## Question 1

a)

\begin{align} f(x,y)&=\frac{ e^{j 2\pi x} \sin(\pi y)}{y} \\ &= \pi \frac{ e^{j 2\pi x} \sin(\pi y)}{\pi y} \\ &= \pi e^{j 2\pi x} sinc(y) \\ &=h(x)g(y) \end{align}

\begin{align} F(x,y)&=H(u)G(v) \\ &= \pi \frac{ e^{j 2\pi x} \sin(\pi y)}{\pi y} \\ &= \pi \delta(u-x)rect(v) \end{align}

## Question 2

a)

\begin{align} y[m,n] =& -\frac{1}{8}x[m+1,n-1] + \frac{1}{2}x[m,n-1] - \frac{1}{8}x[m-1,n-1] \\ & -\frac{1}{4}x[m+1,n] + x[m,n] -\frac{1}{4}x[m,n-1] \\ & -\frac{1}{8}x[m+1,n+1] + \frac{1}{2}x[m,n+1] -\frac{1}{8}x[m-1,n+1] \end{align}

b) Yes. The coefficient matrix of h[m,n] can be written as product of two vectors.

$\begin{pmatrix} -\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \\ -\frac{1}{4} & 1 & -\frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ 1 \\ \frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} -\frac{1}{4} & 1 & -\frac{1}{4} \end{pmatrix}$

Therefore the filter can be separate into two 1-D filters.

$h_1[m] = -\frac{1}{4}\delta[m+1] + \delta[m] -\frac{1}{4}\delta[m-1]$

$h_2[n] = \frac{1}{2}\delta[n+1] + \delta[n] +\frac{1}{2}\delta[n-1]$

c)

$H_1(\mu) = DTFT\{h_1[m]\} = -\frac{1}{4}e^{-j\mu(-1)} + e^{-j\mu(0)} -\frac{1}{4}e^{-j\mu(1)} = 1-\frac{1}{2}cos\mu$

$H_2(\nu) = DTFT\{h_2[n]\} = \frac{1}{2}e^{-j\nu(-1)} + e^{-j\nu(0)} +\frac{1}{2}e^{-j\nu(1)} = 1+cos\nu$

Using the separability,

$H(\mu, \nu) = DSFT\{ h[m,n]\} = H_1(\mu)\cdot H_2(\nu) = (1-\frac{1}{2}cos\mu)(1+cos\nu)$

$H(\mu, 0) = 2(1-\frac{1}{2}cos\mu)$

$H(0, \nu) = 2(1+cos\nu)$

d)

$\begin{array}{cccccccccccc} 0 & 0 & 0 & 0 & -\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} & 0 & 0 & 0& 0 \\ 0 & 0 & 0 & -\frac{1}{8} & \frac{1}{8} & \frac{10}{8} & \frac{1}{8} & -\frac{1}{8} & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{8} & \frac{1}{8} & \frac{7}{8} & \frac{10}{8} & \frac{7}{8} & \frac{1}{8} & -\frac{1}{8} & 0 & 0 \\ 0 & -\frac{1}{8} & \frac{1}{8} & \frac{7}{8} & \frac{9}{8} & 1 & \frac{9}{8} & \frac{7}{8} & \frac{1}{8} & -\frac{1}{8} & 0 \\ -\frac{1}{8} & \frac{1}{8} & \frac{7}{8} & \frac{9}{8} & 1 & 1 & 1 & \frac{9}{8} & \frac{7}{8} & \frac{1}{8} & -\frac{1}{8} \\ -\frac{3}{8} & 1 & \frac{9}{8} & 1 & 1 & 1 & 1 & 1 & \frac{9}{8} & 1 & -\frac{3}{8} \\ -\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\ -\frac{3}{8} & \frac{9}{8} & \frac{6}{8} & \frac{6}{8} & \frac{6}{8} & \frac{6}{8} & \frac{6}{8} & \frac{6}{8} & \frac{6}{8} & \frac{9}{8} & -\frac{3}{8} \\ -\frac{1}{8} & \frac{3}{8} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{3}{8} & -\frac{1}{8} \end{array}$

## Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal