京都大学 情報学研究科 知能情報学専攻 2020年8月実施 専門科目 S-5

标签：

• Kyoto-University
• Discrete-Fourier-Transform

Author

realball

Description

The Fourier spectrum of a continuous-time signal $f(t)$ is given by $F(\omega) = \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt$, where $j$ denotes the imaginary unit. Let $X[k]$ denote discrete Fourier transform of a finite-length discrete signal of length $N$, $x[n] \ (n = 0, \ldots, N-1)$. Answer the following questions.

Q.1

Suppose that $f(t)$ is an even function, that is $f(x) = f(-x).$ The Fourier spectrum of $f(t)$ is given by

$$F(\omega) = 2 \int_0^{\infty} f(t) \boxed{\ (A)\ }dt.$$

Answer $\boxed{\ (A)\ }$. Calculation procedure must also be included in the answer.

Q.2

Let $x_s[n]$ be the circular shifted version of $x[n]$ by $s$,

$$x_s[n] = \begin{cases} x[N+n-s] &(n<s) \\ x[n-s] &(n \geq s) \end{cases}$$

The discrete Fourier transform of $x_s[n]$ is given by $\boxed{\ (B)\ }X[k]$.

Answer $\boxed{\ (B)\ }$. Calculation procedure must also be included in the answer.

Q.3

Let $y[n]$ be a finite-length discrete signal of length $2N$,

$$y[n] = \begin{cases} x[n] &(0 \leq n < N) \\ x[2N-1-n] &(N \leq n < 2N). \end{cases}$$

The discrete Fourier transform of $y[n]$ is given by

$$Y[k] = 2 \boxed{\ (C)\ } \sum_{n=0}^{N-1} x[n] \cos (\boxed{\ (D)\ }).$$

Answer $\boxed{\ (C)\ }$ and $\boxed{\ (D)\ }$. Calculation procedure must also be included in the answer.

Q.4

The transform of $x[n]$,

$$X_{DCT}[k] = \alpha_k \sum_{n=0}^{N-1} x[n] \cos (\boxed{\ (D)\ }), \alpha_0 = 1/\sqrt{N}, \alpha_k = \sqrt{2/N}$$

is known as a discrete cosine tranform (DCT-II), and is used in data compression such as JPEG.

Explain an advantage of the discrete cosine transform compared to the discrete Fourier transform in terms of data compression.

题目描述

连续时间信号 $f(t)$ 的 Fourier 变换为 $F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,dt$。长度 $N$ 的离散信号为 $x[n]$（$n=0,\ldots,N-1$），其 DFT 记为 $X[k]$。回答各空并写出推导。

1. 若 $f(t)$ 为偶函数，则
   $$F(\omega)=2\int_0^\infty f(t)\boxed{(A)}\,dt.$$
   求 $(A)$。
2. 将 $x[n]$ 循环右移 $s$ 位：
   $$x_s[n]= \begin{cases} x[N+n-s],&n$$
   其 DFT 为 $\boxed{(B)}X[k]$。求 $(B)$。
3. 构造长度 $2N$ 的偶对称延拓
   $$y[n]= \begin{cases} x[n],&0\le n<N,\\ x[2N-1-n],&N\le n<2N. \end{cases}$$
   其 DFT 写为
   $$Y[k]=2\boxed{(C)} \sum_{n=0}^{N-1}x[n]\cos\bigl(\boxed{(D)}\bigr).$$
   求 $(C),(D)$。
4. 定义 DCT-II
   $$X_{\mathrm{DCT}}[k]=\alpha_k \sum_{n=0}^{N-1}x[n]\cos\bigl(\boxed{(D)}\bigr), \quad \alpha_0=\frac1{\sqrt N},\quad \alpha_k=\sqrt{\frac2N}.$$
   从数据压缩角度说明 DCT 相比 DFT 的一个优势，例如其在 JPEG 中的使用。

考点

• 偶信号 Fourier 变换：利用正弦项奇对称消失，将变换化为实余弦积分。
• DFT 循环移位性质：通过索引代换推导时域循环移位对应的频域线性相位因子。
• 偶延拓与 DCT-II：把 $2N$ 点 DFT 配对为余弦和，确定相位因子与余弦角度。
• 变换编码能量集中：说明实信号 DCT 的边界延拓减少不连续，常使低频少量系数集中更多能量且无需复数系数。

Kai

Q.1

$$\begin{aligned} \mathcal{F}(\omega) &=\int_{-\infty}^{\infty} f(t)e^{-j\omega t}dt\\ &= \int_{-\infty}^{\infty}f(t)\left[\cos(\omega t)-j \sin(\omega t)dt \right]\\ &= \int_{-\infty}^{\infty}f(t)\cos(\omega t)dt - \int_{-\infty}^{\infty}f(t)j \sin(\omega t)dt\\ &= \int_{-\infty}^{\infty}f(t)\cos(\omega t)dt \end{aligned}$$

Thus, blank (A) is $\cos(\omega t)$.

Q.2

$$\begin{aligned} X_s[k]&= \sum_{n=0}^{N-1}x_s[n]e^{-j\frac{2\pi}{N}kn}\\ &=\sum_{n=0}^{S-1}x[N+n-s]e^{-j\frac{2\pi}{N}kn}+\sum_{n=s}^{N-1}x[n-s]e^{-j\frac{2\pi}{N}kn}\\ &\text{assume \textcircled{1}:} m=N+n-s;n=m+s-N\\ &\text{assume \textcircled{2}:} m=n-s;n=m+s\\ &=\sum_{m=N-s}^{N-1}x[m]e^{-j\frac{2\pi}{N}k(m+s-N)}+\sum_{m=0}^{N-S-1}x[m]e^{-j\frac{2\pi}{N}k(m+s)}\\ &=e^{-j\frac{2\pi}{N}ks}\sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km}\\ &\text{Thus: } x_s[k]=X[k]e^{-j\frac{2\pi}{N}ks}\\ &\text{Then: } (B)=e^{-j\frac{2\pi}{N}ks} \end{aligned}$$

Q.3

$$\begin{aligned} Y[k] &= \sum_{n=0}^{2N-1}x[n] e^{-j\frac{2\pi}{2N}kn} \\ &= \sum_{n=0}^{N-1}x[n] e^{-j\frac{\pi}{N}kn} + \sum_{n=N}^{2N-1}x[2N-1-n] e^{-j\frac{\pi}{N}kn} \\ &= \sum_{n=0}^{N-1}x[n] e^{-j\frac{\pi}{N}kn} + \sum_{n=0}^{N-1}x[N-1-n] e^{-j\frac{\pi}{N}kn} \\ &= \sum_{n=0}^{N-1}x[n] e^{-j\frac{\pi}{N}kn} + \sum_{m=0}^{N-1}x[m] e^{-j\frac{\pi}{N}k(N-1-m)} \\ &= \sum_{n=0}^{N-1}x[n] e^{-j\frac{\pi}{N}kn} + \sum_{m=0}^{N-1}x[m] e^{j\frac{\pi}{N}k(m+1)} \\ &= e^{j\frac{\pi}{2N}k}\left( \sum_{n=0}^{N-1}x[n] e^{-j\pi k\frac{2n+1}{2N}} + \sum_{n=0}^{N-1}x[n] e^{j\pi k\frac{2n+1}{2N}} \right) \\ &= 2e^{j\frac{\pi}{2N}k}\left( \sum_{n=0}^{N-1}x[n] \frac{e^{-j\pi k\frac{2n+1}{2N}} + e^{j\pi k\frac{2n+1}{2N}}}{2} \right) \\ &= 2e^{j\frac{\pi}{2N}k}\left( \sum_{n=0}^{N-1}x[n] \cos \frac{2n + 1}{2N}\pi k \right) \end{aligned}$$

Q.4

Better concentrate energy