PID控制器积分项导致极端不稳定

我有一个在机器人上运行的PID控制器，旨在使机器人转向指南针前进。 以20Hz的速率重新计算/应用PID校正。

虽然PID控制器在PD模式下运行良好（IE，积分项为零），但即使是最轻微的积分也会迫使输出不稳定，使得转向执行器被推到左或右极限。

码：

private static void DoPID(object o) { // Bring the LED up to signify frame start BoardLED.Write(true); // Get IMU heading float currentHeading = (float)RazorIMU.Yaw; // We just got the IMU heading, so we need to calculate the time from the last correction to the heading read // *immediately*. The units don't so much matter, but we are converting Ticks to milliseconds int deltaTime = (int)((LastCorrectionTime - DateTime.Now.Ticks) / 10000); // Calculate error // (let's just assume CurrentHeading really is the current GPS heading, OK?) float error = (TargetHeading - currentHeading); LCD.Lines[0].Text = "Heading: "+ currentHeading.ToString("F2"); // We calculated the error, but we need to make sure the error is set so that we will be correcting in the // direction of least work. For example, if we are flying a heading of 2 degrees and the error is a few degrees // to the left of that ( IE, somewhere around 360) there will be a large error and the rover will try to turn all // the way around to correct, when it could just turn to the right a few degrees. // In short, we are adjusting for the fact that a compass heading wraps around in a circle instead of continuing // infinity on a line if (error  180) error = error - 360; // Add the error calculated in this frame to the running total SteadyError = SteadyError + (error * deltaTime); // We need to allow for a certain amount of tolerance. // If the abs(error) is less than the set amount, we will // set error to 0, effectively telling the equation that the // rover is perfectly on course. if (MyAbs(error) < AllowError) error = 0; LCD.Lines[2].Text = "Error: " + error.ToString("F2"); // Calculate proportional term float proportional = Kp * error; // Calculate integral term float integral = Ki * (SteadyError * deltaTime); // Calculate derivative term float derivative = Kd * ((error - PrevError) / deltaTime); // Add them all together to get the correction delta // Set the steering servo to the correction Steering.Degree = 90 + proportional + integral + derivative; // We have applied the correction, so we need to *immediately* record the // absolute time for generation of deltaTime in the next frame LastCorrectionTime = DateTime.Now.Ticks; // At this point, the current PID frame is finished // ------------------------------------------------------------ // Now, we need to setup for the next PID frame and close out // The "current" error is now the previous error // (Remember, we are done with the current frame, so in // relative terms, the previous frame IS the "current" frame) PrevError = error; // Done BoardLED.Write(false); } 

有谁知道为什么会发生这种情况或如何解决它？

Answer 1

看起来你正在将你的时基用于积分三次。 错误已经是自上一个样本以来的累积误差，所以你不需要将deltaTime乘以它。 所以我会将代码更改为以下内容。

SteadyError += error ;

SteadyError是误差的积分或总和。

所以积分应该只是SteadyError * Ki

float integral = Ki * SteadyError;

编辑：

我已经完成了你的代码，除了上面的修复之外，我还会修复其他几个项目。

1）您不希望以毫秒为单位的增量时间。 在正常的采样系统中，delta项将是1，但是对于20Hz的速率，你输入的值为50，这具有通过该因子增加Ki并且还将Kd减小50倍的效果。 如果您担心抖动，则需要将增量时间转换为相对采样时间。 我会用这个公式代替。

float deltaTime = (LastCorrectionTime - DateTime.Now.Ticks) / 500000.0

500000.0是每个样本的预期滴答数，对于20Hz是50ms。

2）将积分项保持在一个范围内。

 if ( SteadyError > MaxSteadyError ) SteadyError = MaxSteadyError; if ( SteadyError < MinSteadyError ) SteadyError = MinSteadyError; 

3）更改以下代码，以便当错误在-180左右时，您只需稍微更改就不会出错。

 if (error < -270) error += 360; if (error > 270) error -= 360; 

4）validationSteering.Degree正在接收正确的分辨率和符号。

5）最后哟可能只是将deltaTime一起丢弃并按以下方式计算差分项。

 float derivative = Kd * (error - PrevError); 

随之而来的是你的代码。

 private static void DoPID(object o) { // Bring the LED up to signify frame start BoardLED.Write(true); // Get IMU heading float currentHeading = (float)RazorIMU.Yaw; // Calculate error // (let's just assume CurrentHeading really is the current GPS heading, OK?) float error = (TargetHeading - currentHeading); LCD.Lines[0].Text = "Heading: "+ currentHeading.ToString("F2"); // We calculated the error, but we need to make sure the error is set // so that we will be correcting in the // direction of least work. For example, if we are flying a heading // of 2 degrees and the error is a few degrees // to the left of that ( IE, somewhere around 360) there will be a // large error and the rover will try to turn all // the way around to correct, when it could just turn to the right // a few degrees. // In short, we are adjusting for the fact that a compass heading wraps // around in a circle instead of continuing infinity on a line if (error < -270) error += 360; if (error > 270) error -= 360; // Add the error calculated in this frame to the running total SteadyError += error; if ( SteadyError > MaxSteadyError ) SteadyError = MaxSteadyError; if ( SteadyError < MinSteadyError ) SteadyError = MinSteadyError; LCD.Lines[2].Text = "Error: " + error.ToString("F2"); // Calculate proportional term float proportional = Kp * error; // Calculate integral term float integral = Ki * SteadyError ; // Calculate derivative term float derivative = Kd * (error - PrevError) ; // Add them all together to get the correction delta // Set the steering servo to the correction Steering.Degree = 90 + proportional + integral + derivative; // At this point, the current PID frame is finished // ------------------------------------------------------------ // Now, we need to setup for the next PID frame and close out // The "current" error is now the previous error // (Remember, we are done with the current frame, so in // relative terms, the previous frame IS the "current" frame) PrevError = error; // Done BoardLED.Write(false); } 

Answer 2

你是在初始化SteadyError （奇怪的名字……为什么不是“积分器”）？ 如果它在启动时包含一些随机值，它可能永远不会返回到接近零（ 1e100 + 1 == 1e100 ）。

您可能会受到积分器饱和的影响 ，通常应该消失，但是如果它需要更长的时间来减少，而不是完成整个旋转（并再次完成积分器）。 如果您的系统需要，尽管有更先进的解决方案 （PDF，879 kB），但平凡的解决方案是对集成商施加限制。

Ki有正确的标志吗？

我强烈反对使用浮点数作为PID参数，因为它们具有任意精度。 使用整数（可能是固定点 ）。 你将不得不进行限制检查，但它比使用浮动更加明智。

Answer 3

积分项已经累积了一段时间，乘以deltaTime将使其以时间平方的速率累积。 事实上，由于已经通过将错误乘以deltaTime错误地计算了SteadyError，这是时间立方的！

在SteadyError中，如果您尝试补偿非周期性更新，则最好修复非周期性。 但是，无论如何，计算都是有缺陷的。 您已经以误差/时间为单位计算，而您只想要误差单位。 如果真的有必要，可以通过以下方式补偿定时抖动的算术正确方法：

 SteadyError += (error * 50.0f/deltaTime); 

如果deltaTime保持为毫秒且标称更新速率为20Hz。 但是deltaTime可以更好地计算为float，如果它是你试图检测的定时抖动，则根本不会转换为毫秒; 你不必要地抛弃精确度。 无论哪种方式，您需要的是通过标称时间与实际时间的比率来修改误差值。

一个好的读取是没有博士学位的PID

Answer 4

我不确定为什么你的代码不起作用，但我几乎是肯定的，你也无法测试它以查明原因。 你可以注入一个计时器服务，这样你就可以嘲笑它，看看发生了什么：

 public interace ITimer { long GetCurrentTicks() } public class Timer : ITimer { public long GetCurrentTicks() { return DateTime.Now.Ticks; } } public class TestTimer : ITimer { private bool firstCall = true; private long last; private int counter = 1000000000; public long GetCurrentTicks() { if (firstCall) last = counter * 10000; else last += 3500; //ticks; not sure what a good value is here //set up for next call; firstCall = !firstCall; counter++; return last; } } 

然后，用GetCurrentTicks()替换对DateTime.Now.Ticks两次调用，您可以单步GetCurrentTicks()代码并查看值的样子。