Easy Way To Tune Pid

[Bob O]

I adjusted Kp and this is what I came up with.
1. < 10
2. = 10
3. > 10

 

[Peter Nachtwey]

Bob, you are guessing.

That isn't right.

 

[Bob O]

I'm not guessing. I put there code into MatLab and adjusted the Kp. Now I maybe doing something wrong since it has been a while.

 

[Peter Nachtwey]

I sent a PM to Bob O with the solution

The solution only requires high school algebra. Matlab isn't required.

 

[parky]

About the post of beware of the so called experts.

Even in my early days of setting PID loops (before I was taught) I never found it too difficult, usually if it could not be tuned it was a problem with the design of the system not the PID controller (I refer mainly to direct steam injection into product in-line one of the hardest to control).

I have worked on many systems including ovens, speed control, heat exchange in-line injection, I worked with an expert in steam injection who designed a well known system, used extensively in food production controlling in-line cooking to +-0.5 deg C, this system used pressure & temperature control on the steam (one of the hardest to set-up).
Just because some people out there have no idea dosen't mean we are numbskulls.
I am currently working in a food envioment where we have 22 plants producing food using probably more than 50 pid control loops, most of these are using PLC based systems, incidently I have set up all of these, so I think even by default I can be called a sort of "Expert".

 

[Peter Nachtwey]

The answer

Values of KP
0 < Kp < 5 overdamped.
Kp = 5 critically damped
Kp > 5 underdamped

The key is to find where Kp is critically damped because that happens at just one point. A critically damped system has all the poles on the negative real axis at the same place. The denominator of the transefer function for proportional control is S^2+10*s+(20+Kp) and this must be equal to a form of (s+x)(s+x) where -x is the location of the poles. (s+x)(s+x) = s^2+2*x*s+x^2 and that is equal to s^2+10*s+(20+Kp). I can from the coefficients for the first power of s that 2*x = 10 therefore x= 5 which means there are two poles at -5. Now I examine the coefficient for the 0th power of s and see that x^2 = 20+Kp. Basically I am just finding the roots of the polynomial. Since x = 5 then Kp must be equal to 5 too. The fact there are a lot of 5s is just how the problem worked out.

I can go on but I won't unless someone expresses some interest. I don't think there are many that understand the web page Ron posted a link to. I do think a lot can be learned from examining that web page closely. If I do I will start a new thread.