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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 semi-definite.

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 semi-definite.

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

## Alumni Liaison

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