YY^t $

To prove it is positive semi-definite:

Let x be an arbitrary vector

$ x^tYY^tx = (Y^tx)^T(Y^tx) \geq 0 $ so the matrix of $ YY^t $ is positive semi-definite.

The proving procedures for $ Y^tY $ are the same

- Derive expressions for $ V $ and $ \Sigma $ in terms of $ T $, and $ D $.

$ Y^tY = (U \Sigma V^t)^T U \Sigma V^t = V\Sigma^2 V^t = TDT^t $ therefore $ V =T $ and $ \Sigma = D^{\frac{1}{2}} $

- Derive an expression for $ U $ in terms of $ Y $, $ T $, $ D $.

$ Y = U\Sigma V^t= UD^{\frac{1}{2}}T^t $

$ \therefore U = Y(D^{\frac{1}{2}}T^t)^{-1} $

- Derive expressions for E in terms of Y, T, and D.

$ YY^t = U\Sigma V^t(U\Sigma V^t)^t= U\Sigma^2 U^t = E\Gamma E^t $

therefore

$ E = U = Y(D^{\frac{1}{2}}T^t)^{-1} $

- If the columns of Y are images from a training database, then what name do we give to the columns of U?

They are called **eigenimages**