PID block algorithm in Simatic ?

[Peter Nachtwey]

Then it is back to the basics

What ever happened to using the basics? This is how I solved these kinds of problems back in college ( 30+ years ago ) before I learned about state space and the z domain. Simple differential equations still work well. Notice the response is the same as my PID response in Mathcad.

Lurkers can cut an paste the program below into their Scilab and run it. This should work with Matlab too with little if any modification because I tried not to use any fancy functions.

// TANK LEVEL CONTROL
Kp=-0.2368;				 // pump gain.  (m^3/m)/CO% 
Tp=1/12;					// plump time constant. minutes
Kt=0.0279;				  // 1/(tank surface area)   1/m^2
Tc=0.25;					// closed loop time constant. minutes
Kc=-605.444;				// controller Gain. percent output per meter of error
Ti=3*Tc;					// integrator time constant   minutes
T=1/120;					// simulation update period  minutes
PV(1)=0;					// fluid level  meters
CO(1)=0;					// control output. percent of full output
ui(1) = 0;				  // integrator contribution to CO
flow_in=20;				 // flow in.  m^3/min
flow_out(1)=0;			  // flow pumped out.  m^3/min

N=round(10/T);			  // CONVERT TO SAMPLE TIMES

for n=1:N;				  // simulate 10 minutes	   
  SP(n)=2;				  // set point  meters
  
  // calculate the rate of change in flow_out.  This uses the simple differential
  // equation Tp*flow_out_dot+flow_outy=Kp*CO(n).  Note flow_out is negative.
  flow_out_dot(n) = -flow_out(n)/Tp + (Kp/Tp)*CO(n);
  // calculate the new flow out by adding the rate of change times T
  flow_out(n+1) = flow_out(n) + flow_out_dot(n)*T;
  // integrate the difference between the flow in and flow out.
  PV(n+1) = PV(n) + Kt*(flow_in+flow_out(n))*T;
  err(n) =  SP(n) - PV(n);  // calculate error
  up = Kc*err(n);		   // calculate the proportional control output percent
  ui = ui + (Kc*T/Ti)*err(n); // calculate the integrator control output percent
  if ui > (100-up) then	 // limit the integrator contribution to the control output
	ui = 100-up;
  elseif ui < 0-up then
	ui = 0-up;
  end
  CO(n+1) = ui + up;		// the control output is the sum of the integrator and proportional terms
 end
 PV($)=[];				  // LIMIT TO 1200 ITEMS
 CO($)=[];
 
// PLOT THE SIMULATION
// CREATE TO PLOTS USING THE SUBPLOTS FUNCTION
// THE TOP PLOT SHOWS THE SP AND PV. THE BOTTOM PLOT SHOWS THE CO 

t=T:T:N*T;		// COMPUTE TIME INDEXES. START A TIME T IN INCREMENTS OF T TO N*T
clf();			// CLEAR OR RESET THE CURRENT GRAPHICS FIGURE

subplot(2,1,1);   // PLOT TEMPERATURES ON THE TOP PLOT
plot(t,[SP PV]);
xtitle('Tank Level Control Simulation','Time In Minutes','Tank Level');
legend("SP","PV");
subplot(2,1,2);   // PLOT CONTROL OUTPUT ON THE BOTTOM PLOT

// PLOT THE CONTROL OUTPUT TERMS

plot(t,[CO]);
xtitle('Tank Level Control Simulation','Time In Minutes','Control Output%');
legend("CO")

 

[Pandiani]

You asked me what I did and how I calculated gains, so here it is...

 

[Peter Nachtwey]

It looks good. That wasn't so hard.

I think you should take the weekend off after you fix your integrator. Then you should have a beer or two. Doesn't this seem like homework?

I was wondering. Did my Scilab program work on your Matlab?

 

[Pandiani]

Of course, it worked in Matlab-
Here's Matlab code (almost without modifications):

%TANK LEVEL CONTROL
clear all
clc

Kp=-0.2368;				 % pump gain.  (m^3/m)/CO% 
Tp=1/12;					% plump time constant. minutes
Kt=0.0279;				  % 1/(tank surface area)   1/m^2
Tc=0.25;					% closed loop time constant. minutes
Kc=-605.444;				% controller Gain. percent output per meter of error
Ti=3*Tc;					% integrator time constant   minutes
T=1/120;					% simulation update period  minutes
PV(1)=0;					% fluid level  meters
CO(1)=0;					% control output. percent of full output
ui(1) = 0;				  % integrator contribution to CO
flow_in=20;				 % flow in.  m^3/min
flow_out(1)=0;			  % flow pumped out.  m^3/min

N=round(10/T);			  % CONVERT TO SAMPLE TIMES

for n=1:N;				  % simulate 10 minutes	   
  SP(n)=2;				  % set point  meters
  
  % calculate the rate of change in flow_out.  This uses the simple differential
  % equation Tp*flow_out_dot+flow_outy=Kp*CO(n).  Note flow_out is negative.
  flow_out_dot(n) = -flow_out(n)/Tp + (Kp/Tp)*CO(n);
  % calculate the new flow out by adding the rate of change times T
  flow_out(n+1) = flow_out(n) + flow_out_dot(n)*T;
  % integrate the difference between the flow in and flow out.
  PV(n+1) = PV(n) + Kt*(flow_in+flow_out(n))*T;
  err(n) =  SP(n) - PV(n);  % calculate error
  up = Kc*err(n);		   % calculate the proportional control output percent
  ui = ui + (Kc*T/Ti)*err(n); % calculate the integrator control output percent
  if ui > (100-up) 	 % limit the integrator contribution to the control output
	ui = 100-up;
  elseif ui < 0-up
	ui = 0-up;
  end
  CO(n+1) = ui + up;		% the control output is the sum of the integrator and proportional terms
 end
 PV=PV(1:length(PV)-1);				  % LIMIT TO 1200 ITEMS
 CO=CO(1:length(CO)-1);
 
% PLOT THE SIMULATION
% CREATE TO PLOTS USING THE SUBPLOTS FUNCTION
% THE TOP PLOT SHOWS THE SP AND PV. THE BOTTOM PLOT SHOWS THE CO 

t=T:T:N*T;		% COMPUTE TIME INDEXES. START A TIME T IN INCREMENTS OF T TO N*T
clf();			% CLEAR OR RESET THE CURRENT GRAPHICS FIGURE

subplot(2,1,1);   % PLOT TEMPERATURES ON THE TOP PLOT
plot(t, SP,'Linewidth', 2);
hold on
plot(t, PV, 'r','LineWidth', 2);
title('Tank Level Control Simulation'),xlabel('Time In Minutes'),ylabel('Tank Level');
legend('SP','PV');
subplot(2,1,2);   % PLOT CONTROL OUTPUT ON THE BOTTOM PLOT

% PLOT THE CONTROL OUTPUT TERMS

plot(t,[CO],'g','LineWidth', 2);
title('Tank Level Control Simulation'),xlabel('Time In Minutes'),ylabel('Control Output%');
legend('CO')

Response is naturally same as yours.
Yes, after I make windup correction I will rest.
I must spend some time trying to figure out exactly why I had problems with transfer functions blocks in Simulin and clasical PID block. I must figure that out also.

Thank you very much "teacher"

 

[Schtiel]

Hi all! I am new to PID and I need help. Can somebody post here an example of usage Simatic PID with two discrete outputs? Is it difficult to program own PID algorithm (with discrete outputs)? Are there any examples?
Thanks.