Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published on Jul 2019)

## Problem 1

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

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

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

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

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

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

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

1. 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

1. 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 $N<<p$.

1. 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.

1. 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 $\Sigma=D^\frac{1}{2}$

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

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

1. Derive expressions for $E$ in terms of $Y$, $T$, and $D$.
1. If the columns of $Y$ are images from a training database, then what name do we give to the columns of $U$?

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