(Created page with "Hello world")
 
 
(51 intermediate revisions by 2 users not shown)
Line 1: Line 1:
Hello world
+
[[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


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published in Jul 2019)

Problem 1

  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 $

  1. 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 $

  1. 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}) $

  1. 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

  1. 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.

  1. 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

  1. 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}} $

  1. 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} $

  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} $

  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

Alumni Liaison

EISL lab graduate

Mu Qiao