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<center>
 
<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^{\farc{1}{2}}</math>
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<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>
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</center>
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#Derive an expression for <math>U</math> in terms of <math>Y</math>, <math>T</math>, <math>D</math>.
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<center>
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<math>Y = U\Sigma V^t= UD^{\frac{1}{2}}T^t</math>
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<math>\therefore U = Y(D^{\frac{1}{2}}T^t)^{-1}</math>
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</center>
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#Derive expressions for E in terms of Y, T, and D.
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<center>
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<math>YY^t = U\Sigma V^t(U\Sigma V^t)^t= U\Sigma^2 U^t = E\Gamma E^t</math>
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therefore
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<math>E = U  = Y(D^{\frac{1}{2}}T^t)^{-1}</math>
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</center>
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#If the columns of Y are images from a training database, then what name do we give to the columns of U?
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<center>
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They are called '''eigenimages'''
 
</center>
 
</center>

Latest revision as of 19:50, 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

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang