EE261 Problem Set 2

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

这次回顾Problem Set 2。

Problem 1

(1)

对于$n\neq 0$

$$\begin{aligned} c_{n} &=\frac{1}{T} \int_{0}^{T} e^{-2 \pi i n t / T} f(t) d t\\ &=\frac{1}{2} \int_{0}^{2} e^{- \pi i n t}t^2 d t\\ &=\frac{1}{2} \left( t^2 \frac{ e^{- \pi i n t}}{- \pi i n }\Big|_{t=0}^{t=2} - \int_{0}^{2} \frac{ e^{- \pi i n t}}{- \pi i n }\times 2td t\right)\\ &=-\frac 2{\pi i n} +\frac 1 { \pi i n}\int_{0}^{2}{ e^{- \pi i n t}}\times td t \\ &= -\frac 2{\pi i n}+\frac 1 { \pi i n} \left(t \frac{ e^{- \pi i n t}}{- \pi i n }\Big|_{t=0}^{t=2} -\int_{0}^{2} \frac{ e^{- \pi i n t}}{- \pi i n }d t\right) \\ &=-\frac 2{\pi i n} +\frac 2{\pi^2n^2} \end{aligned}$$

对于$n=0$

$$\begin{aligned} c_{n} &=\frac{1}{T} \int_{0}^{T} f(t) d t\\ &=\frac{1}{2} \int_{0}^{2} t^2 d t\\ &=\frac 4 3 \end{aligned}$$

(2)

$$t^2=\frac 4 3+\sum_{n=-\infty,n\neq 0}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) e^{\pi int}$$

取$t=0$，得到

$$\begin{aligned} \frac {0+2^2} 2 &= 2\\ &=\frac 4 3 +\sum_{n=-\infty,n\neq 0}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) \\ &=\frac 4 3 +\sum_{n=-\infty}^{-1} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) +\sum_{n=1}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) \\ &=\frac 4 3+ 4\sum_{n=1}^{\infty} \frac1{\pi^2n^2} \end{aligned}$$

所以

$$\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$

取$t=1$，得到

$$\begin{aligned} 1&=\frac 2 3 +\sum_{n=-\infty,n\neq 0}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) (-1)^{n+1}\\ &=\frac 2 3 +\sum_{n=-\infty}^{-1} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) (-1)^{n+1} +\sum_{n=1}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) (-1)^{n+1} \\ &=\frac 2 3 +\sum_{n=1}^{\infty} \left(\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) (-1)^{-n+1} +\sum_{n=1}^{\infty} \left(-\frac 2{\pi i n} +\frac 2{\pi^2n^2}\right) (-1)^{n+1} \\ &=\frac 2 3 +4\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\pi^2n^{2}} \end{aligned}$$

所以

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$$

将之前两个等式相加得到

$$\frac{3 \pi^2}{12}= \sum_{n=1}^{\infty} \frac{1}{n^{2}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} =\sum_{n=0}^{\infty} \frac{2}{(2 n+1)^{2}}$$

所以

$$\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^{2}}=\frac{\pi^{2}}{8}$$

Problem 2

记

$$e_{n}(t)=\frac{1}{\sqrt{T}} e^{2 \pi i n t / T}$$

不难得到

$$\left(e_{n}, e_{m}\right)=\left\{\begin{array}{ll}{1} & {n=m} \\ {0} & {n \neq m}\end{array}\right.$$

以及

$$\left(f, e_{n}\right)=\frac{1}{\sqrt{T}} \int_{0}^{T} f(t) e^{-2 \pi i n t / T} d t ={\sqrt T}c_n$$

所以

$$f=\sum_{n=-\infty}^{\infty}\left(f, e_{n}\right) e_{n}$$

两边取模得到

$$\begin{aligned} \int_{0}^{1}|f(t)|^{2} d t &=\|f\|^{2}\\ &=(f, f) \\ &=\left(\sum_{n=-\infty}^{\infty}\left(f, e_{n}\right) e_{n}, \sum_{m=-\infty}^{\infty}\left(f, e_{m}\right) e_{m}\right) \\ &=\sum_{n, m}\left(f, e_{n}\right) \overline{\left(f, e_{m}\right)}\left(e_{n}, e_{m}\right)\\ &=\sum_{n, m=-\infty}^{\infty}\left(f, e_{n}\right) \overline{\left(f, e_{m}\right)} \delta_{n m} \\ &=\sum_{n=-\infty}^{\infty}\left(f, e_{n}\right)\overline{\left(f, e_{n}\right)}\\ &=\sum_{n=-\infty}^{\infty}\left|\left(f, e_{n}\right)\right|^{2}\\ &=\sum_{n=-\infty}^{\infty}T|c_n|^{2} \end{aligned}$$

因此

$$\frac 1 T \int_{0}^{1}|f(t)|^{2} d t =\sum_{n=-\infty}^{\infty}|c_n|^{2}$$

Problem 3

略过

Problem 4

(a)回顾

$$\Lambda(x)=\left\{\begin{array}{ll}{1-|x|} & {|x| \leq 1} \\ {0} & {\text { otherwise }}\end{array}\right.$$

记

$$\Lambda_a(x)=\Lambda(\frac x a)=\left\{\begin{array}{ll}{1-|\frac x a|} & {|\frac x a| \leq 1} \\ {0} & {\text { otherwise }}\end{array}\right.$$

不难看出上图的函数为

$$2\Lambda_2(x-2) +2.5\Lambda_2(x-4)$$

所以该函数的傅里叶变换为

$$\begin{aligned} 2e^{-2\pi is 2}2\operatorname{sinc}^{2} (2s) +2.5e^{-4\pi is2}2\operatorname{sinc}^{2} (2s) &=\operatorname{sinc}^{2} (2s) \left(4e^{-4\pi is }+5e^{-8\pi is }\right) \end{aligned}$$

(b)同(a)可得，考虑$[kT,(k+2)T]$区间，该区间对应的$\Lambda$函数为

$$f((k+1)T) \Lambda_{T}(x-(k+1)T)$$

所以

$$\begin{aligned} g(x)&=\sum_{k=0}^{n-2}f((k+1)T) \Lambda_{T}(x-(k+1)T)\\ \mathcal{F} g(s) &=\sum_{k=0}^{n-2} f((k+1)T) e^{-2(k+1)T\pi is } T\operatorname{sinc}^{2} (Ts)\\ &=T\operatorname{sinc}^{2} (Ts) \sum_{k=1}^{n-1} f(kT) e^{-2kT\pi is } \end{aligned}$$

Problem 5

(a)

$$\begin{aligned} \mathcal{F} g(s) &=\int_{-\infty}^{\infty} e^{-2\pi ist} f(t) \cos \left(2 \pi s_{0} t\right) dt\\ &=\frac 1 2\int_{-\infty}^{\infty} e^{-2\pi ist} f(t) \left(e^{2 \pi i s_{0} t}+e^{-2 \pi i s_{0} t}\right) dt\\ &= \frac 1 2\int_{-\infty}^{\infty} e^{-2\pi i(s-s_0)t} f(t)dt+ \frac 1 2\int_{-\infty}^{\infty} e^{-2\pi i(s+s_0)t} f(t)dt\\ &= \frac{1}{2} \mathcal{F} f\left(s-s_{0}\right)+\frac{1}{2} \mathcal{F} f\left(s+s_{0}\right) \end{aligned}$$

(b)图形对应的函数为

$$\Lambda_2(x-4)+\Lambda_2(x+4) =\frac 12 \times 2\Lambda_2(x-4)+\frac 12 \times 2\Lambda_2(x+4)$$

注意到

$$\mathcal{F} \operatorname{sinc}^{2}=\Lambda$$

记

$$f(t)= 4\operatorname{sinc}^{2}(2t)$$

那么

$$\begin{aligned} \mathcal{F} \left(f\right)=4\times\frac 12 \Lambda_2=2\Lambda_2 \end{aligned}$$

结合(a)可得

$$g(t) =4\operatorname{sinc}^{2}(2t)\cos \left(8 \pi t\right)$$

Problem 6

$$\begin{aligned} \hat{g}(n) &=\int_{-\frac 1 2}^{\frac 12 }e^{-2\pi int} g(t) dt\\ &=\int_{-\frac 1 2}^{\frac 12 }e^{-2\pi int} \sum_{k=-\infty}^{\infty} f(t-k) dt\\ &= \sum_{k=-\infty}^{\infty}\int_{-\frac 1 2}^{\frac 12 }e^{-2\pi int} f(t-k) dt\\ &=\sum_{k=-\infty}^{\infty} \int_{k-\frac 1 2}^{k+\frac 12 }e^{-2\pi in(t+k)} f(t) dt\\ &=\sum_{k=-\infty}^{\infty} e^{-2\pi ink}\int_{k-\frac 1 2}^{k+\frac 12 }e^{-2\pi int} f(t) dt\\ &= \sum_{k=-\infty}^{\infty} \int_{k-\frac 1 2}^{k+\frac 12 }e^{-2\pi int} f(t) dt\\ &= \int_{-\infty}^{\infty }e^{-2\pi int} f(t) dt\\ &=\mathcal{F} f(n) \end{aligned}$$

Problem 7

(a)因为

$$\lim_{s\to 0} \frac{\sin (2 \pi s)-2 \pi s} {4 \pi^{2} s^{2}} = \lim_{s\to 0} \frac{-\frac 1 6(2\pi s)^3}{4 \pi^{2} s^{2}} =0$$

所以$s=0$不是奇异点。

(b)因为$h(x)$是实，奇函数，所以$\mathcal{F} h(s)$是纯虚数，因此$\angle \mathcal{F} h(s)$为$\frac \pi 2$或$-\frac \pi 2$

(c)

$$\begin{aligned} \int_{-\infty}^{\infty} \mathcal{F} g(s) \cos (\pi s) d s &=\frac 1 2\int_{-\infty}^{\infty} \mathcal{F} g(s) \left(e^{i\pi s} +e^{-i\pi s}\right)d s\\ &=\frac 12 \int_{-\infty}^{\infty} \left(\mathcal{F} g(s)+ \mathcal{F} g(-s)\right) e^{i\pi s}d s \end{aligned}$$

注意到

$$\begin{aligned} \mathcal Fg +(\mathcal Fg)^- &=\mathcal Fg+\mathcal Fg^- \\ &=\mathcal F\left(g+g^-\right)\\ &=\mathcal F h \end{aligned}$$

因此

$$\begin{aligned} \int_{-\infty}^{\infty} \mathcal{F} g(s) \cos (\pi s) d s &=\frac 12 \int_{-\infty}^{\infty} \mathcal F he^{ 2i\pi s \times\frac 1 2}d s\\ &=\frac 12 \mathcal F^{-1}\mathcal F h(\frac 12 )\\ &=\frac 1 2h(\frac 12 )\\ &=\frac 1 4 \end{aligned}$$

(d)

$$\begin{aligned} \int_{-\infty}^{\infty} \mathcal{F} h(s) e^{4i \pi s} d s &= \int_{-\infty}^{\infty} \mathcal{F} h(s) e^{2i \pi s \times 2} d s \\ &=\mathcal F^1\mathcal{F} h(2) \\ &=h(2) \\ &=0 \end{aligned}$$

(e)由对称性可得

$$\begin{aligned} \Re \mathcal{F} g(s) &=\frac 12 \Re \mathcal{F} \Pi(s)\\ &=\frac 12 \operatorname{sinc}^{2}(s) \end{aligned}$$

(f)依然由对称性，我们有

$$\begin{aligned} \Im \mathcal{F} h(s) &=2\Im \mathcal{F} g(s)\\ &=2 \frac{\sin (2 \pi s)-2 \pi s} {4 \pi^{2} s^{2}}\\ &= \frac{\sin (2 \pi s)-2 \pi s} {2 \pi^{2} s^{2}} \end{aligned}$$

(g)

$$\begin{aligned} c_k&=\frac 12 \int_{-1}^1 e^{-\pi ikx} h(x) dx\\ &=\frac 12 \int_{-1}^1 e^{-\pi i2x (\frac k 2)} h(x) dx\\ &=\frac 12 \mathcal Fh(\frac k2)\\ &=i \frac{\sin ( \pi k)- \pi k} {\pi^{2} k^{2}} & \mathcal{F} h(s)是纯虚数\\ &=-\frac i {\pi k} \end{aligned}$$