Matlab 仿真——直流电机速度控制（2）系统分析

上一节我们成功在matlab建立好了电机系统模型，这一节我们来分析这个系统的时域响应

0. 被控对象与设计要求

这里把设计需求和系统转换方程粘贴在这里：设计需求（阶跃响应）：

稳定时间<2s
超调<5%
稳态误差<1%

转换方程：
Matlab 仿真——直流电机速度控制（2）系统分析

1. 开环响应

让我们看看该系统的开环响应如何，是否满足我们对电机的要求。matlab自带了线性系统分析工具。 Matlab运行以下代码

<code class="prism language-python has-numbering"><span class="token operator">%</span>motor parameter
J <span class="token operator">=</span> <span class="token number">0.01</span><span class="token punctuation">;</span>
b <span class="token operator">=</span> <span class="token number">0.1</span><span class="token punctuation">;</span>
K <span class="token operator">=</span> <span class="token number">0.01</span><span class="token punctuation">;</span>
R <span class="token operator">=</span> <span class="token number">1</span><span class="token punctuation">;</span>
L <span class="token operator">=</span> <span class="token number">0.5</span><span class="token punctuation">;</span>
<span class="token operator">%</span>motor tf function
s <span class="token operator">=</span> tf<span class="token punctuation">(</span><span class="token string">'s'</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
P_motor <span class="token operator">=</span> K<span class="token operator">/</span><span class="token punctuation">(</span><span class="token punctuation">(</span>J<span class="token operator">*</span>s<span class="token operator">+</span>b<span class="token punctuation">)</span><span class="token operator">*</span><span class="token punctuation">(</span>L<span class="token operator">*</span>s<span class="token operator">+</span>R<span class="token punctuation">)</span><span class="token operator">+</span>K<span class="token operator">^</span><span class="token number">2</span><span class="token punctuation">)</span>
<span class="token operator">%</span>求系统P_motor的阶跃响应，时间为<span class="token number">0</span>到<span class="token number">5</span>秒，步长<span class="token number">0.</span>1s
linearSystemAnalyzer<span class="token punctuation">(</span><span class="token string">'step'</span><span class="token punctuation">,</span>P_motor<span class="token punctuation">,</span><span class="token number">0</span><span class="token punctuation">:</span><span class="token number">0.1</span><span class="token punctuation">:</span><span class="token number">5</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
</code>

输出
在这里插入图片描述在窗口上右击选择要显示的数据，我们得到系统的稳定时间为2.07秒，终值为0.0999。稳定时间不满足我们的要求。

2. LTI 模型特征

我们选择该系统的零极点图来观察
在这里插入图片描述
发现该系统有两个实极点，一个在-2处，一个在-10处。也因为极点没有虚部，所以系统不存在震荡。根据现代控制系统的知识，我们推测-2处的极点主导了系统的响应。为了验证这个想法，我们找到一个单极点的系统进行比较 rP_motor = 0.1/(0.5*s+1)。代码如下

<code class="prism language-csharp has-numbering"><span class="token operator">%</span>motor parameter
J <span class="token operator">=</span> <span class="token number">0.01</span><span class="token punctuation">;</span>
b <span class="token operator">=</span> <span class="token number">0.1</span><span class="token punctuation">;</span>
K <span class="token operator">=</span> <span class="token number">0.01</span><span class="token punctuation">;</span>
R <span class="token operator">=</span> <span class="token number">1</span><span class="token punctuation">;</span>
L <span class="token operator">=</span> <span class="token number">0.5</span><span class="token punctuation">;</span>
<span class="token operator">%</span>motor tf function
s <span class="token operator">=</span> <span class="token function">tf</span><span class="token punctuation">(</span><span class="token string">'s'</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
P_motor <span class="token operator">=</span> K<span class="token operator">/</span><span class="token punctuation">(</span><span class="token punctuation">(</span>J<span class="token operator">*</span>s<span class="token operator">+</span>b<span class="token punctuation">)</span><span class="token operator">*</span><span class="token punctuation">(</span>L<span class="token operator">*</span>s<span class="token operator">+</span>R<span class="token punctuation">)</span><span class="token operator">+</span>K<span class="token operator">^</span><span class="token number">2</span><span class="token punctuation">)</span>
rP_motor <span class="token operator">=</span> <span class="token number">0.1</span><span class="token operator">/</span><span class="token punctuation">(</span><span class="token number">0.5</span><span class="token operator">*</span>s<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
<span class="token operator">%</span>求系统P_motor的阶跃响应，时间为<span class="token number">0</span>到<span class="token number">5</span>秒，步长<span class="token number">0.1</span>s
<span class="token function">linearSystemAnalyzer</span><span class="token punctuation">(</span><span class="token string">'step'</span><span class="token punctuation">,</span>P_motor<span class="token punctuation">,</span><span class="token number">0</span><span class="token punctuation">:</span><span class="token number">0.1</span><span class="token punctuation">:</span><span class="token number">5</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
</code>

在弹出的窗口里面我们导入新的系统阶跃响应
在这里插入图片描述

可以发现系统阶跃响应很接近，这印证了我们的猜想。

3. 其他输入信号的响应

除了阶跃响应，有时我们也想看看系统输入为其他输入信号时的情况。你可以直接导入你自己定义好的输入信号，或者用Matlab为你准备好的信号模板定制输入信号。接下来我们演示如何用模板信号。右击，选择Plot Types——Linear Simulation
在这里插入图片描述定义仿真时间，步长，然后点击设计信号
填写信号参数，插入，然后点击仿真