📂Fourier Analysis

Sampling Theorem

Buildup[^1]

Consider a physical signal $f$ being measured over time $t_{1} < t_{2} < t_{3} < \cdots$. Even if we know $f(t_{1}), f(t_{2}), \dots$, we generally cannot know the values for arbitrary $f(t)$. However, let’s assume that signal $f$ contains only frequencies within a certain range. That is, we consider a signal $f$ that contains only frequencies smaller than some constant $\Omega$, which is referred to as a band-limited signal.

In the language of Fourier analysis, this is the same as saying that the function values are all $0$ in the region where $\hat{f}(\omega)$ is $|\omega| \ge \Omega$. Therefore, the condition that signal $f$ is band-limited is equivalent to the condition $\hat{f} (\omega) = 0\ \text{for } | \omega | \ge \Omega$, meaning $\hat{f} \in L^{1}$. Under such conditions, the following powerful theorem holds.

Theorem

Assuming that $f\in L^{2}$ and $\hat{f} (\omega) = 0\ \text{for } | \omega | \ge \Omega$, then $f(t)$ is determined by the values in $n\pi / \Omega (n=0, \pm 1, \pm 2, \dots)$. That is, the following applies.

$$ f(t) = \sum \limits_{n = -\infty}^{\infty} f \left( \dfrac{n\pi}{\Omega}\right) \dfrac{\sin (\Omega t -n\pi )}{\Omega t - n\pi} $$

Explanation

This is referred to as the sampling theorem. The sampling theorem provides the condition under which all $f(t)$ for all $t$ can be determined using a countable number of function values $f \left( \dfrac{n\pi}{\Omega}\right)$.

From the perspective of signal analysis, the condition $\hat{f} (\omega) = 0\ \text{for } | \omega | \ge \Omega$ means that the frequency of signal $f$ is limited.

Also, a similar formula applies to the frequency function $\hat{f} \in L^{2}$ and time-limited signal $f(t)$.

Frequency Sampling Theorem

Assume that $\hat{f} \in L^{2}$ and $f (t) = 0\ \text{for } | t | \ge L$. Then $\hat{f}(\omega)$ is determined by the values in $n\pi / L(n=0, \pm 1, \pm 2, \dots)$. That is, the following applies.

$$ \hat{f}(\omega) = \sum \limits_{n = -\infty}^{\infty} \hat{f} \left( \dfrac{n\pi}{L}\right) \dfrac{\sin (L \omega -n\pi )}{L \omega - n\pi} $$

The proof is similar to that of the time sampling theorem.

Proof

Since $\hat{f} \in L^{1}$, $\hat{f}$ can be expressed as a Fourier series. $\hat{f} (\omega)$ is defined at $[-\Omega, \Omega]$, so the complex Fourier series is as follows.

$$ \begin{equation} \hat{f}(\omega)=\sum_{-\infty}^{\infty} c_{-n} e^{-i n \pi \omega / \Omega} \quad(|\omega| \leq \Omega) \label{eq1} \end{equation} $$

For convenience in the proof process, $n$ is denoted by $-n$. The coefficient $c_{-n}$ is as follows.

$$ \begin{align*} c_{-n} &= \dfrac{1}{2 \Omega} \int_{-\Omega}^{\Omega} \hat{f}(\omega) e^{i n \pi \omega / \Omega} d \omega \\ &= \dfrac{1}{2 \Omega} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i n \pi \omega / \Omega} d \omega \\ &= \dfrac{\pi}{\Omega} \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega (n \pi / \Omega)} d\omega \end{align*} $$

Since $\hat{f} (\omega) = 0\ for\ | \omega | \ge \Omega$, then $\hat{f} (\omega) \in L^{2}$, and by the Fourier inverse transform theorem, the following applies.

$$ \begin{equation} f(t) =\dfrac{1}{2\pi} \int _{-\infty} ^{\infty}\hat{f}(\omega) e^{i\omega t}d\omega \label{eq2} \end{equation} $$

Therefore, one obtains the following equation.

$$ \begin{align*} c_{-n} &= \dfrac{\pi}{\Omega} \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega (n \pi / \Omega)} d\omega \\ &= \dfrac{\pi}{\Omega} f \left( \dfrac{n\pi}{\Omega}\right) \end{align*} $$

Inserting this into $\eqref{eq1}$ yields the following.

$$ \hat{f}(\omega) = \sum_{-\infty}^{\infty} c_{-n} e^{-i n \pi \omega / \Omega} \quad(|\omega| \leq \Omega) = \sum_{-\infty}^{\infty} \dfrac{\pi}{\Omega} f \left( \dfrac{n\pi}{\Omega}\right) e^{-i n \pi \omega / \Omega} \quad(|\omega| \leq \Omega) $$

Again inserting this into $\eqref{eq2}$ yields the equation below.

$$ \begin{align*} f(t) &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega t} d \omega = \frac{1}{2 \pi} \int_{-\Omega}^{\Omega} \hat{f}(\omega) e^{i \omega t} d \omega \\ &=\frac{1}{2 \Omega} \int_{-\Omega}^{\Omega} \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega \end{align*} $$

In the function space, the inner product is defined as a definite integral, and since the inner product is continuous, the limit can be taken out, so the equation is as follows.

$$ \begin{align*} f(t) &= \frac{1}{2 \Omega} \int_{-\Omega}^{\Omega} \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega \\ &= \frac{1}{2 \Omega} \left\langle \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) e^{-i n \pi \omega / \Omega}, e^{-i \omega t} \right\rangle \\ &= \frac{1}{2 \Omega} \sum_{-\infty}^{\infty} \left\langle f\left(\frac{n \pi}{\Omega}\right) e^{-i n \pi \omega / \Omega}, e^{-i \omega t} \right\rangle \\ &= \frac{1}{2 \Omega} \sum_{-\infty}^{\infty} \int_{-\Omega}^{\Omega} f\left(\frac{n \pi}{\Omega}\right) e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega \\ &= \frac{1}{2 \Omega} \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) \int_{-\Omega}^{\Omega} e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega \end{align*} $$

Calculating the integral above yields the following.

$$ \begin{align*} \int_{-\Omega}^{\Omega} e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega &= \int_{-\Omega}^{\Omega} e^{i(\Omega t- n \pi )\omega/\Omega} d \omega \\ &= \left. \frac{e^{i(\Omega t-n \pi) \omega / \Omega}}{i(\Omega t-n \pi) / \Omega}\right|_{-\Omega} ^{\Omega} \\ &= \frac{1}{i(\Omega t-n \pi) / \Omega}\left[ e^{i(\Omega t-n \pi)} - e^{-i(\Omega t-n \pi)} \right] \\ &= \frac{2}{(\Omega t-n \pi) / \Omega}\dfrac{ e^{i(\Omega t-n \pi)} - e^{-i(\Omega t-n \pi)}}{2i} \\ &= \frac{2 \sin (\Omega t-n \pi)}{(\Omega t-n \pi) / \Omega} \end{align*} $$

Substituting and rearranging yields the following.

$$ \begin{align*} f(t) &= \frac{1}{2 \Omega} \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) \int_{-\Omega}^{\Omega} e^{-i n \pi \omega / \Omega} e^{i \omega t} d \omega \\ &= \frac{1}{2 \Omega} \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) \frac{2 \sin (\Omega t-n \pi)}{(\Omega t-n \pi) / \Omega} \\ &= \sum_{-\infty}^{\infty} f\left(\frac{n \pi}{\Omega}\right) \frac{2 \sin (\Omega t-n \pi)}{\Omega t-n \pi} \end{align*} $$

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