数据的维度不得超过 2 个！

发布于 2024-11-10 00:40:16 字数 773 浏览 3 评论 0原文

我正在尝试在某些控制系统中进行开环分析。

首先，我分析了PT1，效果很好！

% Assigning the variables
Ksys = 0.8667;
T1 = 1.65;

% PT1 modeling
num = [0 Ksys];
den = [T1 1];
PT1 = tf(num, den);

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PT1, t);

% Plotting
plot(t, y);
ylim([0, 6]);

但是当我分析高阶工厂时：

% Assigning the variables
Ksys = 0.8667;
Tc = 1.65;
Td = 0.25;

% PTn modeling
num = [0 Ksys];
den = [Tc 1];
PT1 = tf(num, den);
sh = pade(Td, 2);
PTn = PT1 * sh;

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PTn, t);

% Plotting
plot(t, y);
ylim([0, 6]);

它给出以下错误：

??? Error using ==> plot
Data may not have more than 2 dimensions

Error in ==> TestProject at 25
plot(t, y);

我该如何解决这个问题？

原文

I am trying to do open-loop analysis in some control system.

First, I analyze PT1, and it works fine!

% Assigning the variables
Ksys = 0.8667;
T1 = 1.65;

% PT1 modeling
num = [0 Ksys];
den = [T1 1];
PT1 = tf(num, den);

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PT1, t);

% Plotting
plot(t, y);
ylim([0, 6]);

But when I analyze the higher order plant:

% Assigning the variables
Ksys = 0.8667;
Tc = 1.65;
Td = 0.25;

% PTn modeling
num = [0 Ksys];
den = [Tc 1];
PT1 = tf(num, den);
sh = pade(Td, 2);
PTn = PT1 * sh;

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PTn, t);

% Plotting
plot(t, y);
ylim([0, 6]);

it gives the following error:

??? Error using ==> plot
Data may not have more than 2 dimensions

Error in ==> TestProject at 25
plot(t, y);

How can I solve this issue ?

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街角卖回忆 2024-11-17 00:40:16

plot 给出错误的原因是 y 是一个 3D 数组，并且它需要一个 2D 矩阵。 y 中的第二个维度是单一维度，出于各种目的，可以将其删除。使用函数 squeeze 消除单一维度。请尝试以下操作：

plot(t',squeeze(y))

The reason plot gives you an error is because y is a 3D array and it expects a 2D matrix. The second dimension in y is a singleton dimension and for all purposes, can be removed. Use the function squeeze to get rid of the singleton dimension. Try the following:

plot(t',squeeze(y))

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绝不放开 2024-11-17 00:40:16

好吧，我找到了问题所在，它在 pade() 函数中。这是最终的代码：

% Assigning the variables
Ksys = 0.8667;
Tc = 1.65;
Td = 0.25;

% PTn modeling
num = [0 Ksys];
den = [Tc 1];
PT1 = tf(num, den);
[nums, dens] = pade(Td, 2);
sh = tf(nums, dens);
PTn = PT1 * sh;

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PTn, t);

% Plotting
plot(t, y);
ylim([0, 6]);

Well, I figured out the probelm, it was in pade() function. Here is the final code:

% Assigning the variables
Ksys = 0.8667;
Tc = 1.65;
Td = 0.25;

% PTn modeling
num = [0 Ksys];
den = [Tc 1];
PT1 = tf(num, den);
[nums, dens] = pade(Td, 2);
sh = tf(nums, dens);
PTn = PT1 * sh;

% Step Response
t = 0:0.01:10;
y = 4.5 * step(PTn, t);

% Plotting
plot(t, y);
ylim([0, 6]);

回复收藏 0 原文

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