Question

英文文献翻译 高分悬赏

The first thing to note about the process of taking
one step is that the maximum available torque is at
a minimum when the rotor is halfway from one step
to the next. This minimum determines the running
torque, the maximum torque the motor can drive as it
steps slowly forward. For common two-winding permanent
magnet motors with ideal sinusoidal torque
versus position curves and holding torque h, this will
be h/(20.5). If the motor is stepped by powering two
windings at a time, the running torque of an ideal
two-winding permanent magnet motor will be the
same as the single-winding holding torque.
It should be noted that at higher stepping speeds, the
running torque is sometimes defined as the pull-out
torque. The resulting trajectory may resemble the one
shown in Fig. 5.
The resonant frequency of the motor rotor depends
on the amplitude of the oscillation; but as the
amplitude decreases, the resonant frequency rises to
a well-defined small-amplitude frequency. Formally,
the small-amplitude resonance can be computed as
follows. First, recall Newton’s law for angular acceleration:
T = μA
where T torque applied to rotor, μ moment of inertia
of rotor and load, and A angular acceleration, in radians
per second. We assume that, for small amplitudes,
the torque on the rotor can be approximated as a linear
function of the displacement from the equilibrium
position. Therefore, Hooke’s law applies:
T = −kΘ
where k is the “spring constant” of the system, in
torque units per radian, Θ the angular position of
rotor, in radians.
We can equate the two formulas for the torque to
get: μA = −kΘ. Note that acceleration is the second
derivative of position with respect to time: A = d2Θ/dt2, so we can rewrite this the above in differential
equation form: d2Θ/dt2 = −(k/μ)Θ. To solve
this, recall that, for: f(t) = a sin bt.
The derivatives are:
df(t)
dt = ab cos bt
d2f(t)
dt2 = −ab2 sin bt = −b2f(t)
Note that, throughout this discussion, we assumed that
the rotor is resonating. Therefore, it has an equation
of motion something like:
Θ = a sin(2πft)
a is the angular amplitude of resonance; f the resonant
frequency. This is an admissible solution to the above
differential equation if we agree that:
b = 2πf
b2 =
k
μ
In practice, this oscillation can cause significant problems
when the stepping rate is anywhere near a resonant
frequency of the system; the result frequently
appears as random and uncontrollable motion.
The fuzzy logic controller provides an algorithm,
which converts the linguistic control, based on expert
knowledge into an automatic control strategy [13].
Therefore, the fuzzy logic algorithm is much closer
in spirit to human thinking than traditional logical
systems [5,14]. The main problem with fuzzy logic
controller generation is related to the choice of the
regulator parameters [11].
公式看不清楚可以不要翻译！谢谢

Answer 1

首先要注意的过程索取

第一步是,最大限度地利用现有的转矩的所在

当转子至少一半的从一步

到下一个。这个最低决定了跑步

扭矩,最大转矩马达开车送了

慢慢向前步骤。对普通的two-winding永久

磁铁电动机和理想的正弦扭矩

与位置曲线和控股扭矩h的时候,这个意愿

是氢/(7)。如果电动机驱动的了两个

绕组,这对理想的运行扭矩

two-winding永磁电机将

single-winding一样持有扭矩。

值得注意的是,在较高的步进速度

有时可定义为运行扭矩的更远

扭矩。由此产生的轨迹类似于

显示在图5。

谐振频率的电机转子视情况而定

在幅值的振荡;而作为

振幅减少,谐振频率上升到

一个定义明确的方法的频率。正式,

就可以计算方法的共鸣

随之而来。首先,回忆牛顿定律为角加速度:

μA T =

在应用于转子,T扭矩μ惯性矩的吗

转子和负载和角加速度、按弧度

/每秒。我们假定对小振幅,

转矩对转子近似为一个线性

功能的位移平衡

位置。因此,虎克定律运用:

kΘ−T =

在k的“弹簧不变的”的基础上,在吗

扭矩单元每弧度,Θ角位置

转子,按弧度。

我们可以把大两公式

μA =−得到:kΘ。注意,加速度为第二阶段

衍生的位置相对于时间:d2Θ/ dt2),因此我们可以重写这上面的鉴别

方程形式:d2Θ/ dt2 =−(k /μ)Θ。解决

这,回忆,对于:女(t)=赎罪bt。

衍生金融工具的是:

df(t)

dt = ab因为bt

d2f(t)

dt2 =−ab2−b2f罪bt =(t)

注意到,在这个讨论中,我们假设

转子共鸣。因此,它拥有一个等式

运动类似:

罪(=Θ2πft)

是角振幅的共振;如果共振

频率。这是一个允许的解决以上

如果我们同意微分方程。

b = 2πf

b2 =

凯西

μ

在实践中,这振荡会导致严重的问题

当步进率是一个共振附近的什么地方

频率的系统;结果频繁

表现为随机和不可控制的运动。

模糊逻辑控制器提供了一种运算,

它把语言控制,基于专家吗

知识转化为自动控制策略[13]。

因此,模糊逻辑算法是近多了

精神上的比传统人类思维的逻辑上的

系统[5,14]。模糊逻辑的主要问题

控制器产生相关的选择

[11]调节器参数。

Answer 2

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