ee261 problem set 9

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

取逆变换可得

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

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

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

(a)

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

(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

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

从而

$$begin{aligned} mathcal F e^{-ar^2} &= mathcal F e^{-a(x^2+y^2)}\ &=mathcal F e^{-ax^2}mathcal F e^{-ay^2}\ &=frac pi a e^{-axi_1^2}e^{-axi_2^2} end{aligned}$$

(c)

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

所以

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

(d)

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

取傅里叶变换可得

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

(a)

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

(b)

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

(c)

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

(d)

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

(a)考虑$f$第$m$行$f_m[l]$，我们有

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

那么

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

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)

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

(c)

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

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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：

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function  = 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;