Why are there negative PID coefficients?

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).

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

Answer 1

[Sam Chak]

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.