Kyoto-University
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
Author realball
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 \(T\) .
Kai Q.1
\[
\begin{aligned}
\mathcal{F}^{-1}[P_\Omega(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}P_{\Omega}(w)e^{jwt} d\omega\\
&=\frac{1}{2\pi}\int_{-\Omega}^{\Omega}e^{j\omega t}d\omega=\frac{1}{2\pi}\cdot\frac{e^{j\omega}}{j\omega}\mid_{-\omega}^{\omega}=\frac{1}{2\pi jt}\cdot 2j\sin \Omega t\\
&=\frac{1}{\pi t}\cdot\sin \Omega t
\end{aligned}
\]
Q.2
\[
S_{T}(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\frac{2\pi}{T}t}
\]
\[
{{a_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\delta_{T}(t)e^{-jt\frac{2\pi}{T}t} dt=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=-\infty}^{\infty}\delta(t-nT)e^{-j\frac{2\pi}{T}t}dt=\frac{1}{T}}}
\]
\[
\delta_T(t)=\frac{1}{T}\sum_{k=-\infty}^{\infty}e^{jk\frac{2\pi}{T}t}
\]
\[
\begin{aligned}
\mathcal{F}\left[\delta_{T}(t)\right]&=\sum_{k=-\infty}^{\infty}\frac{2\pi}{T}\delta(\omega-k\frac{2\pi}{T})\\
&=\frac{2\pi}{T}\delta_\frac{2\pi}{T}(\omega)
\end{aligned}
\]
Q.3
\[
\begin{aligned}\mathcal{F}[x_{s}(t)]=X_{s}(j\omega)
&=\frac1{2\pi}\int_{-\infty}^{\infty}X\left[j{\theta}\cdot\mathcal{F}[\delta_T(t)]\right]\\
&=\frac1{2\pi}\int_{-\infty}^{\infty}X(j\theta)\cdot\sum_{k=-\infty}^{\infty}\frac{2\pi}{T}\delta(\omega-\theta-\frac{2\pi k}{T})d\theta\\
&=\frac1{2\pi}\cdot\frac{2\pi}{T}\sum_{k=\infty}^{\infty}\int_{-\infty}^{\infty}X(j\theta)\delta(w-\theta-\frac{2\pi k}T)d\theta\\
&\text{when }\theta=\omega+\frac{2\pi k}T, \delta(\omega-\theta-\frac{2\pi k}T)\not =0\\
&=\frac1T\sum_{k=-\infty}^{\infty}X(j(w+\frac{2\pi k}T))
\end{aligned}
\]