Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published on Jul 2019)

## Problem 1

- Calculate an expression for $ \lambda_n^c $, the X-ray energy corrected for the dark current.

$ \lambda_n^c=\lambda_n^b-\lambda_n^d $

- Calculate an expression for $ G_n $, the X-ray attenuation due to the object's presence.

$ G_n=-\mu(x,y_0+n*\Delta d)\lambda_n $

- Calculate an expression for $ \hat{P}_n $, an estimate of the integral intensity in terms of $ \lambda_n $, $ \lambda_n^b $, and $ \lambda_b^d $.

$ \lambda_n=(\lambda_n^b-\lambda_n^d)e^{-\int_0^x \mu(t)dt} $

$ \hat{P}_n=\int_0^x \mu(t)dt=-log\frac{\lambda_n}{\lambda_n^b-\lambda_n^d} $

- For this part, assume that the object is of constant density with $ \mu(x,y)=\mu_0 $. Then sketch a plot of $ \hat{P}_n $ versus the object thickness, $ T_n $, in $ mm $, for the $ n^{th} $ detector. Label key features of the curve such as its slope and intersection.

## Problem 2

- Specify the size of $ YY^t $ and $ Y^tY $. Which matrix is smaller?

$ Y $ is of size $ p\times N $, so the size of $ YY^t $ is $ p\times p $.

$ Y $ is of size $ p\times N $, so the size of $ Y^tY $ is $ N\times N $.

Obviously, the size of $ Y^tY $ is much smaller, since $ N<<p $.

- Prove that both $ YY^t $ and $ Y^tY $ are both symmetric and positive semi-definite matrices.

To prove it is symmetric:

$ (YY^t)^t=YY^t $

To prove it is positive semi-definite:

Let $ x $ be an arbitrary vector

$ x^tYY^tx=(Y^tx)^T(Y^tx)\geq 0 $ So the matrix $ YY^t $ is positive semi-definite.

The proving procedures for $ Y^tY $ are the same.

- Derive expressions for $ V $ and $ \Sigma $ in terms of $ T $, and $ D $.

$ Y^tY=(U\Sigma V^t)^tU\Sigma V^t=V\Sigma^2V^t=TDT^t $ therefore $ V=T $ and $ \Sigma=D^\frac{1}{2} $

- Drive expressions for $ U $ in terms of $ Y $, $ T $, and $ D $.

$ Y=U\Sigma V^t=UD^\frac{1}{2}T^t $

$ \therefore U=Y(D^\frac{1}{2}T^t)^{-1} $

- Derive expressions for $ E $ in terms of $ Y $, $ T $, and $ D $.

$ YY^t=U\Sigma V^t(U\Sigma V^t)^t=U\Sigma^2U^t=E\Gamma E^t $

therefore

$ E=U=Y(D^\frac{1}{2}T^t)^{-1} $

- If the columns of $ Y $ are images from a training database, then what name do we give to the columns of $ U $?

They are called eigenimages.