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<math></math>
 
<math></math>
 
  
 
==Problem 2==
 
==Problem 2==
  
a)Since U is <math>p \times N</math>, <math>\Sigma</math> and V are <math>N \times N</math>
+
a)Since U is <math>p \times N</math>, <math>\Sigma</math> and V are <math>N \times N</math>]
 +
 
 +
<math>Y = U \Sigma V^t = p \times N</math>
 +
 
 +
<math>YY^t = (p\times N)(N\times p) = p \times p</math>
 +
 
 +
<math>Y^tY = (N\times p)(p\times N) = N \times N</math>

Revision as of 19:25, 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 $

b) $ G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\Delta d)\lambda_n^c $

c) $ \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}) $

d) $ \hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n $

  A straight line with slope $ \mu_0 $


Problem 2

a)Since U is $ p \times N $, $ \Sigma $ and V are $ N \times N $]

$ Y = U \Sigma V^t = p \times N $

$ YY^t = (p\times N)(N\times p) = p \times p $

$ Y^tY = (N\times p)(p\times N) = N \times N $

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