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==Problem 1==
 
==Problem 1==
  
# Calculate an expression for <math>\lambda_n^c</math>, the X-ray energy corrected for the dark current.
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a) <math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math>
 
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<center>
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<math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math>
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</center>
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# Calculate an expression for <math>G_n</math>, the X-ray attenuation due to the object's presence.
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<center>
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<math>
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G_n=-\mu(x,y_0+n*\Delta d)\lambda_n
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</math>
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</center>
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# 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_b^d</math>.
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<center>
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<math>
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\lambda_n=(\lambda_n^b-\lambda_n^d)e^{-\int_0^x \mu(t)dt}
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</math>
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</center>
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<center>
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<math>
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\hat{P}_n=\int_0^x \mu(t)dt=-log\frac{\lambda_n}{\lambda_n^b-\lambda_n^d}
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</math>
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</center>
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# 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 <math>mm</math>, for the <math>n^{th}</math> detector. Label key features of the curve such as its slope and intersection.
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==Problem 2==
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# Specify the size of <math>YY^t</math> and <math>Y^tY</math>. Which matrix is smaller?
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<center>
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<math>Y</math> is of size <math>p\times N</math>, so the size of <math>YY^t</math> is <math>p\times p</math>.
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<math>Y</math> is of size <math>p\times N</math>, so the size of <math>Y^tY</math> is <math>N\times N</math>.
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Obviously, the size of <math>Y^tY</math> is much smaller, since <math>N<<p</math>.
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</center>
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# Prove that both <math>YY^t</math> and <math>Y^tY</math> are both symmetric and positive semi-definite matrices.
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<center>
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To prove it is symmetric:
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<math>
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(YY^t)^t=YY^t
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</math>
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To prove it is positive semi-definite:
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Let <math>x</math> be an arbitrary vector
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<math>
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x^tYY^tx=(Y^tx)^T(Y^tx)\geq 0
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</math>
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So the matrix <math>YY^t</math> is positive semi-definite.
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The proving procedures for <math>Y^tY</math> are the same.
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</center>
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# Derive expressions for <math>V</math> and <math>\Sigma</math> in terms of <math>T</math>, and <math>D</math>.
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<center>
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<math>
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Y^tY=(U\Sigma V^t)^tU\Sigma V^t=V\Sigma^2V^t=TDT^t
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</math>
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therefore <math>V=T</math> and <math>\Sigma=D^\frac{1}{2}</math>
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</center>
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# Drive expressions for <math>U</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
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<center>
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<math>
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Y=U\Sigma V^t=UD^\frac{1}{2}T^t
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</math>
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</center>
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<center>
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<math>
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\therefore U=Y(D^\frac{1}{2}T^t)^{-1}
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</math>
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</center>
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# Derive expressions for <math>E</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
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<center>
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<math>
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YY^t=U\Sigma V^t(U\Sigma V^t)^t=U\Sigma^2U^t=E\Gamma E^t
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</math>
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therefore
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<math>
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E=U=Y(D^\frac{1}{2}T^t)^{-1}
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</math>
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</center>
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# 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>?
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<center>
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They are called '''eigenimages'''.
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</center>
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Revision as of 19:14, 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

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

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009