From 92dace2416473fd793c9e5b351a047ed2825a1d8 Mon Sep 17 00:00:00 2001 From: myyura Date: Tue, 7 Jan 2025 18:15:19 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BA=AC=E9=83=BD=E5=A4=A7=E5=AD=A6=20?= =?UTF-8?q?=E6=83=85=E5=A0=B1=E5=AD=A6=E7=A0=94=E7=A9=B6=E7=A7=91=20?= =?UTF-8?q?=E7=9F=A5=E8=83=BD=E6=83=85=E5=A0=B1=E5=AD=A6=E5=B0=82=E6=94=BB?= =?UTF-8?q?=202023=E5=B9=B48=E6=9C=88=E5=AE=9F=E6=96=BD=20=E5=B0=82?= =?UTF-8?q?=E9=96=80=E7=A7=91=E7=9B=AE=20S-5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- docs/kakomonn/kyoto_university/index.md | 1 + .../informatics/ist_202308_senmon_s_5.md | 258 ++++++++++++++++++ mkdocs.yml | 3 +- 3 files changed, 261 insertions(+), 1 deletion(-) create mode 100644 docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_5.md diff --git a/docs/kakomonn/kyoto_university/index.md b/docs/kakomonn/kyoto_university/index.md index c7ea1f8b1..680955a39 100644 --- a/docs/kakomonn/kyoto_university/index.md +++ b/docs/kakomonn/kyoto_university/index.md @@ -81,6 +81,7 @@ tags: - [専門科目 S-2](informatics/ist_202308_senmon_s_2.md) - [専門科目 S-3](informatics/ist_202308_senmon_s_3.md) - [専門科目 S-4](informatics/ist_202308_senmon_s_4.md) + - [専門科目 S-5](informatics/ist_202308_senmon_s_5.md) - 2023年度: - [情報学基礎 F1-1](informatics/ist_202208_kiso_f1_1.md) - [情報学基礎 F1-2](informatics/ist_202208_kiso_f1_2.md) diff --git a/docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_5.md b/docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_5.md new file mode 100644 index 000000000..04b50497e --- /dev/null +++ b/docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_5.md @@ -0,0 +1,258 @@ +--- +comments: false +title: 京都大学 情報学研究科 知能情報学専攻 2023年8月実施 専門科目 S-5 +tags: + - Kyoto-University +--- +# 京都大学 情報学研究科 知能情報学専攻 2023年8月実施 専門科目 S-5 + +## **Author** +祭音Myyura + +## **Description** +### 設問1 +2次元信号 $f(x, y)$ の2次元フーリエ変換を + +$$ +F(u, v) = \iint_{-\infty}^{\infty} f(x, y) e^{-j(ux+vy)} dxdy +$$ + +とする。ただし $j$ は虚数単位である。 +また $f(x, y)$ のある軸 $l$ への投影を、軸 $l$ 上の各点における、$l$ に垂直な直線に沿った $f(x, y)$ の線積分とする。以下の問いに答えよ。 + +(1) $f(x, y)$ を $x$ 軸に投影した信号 $p(x)$ の1次元フーリエ変換を、$F(u, v)$ を用いて表せ。 + +(2) 原点を中心として $x$ 軸を反時計回りに角度 $\theta$ 回転して得られた $s$ 軸上に $f(x, y)$ を投影した信号を $p_\theta(s)$ とする。 +$p_\theta(s)$ の $s$ についての1次元フーリエ変換を $F(u, v)$ を用いて表せ。 + +### 設問2 +長さ $N$ の離散時間信号 $x[n]$ の $N$ 点離散フーリエ変換 $X[k]$ を + +$$ +X[k] = \sum_{n=0}^{N-1} x[n] W_N^{kn}, \quad W_N = e^{-j\frac{2\pi}{N}} +$$ + +とする。ただし $j$ は虚数単位、$n, k = 0, \ldots, N-1$ であり、$N$ は正の偶数とする。以下の問いに答えよ。 + +(1) 観測系列 $x_0[n] = \{x_0[0], x_0[1], x_0[2], x_0[3]\} = \{1, 2, 1, -2\}$ を、ある信号を 4000Hz で等間隔にサンプリングすることで得たとする。 +$x_0[n]$ の４点離散フーリエ変換を計算し、周波数（Hz）に対応する振幅スペクトルおよび位相スペクトルを図示せよ。 + +(2) ２つの要素数 $N$ の実数値系列 $x_1[n]$ および $x_2[n]$ の $N$ 点離散フーリエ変換を、１回の $N$ 点離散フーリエによって計算する方法を導出せよ。 + +(3) 要素数 $2N$ の実数値系列の $2N$ 点離散フーリエ変換を、１回の $N$ 点離散フーリエ変換によって計算する方法を導出せよ。 + +## **Kai** +### 設問1 +#### (1) +By the definition of projection, we have + +$$ +p(x) = \int_{-\infty}^{\infty}f(x,y)dy +$$ + +hence the 1D Fourier transform of $p(x)$ is + +$$ +\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(x,y)dy\right)e^{-jux}dx += \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-j(ux+0\cdot y)}dxdy += F(u, 0) +$$ + +#### (2) +Let $(s,t)$ denote the coordinates obtained by rotating $(x, y)$ counterclockwise by an angle $\theta$. Then we have + +$$ +\begin{pmatrix} +s\\ +t +\end{pmatrix} += +\begin{pmatrix} +\cos\theta&-\sin\theta\\ +\sin\theta&\cos\theta +\end{pmatrix} +\begin{pmatrix} +x\\ +y +\end{pmatrix} \Rightarrow +\begin{cases} +x = s\cos\theta-t\sin\theta\\ +y = s\sin\theta+t\cos\theta +\end{cases} +$$ + +by calculating the Jacobian determinant + +$$ +J = +\begin{vmatrix} +\cos\theta&-\sin\theta\\ +\sin\theta&\cos\theta +\end{vmatrix} += \cos^{2}\theta+\sin^{2}\theta = 1 +$$ + +we know that $f(x,y)dxdy{=}f(s,t)dsdt$. Hence the 1D Fourier transform of $p_{\theta}(x)$ is + +$$ +\begin{aligned} +\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s,t)dt\right)e^{-jus}ds +&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s,t)e^{-j(us+0\cdot t)}dsdt\\ +&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-j(u\cos\theta x+(-u\sin\theta)y)}dxdy\\ +&= F(u\cos\theta, -u\sin\theta) +\end{aligned} +$$ + +### 設問2 +#### (1) +The 4-point discrete Fourier transform of $x_0[n]$: + +$$ +\begin{pmatrix} +X[0]\\ +X[1]\\ +X[2]\\ +X[3] +\end{pmatrix} += +\begin{pmatrix} +W_{4}^{0}&W_{4}^{0}&W_{4}^{0}&W_{4}^{0}\\ +W_{4}^{0}&W_{4}^{1}&W_{4}^{2}&W_{4}^{3}\\ +W_{4}^{0}&W_{4}^{2}&W_{4}^{4}&W_{4}^{6}\\ +W_{4}^{0}&W_{4}^{3}&W_{4}^{6}&W_{4}^{9} +\end{pmatrix} +\begin{pmatrix} +x[0]\\ +x[1]\\ +x[2]\\ +x[3] +\end{pmatrix} += +\begin{pmatrix} +1&1&1&1\\ +1&-j&-1&j\\ +1&-1&1&-1\\ +1&j&-1&-j +\end{pmatrix} +\begin{pmatrix} +1\\ +2\\ +1\\ +-2 +\end{pmatrix} += +\begin{pmatrix} +2\\ +-4j\\ +2\\ +4j +\end{pmatrix} +$$ + +#####
Fig. magnitude and phase spectra + +
+ +
+ +#### (2) + +$$ +\begin{aligned} +X[k] +&= \sum_{n=0}^{N-1}x[n]W_{N}^{kn}\\ +&= \sum_{n=0}^{N-1}x[n]\cos\left(2\pi-\frac{2\pi kn}{N}\right) ++ j\sum_{n=0}^{N-1}x[n]\sin\left(2\pi-\frac{2\pi kn}{N}\right)\\ +&= \sum_{n=0}^{N-1}x[n]\cos\frac{2\pi n(N-k)}{N} ++ j\sum_{n=0}^{N-1}x[n]\sin\frac{2\pi n(N-k)}{N}\\ +&= \text{Re } X[N-k]-j\text{Im } X[N-k] +\end{aligned} +$$ + +which implies that + +$$ +\begin{align} +\begin{cases} +\text{Re } X[k] = \text{Re } X[N-k]\\ +\text{Im } X[k] = -\text{Im } X[N-k] +\end{cases} \tag{j} +\end{align} +$$ + +Let $y[n] = x_{1}[n]+j x_{2}[n]$. Let $X_{1}[k],X_{2}[k],Y[k]$ denote the discrete Fourier transform of $x_{1}[n],x_{2}[n],y_{n}$, respectively. Then, + +$$ +\begin{align} +Y[k] +&= \sum_{n=0}^{N-1}(x_{1}[n]+j x_{2}[n])W_{N}^{kn} \nonumber \\ +&= \sum_{n=0}^{N-1}x_{1}[n]W_{N}^{kn}+j\sum_{n=0}^{N-1}x_{2}[n]W_{N}^{kn} \nonumber \\ +&= X_{1}[k]+iX_{2}[k] \nonumber \\ +&= (\text{Re } X_{1}[k]+j\text{Im } X_{1}[k])+j(\text{Re } X_{2}[k]+j \text{Im } X_{2}[k]) \nonumber \\ +&= (\text{Re } X_{1}[k]-\text{Im } X_{2}[k])+j(\text{Im } X_{1}[k]+\text{Re } X_{2}[k]) \tag{ii} +\end{align} +$$ + +By (i) we know that + +$$ +\begin{align} +Y[N-k] +&= (\text{Re } X_{1}[N-k]-\text{Im } X_{2}[N-k])+j(\text{Im } X_{1}[N-k]+\text{Re } X_{2}[N-k]) \nonumber \\ +&= (\text{Re } X_{1}[k]+\text{Im } X_{2}[k])+j(-\text{Im } X_{1}[k]+\text{Re } X_{2}[k]) \tag{iii} +\end{align} +$$ + +and by (ii), (iii) we have + +$$ +\begin{align} +\begin{cases} +\displaystyle +X_{1}[k] = \frac{\text{Re } Y[k]+\text{Re } Y[N-k]}{2}+j\frac{\text{Im } Y[k]-\text{Im } Y[N-k]}{2}\\ +\displaystyle +X_{2}[k] = \frac{\text{Im } Y[k]+\text{Im } Y[N-k]}{2}-j\frac{\text{Re } Y[k]-\text{Re } Y[N-k]}{2} +\end{cases} \tag{iv} +\end{align} +$$ + +#### (3) +By definition we know that $W_{N}^{2kn}{=}W_{N/2}^{kn}$ and $W_{N}^{kN}{=}1$, hence we have + +$$ +\begin{aligned} +X[2k] +&= \sum_{n=0}^{N-1}x[n]W_{2N}^{2kn}+\sum_{n=N}^{2N-1}x[n]W_{2N}^{2kn}\\ +&= \sum_{n=0}^{N-1}x[n]W_{2N}^{2kn}+\sum_{n=0}^{N-1}x[n+N]W_{2N}^{2k(n+N)}\\ +&= \sum_{n=0}^{N-1}x[n]W_{N}^{kn}+\sum_{n=0}^{N-1}x[n+N]W_{N}^{k(n+N)}\\ +&= \sum_{n=0}^{N-1}x[n]W_{N}^{kn}+\sum_{n=0}^{N-1}x[n+N]W_{N}^{kn}\\ +&= \sum_{n=0}^{N-1}(x[n]+x[n+N])W_{N}^{kn}\\ +\end{aligned} +$$ + +Similarly, since $W_{2N}^{N}{=}{-}1$, we have + +$$ +\begin{aligned} +X[2k+1] +&= \sum_{n=0}^{N-1}x[n]W_{2N}^{(2k+1)n}+\sum_{n=0}^{N-1}x[n+N]W_{2N}^{(2k+1)(n+N)}\\ +&= \sum_{n=0}^{N-1}\left(x[n]+x[n+N]W_{2N}^{(2k+1)N}\right)W_{2N}^{(2k+1)n}\\ +&= \sum_{n=0}^{N-1}\left(x[n]+x[n+N]W_{2N}^{N}\right)W_{2N}^{2kn}W_{2N}^{n}\\ +&= \sum_{n=0}^{N-1}\left(x[n]-x[n+N]\right)W_{2N}^{n}W_{N}^{kn} +\end{aligned} +$$ + +Therefore, let $y[n]{=}x[n]{+}x[n+N]$ and $z[n]{=}x[n]{-}x[n+N]$, we have + +$$ +\begin{aligned} +\begin{cases} +X[2k] = \sum_{n=0}^{N-1}y[n]W_{N}^{kn}\\ +X[2k] = \sum_{n=0}^{N-1}y[n]W_{2N}^{n}W_{N}^{kn} +\end{cases} +\end{aligned} +$$ + +which implies that $2N$-point discrete Fourier transforms can be obtained using two executions of the $N$-point Fourier transform. + +By using the result from the previous question (2), it has been demonstrated that the $N$-point discrete Fourier transforms of two different sequences can be obtained using a single execution of the $N$-point Fourier transform, thereby showing that a $2N$-point discrete Fourier transform can be obtained using a single execution of the $N$-point Fourier transform. diff --git a/mkdocs.yml b/mkdocs.yml index 19a0812ec..c0c07a3b0 100644 --- a/mkdocs.yml +++ b/mkdocs.yml @@ -450,7 +450,8 @@ nav: - 情報学基礎 F2-2: kakomonn/kyoto_university/informatics/ist_202308_kiso_f2_2.md - 専門科目 S-2: kakomonn/kyoto_university/informatics/ist_202308_senmon_s_2.md - 専門科目 S-3: kakomonn/kyoto_university/informatics/ist_202308_senmon_s_3.md - - 専門科目 S-4: kakomonn/kyoto_university/informatics/ist_202308_senmon_s_4.md + - 専門科目 S-4: kakomonn/kyoto_university/informatics/ist_202308_senmon_s_4.md + - 専門科目 S-5: kakomonn/kyoto_university/informatics/ist_202308_senmon_s_5.md - 2023年度: - 情報学基礎 F1-1: kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_1.md - 情報学基礎 F1-2: kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_2.md