HW6

Problems:
#5.91, 94, 97, 100

Problems: [click here]
Due: May 9 (Thursday)
Submit: To the TA 

Project Final Report

Due: May 9 (Thursday) 1159pm

The requirement for the final report is no more than 
(1) eight (8) pages for individual project,
(2) ten (10) pages for group of 2 students,
(3) twelve (12) pages for group of 3 students.

Your project report should contain some background on the topic, the research methodology and your findings with enough supports. It will be graded based on, but not limited to, the following criteria (max 3% each)
1) accuracy;
2) execution;
3) presentation;
4) creativeness.

Any evidence of word-for-word copying from books or webpage will result in a lowered grade. You should research ideas for your group in the library and on the world-wide-web. Your paper should reference papers, books, etc. that you use in your work. 

Please email me a zip file of the final report and any code developed.

Lecture 19 (Apr 15)

Image Denoising

Ideal low pass filter:

$$

\hat{h}_{p,q}= \left\{

\begin{array}{cc}

1 & \mbox{ if $D(p,q)\le D_0$} \\

0 & \mbox{ otherwise.}

\end{array}

\right.

$$

Butterworth low pass filter

$$

\hat{h}_{p,q}=\frac{1}{1+[D(p,q)/D_0]^{2n}} \, .

$$

Gaussian low pass filter

$$

\hat{h}_{p,q}=e^{-D(p,q)^2}{2\sigma^2} \, .

$$

Image Edge Detection

Ideal high pass filter:

$$

\hat{h}_{p,q}= \left\{

\begin{array}{cc}

0 & \mbox{ if $D(p,q)\le D_0$} \\

1 & \mbox{ otherwise.}

\end{array}

\right.

$$

Butterworth high pass filter

$$

\hat{h}_{p,q}=1-\frac{1}{1+[D(p,q)/D_0]^{2n}}=\frac{1}{1+[D_0/D(p,q)]^{2n}} \, .

$$

Gaussian low pass filter

$$

\hat{h}_{p,q}=1-e^{-D(p,q)^2/(2\sigma^2)} \, .

$$

Deconvolution

$g=h*f$, given both $g$ and $h$, want to determine $f$.

If only $g$ is given, we have a blind deconvolution problem.

$$
f=\mathcal{F}^{-1}\left[ \frac{\mathcal{F}(g)}{\mathcal{F}(h)} \right] \, .
$$

Reading materials

MATLAB demo files for the ideal low pass filter: [Driver.m] [ideal_low.m]
Lecture notes: p.94-100

Lecture 18 (Apr 12)

Two dimensional generalizations

1) Fourier Transform
$$
f(x,y)=\int \int \hat{f}(u,v) e^{i2\pi(ux+vy)} du dv \\
\hat{f}(u,v)=\int \int f(x,y) e^{-i2\pi(ux+vy)} dx dy \, .
$$
2) Discrete Fourier Transform
$$
f_{m,n}= \frac{1}{MN} \sum_{p=0}^{M-1} \sum_{q=0}^{N-1} \hat{f}_{p,q} e^{i 2\pi \left( \frac{pm}{M}+\frac{qn}{N} \right)} \\
\hat{f}_{p,q}=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} f_{m,n} e^{-i 2\pi \left( \frac{pm}{M}+\frac{qn}{N} \right)} \, .
$$

Some properties

1) Periodicity:
$$
\hat{f}_{p,q}=\hat{f}_{p+k_1 M,q}=\hat{f}_{p,q+k_2 N}=\hat{f}_{p+k_1 M,q+k_2 N}
$$
for any integer $k_1$ and $k_2$.
2) Translation:
$$
\mathcal{F} \left[
f_{m,n} e^{i 2\pi \left( \frac{p^*m}{M}+\frac{q^*n}{N} \right)} \right] = \hat{f}_{p-p^*,q-q^*}
$$
3) From (2), take $p^*=M/2$ and $q^*=N/2$, we have
$$
\mathcal{F} \left[
f_{m,n} (-1)^{m+n} \right] = \hat{f}_{p-M/2,q-N/2} \, .
$$

Image Enhancement

Design $\hat{h}$ such that $\tilde{f}=\mathcal{F}^{-1}(\hat{h} \cdot \hat{g})$ achieves some properties.

Reading Materials:

Lecture Notes: p.94

HW5

Problems:
#5.60, 61, 63, 67, 75
#6.10 [Driver_FFT1.m] [Driver_FFT2.m], 11 (Use the expressions in Sec 3.5 and 3.6) [Fig4.11(a)]

Problems: [click here]
Due: May 3 (Friday)
Submit: To the TA in the tutorial

Lecture 17 (Apr 8)

Discrete Fourier Transform

$$

\hat{f}_n = \sum_{m=0}^{N-1} f_m e^{-2\pi i mn/N}

$$

Inverse Discrete Fourier Transform

$$
f_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{f}_n e^{2\pi i mn/N}
$$

Fourier Matrix: $F_N=[\omega^{mn}]$ where $\omega=e^{-2\pi i/N}$.

The inverse of the Fourier matrix: $F_N^{-1}=\frac{1}{N} \bar{F}_N$.

Fast Fourier Transform

Reading Materials

Lecture Notes: p.92-94

Lecture 16 (Apr 1)

Convolution Theorem

$\mathcal{F}(f*g)(s)= \hat{f}(s) \cdot \hat{g}(s)$.

Discrete Fourier Transform

$$
\hat{f}_n = \sum_{m=0}^{N-1} f_m e^{-2\pi i mn/N}
$$

Inverse Discrete Fourier Transform

$$
f_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{f}_n e^{2\pi i mn/N}
$$

Fourier Matrix: $F_N=[\omega^{mn}]$ where $\omega=e^{-2\pi i/N}$.

The inverse of the Fourier matrix: $F_N^{-1}=\frac{1}{N} \bar{F}_N$.

Reading Materials

Lecture Notes: p.90-92