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)

\[ F[f_1(x)]=\frac{1}{\sqrt{2\pi}} \left( \frac{e^{-2i k} - 1}{i k} \right) \]

(2)

\[ F[f_2(x)]=\frac{\sqrt{2\pi}}{2}\delta(-k)+\frac{\sqrt{2\pi}}{4}\delta(2\omega-k)+\frac{\sqrt{2\pi}}{4}\delta(-2\omega-k) \]

Q.2

(1)

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

(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)

\[ \mathcal{F}[f_3(x)]=\frac{2\sin^2k}{-\sqrt{2\pi}k^2} \]