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a) <math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math>
 
a) <math>\lambda_n^c=\lambda_n^b-\lambda_n^d</math>
  
b) <math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\delta d)\lambda_n^c</math>
+
b) <math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\Delta d)\lambda_n^c</math>
  
c)
+
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>
 +
 
 +
d) <math>\hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n<\math>
 +
  A straight line with slope <math>\mu_0<\math>
 +
 
 +
<math><\math>
 +
 
 +
 
 +
==Problem 2==
 +
 
 +
a)Since U is $p \times N$, $\Sigma$ and V are $N \times N$\\

Revision as of 19:21, 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})<\math> d) <math>\hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n<\math> A straight line with slope <math>\mu_0<\math> <math><\math> ==Problem 2== a)Since U is $p \times N$, $\Sigma$ and V are $N \times N$\\ $

Alumni Liaison

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

Buyue Zhang