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A new variable step-size LMS adaptive filtering algorithm is presented, which makes use of the intercorrelation function of the relative output error signal to control the new variable step-size.

上述问题,提种新步长LMS自适滤波算法,它采用相差互相关函数来控制步长更新。

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妙想物语

What does it mean to improve your stride length?

提高步长意味着什

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妙想物语

Your stride is made up of two components: stride length (L) and rhythm (Y) .

步长(L)和节奏(Y)。

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妙想物语

Use deliberate practice to enhance your skills, and subsequently, your stride length.

通过刻意练习来提高你的技能,进而提高你的步长

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妙想物语

So how do you improve your stride length?

要如何提高步长

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妙想物语

Stride length is the distance between each step, and rhythm is how often you take each step.

步长是指每一步之间的距离,节奏是你每一步的频

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妙想物语

Stride length is a measure of how effective the action you take is with respect to your goal.

步长是衡量你所采取的行动对你的目标有多大成效的一种方法。

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妙想物语

Now, after stride length, the final component is rhythm: how often you take a stride.

现在,在步长之后,最后一个组成部分是节奏:你迈出步幅的频

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科学60秒-科学美国人 2021年5月合集

And we scaled it up to the size of our T. rex. And that gives you step length.

放大到霸王龙的大小。这给了你步长

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文献阅读

Meanwhile, an improved self-adaptive bat algorithm with step-controlled and mutation mechanisms is proposed herein.

同时,本文提出了一种改进的自适应蝙蝠算法,该算法结合了步长控制与变异机制。

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TED-Ed(视频版)

The answer is given by this formula: distance equals step size times the square root of the number of steps.

这个公式给出了答案:距离等于步长乘以步数的平方根。

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科学60秒-科学美国人 2021年5月合集

Intagliata: Multiply step length by step rhythm, and you get a speed of 1.28 meters per second, or about 2.9 miles per hour.

Intagliata:将步长乘以步速,得到每秒 1. 28 米的速度, 或约每小时 2. 9 英里。

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高等数学之美

And instead of describing the step size as delta theta, a  concrete finite amount, we instead describe it as " d" theta, which I like to think of as signaling  the fact that some kind of limit is being taken.

没有将步长描述为 delta theta(具体的有限量),而是将其描述为“d”theta,我喜欢将其视为表示正在采取某种限制的事实。

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高等数学之美

This alternate solution gives a richer understanding that can let you answer other questions about the setup, and it's also core to how I coded accurate simulations of these blocks without requiring absurdly small time steps and huge computation time.

这种替代解决方案提供了更丰富的理解, 可以让您回答有关设置的其他问题,也是我如何对这些块进行精确模拟编码的核心, 而不需要极小的时间步长和大量的计算时间。

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