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==Problem 1==
 
==Problem 1==
  
a) <math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math>
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b) <math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\Delta d)\lambda_n^c</math>
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c) <math>\lambda_n = \lambda_n^c e^{-\int_{0}^{x}\mu(t)dt} \Longrightarrow \hat{P}_n = \int_{0}^{x}\mu(t)dt= -ln(\frac{\lambda_n}{\lambda_n^c}) = -ln(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d})</math>
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d) <math>\hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n</math>
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  A straight line with slope <math>\mu_0</math>
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==Problem 2==
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a)Since U is <math>p \times N</math>, <math>\Sigma</math> and V are <math>N \times N</math>]
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<math>Y = U \Sigma V^t = p \times N</math>
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<math>YY^t = (p\times N)(N\times p) = p \times p</math>
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<math>Y^tY = (N\times p)(p\times N) = N \times N</math>
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b) <math>YY^t = U \Sigma V^t V \Sigma U^t = U\Sigma^2 U^t</math> and <math>(YY^t)^t = (U\Sigma^2 U^t)^t = U\Sigma^2 U^t = YY^t </math>. Therefore, <math>YY^t</math> is symmetric
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For an arbitrary x, <math>x^tYY^tx = x^t U\Sigma \Sigma U^t x=(\Sigma U^t x)^t\Sigma U^t x=\|\Sigma U^t x\|^2 \geq 0</math>. Therefore, <math>YY^t</math> is positive semi-definite.
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Similarly, <math>Y^tY = V \Sigma^2 V^t</math> and <math>(Y^tY)^t = (V \Sigma^2 V^t)^t = V \Sigma^2 V^t = Y^tY</math>, <math>Y^tY</math> is symmetric
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For an arbitrary x, <math>x^tY^tYx = x^t V\Sigma \Sigma V^t x=(\Sigma V^t x)^t\Sigma V^t x=\|\Sigma V^t x\|^2 \geq 0</math>. Therefore, <math>Y^tY</math> is positive semi-definite.
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c) From B, obtain that <math>Y^tY = V\Sigma^2 V^t </math> while <math>Y^tY = TDT^t</math>. <math>V = T</math> and <math>\Sigma = D^{1/2}</math>
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d)<math>U\Sigma V^t=Y \Longrightarrow U = Y(\Sigma V^t)^{-1} = Y(D^{1/2}T^t)^{-1}</math>
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e)<math>YY^t = U\Sigma U^t = E\Sigma E^t</math>
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<math>E = U = Y(D^{1/2}T^t)^{-1}</math>
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f)The name we give to the column of U is eigenimages
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Revision as of 20:01, 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

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Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics