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

Author

realball

Description

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

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,

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

(2)

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

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

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

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)

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

(2)

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

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

(3)

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