京都大学 情報学研究科 知能情報学専攻 2018年8月実施 専門科目 T-2

标签：

• Kyoto-University
• Continuous-Time-Fourier-Transform
• Dirac-Comb-Fourier-Transform
• Sampling-Theorem-and-Aliasing

Author

realball, 祭音Myyura

Description

The Fourier spectrum of a continuous-time signal $x(t)$ is given by

$$X(\omega) = \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t)e^{-j\omega t}dt,$$

where $j$ denotes the imaginary unit and $\mathcal{F}[]$ denotes Fourier transform. Let

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

be the inverse Fourier transform.

Q.1

Find the continuous-time signal $\mathcal{F}^{-1}[P_\Omega(\omega)]$ corresponding to the Fourier spectrum $P_\Omega(\omega)$, where $P_\Omega(\omega)$ denotes a rectangular function of width $2\Omega$ and is given by

$$P_\Omega(\omega) = \begin{cases} 1 & |\omega| < \Omega, \\ 0 & |\omega| \geq \Omega. \end{cases}$$

Q.2

Let $\delta_T(t)$ be a comb function whose time period is $T$,

$$\delta_T(t) = \sum_{n=-\infty}^\infty \delta(t - nT).$$

Show that

$$\mathcal{F}[\delta_T(t)] = \frac{2\pi}{T} \delta_{\frac{2 \pi}{T}}(\omega),$$

where $\delta(t)$ is a function that satisfies

$$\delta(t) = \begin{cases} \infty & t = 0, \\ 0 & t \neq 0, \end{cases}$$

and

$$\int_{-\infty}^\infty \delta(t)dt = 1.$$

You may use $\mathcal{F}[e^{j\omega_0t}] = \int_{-\infty}^\infty e^{j\omega_0t}e^{-j\omega t}dt = 2\pi \delta(\omega - \omega_0)$ for any real number $\omega_0$.

Q.3

Let $x_s(t) = x(t)\delta_T(t)$ be a continuous-time signal sampled from $x(t)$ with a sampling period $T$. Describe $\mathcal{F}[x_s(t)]$ with $X(\omega)$ and $T$.

题目描述

连续时间信号 $x(t)$ 的 Fourier 变换与逆变换定义为

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt, \qquad \mathcal F^{-1}[X(\omega)] =\frac1{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}\,dt,$$

其中 $j$ 为虚数单位。

1. 设宽度为 $2\Omega$ 的矩形频谱

   $$P_\Omega(\omega)= \begin{cases} 1,&|\omega|<\Omega,\\ 0,&|\omega|\ge\Omega. \end{cases}$$

   求对应的连续时间信号 $\mathcal F^{-1}[P_\Omega(\omega)]$。

2. 设周期为 $T$ 的冲激梳

   $$\delta_T(t)=\sum_{n=-\infty}^{\infty}\delta(t-nT).$$

   证明

   $$\mathcal F[\delta_T(t)] =\frac{2\pi}{T}\delta_{2\pi/T}(\omega).$$

   其中 Dirac 冲激满足在 $t=0$ 为无穷、其余处为零且 $\int_{-\infty}^{\infty}\delta(t)\,dt=1$。可使用 $\mathcal F[e^{j\omega_0t}]=2\pi\delta(\omega-\omega_0)$。

3. 令采样信号 $x_s(t)=x(t)\delta_T(t)$。用 $X(\omega)$ 与 $T$ 表示 $\mathcal F[x_s(t)]$。

Kai

Q.1

$$\begin{aligned} \mathcal{F}^{-1}[P_\Omega(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}P_{\Omega}(\omega)e^{j\omega t} d\omega\\ &=\frac{1}{2\pi}\int_{-\Omega}^{\Omega}e^{j\omega t}d\omega =\frac{e^{j\Omega t}-e^{-j\Omega t}}{2\pi jt} =\frac{1}{2\pi jt}\cdot 2j\sin \Omega t\\ &=\frac{1}{\pi t}\cdot\sin \Omega t \end{aligned}$$

At $t=0$, its continuous extension is $\Omega/\pi$.

Q.2

$$\delta_{T}(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\frac{2\pi}{T}t}$$

$$a_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\delta_{T}(t)e^{-jk\frac{2\pi}{T}t}dt =\frac{1}{T}.$$

$$\delta_T(t)=\frac{1}{T}\sum_{k=-\infty}^{\infty}e^{jk\frac{2\pi}{T}t}$$

$$\begin{aligned} \mathcal{F}\left[\delta_{T}(t)\right]&=\sum_{k=-\infty}^{\infty}\frac{2\pi}{T}\delta(\omega-k\frac{2\pi}{T})\\ &=\frac{2\pi}{T}\delta_\frac{2\pi}{T}(\omega) \end{aligned}$$

Q.3

$$\begin{aligned}\mathcal{F}[x_{s}(t)]=X_{s}(\omega) &=\frac1{2\pi}\left(X*\mathcal{F}[\delta_T]\right)(\omega)\\ &=\frac1{2\pi}\int_{-\infty}^{\infty}X(\theta) \sum_{k=-\infty}^{\infty}\frac{2\pi}{T} \delta\left(\omega-\theta-\frac{2\pi k}{T}\right)d\theta\\ &=\frac1T\sum_{k=-\infty}^{\infty}X\left(\omega-\frac{2\pi k}{T}\right). \end{aligned}$$