EE261 Problem Set 9

课程主页：https://see.stanford.edu/Course/EE261

这次回顾Problem Set 9。

Problem 1

$$\begin{aligned} \mathcal F g(x,y) &=\mathcal F \left(\Pi(x) \Pi(y) \right) \mathcal F \left(\Pi(x) \Pi(y) \right)\\ &= \mathcal F^2 (\Pi(x))\mathcal F^2 (\Pi(y))\\ &=\text{sinc}^2(\xi)\text{ sinc}^2(\eta)\\ &=\mathcal F(\Lambda(x)\Lambda(y)) \end{aligned}$$

取逆变换可得

$$g(x,y) = \Lambda(x)\Lambda(y)$$

Problem 2

注意汉高变换为

$$G(\rho)=2 \pi \int_{0}^{\infty} g(r) J_{0}(2 \pi r \rho) r d r$$

考虑$g(ar)$的汉高变换

$$\begin{aligned} 2 \pi \int_{0}^{\infty} g(a r) J_{0}(2 \pi r \rho) r d r &=\frac{2 \pi}{a} \int_{0}^{\infty} g(r') J_{0}(2 \pi (r' /a) \rho) r'/ a d r'\\ &=\frac{2 \pi}{a^2} \int_{0}^{\infty} g(r') J_{0}(2 \pi r'(\rho /a) ) r' d r'\\ &=\frac{1}{|a|^{2}} G\left(\frac{\rho}{a}\right) \end{aligned}$$

Problem 3

(a)

$$\begin{aligned} \mathcal F\left(\sin 2 \pi a x_{1} \sin 2 \pi b x_{2}\right) &= \mathcal F(\sin 2 \pi a x_{1})\mathcal F(\sin 2 \pi b x_{2})\\ &=\frac{1}{2 i}\left(\delta(\xi_1-a)-\delta(\xi_1+a)\right) \frac{1}{2 i}\left(\delta(\xi_2 -b)-\delta(\xi_2+b)\right)\\ &=-\frac 1 4 \left( \delta(\xi_1-a, \xi_2-b) -\delta(\xi_1+a, \xi_2-b) -\delta(\xi_1-a, \xi_2+b)+\delta(\xi_1+a, \xi_2+b) \right) \end{aligned}$$

(b)注意到我们有

$$\begin{aligned} \mathcal F e^{-\pi x^2} =\mathcal F e^{-\pi s^2} \end{aligned}$$

所以

$$\begin{aligned} \mathcal F e^{-ax^2} &=\mathcal F e^{-\pi \left(\frac{\sqrt a}{\sqrt \pi} x\right)^2 }\\ &= \sqrt{\frac \pi a} e^{-as^2} \end{aligned}$$

从而

$$\begin{aligned} \mathcal F e^{-ar^2} &= \mathcal F e^{-a(x^2+y^2)}\\ &=\mathcal F e^{-ax^2}\mathcal F e^{-ay^2}\\ &=\frac \pi a e^{-a\xi_1^2}e^{-a\xi_2^2} \end{aligned}$$

(c)

$$\cos (2 \pi c x) =\frac{1}{2}\left(e^{2\pi icx}+e^{-2\pi icx}\right)$$

所以

$$\begin{aligned} \mathcal F\left(e^{-2 \pi i(a x+b y)} \cos (2 \pi c x)\right) &=\mathcal F\left(e^{-2 \pi i(a x+b y)}\frac{1}{2}\left(e^{2\pi icx}+e^{-2\pi icx}\right)\right) \\ &=\frac 12 \left(\mathcal F (e^{-2 \pi i((a-c) x+b y)})+e^{-2 \pi i((a+c) x+b y)}\right)\\ &=\frac 12 \left(\mathcal F (e^{-2 \pi i(a-c) x}) \mathcal F (e^{-2 \pi ib y})+ \mathcal F (e^{-2 \pi i(a+c) x}) \mathcal F (e^{-2 \pi ib y}) \right)\\ &=\frac 12 \left(\delta(\xi_1 + a-c) \delta(\xi_2 +b) + \delta(\xi_1 + a+c) \delta(\xi_2 +b)\right)\\ &=\frac 12 \left(\delta(\xi_1 +a-c, \xi_2 +b)+\delta(\xi_1 +a+c, \xi_2 +b)\right) \end{aligned}$$

(d)

$$\begin{aligned} \cos (2 \pi(a x+b y)) &= \cos (2\pi a x) \cos(2\pi by) -\sin (2\pi a x) \sin(2\pi by) \end{aligned}$$

取傅里叶变换可得

$$\begin{aligned} \mathcal F\left(\cos (2 \pi(a x+b y))\right) &= \mathcal F \left(\cos (2\pi a x) \cos(2\pi by)\right) -\mathcal F \left(\sin (2\pi a x) \sin(2\pi by)\right)\\ &=\frac 1 4 \left(\left( \delta(\xi_1 -a)+\delta(\xi_1 +a)\right)\left( \delta(\xi_2 -b)+\delta(\xi_2 +b)\right) \right) \\ &+\frac 1 4 \left(\left( \delta(\xi_1 -a)-\delta(\xi_1 +a)\right)\left( \delta(\xi_2 -b)-\delta(\xi_2 +b)\right) \right)\\ &= \frac 1 2 \left(\delta(\xi_1 -a, \xi_2 -b) +\delta(\xi_1 +a, \xi_2 +b) \right) \end{aligned}$$

Problem 4

(a)

$$\begin{aligned} \mathcal F g(\xi_1, \xi_2) &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} g(x_1,x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} f(-x_1,x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (-\xi_1 x_1+\xi_2 x_2)} f(x_1,x_2) dx_1 dx_2\\ &=\mathcal F f(-\xi_1, \xi_2) \end{aligned}$$

(b)

$$\begin{aligned} \mathcal F h(\xi_1, \xi_2) &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} h(x_1,x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} f(x_1,-x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1-\xi_2 x_2)} f(x_1,x_2) dx_1 dx_2\\ &=\mathcal F f(\xi_1, -\xi_2) \end{aligned}$$

(c)

$$\begin{aligned} \mathcal F k(\xi_1, \xi_2) &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} k(x_1,x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} f(x_2,x_1) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_2 x_1+\xi_1 x_2)} f(x_1,x_2) dx_1 dx_2\\ &=\mathcal F f(\xi_2, \xi_1) \end{aligned}$$

(d)

$$\begin{aligned} \mathcal F m(\xi_1, \xi_2) &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} m(x_1,x_2) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (\xi_1 x_1+\xi_2 x_2)} f(-x_2,-x_1) dx_1 dx_2\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2\pi (-\xi_2 x_1-\xi_1 x_2)} f(x_1,x_2) dx_1 dx_2\\ &=\mathcal F f(-\xi_2, -\xi_1) \end{aligned}$$

Problem 5

(a)考虑$f$第$m$行$f_m[l]$，我们有

$$\begin{aligned} \mathcal F f_m[l] &=\sum_{n=0}^{N-1} f_{m}[n]\omega_{N}^{-l n}\\ &=\sum_{n=0}^{N-1}f[m,n]\omega_{N}^{-l n} \end{aligned}$$

那么

$$\begin{aligned} \mathcal{F} f[k, l] &=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} f[m, n] \omega_{N}^{-l n} \omega_{M}^{-k m}\\ &= \sum_{m=0}^{M-1} \mathcal F f_m[l]\omega_{M}^{-k m} \end{aligned}$$

所以可以先利用FFT计算$\mathcal F f_m[l] $，然后再利用FFT计算二维的结果。

(b)

$$\begin{aligned} \mathcal{F} g[k, l] &=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} g[m, n] \omega_{N}^{-l n} \omega_{M}^{-k m}\\ &=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} f[m, n] \omega_{M}^{m k_{0}} \omega_{N}^{n l_{0}} \omega_{N}^{-l n} \omega_{M}^{-k m}\\ &=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} f[m, n] \omega_{N}^{n (l_{0}-l)} \omega_{M}^{m (k_{0}-k)}\\ &=\mathcal F [ k-k_0,l-l_0] \end{aligned}$$

(c)

$$\begin{aligned} \mathcal F(f * g)[k, l] &= \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} (f * g)[m, n] \omega_{N}^{-l n} \omega_{M}^{-k m}\\ &= \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} \sum_{u=0}^{M-1} \sum_{v=0}^{N-1} f[u, v] g[m-u, n-v]\omega_{N}^{-l n} \omega_{M}^{-k m}\\ &= \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} \sum_{u=0}^{M-1} \sum_{v=0}^{N-1} f[u, v]\omega_{N}^{-l u} \omega_{M}^{-k v} g[m-u, n-v] \omega_{N}^{-l (n-u)} \omega_{M}^{-k(m-v)}\\ &=\left(\sum_{u=0}^{M-1} \sum_{v=0}^{N-1} f[u, v]\omega_{N}^{-l u} \omega_{M}^{-k v}\right) \left( \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} g[m-u, n-v] \omega_{N}^{-l (n-u)} \omega_{M}^{-k(m-v)}\right)\\ &= \mathcal F f[k, l]\times \mathcal F g[k, l] \end{aligned}$$

Problem 6

%(a)
data = imread("dog.jpg");
Max = 255;
data = im2double(data);
imshow(data);

%(b)
data_new = treat(data);
figure(1)
imshow(data_new);

%(c)
i = 2;
F = 0.5: -0.05: 0.1;
[Xmax ,Ymax] = size(data);
for f = 0.5: -0.05: 0.1
    h = LP_filter(Xmax, Ymax, f);
    data1 = real(ifft2(fft2(data) .* h));
    data_new = treat(data1);
    figure(i);
    suptitle(sprintf('Initial Image with Object Boundaries (f=%g)', f));
    imshowpair(data1, data_new, 'montage');
    i = i + 1;
end

函数treat：

function data_new = treat(data)

Max = 255;
Bx = [0, 0, 0; 1, -1, 0; 0, 0, 0];
By = [0, 1, 0; 0, -1, 0; 0, 0, 0];
datax = Max * conv2(data, Bx, 'same');
datay = Max * conv2(data, By, 'same');
data_new = data;
data_new(abs(datax) > 10) = 1;
data_new(abs(datay) > 10) = 1;