京都大学 情報学研究科 知能情報学専攻 2024年8月実施 専門科目 S-4

标签：

• Kyoto-University
• Convolution-Theorem
• Sampling-Theorem-and-Aliasing
• Fourier-Analysis

Author

itsuitsuki

Description

大学公表の原題

Let us define the Fourier transform $\mathcal{F}[x(t)]$ of a real function $x(t)$ and the inverse Fourier transform $\mathcal{F}^{-1}[X(\omega)]$ of a function $X(\omega)$ with the following formulas, where $t$ and $\omega$ denote real numbers, and $j = \sqrt{-1}$.

$$\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} \mathrm d t \\ \mathcal{F}^{-1}[X(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j\omega t} \mathrm d\omega$$

Answer the following questions, where $T_0, \omega_0$, and $T$ denote positive constants.

Q.1

Prove that the following equation holds for real functions $f(t)$ and $g(t)$, where $*$ denotes convolution.

$$\mathcal{F}[f(t)g(t)] = \frac{1}{2\pi} \mathcal{F}[f(t)] * \mathcal{F}[g(t)]$$

Q.2

Compute the Fourier transform of the functions given below. (1) $x_1(t) = \frac{1}{2}(\text{sgn}(T_0 - t) + \text{sgn}(T_0 + t))$,

$$\text{where } \text{sgn}(t) = \begin{cases} -1 & (t < 0) \\ 0 & (t = 0) \\ 1 & (t > 0) \end{cases}$$

(2) $x_2(t) = \begin{cases} \frac{\sin \omega_0 t}{\pi t} & (t \neq 0) \\ \frac{\omega_0}{\pi} & (t = 0) \end{cases}$

Q.3

Let $x_s(t, T) = x_2(t)\delta_T(t)$ be a signal sampled from $x_2(t)$ in Q.2 using a comb function $\delta_T(t) = \sum_{k=-\infty}^{\infty} \delta(t - kT)$, where $\delta(t)$ denotes the Dirac delta function. Answer the following questions. You may use that $\mathcal{F}[\delta_T(t)] = \frac{1}{T} \sum_{k=-\infty}^{\infty} \delta(\omega - \frac{k}{T})$ holds.

(1) Draw the graph of $\mathcal{F}[x_s(t, \frac{1}{3\omega_0})]$ in the range of $|\omega| \leq 3\omega_0$.

(2) Show the condition for $T$ to satisfy $\mathcal{F}[x_2(t)] = \mathcal{F}[x_s(t, T)]$ in the range of $|\omega| \leq \omega_0$.

(3) Draw the graph of $\mathcal{F}[x_s(t, \frac{2}{3\omega_0})]$ in the range of $|\omega| \leq 3\omega_0$.

(4) Draw the graph of $\mathcal{F}^{-1}[X_s(\omega)]$ in the range of $|t| \leq \frac{\pi}{\omega_0}$. $X_s(\omega)$ is given below.

$$X_s(\omega) = \begin{cases} \mathcal{F}[x_s(t, \frac{2}{3\omega_0})] & (|\omega| \leq \omega_0) \\ 0 & (|\omega| > \omega_0) \end{cases}$$

题目描述

连续时间 Fourier 变换及其逆变换定义为

$$\mathcal{F}[x(t)]=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt,\qquad \mathcal{F}^{-1}[X(\omega)]=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}\,d\omega.$$

1. 证明时域乘积的 Fourier 变换满足

   $$\mathcal{F}[f(t)g(t)] =\frac{1}{2\pi}\bigl(\mathcal{F}[f(t)]*\mathcal{F}[g(t)]\bigr),$$

   其中 $*$ 表示卷积。

2. 求下列两个信号的 Fourier 变换：

   $$x_1(t)=\frac12\{\operatorname{sgn}(T_0-t)+\operatorname{sgn}(T_0+t)\},$$

   其中符号函数 $\operatorname{sgn}(\cdot)$ 按题中定义取值；以及

   $$x_2(t)= \begin{cases} \dfrac{\sin\omega_0t}{\pi t} & (t\ne0),\\[4pt] \dfrac{\omega_0}{\pi} & (t=0). \end{cases}$$

3. 用冲激梳

   $$\delta_T(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)$$

   对 $x_2(t)$ 采样，令 $x_s(t,T)=x_2(t)\delta_T(t)$。可使用

   $$\mathcal{F}[\delta_T(t)] =\frac1T\sum_{k=-\infty}^{\infty}\delta\left(\omega-\frac{k}{T}\right).$$

   （1）在 $|\omega|\leq3\omega_0$ 范围内画出 $\mathcal{F}[x_s(t,\frac{1}{3\omega_0})]$；（2）求使得在 $|\omega|\leq\omega_0$ 上 $\mathcal{F}[x_2(t)]=\mathcal{F}[x_s(t,T)]$ 成立的 $T$ 的条件；（3）在 $|\omega|\leq3\omega_0$ 范围内画出 $\mathcal{F}[x_s(t,\frac{2}{3\omega_0})]$；（4）定义

   $$X_s(\omega)= \begin{cases} \mathcal{F}[x_s(t,\frac{2}{3\omega_0})] & (|\omega|\leq\omega_0),\\ 0 & (|\omega|>\omega_0), \end{cases}$$

   在 $|t|\leq\frac{\pi}{\omega_0}$ 范围内画出 $\mathcal{F}^{-1}[X_s(\omega)]$。

Kai

Q.1

Assume, for example, that $f,g$ are Schwartz functions. Substituting the inverse transform of $g$ and applying Fubini's theorem gives

$$\begin{aligned} \mathcal F[fg](\omega) &=\int_{\mathbb R}f(t)e^{-j\omega t} \left(\frac1{2\pi}\int_{\mathbb R}G(\nu)e^{j\nu t}\,d\nu\right)dt\\ &=\frac1{2\pi}\int_{\mathbb R}G(\nu)F(\omega-\nu)\,d\nu =\frac1{2\pi}(F*G)(\omega). \end{aligned}$$

Q.2

(1)

The function is $1$ for $|t|<T_0$, $0$ for $|t|>T_0$, and $1/2$ at the two endpoints. These endpoint values do not affect its integral. Hence

$$X_1(\omega)=\int_{-T_0}^{T_0}e^{-j\omega t}\,dt =\begin{cases}2\sin(\omega T_0)/\omega,&\omega\ne0,\\2T_0,&\omega=0.\end{cases}$$

(2)

The inverse transform of the frequency rectangle is

$$\frac1{2\pi}\int_{-\omega_0}^{\omega_0}e^{j\omega t}\,d\omega =\frac{\sin(\omega_0t)}{\pi t},$$

with the continuous value $\omega_0/\pi$ at $t=0$. Thus, with symmetric improper integration at the jump points,

$$X_2(\omega)=R(\omega):= \begin{cases} 1,&|\omega|<\omega_0,\\ 1/2,&|\omega|=\omega_0,\\ 0,&|\omega|>\omega_0. \end{cases}$$

Q.3

For the angular-frequency convention $e^{-j\omega t}$, the distributional Fourier-series identity is

$$\delta_T(t)=\frac1T\sum_{m\in\mathbb Z}e^{j2\pi mt/T}, \qquad \mathcal F[\delta_T](\omega) =\frac{2\pi}{T}\sum_{m\in\mathbb Z}\delta\left(\omega-\frac{2\pi m}{T}\right).$$

Using Q.1, the sampled spectrum is therefore

$$\mathcal F[x_s](\omega)=\frac1T\sum_{m\in\mathbb Z} R\left(\omega-\frac{2\pi m}{T}\right). \tag{*}$$

The following answers use this transform convention and the half-height values of $R$ at its two discontinuities.

(1)

For $T=1/(3\omega_0)$, neighboring replicas are separated by $6\pi\omega_0$. Only the central rectangle intersects $|\omega|\le3\omega_0$, so

$$\mathcal F[x_s](\omega)=3\omega_0 R(\omega) \qquad(|\omega|\le3\omega_0).$$

(2)

In the absence of spectral overlap, $2\pi/T>2\omega_0$ and (*) equals $X_2/T$ on the baseband. Thus exact equality of the unscaled spectra throughout the closed interval $|\omega|\le\omega_0$ requires

$$\boxed{T=1\quad\text{and}\quad0<\omega_0<\pi}.$$

To see that overlap cannot provide another solution with these endpoint values, note that (*) is periodic with period $a=2\pi/T$. If $a<2\omega_0$, periodicity makes its value at $-\omega_0$ equal its value at the interior point $-\omega_0+a$, whereas $R$ takes the unequal values $1/2$ and $1$ there. If $a=2\omega_0$, two half-height replicas sum to one at either boundary, again preventing equality with $R$ after matching the interior amplitude.

For amplitude-normalized sampling, the usual no-aliasing identity is instead

$$T\mathcal F[x_s](\omega)=X_2(\omega), \qquad T<\frac{\pi}{\omega_0},\quad |\omega|\le\omega_0.$$

The non-strict bound $T\le\pi/\omega_0$ suffices if equality is required only almost everywhere and boundary values are disregarded.

(3)

For $T=2/(3\omega_0)$, replicas are separated by $3\pi\omega_0$. Again only the central rectangle intersects the indicated range, giving

$$\mathcal F[x_s](\omega)=\frac{3\omega_0}{2}R(\omega) \qquad(|\omega|\le3\omega_0).$$

(4)

The baseband restriction keeps this central rectangle. Consequently

$$\mathcal F^{-1}[X_s](t) =\frac{3\omega_0}{2}\,x_2(t) =\begin{cases} \dfrac{3\omega_0\sin(\omega_0t)}{2\pi t},&t\ne0,\\[4pt] \dfrac{3\omega_0^2}{2\pi},&t=0. \end{cases}$$

It is even, nonnegative on the requested interval, maximal at zero, and zero at $t=\pm\pi/\omega_0$.