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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})< | + | 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< | + | d) <math>\hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n</math> |
| − | A straight line with slope <math>\mu_0< | + | A straight line with slope <math>\mu_0</math> |
<math><\math> | <math><\math> | ||
Revision as of 18:22, 9 July 2019
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 $N \times N$\\ $
