신호의 자기상관함수

Definition¹

Analog Signal
For an energy signal $f \in L^{2}(\mathbb{R})$, $R_{f}$ defined as follows is called the auto-correlation function.
$$ R_{f}(\tau) := \int_{-\infty}^{\infty} \overline{f(t)} f(t + \tau) dt $$
Here $\overline{f(t)}$ is the complex conjugate of $f(t)$.
Digital Signal
The auto-correlation function for the energy signal $\left\{ x_{n} \right\} \in \ell^{2}$ is defined as follows.
$$ R_{x}(m) := \sum\limits_{n \in \mathbb{N}} \overline{x_{n}}x_{n+m} $$

Explanation

Expressed in terms of inner product, it can be written as follows. We simplify $f$’s translation as $f_{-\tau} = T_{-\tau}f$,

$$ R_{f}(\tau) = \braket{f, T_{-\tau}f} = \braket{f, f_{-\tau}} $$

The auto-correlation function is a measure of the similarity between signal $f$ and $f_{-\tau}$. If we define the distance in space $L^{2}$ as $d(f,g) = \left\| f - g \right\|_{2} = \sqrt{\braket{f-g, f-g}}$,

$$ \begin{align*} d(f, f_{-\tau}) ^{2} &= \braket{f-f_{-\tau}, f-f_{-\tau}} \\ &= \braket{f, f} - \braket{f, f_{-\tau}} - \braket{f_{-\tau}, f} + \braket{f_{-\tau}, f_{-\tau}} \\ &= \left\| f \right\|_{2} - \braket{f, f_{-\tau}} - \overline{\braket{f, f_{-\tau}}} + \left\| f_{-\tau} \right\| \\ &= 2\left\| f \right\|_{2} - 2 \Re \braket{f, f_{-\tau}} \\ &= 2\left\| f \right\|_{2} - 2 \Re \left( R_{f}(\tau) \right) \end{align*} $$

Therefore, when the value of $R_{f}(\tau)$ increases, the disparity between $f$ and $f_{-\tau}$ decreases, and when the value decreases, the disparity increases.

Theorem

For an analog signal $f \in L^{2}(\mathbb{R})$, the following holds:

$$ S_{f}(\omega) = \int_{-\infty}^{\infty} R_{f}(\tau) e^{-i\omega \tau} d\tau = \hat{R_{f}}(\omega) $$

Here, $S_{f}(\omega)$ is the energy spectrum density of $f$, and $\hat{R_{f}}$ is the Fourier transform of $R_{f}$.

Proof

By appropriately substituting variables, this can be easily shown. $$ \begin{align*} \hat{R_{f}}(\omega) &= \int_{-\infty}^{\infty} R_{f}(\tau) e^{-i\omega \tau} d\tau \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \overline{f(t)} f(t + \tau) dt e^{-i\omega \tau} d\tau \\ &= \int_{-\infty}^{\infty} \overline{f(t)} \int_{-\infty}^{\infty} f(t + \tau) e^{-i\omega \tau} d\tau dt \\ &= \int_{-\infty}^{\infty} \overline{f(t)} \int_{-\infty}^{\infty} f(\tau^{\prime}) e^{-i\omega (\tau^{\prime} - t)} d\tau^{\prime} dt \qquad (\tau^{\prime} = t + \tau) \\ &= \int_{-\infty}^{\infty} \overline{f(t)} \int_{-\infty}^{\infty} f(\tau^{\prime}) e^{-i\omega \tau^{\prime}} d\tau^{\prime} e^{i\omega t}dt \\ &= \int_{-\infty}^{\infty} \overline{f(t)} \hat{f}(\omega) e^{i\omega t}dt \\ &= \hat{f}(\omega) \int_{-\infty}^{\infty} \overline{f(t)} e^{i\omega t}dt \\ &= \hat{f}(\omega) \overline{\int_{-\infty}^{\infty} f(t) e^{-i\omega t}dt} \\ &= \hat{f}(\omega) \overline{\hat{f}(\omega)} \\ &= | \hat{f}(\omega) |^{2} \\ &= S_{f}(\omega) \end{align*} $$

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