m
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\therefore U=Y(D^\frac{1}{2}T^t)^{-1}
 
\therefore U=Y(D^\frac{1}{2}T^t)^{-1}
 
</math>
 
</math>
 +
</center>
  
 
# Derive expressions for <math>E</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
 
# Derive expressions for <math>E</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
 +
 +
<center>
 +
<math>
 +
YY^t=U\Sigma V^t(U\Sigma V^t)^t=U\Sigma^2U^t=E\Gamma E^t
 +
</math>
 +
 +
therefore
 +
 +
<math>
 +
E=U=Y(D^\frac{1}{2}T^t)^{-1}
 +
</math>
 +
</center>
  
 
# If the columns of <math>Y</math> are images from a training database, then what name do we give to the columns of <math>U</math>?
 
# If the columns of <math>Y</math> are images from a training database, then what name do we give to the columns of <math>U</math>?

Revision as of 02:07, 7 July 2019


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published on Jul 2019)

Problem 1

  1. Calculate 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=-\mu(x,y_0+n*\Delta d)\lambda_n $

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

$ \lambda_n=(\lambda_n^b-\lambda_n^d)e^{-\int_0^x \mu(t)dt} $

$ \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 $ 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)^tU\Sigma V^t=V\Sigma^2V^t=TDT^t $ therefore $ V=T $ and $ \Sigma=D^\frac{1}{2} $

  1. Drive expressions for $ U $ in terms of $ Y $, $ T $, and $ 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^2U^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 $?

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett