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Revision as of 18:34, 7 July 2019
Communication, Networking, Signal and Image Processing (CS)
Question 5: Image Processing
August 2016 (Published in 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.
