EE261 Problem Set 6

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

这次回顾Problem Set 6。

Problem 1

只要没有重合即可，考虑如下两种极限情形：

所以结果为

$$B_2<s <2B_1$$

Problem 2

(a)利用$\text{III}$函数对原式进行化简可得

$$\sum_{k=-\infty}^{\infty} T p(t-k T) =T p(t) * \text{III}_T (t)$$

取傅里叶变换可得

$$\begin{aligned} \mathcal F g(s) &= \mathcal Ff(s) * \left(T\mathcal Fp(s) \frac 1 T \text{III}_{\frac 1T} (t) \right)\\ &=\mathcal Ff(s) * \left( \mathcal Fp(s) \text{ III}_{\frac 1T} (s) \right)\\ &= \mathcal Ff(s) * \left( \sum_{k=-\infty}^{\infty} \mathcal Fp(s) \delta \left(s- \frac k T \right) \right)\\ &= \mathcal Ff(s) * \left( \sum_{k=-\infty}^{\infty} \mathcal Fp \left( \frac k T \right) \delta \left(s- \frac k T \right) \right)\\ &= \sum_{k=-\infty}^{\infty} \mathcal Fp \left( \frac k T \right) \left(\mathcal Ff(s) * \delta \left(s- \frac k T \right)\right) \\ &=\sum_{k=-\infty}^{\infty} \mathcal Fp \left( \frac k T \right) \mathcal Ff\left( s-\frac k T \right) \end{aligned}$$

取逆变换可得

$$g(s)=\sum_{k=-\infty}^{\infty} \mathcal Fp \left( \frac k T \right) f\left( s-\frac k T \right)$$

对比采样定理

$$f(t)=\sum_{k=-\infty}^{\infty} f\left(\frac{k}{p}\right) \operatorname{sinc} p\left(t-\frac{k}{p}\right)$$

只要利用第一个式子求解出第二个式子的系数即可，其条件为采样间隔大于带宽，即$\frac 1 T >2 B$

Problem 3

因为

$$f(x) = \frac{1}{2} \left( e^{-2\pi ix} + e^{2\pi ix}\right)$$

取傅里叶变换可得

$$\mathcal F f(s) = \frac{1}{2}\left(\delta (s-1)+\delta(s+1)\right)$$

按$\frac 2 3$采样即使用$\text{III}_{\frac 2 3}$，所以

$$\begin{aligned} \mathcal F f(s) * \text{III}_{\frac 2 3} &= \frac{1}{2}\left(\delta(s-1)+\delta(s+1)\right) * \sum_{i=1}^\infty \delta \left(x- \frac 2 3k \right) \\ &= \frac 12 \sum_{i=1}^\infty \delta \left(s- \frac 2 3k -1 \right) +\frac 12 \sum_{i=1}^\infty \delta \left(s- \frac 2 3k +1\right) \end{aligned}$$

按照$\frac 2 3 $赫兹截断，那么

$$\begin{aligned} \Pi_{\frac 23 } \left(\mathcal F f(s) * \text{III}_{\frac 2 3} \right) &=\frac 12 \left(\delta \left(s -\frac 13 \right)+ \delta \left(s +\frac 13 \right) \right)+ \frac 12\left(\delta \left(s -\frac 13 \right)+ \delta \left(s +\frac 13 \right) \right)\\ &=\delta \left(s -\frac 13 \right)+ \delta \left(s +\frac 13 \right) \end{aligned}$$

取傅里叶逆变换可得

$$e^{-2\pi it \frac 13 } + e^{2\pi it \frac 13 } =2\cos \left(\frac 2 3 \pi t \right)$$

Problem 4

由采样定理，我们得到

$$g(t)=\sum_{k=-\infty}^{\infty} g\left(t_{k}\right) \operatorname{sinc} p'\left(t-t_{k}\right)$$

其中

$$p' \ge p$$

对上式积分

$$\begin{aligned} \int_{-\infty}^{\infty} g(t) d t &= \int_{-\infty}^{\infty} \sum_{k=-\infty}^{\infty} g\left(t_{k}\right) \operatorname{sinc} p'\left(t-t_{k}\right)d t \\ &= \sum_{k=-\infty}^{\infty} g\left(t_{k}\right) \int_{-\infty}^{\infty} \operatorname{sinc} p'\left(t-t_{k}\right)d t \end{aligned}$$

现在计算

$$\int_{-\infty}^{\infty} \operatorname{sinc} p'\left(t-t_{k}\right)d t$$

我们有

$$\begin{aligned} \mathcal F \operatorname{sinc} p'\left(t-t_{k}\right) &= e^{-2\pi i p't_k s} \mathcal F\operatorname{sinc} p' t \\ &= e^{-2\pi i p't_k s} \frac 1 {p'} \Pi_{ {p'} } (s) \end{aligned}$$

注意到上述积分为傅里叶变换在$s=0$处的值，所以

$$\int_{-\infty}^{\infty} \operatorname{sinc} p'\left(t-t_{k}\right)d t = \frac 1 {p'}$$

因此

$$\begin{aligned} \int_{-\infty}^{\infty} g(t) d t &= \sum_{k=-\infty}^{\infty} g\left(t_{k}\right) \int_{-\infty}^{\infty} \operatorname{sinc} p'\left(t-t_{k}\right)d t \\ &=\frac 1 {p'}\sum_{k=-\infty}^{\infty} g\left(t_{k}\right) \end{aligned}$$

Problem 5

% (a)
d1 = imread("man.gif");
%转换类型
%data = im2double(data);
d2 = d1(:);
d3 = double(d2);
d4 = d3 / max(d3);
n = length(d4);

% (b)
x = fft(d4);
l = length(x);
% 中心化
x_center = [x(l/2 + 1: end); x(1: l/2)];
plot([-l/2: l / 2 - 1], abs(x_center))

% (c)
E_total = norm(x_center) ^ 2;
E_partial = zeros(1, l / 2);
E_partial(1) = abs(x(1)) ^ 2;
for i = 2: (l / 2 - 1)
    E_partial(i) = E_partial(i - 1) + 2 * abs(x(i)) ^ 2;
end
%比例
ratio = E_partial / E_total;

Alpha = [0.9, 0.95, 0.99];
%Type = ['nearest', 'linear'];
cnt = 1;
for i = 1: 2
    for j = 1: 3
        %type = Type(i);
        if i == 1
            type = 'nearest';
        else
            type = 'linear';
        end
        alpha = Alpha(j);
        %计算p
        p = double(find_p(ratio, alpha));
        %采样频率
        rate = floor(n / (2 * p));
        %索引
        index = 1: rate: n;
        %对应的值
        x_res = d4(index);
        %插值
        res = interp1(index, x_res, 1: n, type);
        %恢复图像
        res = max(d3) * res;
        res = reshape(res, 256, 256);
        %作图
        figure(cnt);
        imagesc(res); 
        colormap('gray');
        title(['\alpha =' num2str(alpha) ', ', type]);
        cnt = cnt + 1;
    end
end

结果如下：