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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 $
b) $ YY^t = U \Sigma V^t V \Sigma U^t = U\Sigma^2 U^t $ and $ (YY^t)^t = (U\Sigma^2 U^t)^t = U\Sigma^2 U^t = YY^t $. Therefore, $ YY^t $ is symmetric
For an arbitrary x, $ 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 $. Therefore, $ YY^t $ is positive semidefinite.
Similarly, $ Y^tY = V \Sigma^2 V^t $ and $ (Y^tY)^t = (V \Sigma^2 V^t)^t = V \Sigma^2 V^t = Y^tY $, $ Y^tY $ is symmetric
For an arbitrary x, $ 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 $. Therefore, $ Y^tY $ is positive semidefinite.
c) From B, obtain that $ Y^tY = V\Sigma^2 V^t $ while $ Y^tY = TDT^t $. $ V = T $ and $ \Sigma = D^{1/2} $
d)$ U\Sigma V^t=Y \Longrightarrow U = Y(\Sigma V^t)^{1} = Y(D^{1/2}T^t)^{1} $
e)$ YY^t = U\Sigma U^t = E\Sigma E^t $
$ E = U = Y(D^{1/2}T^t)^{1} $
f)The name we give to the column of U is eigenimages