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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>
 
b) <math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\Delta d)\lambda_n^c</math>
  
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>
+
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>
+
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>
+
   A straight line with slope <math>\mu_0</math>
  
 
<math><\math>
 
<math><\math>

Revision as of 19:22, 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 $

$ <\math> ==Problem 2== a)Since U is $p \times N$, $\Sigma$ and V are $N \times N$\\ $

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