Why are there negative PID coefficients?
显示 更早的评论
Hi all
The transfer function I have is a type 1 transfer function so it doens't respond to step response and has 0 steady state error. 
. (The transfer function).
. (The transfer function).However, when I use PID controller only negative gains can stabilise it and not positive, which is unusual and something which perplexes me. I would appreciate any insight as to why is that so and what that tells us about the system. Any help would be appreciated
采纳的回答
Hi @Maaz Madha
You have to look into the plant transfer function first. On the denominator, the 's' can be factored out. On the numerator, the last term is a 'minus' sign. Thus, the step response of the plant will go into the negative direction.
s = tf('s');
Gp = 13.11*(s + 37.85)*(s - 35.58)*(s + 0.03922)/(s*(s^2 + 0.06348*s + 0.1765)*(s^2 + 19.91*s + 148.7))
step(Gp, 1e3)
Gp =
13.11 s^3 + 30.27 s^2 - 1.765e04 s - 692.4
-------------------------------------------------
s^5 + 19.97 s^4 + 150.1 s^3 + 12.95 s^2 + 26.25 s
tem To make the closed-loop system goes in the positive direction, naturally, the polarity of the proportional control gain has to be reversed. In this example, only the proportional controller is used to demonstrate the concept, Gc = -2.5827e-4.
Gc = -2.5827e-4;
Gcl = feedback(Gp*Gc, 1)
step(Gcl)
Gcl =
-0.003386 s^3 - 0.007819 s^2 + 4.56 s + 0.1788
----------------------------------------------------------
s^5 + 19.97 s^4 + 150.1 s^3 + 12.95 s^2 + 30.81 s + 0.1788
Applying the final value theorem, your can see that the step response of the closed-lopp control system will converge to 
, when 
. The direction of the system response is successful inverted. Note that PI and PID controllers are not needed to stabilize the plant.
, when . The direction of the system response is successful inverted. Note that PI and PID controllers are not needed to stabilize the plant.
2 个评论
Muhammad Daffa Pratama
2024-6-19
Hey I have the same problem but solved thanks to your explanation. But another problem arose. That negative proportional gain, when I check to its root locus, is in an unstable region. But my system is stable. How can that happen?
Sam Chak
2024-6-26
Can you post a New Question on MATLAB Answers and provide the transfer function or state-space of the system?
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Classical Control Design 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
