京都大学 情報学研究科 知能情報学専攻 2022年8月実施 専門科目 S-5

标签：

• Kyoto-University
• Convolution
• Signal-Processing

Author

realball, 祭音Myyura

Description

大学公表の原題 Suppose that the Fourier transform $\mathcal{F}[f(x)]$ of a function $f(x)$ and the Fourier integral representation of the Dirac delta function $\delta(x)$ are given by the following formulae, where $x$ and $k$ are real numbers, and $i=\sqrt{-1}$. Answer the following questions.

$$\begin{align} \mathcal{F}[f(x)]&=F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\mathrm{e}^{-ikx}\text{d}x \tag{i}\\ \delta(x)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{e}^{ikx}\text{d}k \tag{ii} \end{align}$$

Q.1

Compute the Fourier transform of the function given below, where $\omega$ is a real number.

(1) $f_1(x)=\left\{\begin{array}{ll}0&(x<0) \\ 1&(0\le x\le2)\\0&(x>2)\end{array}\right.$

(2) $f_{2}(x)=\cos^{2}\omega x$

Q.2

Compute the Fourier transform of function $f_3(x)$ by following the steps below.

$$f_3(x)=\left\{\begin{array}{l}0 \ \ (x<-2)\\ x+2 \ \ (-2\leq x<0)\\ 2-x \ \ (0\leq x<2)\\ 0 \ \ (x\geq2)\end{array}\right.$$

(1) Derive the following equation concerning convolution operation.

$$\mathcal{F}[f(x)*g(x)]=\mathcal{F}\left[\int_{-\infty}^{\infty}f(\tau)g(x-\tau)\text{d}\tau\right]=\sqrt{2\pi}\mathcal{F}[f(x)]\mathcal{F}[g(x)]$$

(2) Find function $f_4(x)$ whose convolution with the above $f_1(x)$ satisfies $f_3(x)=f_1(x) * f_4(x)$, and explain how the convolution gives $f_3(x)$.

(3) Compute $\mathcal{F}[f_3(x)]$.

题目描述

采用归一化

$$\mathcal F[f](k)=\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx, \qquad \delta(x)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{ikx}\,dk,$$

其中 $x,k\in\mathbb R$、$i=\sqrt{-1}$。

1. 求 Fourier 变换：

   1. $f_1(x)=1$（$0\le x\le2$），区间外为 0；
   2. $f_2(x)=\cos^2(\omega x)$，$\omega\in\mathbb R$。

2. 三角函数

   $$f_3(x)= \begin{cases} 0,&x<-2,\\ x+2,&-2\le x<0,\\ 2-x,&0\le x<2,\\ 0,&x\ge2 \end{cases}$$

   按以下步骤求变换：

   1. 推导卷积定理

      $$\mathcal F[f*g] =\mathcal F\!\left[\int f(\tau)g(x-\tau)\,d\tau\right] =\sqrt{2\pi}\,\mathcal F[f]\mathcal F[g].$$

   2. 求 $f_4$ 使 $f_3=f_1*f_4$，并说明卷积如何得到 $f_3$。

   3. 求 $\mathcal F[f_3]$。

Kai

Q.1

(1)

$$\begin{aligned} \mathcal{F}\left[f_{1}(x)\right] &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\\ &= \frac{1}{\sqrt{2\pi}}\int_{0}^{2}e^{-ikx}dx\\ &= -\frac{1}{ik\sqrt{2\pi}}\left[e^{-ikx}\right]_{0}^{2}\\ &= \frac{1}{ik\sqrt{2\pi}}\left(1-e^{-2ik}\right) \end{aligned}$$

applying the Euler formula,

$$\begin{aligned} \mathcal{F}\left[f_{1}(x)\right] &= e^{-ik}\frac{2}{k\sqrt{2\pi}}\frac{e^{ik}-e^{-ik}}{2i}\\ &= e^{-ik}\sqrt{\frac{2}{\pi}}\frac{\sin k}{k} \end{aligned}$$

At $k=0$, the value is understood by continuity as $\sqrt{2/\pi}$.

(2)

The Fourier transform here is interpreted in the sense of tempered distributions.

$$\begin{aligned} \mathcal{F}(f_{2}(x)) &= \mathcal{F}\left(\cos^{2}\omega x\right)\\ &= \frac{1}{2}\mathcal{F}\left(1+\cos 2\omega x\right)\\ &= \frac{1}{2}\left(\mathcal{F}(1) + \mathcal{F}\left(\cos 2\omega x\right)\right) \end{aligned}$$

$$\sqrt{2\pi}\delta(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}dx \Rightarrow \mathcal{F}(1) = \sqrt{2\pi}\delta(k)$$

$$\begin{aligned} \mathcal{F}\left[\cos 2\omega x\right] &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\cos 2\omega x)e^{-ikx}dx\\ &= \frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(e^{i2\omega x}+e^{-i2\omega x}\right) e^{-ikx}dx\\ &= \frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(e^{-i(k-2\omega)x}+e^{-i(k+2\omega)x}\right)dx\\ &= \frac{1}{2}\cdot\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i(k-2\omega)x}dx+\frac{1}{2}\cdot\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i(k+2\omega)x}dx\\ &= \frac{1}{2}\cdot \sqrt{2\pi}\delta(k-2\omega)+\frac{1}{2}\cdot \sqrt{2\pi}\delta(k+2\omega)\\ &= \frac{\sqrt{2\pi}}{2}\left(\delta(k-2\omega)+\delta(k+2\omega)\right) \end{aligned}$$

Hence

$$\mathcal{F}(f_{2}(x)) = \frac{\sqrt{2\pi}}{4}\left(\delta(k-2\omega)+2\delta(k)+\delta(k+2\omega)\right)$$

Q.2

(1)

For integrable $f,g$, Fubini's theorem justifies exchanging the integrals below.

$$\begin{aligned} \mathcal{F}\left[f(x)\ast g(x)\right] &= \mathcal{F}\left[\int_{-\infty}^{\infty}f(\tau)g(x-\tau)d\tau\right]\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(\tau)g(x-\tau)d\tau\right)e^{-ikx}dx\\ &= \int_{-\infty}^{\infty}f(\tau)\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(x-\tau)e^{-ikx}dx\right)d\tau\\ &= \int_{-\infty}^{\infty}f(\tau)e^{-ik\tau}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(y)e^{-iky}dy\right)d\tau\\ &= \sqrt{2\pi}\mathcal{F}\left[f(x)\right]\mathcal{F}\left[g(x)\right]. \end{aligned}$$

(2)

$$f_4(x)=\begin{cases} 0 & x < -2 \\ 1 & -2\leq x \leq 0\\ 0 & x > 0 \end{cases}$$

$$f_3(x)=f_1(x)\ast f_4(x) =\int_{0}^{2}f_4(x-\tau)d\tau$$

Indeed, the integral is the length of $[0,2]\cap[x,x+2]$, namely $x+2$ for $-2\leq x<0$, $2-x$ for $0\leq x<2$, and $0$ otherwise.

(3)

$$\begin{aligned} \mathcal{F}\left[f_{3}(x)\right] &= \mathcal{F}\left[f_{1}(x)\ast f_{4}(x)\right]\\ &= \mathcal{F}\left[f_{1}(x)\ast f_{1}(x+2)\right]\\ &= \sqrt{2\pi}\mathcal{F}\left[f_{1}(x)\right]\mathcal{F}\left[f_{1}(x+2)\right]\\ &= \sqrt{2\pi}\mathcal{F}\left[f_{1}(x)\right]\left(e^{2ik}\mathcal{F}\left[f_{1}(x)\right]\right)\\ &= \sqrt{2\pi}e^{2ik}\left(\mathcal{F}\left[f_{1}(x)\right]\right)^{2}\\ &= \sqrt{2\pi}e^{2ik}\left(e^{-ik}\sqrt{\frac{2}{\pi}}\frac{\sin k}{k}\right)^{2}\\ &= 2\sqrt{\frac{2}{\pi}}\left( \frac{\sin k}{k} \right)^{2} \end{aligned}$$

At $k=0$, the last expression has the continuous value $2\sqrt{2/\pi}$, consistent with the area $\int f_3(x)\,dx=4$ and the Fourier normalization.