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(u,v)&=H(u)G(v) \\ &= \pi \frac{ e^{j 2\pi x} \sin(\pi y)}{\pi y} \\ &= \pi \delta(u-x)rect(v) \end{align} $
b)
$ \begin{align} f(x,y)&=rect(x-x_0) \\ &= rect(x-x_0) \cdot 1\\ &=h(x)g(y) \end{align} $
$ \begin{align} F(u,v)&=H(u)G(v) \\ &= sinc(u) e^{-j 2\pi x_0} \delta(v)\\ \end{align} $
c)
$ \begin{align} f(x,y)&=cos(\pi x) \\ &= cos(\pi x) \cdot 1\\ &=h(x)g(y) \end{align} $
$ \begin{align} F(u,v)&=H(u)G(v) \\ &= \frac{1}{2} [\delta (u-\frac{1}{2})+\delta (u+\frac{1}{2})] \delta(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) = \frac{1}{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} $
