m (Corrected from H(0,v) = 2(1+cos(v)) to 1/2(1+cos(v)))
 
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[[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]]
 
[[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]][[Category:HW11ECE438F13]]
  
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=Hw11_ECE438F13sln=
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==Question 1==
  
  
 +
a)
  
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Put your content here . . .
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<math>\begin{align}
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f(x,y)&=\frac{ e^{j 2\pi x} \sin(\pi y)}{y} \\
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&= \pi \frac{ e^{j 2\pi x} \sin(\pi y)}{\pi y} \\
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&= \pi  e^{j 2\pi x} sinc(y) \\
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&=h(x)g(y)
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\end{align}</math>
  
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<math>\begin{align}
 +
F(u,v)&=H(u)G(v) \\
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&= \pi \frac{ e^{j 2\pi x} \sin(\pi y)}{\pi y} \\
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&= \pi \delta(u-x)rect(v)
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\end{align}</math>
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 +
b)
 +
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<math>\begin{align}
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f(x,y)&=rect(x-x_0) \\
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&= rect(x-x_0) \cdot 1\\
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&=h(x)g(y)
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\end{align}</math>
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 +
<math>\begin{align}
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F(u,v)&=H(u)G(v) \\
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&= sinc(u) e^{-j 2\pi x_0} \delta(v)\\
 +
\end{align}</math>
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 +
c)
 +
 +
<math>\begin{align}
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f(x,y)&=cos(\pi x) \\
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&= cos(\pi x) \cdot 1\\
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&=h(x)g(y)
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\end{align}</math>
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 +
<math>\begin{align}
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F(u,v)&=H(u)G(v) \\
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&= \frac{1}{2} [\delta (u-\frac{1}{2})+\delta (u+\frac{1}{2})] \delta(v)\\
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\end{align}</math>
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----
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==Question 2==
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a)
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 +
<math>
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\begin{align}
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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] \\
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& -\frac{1}{4}x[m+1,n] + x[m,n] -\frac{1}{4}x[m,n-1] \\
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& -\frac{1}{8}x[m+1,n+1] + \frac{1}{2}x[m,n+1] -\frac{1}{8}x[m-1,n+1]
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\end{align}</math>
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 +
b) Yes. The coefficient matrix of h[m,n] can be written as product of two vectors.
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 +
<math>
 +
\begin{pmatrix}
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-\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \\
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-\frac{1}{4} & 1 & -\frac{1}{4} \\
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-\frac{1}{8} & \frac{1}{2} & -\frac{1}{8}
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\end{pmatrix} =
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\begin{pmatrix}
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\frac{1}{2}  \\
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1 \\
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\frac{1}{2}
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\end{pmatrix} \cdot
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\begin{pmatrix}
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-\frac{1}{4} & 1 & -\frac{1}{4}
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\end{pmatrix}
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</math>
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 +
Therefore the filter can be separate into two 1-D filters.
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<math>h_1[m] = -\frac{1}{4}\delta[m+1] + \delta[m] -\frac{1}{4}\delta[m-1]</math>
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<math>h_2[n] = \frac{1}{2}\delta[n+1] + \delta[n] +\frac{1}{2}\delta[n-1]</math>
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c)
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 +
<math>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</math>
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 +
<math>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</math>
 +
 +
Using the separability,
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 +
<math>H(\mu, \nu) = DSFT\{ h[m,n]\} = H_1(\mu)\cdot H_2(\nu) = (1-\frac{1}{2}cos\mu)(1+cos\nu)</math>
 +
 +
<math>H(\mu, 0) = 2(1-\frac{1}{2}cos\mu)</math>
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[[Image:HW8Q1fig1.jpg]]
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 +
<math>H(0, \nu) = \frac{1}{2}(1+cos\nu)</math>
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 +
[[Image:HW8Q1fig2.jpg]]
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 +
d)
 +
 +
<math>\begin{array}{cccccccccccc}
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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 \\
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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 \\
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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 \\
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-\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} \\
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-\frac{3}{8} & 1 & \frac{9}{8} & 1 & 1 & 1 & 1 & 1 & \frac{9}{8} & 1 & -\frac{3}{8} \\
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-\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\
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-\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\
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-\frac{1}{2} & \frac{3}{2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{3}{2} & -\frac{1}{2} \\
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-\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} \\
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-\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}</math>
 +
----
  
  
  
 
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Latest revision as of 23:11, 10 December 2013


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) $

HW8Q1fig1.jpg

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

HW8Q1fig2.jpg

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} $


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva