卷积、互相关、自相关可视化比较

$g(\tau )$ is an even function， $(g(-\tau)=g(\tau) )$ , so convolution is equivalent to correlation.

convolution $(f∗g)(t) = \int_{-∞}^{∞} f (τ) g (t-τ) dτ\,.$
cross-correlation $(f \star g)(τ) = \int_{-∞}^{∞} \overline{f (t)} g (t+τ) dt\,.$


1.Express each function in terms of a dummy variable $\tau$ . 2.Reflect one of the functions: $g(\tau )$ → $g(-\tau ).$ 3.Add a time-offset, t, which allows $g(t-\tau )$ to slide along the $\tau$ -axis. 4.Start t at −∞ and slide it all the way to +∞. Wherever the two functions intersect, find the integral of their product. In other words, compute a sliding, weighted-sum of function $f(\tau )$ , where the weighting function is $g(-\tau ).$
In this example, the red-colored “pulse”, $g(\tau )$ ,is an even function $(g(-\tau)=g(\tau) )$ , so convolution is equivalent to correlation. A snapshot of this “movie” shows functions $g(t-\tau )$ and $f(\tau )$ (in blue) for some value of parameter $t$ , which is arbitrarily defined as the distance from the $\tau =0$ axis to the center of the red pulse. The amount of yellow is the area of the product $f(\tau )\cdot g(t-\tau )$ , computed by the convolution/correlation integral. The movie is created by continuously changing $t$ and recomputing the integral. The result (shown in black) is a function of $t$ , but is plotted on the same axis as $\tau$ , for convenience and comparison.
In this depiction, $f(\tau )$ could represent the response of an RC circuit to a narrow pulse that occurs at $\tau =0$ . In other words, if $g(\tau )=\delta (\tau )$ , the result of convolution is just $f(t)$ . But when $g(\tau )$ is the wider pulse (in red), the response is a “smeared” version of $f(t)$ . It begins at $t=-0.5$ , because we defined $t$ as the distance from the $\tau =0$ axis to the center of the wide pulse (instead of the leading edge).

1.Express each function in terms of a dummy variable $\tau$ .

2.Reflect one of the functions: $g(\tau )$ → $g(-\tau ).$

3.Add a time-offset, t, which allows $g(t-\tau )$ to slide along the $\tau$ -axis.

4.Start t at −∞ and slide it all the way to +∞. Wherever the two functions intersect, find the integral of their product. In other words, compute a sliding, weighted-sum of function $f(\tau )$ , where the weighting function is $g(-\tau ).$

In this example, the red-colored “pulse”, $g(\tau )$ ,is an even function $(g(-\tau)=g(\tau) )$ , so convolution is equivalent to correlation. A snapshot of this “movie” shows functions $g(t-\tau )$ and $f(\tau )$ (in blue) for some value of parameter $t$ , which is arbitrarily defined as the distance from the $\tau =0$ axis to the center of the red pulse. The amount of yellow is the area of the product $f(\tau )\cdot g(t-\tau )$ , computed by the convolution/correlation integral. The movie is created by continuously changing $t$ and recomputing the integral. The result (shown in black) is a function of $t$ , but is plotted on the same axis as $\tau$ , for convenience and comparison.

In this depiction, $f(\tau )$ could represent the response of an RC circuit to a narrow pulse that occurs at $\tau =0$ . In other words, if $g(\tau )=\delta (\tau )$ , the result of convolution is just $f(t)$ . But when $g(\tau )$ is the wider pulse (in red), the response is a “smeared” version of $f(t)$ . It begins at $t=-0.5$ , because we defined $t$ as the distance from the $\tau =0$ axis to the center of the wide pulse (instead of the leading edge).

Convolution https://en.wikipedia.org/wiki/Convolution
Cross-correlation https://en.wikipedia.org/wiki/Cross-correlation

g(τ)g(\tau )g(τ) is an even function， (g(−τ)=g(τ))(g(-\tau)=g(\tau) )(g(−τ)=g(τ)), so convolution is equivalent to correlation.

$g(\tau )$ is an even function， $(g(-\tau)=g(\tau) )$ , so convolution is equivalent to correlation.