EE261 Problem Set 4

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

这次回顾Problem Set 4。

Problem 1

(a)$f \star g$很大说明$g$平移后和$f$很接近；很小说明不接近；正表示平移后增长趋势接近；负表示平移后增长趋势不同

(b)首先

$$\begin{aligned} (f \star g)(x) &=\int_{-\infty}^{\infty} f(y) g(x+y) d y \\ &= \int_{-\infty}^{\infty} f^{-}(-y) g(x+y) d y \\ &= (f^{-} * g)(x) \end{aligned}$$

另一方面

$$\begin{aligned} (f \star g)(x) &=\int_{-\infty}^{\infty} f(y) g(x+y) d y \\ &= \int_{-\infty}^{\infty} f(y) g^{-}(-x-y) d y \\ &= \left(f * g^{-}\right)(-x)\\ &=\left(f * g^{-}\right)^{-}(x) \end{aligned}$$

注意到

$$\begin{aligned} (g \star f)(x) &=\left(g * f^{-}\right)^{-} \\ &=\left( f^{-} * g\right)^{-} \\ &\neq (f^{-} * g) \end{aligned}$$

所以

$$f \star g\neq g \star f$$

(c)

$$\begin{aligned} (f \star \left(\tau_{b} g\right))(x) &=\int_{-\infty}^{\infty} f(y) g(x+y-b) d y \\ &= \int_{-\infty}^{\infty} f(y) g(x-b+y) d y \\ &= (f \star g)(x-b) \\ &=(\tau_{b}(f \star g))(x) \end{aligned}$$

所以

$$f \star\left(\tau_{b} g\right)=\tau_{b}(f \star g)$$

另一方面

$$\begin{aligned} ( \left(\tau_{b} f\right) \star g)(x) &=\int_{-\infty}^{\infty} f(y-b) g(x+y) d y \\ &= \int_{-\infty}^{\infty} f(y-b) g(x+b+y-b) d y \\ &=\int_{-\infty}^{\infty} f(y) g(x+b+y) d y \\ &=(f \star g)(x+b)\\ &=(\tau_{-b}(f \star g))(x) \end{aligned}$$

Problem 2

(a)

$$\begin{aligned} \Pi \star \Pi &=\Pi^{-} * \Pi\\ &=\Pi * \Pi\\ &=\Lambda \end{aligned}$$

(b)只要证明如下事实即可

$$\int_{-\infty}^{\infty} f(y) f(x+y) d y \le \int_{-\infty}^{\infty} f(y) f(y) d y=\int_{-\infty}^{\infty} f(y)^2 d y$$

回顾柯西不等式

$$\int_{-\infty}^{\infty} f(y) g(y) d y \leq\left\{\int_{-\infty}^{\infty} f(y)^{2} d y\right\}^{1 / 2}\left\{\int_{-\infty}^{\infty} g(y)^{2} d y\right\}^{1 / 2}$$

取

$$\begin{aligned} f(y) & = f(y)\\ g(y)&= f(x+y) \end{aligned}$$

那么

$$\begin{aligned} \int_{-\infty}^{\infty} f(y) f(x+y) d y & \le \left\{\int_{-\infty}^{\infty} f(y)^{2} d y\right\}^{1 / 2}\left\{\int_{-\infty}^{\infty} f(x+y)^{2} d y\right\}^{1 / 2}\\ &=\int_{-\infty}^{\infty} f(y)^2 d y \end{aligned}$$

等号成立当且仅当

$$f(y) = k f(x+y)$$

(c)

$$\begin{aligned} \mathcal{F}(f \star f) &=\mathcal{F}(f^{-} * f) \\ &=\mathcal{F}f^{-} \mathcal{F}f \\ &=(\mathcal{F}f)^{-} \mathcal{F}f \\ &=\overline {\mathcal F f} \mathcal F f\\ &=|\mathcal{F} f|^{2} \end{aligned}$$

(d)

$$\begin{aligned} (f \star f_{r})(t) &= \left(f \star (\alpha\left(\tau_{2 T} f\right)+n)\right)(t)\\ &=\alpha (f\star (\tau_{2 T} f)) (t) +(f \star n) t\\ &=\alpha \tau_{2 T}(f\star f)(t) +C \\ &= \alpha (f\star f)(t-2T) +C \\ &\le \alpha (f\star f)(0)+C \end{aligned}$$

当满足如下条件时取极大值

$$t_0- 2T = 0 \Rightarrow T=\frac{t_0}{2}$$

Problem 3

Rectangle Window

$$\mathcal F\Pi = \text{sinc}$$

Triangular Window

记

$$g(t)= 2 \Lambda(2 t)$$

那么

$$\mathcal F g (s)= 2 \times \frac 1 2 \times \text{sinc}^2 \left(\frac s 2 \right)=\text{sinc}^2 \left(\frac s 2 \right)$$

Hamming Window

$$\begin{aligned} \mathcal F w( s) &=\int_{-\frac 12 }^{\frac 12 } e^{-2\pi i st}\cos^2(\pi t) dt \\ &=\int_{-\frac 12 }^{\frac 12 } e^{-2\pi i st} \left( \frac{e^{-i\pi t}+e^{i\pi t}}{2}\right)^2 dt \\ &=\frac 1 4\int_{-\frac 12 }^{\frac 12 } e^{-2\pi i st} \left(e^{-2\pi i t}+e^{2\pi i t} +2 \right) dt \\ &=\frac 1 4 \int_{-\frac 12 }^{\frac 12 } \left(e^{-2\pi it(1+s)}+e^{2\pi it(1-s)} +2e^{-2\pi i st} \right) dt \\ &=\frac 1 4 \left(\frac{\sin (\pi (1+s))}{\pi (1+s)}+ \frac{\sin (\pi (1-s))}{\pi (1-s)} +2 \frac{\sin (\pi s)}{\pi s}\right)\\ &=\frac 1 4 \text{sinc} (1+s) +\frac 1 4 \text{sinc} (s-1) +\frac 1 2 \text{sinc} (s) \end{aligned}$$

Problem 4

现在对方程两边关于$x$做傅里叶变换，首先对右边做傅里叶变换得到

$$D\mathcal{F} f_{x x}(s, t)=D(2 \pi i s)^{2} \mathcal{F} f(s, t)=-4D \pi^{2} s^{2} \mathcal{F} f(s, t)$$

其次对左边做傅里叶变换得到

$$\begin{aligned} \mathcal{F} f_{t}(s, t) &=\int_{-\infty}^{\infty} f_{t}(x, t) e^{-2 \pi i s x} d x \quad \text { (Fourier transform in } x ) \\ &=\int_{-\infty}^{\infty} \frac{\partial}{\partial t} f(x, t) e^{-2 \pi i s x} d x \\ &=\frac{\partial}{\partial t} \int_{-\infty}^{\infty} f(x, t) e^{-2 \pi i s x} d x\\ &=\frac{\partial}{\partial t} \hat{f}(s, t) \end{aligned}$$

所以原方程可以化为

$$\frac{\partial \mathcal{F} f(s, t)}{\partial t}=-4D \pi^{2} s^{2} \mathcal{F} f(s, t)$$

因此

$$\mathcal{F}f(s, t)=\mathcal{F} f(s, 0) e^{-4 D\pi^{2} s^{2} t}$$

最后计算初值$\mathcal{F} f(s, 0)$：

$$\begin{aligned} \mathcal{F} f(s, 0) &=\mathcal{F}\delta (s)\\ &=1\end{aligned}$$

从而

$$\mathcal{F} f(s, t)=\mathcal{F}\delta (s) e^{-4 D\pi^{2} s^{2} t} = e^{-4 D\pi^{2} s^{2} t}$$

设

$$h(x)= e^{-\pi x^2}$$

回顾之前的结论，我们有

$$\mathcal F h(s) = h(s)$$

设

$$g(x, t) = a h\left(bx\right)$$

从而

$$\begin{aligned} \mathcal{F} g(s, t) & =\mathcal{F} a h\left(bx\right)\\ &=\frac a {|b |} e^{-\frac{\pi s^2}{b^2} } \end{aligned}$$

因此对于

$$e^{-4 D\pi^{2} s^{2} t}$$

我们有

$$\begin{aligned} 4 D\pi^{2} s^{2} t& =\frac{\pi s^2}{b^2} \\ \frac a {|b |} & =1 \end{aligned}$$

即

$$\begin{aligned} b &=\frac {1}{2 \sqrt{ D \pi t}} \\ a&=\frac {1}{2 \sqrt{ D \pi t}} \end{aligned}$$

所以

$$g(x, t) = \frac {1}{2 \sqrt{ D \pi t}} h\left(\frac {x}{2 \sqrt{ D \pi t}} \right) =\frac {1}{2 \sqrt{ D \pi t}} e^{- \frac{x^2}{4D t}}$$

因此

$$f(x, t) = \frac {1}{2 \sqrt{ D \pi t}} e^{- \frac{x^2}{4D t}}$$

(b)期望为

$$0$$

方差为

$$2Dt$$

(c)将

$$D=\frac{k T}{6 \pi \eta R}$$

代入

$$f(x, t) = \frac {1}{2 \sqrt{ D \pi t}} e^{- \frac{x^2}{4D t}}$$

可得

$$f(x, t) = \frac{ \sqrt{3\eta R}} {\sqrt{2KTt}}e^{- \frac{3\pi \eta R}{2KTt}x^2}$$

从上式可以可以看出，温度越高，例子波动幅度越大；半径越大，粒子波动幅度越小。

Problem 5

这部分参考了解答的代码。

% C B D A
% | | | |
% A B C D
[y, Fs] = audioread("PS-4-scramble.wav");
%变换为频域
z = fft(y);
plot(abs(z))
%plot(imag(z))
%步长，注意有对称性
n = floor(length(y) / 8);
C = z(1: n);
B = z(n + 1: 2 * n);
D = z(2 * n + 1: 3 * n);
A = z(3 * n + 1: 4 * n);
res = [A; B; C; D; flipud([conj(A);conj(B);conj(C);conj(D)])];
%取实部
r = real(ifft(res));
audiowrite('res.wav', r, Fs);