Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published in Jul 2019)

## Problem 1

1. Calcualte 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

$G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n * \Delta d)\lambda_n^c$

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

$\lambda_n = (\lambda_n^b-\lambda_n^d) e^{-\int_{0}^{x}\mu(t)dt}d)\lambda_n^c$

$\hat{P}_n = \int_{0}^{x}\mu(t)dt= -log(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d})$

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 of $YY^t$ 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 $D$.

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

1. Derive an expression for $U$ in terms of $Y$, $T$, $D$.

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

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

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

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

therefore

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

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

## Alumni Liaison

Sees the importance of signal filtering in medical imaging Dhruv Lamba, BSEE2010