EE261 Problem Set 3

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

这次回顾Problem Set 3。

Problem 1

因为

$\mathcal{F} f(s)=\int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) d t$

所以

$\begin{aligned} W_{\mathcal Ff} &=\frac{1}{\mathcal{F} f(0)} \int_{-\infty}^{\infty} \mathcal{F} f(s) d s\\ &=\frac 1 {\int_{-\infty}^{\infty} f(t) d t} \mathcal F(\mathcal F f)(0)\\ &= \frac 1 {\int_{-\infty}^{\infty} f(t) d t} f^{-}(0)\\ &= \frac 1 {\int_{-\infty}^{\infty} f(t) d t} f(0)\\ &=\frac 1 {W_f} \end{aligned}$

Problem 2

因为

$\sin c(x) =\frac {\sin \pi x}{ \pi x}$

所以图像中的函数为

$f(x)=a\sin c\left(b(x- c)\right)$

其傅里叶变换为

$\mathcal F f= ae^{2\pi i s(- c)} \frac 1 {|b|} \Pi\left(\frac s b\right) =\frac{ae^{-2\pi ics}}{|b|}\Pi\left(\frac s b\right)$

Problem 3

(a)

$g_1(t)= f(-t)$

所以

$\mathcal F g_1=\mathcal F f^{-}=(\mathcal{F} f)^{-} \Rightarrow \mathcal F g_1(s)=F(-s)$

(b)

$g_2(t)=g_1(t-2)$

所以

$\mathcal F g_2(s)= e^{-4\pi is } \mathcal Fg_1(s) \Rightarrow \mathcal F g_2(s)=e^{-4\pi is } F(-s)$

(c)

$g_3(t)=f(t-1)$

所以

$\mathcal F g_3(s)= e^{-2\pi is} \mathcal Ff(s) \Rightarrow \mathcal F g_3(s)=e^{-2\pi is } F(s)$

(d)

$g_4(t)=f\left(\frac t 2\right)$

所以

$\mathcal F g_4(s)= 2 \mathcal Ff(2s) \Rightarrow \mathcal F g_4(s)=2 F(2s)$

(e)

$g_5(t)=f(t)+f(t+2)$

所以

$\mathcal F g_5(s)= \mathcal Ff(s)+e^{-2\pi is (-2)} \mathcal Ff(s) \Rightarrow \mathcal F g_5(s)= (1+e^{4\pi is} ) F(s)$

(f)

$g_6(t)= g_1(t)+f(t)$

所以

$\mathcal F g_6(s)= \mathcal Ff(s)+\mathcal Ff(-s) \Rightarrow \mathcal F g_6(s)= F(s) +F(-s)$

Problem 4

(a)因为

$\Pi_a(t)=\left\{\begin{array}{ll}{1} & {|t|<a / 2} \\ {0} & {|t| \geq a / 2}\end{array}\right.$

所以如果$0<t<a$，那么

$\begin{aligned} (\Pi_{a} * \Pi_{a}) (t) &=\int_{-\infty}^{\infty} \Pi_{a} (t-x) \Pi_{a} (x) d x\\ &=\int_{t-\frac a 2}^{\frac a 2} dx \\ &=a-t \end{aligned}$

如果$-a<t<0$，那么

$\begin{aligned} (\Pi_{a} * \Pi_{a}) (t) &=\int_{-\infty}^{\infty} \Pi_{a} (t-x) \Pi_{a} (x) d x\\ &=\int_{-\frac a 2}^{t+\frac a 2} dx \\ &=a+t \end{aligned}$

其余情形都有

$(\Pi_{a} * \Pi_{a}) (t) =0$

所以

$(\Pi_{a} * \Pi_{a}) (t)=\left\{\begin{array}{ll}{a-|t|} & {|t|<a} \\ {0} & {|t| \geq a}\end{array}\right.$

(b)因为

$\begin{aligned} (f * f) (-t) &=\int_{-\infty}^{\infty}f (-t-x) f (x) d x\\ &=\int_{-\infty}^{\infty}f (t+x) f (-x) d x &偶函数 \\ &=\int_{\infty}^{-\infty}f (t-x) f (x) d(-x) &令x=-x \\ &= \int_{-\infty}^{\infty}f (t-x) f (x) d x\\ &=(f * f) (t) \end{aligned}$

即$(f * f) (t)$是偶函数，所以只需讨论$t\ge 0$的情形。

如果$t\ge 0$，那么

$\begin{aligned} (f * f) (t) &=\int_{-\infty}^{\infty}f (t-x) f (x) d x\\ &=\int_{-\infty}^{\infty} e^{-|t-x|-|x|} d x\\ &=\int_{-\infty}^{0} e^{-t+2x} d x+\int_{0}^{t} e^{-t} d x+\int_{t}^{\infty} e^{t-2x} d x\\ &=e^{-t} \frac 1 2 + e^{-t}t + e^t \frac 12 e^{-2t}\\ &= e^{-t}(t+1) \end{aligned}$

利用对称性可得当$t<0$时，

$(f * f) (t)=(f * f) (-t)= e^{t}(-t+1)$

因此

$(f * f) (t)=e^{-|t|}(|t|+1)$

(c)

$\begin{aligned} (g * g) (t) &=\int_{-\infty}^{\infty}g (t-x) g (x) d x\\ &=\int_{-\infty}^{\infty} e^{-\pi (t-x)^{2} } e^{-\pi x^2} d x\\ &=\int_{-\infty}^{\infty} e^{-\pi (2x^2-2tx +t^2) } d x\\ &=\int_{-\infty}^{\infty} e^{-\pi \left(2(x-\frac t 2)^2 +\frac {t^2}2 \right)} d x\\ &=e^{-\frac {\pi t^2}{2}} \int_{-\infty}^{\infty} e^{-2\pi(x-\frac t 2)^2 } d x\\ &=\frac 1 {\sqrt{2}} e^{-\frac {\pi t^2}{2}} \end{aligned}$

(e)推广：

$(g * g * \ldots * g)(t) =\frac 1 {\sqrt n} e^{-\frac {\pi t^2}{n}}$

Problem 5

(a)

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

所以

$(f^{-} * g^{-}) =(f* g)^{-}$

另一方面

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

所以

$(f * g^{-})=(f^{-} * g)^{-}$

(b)

$\begin{aligned} (\left(\tau_{b} f\right) * g)(t) &=\int_{-\infty}^{\infty}\tau_{b} f (t-x) g(x) dx\\ &=\int_{-\infty}^{\infty} f(t-x-b) g(x) dx\\ &= \int_{-\infty}^{\infty} f(t-b-x) g(x) dx\\ &= (f * g)(t-b)\\ &= (\tau_{b}(f * g))(t)\\ &=\int_{-\infty}^{\infty} f(t-x) g(x-b) dx & x=x+b\\ &=\int_{-\infty}^{\infty} f(t-x) \tau_{b} g (x) dx \\ &=(f *\left(\tau_{b} g\right))(t) \end{aligned}$

等式的含义为位移函数的卷积等于卷积的位移。

只需证明$f$的周期为$T$的情形：

$\begin{aligned} (f*g)(t-T) &=\tau_T(f*g)(t)\\ &=(\left(\tau_{T} f\right) * g)(t)\\ &= (f*g)(t) \end{aligned}$

(c)

$\begin{aligned} (\left(\sigma_{a} f\right) * g)(t) &= \int_{-\infty}^{\infty} f(at-ax) g(x)dx\\ &= \int_{-\infty}^{\infty} f(at-ax) \sigma_{1/a} g(ax) dx\\ &= \frac 1{|a|}\int_{-\infty}^{\infty} f(at-x) \sigma_{1/a} g(x) dx & x=ax\\ &=\frac 1{|a|} \sigma_{a}\left(f *\left(\sigma_{1 / a} g\right)\right)(t) \end{aligned}$

所以

$\left(\sigma_{a} f\right) * g=\frac{1}{|a|} \sigma_{a}\left(f *\left(\sigma_{1 / a} g\right)\right)$

另一方面

$\begin{aligned} (\left(\sigma_{a} f\right) *\left(\sigma_{a} g\right))(t) &= \int_{-\infty}^{\infty} f(at-ax) g(ax)dx\\ &= \frac 1{|a|}\int_{-\infty}^{\infty} f(at-x) g(x) dx & x=ax\\ &=\frac 1{|a|} \sigma_{a}\left(f *g\right)(t) \end{aligned}$

所以

$\left(\sigma_{a} f\right) *\left(\sigma_{a} g\right)=\frac{1}{|a|} \sigma_{a}(f * g)$

Problem 6

首先计算卷积

$\begin{aligned} \int_{-\infty}^{\infty} \sin 2\pi(t-x) \sin 2\pi x dx &= \frac 1 2 \int_{-\infty}^{\infty} \left( \cos(2\pi(t-2x)) +\cos(2\pi t)\right) dx \end{aligned}$

所以上述积分并不存在，上述例子说明两个周期函数的卷积并不存在。

Problem 7

(a)

$p_n=\sum_{i+j+k=n} p_ip_jp_k$

代码如下：

import numpy as np
import matplotlib.pyplot as plt

X = np.ones(6) / 6
Y = np.array([1, 1, 2, 2, 3, 3]) / 12
Z = np.array([4, 3, 2, 1, 1, 1]) / 12
data = np.arange(1, 7)
N = 6

####(a)
def P(n):
    p = 0
    for i in range(N):
        for j in range(N):
            for k in range(N):
                if i + j + k + 3 == n:
                    p += X[i] * Y[j] * Z[k]
    
    return p

res = [7, 8, 12]
p = 0
for n in res:
    p += P(n)
print(p)

0.2997685185185185

(b)注意到

$\begin{aligned} \mathbb E[X_i] &=\frac1 6 \sum_{i=1}^6 i\\ &=\frac 7 2\\ \mathbb E[Y_i] &= \frac {1\times1 +2\times 1+3\times 2 +4\times 2 +5\times 3+ 6\times 3} {12}\\ &=\frac {50}{12}\\ &=\frac {25}{6}\\ \mathbb E[Z_i] &= \frac {1\times 4 +2\times 3+3\times 2 +4\times 1 +5\times 1+ 6\times 1} {12}\\ &=\frac {31}{12} \end{aligned}$

由大数定律可得

$\begin{aligned} \lim_{N\to\infty} \frac{1}{3 N} \sum_{i=1}^{N}\left(X_{i}+Y_{i}+Z_{i}\right) &=\mathbb E[X_i]+\mathbb E[Y_i]+\mathbb E[Z_i]\\ &=\frac 1 3\left(\frac 7 2 + \frac {25}{6}+\frac {31}{12} \right)\\ &=\frac {41}{12} \end{aligned}$

(c)代码如下：

####(c)
def Print(n, M=100000):
    x = np.random.choice(data, p=X, size=(M, n))
    y = np.random.choice(data, p=Y, size=(M, n))
    z = np.random.choice(data, p=Z, size=(M, n))
    
    res = np.mean(x + y + z, axis=1) / 3
    plt.hist(res)
    plt.title("N={}".format(n))
    plt.show()
    
N = [2, 10, 100, 1000]
for n in N:
    Print(n)