Trending New Products Control Systems Process Manufacturing Discrete Manufacturing System Integration Networking & Security Info Manageme Forgot password? Username ******* Subscribe Manage Account Maga Newsletters Webcasts Research Sponsor May 06, 2013 Share Real World Engineering This is a blog from the trenches—written by engineers at Maverick Technologies who are implementing and upgrading control systems every day across every industry. This isn’t what they teach you in engineering school. These are lessons learned from years on the job, encountering the obstacles and issues that are part of the real world of control and process engineering. PID math demystified, part 1 You’ve seen the equations, but have you thought about how those elements work together? Part 1: The basic concepts and proportional control. Scott Hayes Most process control engineers have been exposed to the basic equation in a form that looks something like this: More than you want to swallow in one bite? Let’s break this down into the major components: Output: u(t) is the output of the controller at the end of the scan. If the output of the controller is a valve, then the output is the valve position that the controller is requesting after it has seen the inputs. In most controllers, this is actually the change in output from 50%. So if u(t) = 0 then the valve output is 50%; if u(t) = 1 then the valve output is 51%; and if u(t) = 2 then the valve output is 48%. You get the idea. But what’s important is that it’s not a change in output from the previous scan, but a new output. Proportional K p e(t) is the proportional component, the P in PID. If you have a controller configured as proportional only, this is it. So let’s look at how this works. Let’s start with my own misconception of how I thought it worked. When I imagine a controller, I picture myself turning a valve while watching a gage. I look at the gage, decide if I need more or less, turn that valve a little more or a little less, and then repea the process until the gage shows the value I want. That sounds fundamentally logical, but it is not how a proportional only controller works. It’s more like if I were to look at the gage, subtract what it reads from what I want it to read, and then take that error over to a chart to look up a new value for the valve. e(t) usually called error, is simply the difference between the setpoint and the process variable. It is the difference between where you are and where you want to be, right now, at this instant. K p gain, is a factor that is multiplied by the error to give you the new output, the new valve position. It’s that simple. The error at that instant of the scan is calculated and the new output is calculated. Let’s look at an example of pseudo code to explore how this works: Error = Setpoint ProcessValue; Output = K * Error; International Editions
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3/17/2015 PID math demystified, part 1 | Control Engineering
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Real World Engineering
This is a blog from the trenches—written by engineers at Maverick Technologies who are implementing andupgrading control systems every day across every industry. This isn’t what they teach you in engineeringschool. These are lessons learned from years on the job, encountering the obstacles and issues that arepart of the real world of control and process engineering.
PID math demystified, part 1You’ve seen the equations, but have you thought about how those elements work together? Part 1: The basic concepts andproportional control.Scott Hayes
Most process control engineers have been exposed to the basic equation in a form that looks something like this:
More than you want to swallow in one bite? Let’s break this down into the major components:
Output: u(t) is the output of the controller at the end of the scan. If the output of the controller is a valve, then the output is thevalve position that the controller is requesting after it has seen the inputs. In most controllers, this is actually the change in
output from 50%. So if u(t) = 0 then the valve output is 50%; if u(t) = 1 then the valve output is 51%; and if u(t) = 2 then the valveoutput is 48%. You get the idea. But what’s important is that it’s not a change in output from the previous scan, but a new output.
Proportional
Kpe(t) is the proportional component, the P in PID. If you have a controller configured as proportional only, this is it. So let’s lookat how this works.
Let’s start with my own misconception of how I thought it worked. When I imagine a controller, I picture myself turning a valvewhile watching a gage. I look at the gage, decide if I need more or less, turn that valve a little more or a little less, and then repeatthe process until the gage shows the value I want. That sounds fundamentally logical, but it is not how a proportional onlycontroller works. It’s more like if I were to look at the gage, subtract what it reads from what I want it to read, and then take thaterror over to a chart to look up a new value for the valve.
e(t) usually called error, is simply the difference between the setpoint and the process variable. It is the difference between whereyou are and where you want to be, right now, at this instant.
Kp gain, is a factor that is multiplied by the error to give you the new output, the new valve position. It’s that simple. The error atthat instant of the scan is calculated and the new output is calculated.
Let’s look at an example of pseudo code to explore how this works:
Error = Setpoint ProcessValue;
Output = K * Error;
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3/17/2015 PID math demystified, part 1 | Control Engineering
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This control algorithm is deceptively simple, yet it gives an immediate response to a setpoint change or a disturbance in theprocess. And if K is set correctly, will quickly move the process toward the setpoint. But, it won’t get the process to the setpointbecause there has to be some error if the output is anything other than 50%.
Note: On some systems (though not often in modern systems), gain is expressed as proportional band. Proportional band isdefined as the amount of change in the controlled variable required to drive the loop output from 0 to 100%. To convert betweenthe two, gain = 100/PB.
Next week: Understanding I and D.
This post was written by Scott Hayes. Scott is a senior engineer at MAVERICKTechnologies, a leading system integrator providing industrial automation, operationalsupport, and control systems engineering services in the manufacturing and processindustries. MAVERICK delivers expertise and consulting in a wide variety of areasincluding industrial automation controls, distributed control systems, manufacturingexecution systems, operational strategy, and business process optimization. Thecompany provides a full range of automation and controls services – ranging from PIDcontroller tuning and HMI programming to serving as a main automation contractor.Additionally MAVERICK offers industrial and technical staffing services, placing onsite automation, instrumentation and controls engineers.
Related News: PID math demystified, part 4 10.06.2014 08:00 PID math demystified, part 3: More on derivative control 21.10.2013 15:00 Evolving PID tuning rules 13.03.2013 14:33 Fixing PID 30.11.2012 16:23
3/17/2015 PID math demystified, part 2 | Control Engineering
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This is a blog from the trenches—written by engineers at Maverick Technologies who are implementing andupgrading control systems every day across every industry. This isn’t what they teach you in engineeringschool. These are lessons learned from years on the job, encountering the obstacles and issues that arepart of the real world of control and process engineering.
PID math demystified, part 2You’ve see the equations, but have you thought about how those elements work together? Part 2: Adding integral and derivative tothe mix.Scott Hayes
Last week we started with proportional. (Read Part 1) Now let’s look at the next part of the equation, the integral component:
The most striking (and scariest) part of this equation is the big integral sign in the middle. Ifyou’ve had high school calculus, you think to yourself, “I’ve got this. Integrals don’t scareme. I just need to find the area under the curve from time zero to time t of the errorfunction.”
But, this is the real world. What is time zero? How do I integrate an error function? The good news is that the real definition ismuch simpler than calculus. What the PID function does is take a portion of the error and adds it to a running total. This runningtotal, sometimes called reset, is added to the output. Since reset increases or decreases a little at a time, it adjusts the output ofthe valve incrementally each scan.
For a PI controller, the two factors that we have covered so far are Kp and Ki, but if you look at the faceplate for most industrialsystems, there is only one K (gain) that has no units, and a τi (integral time constant) designated as seconds or minutes perrepeat. So, a little translation is required. Most industrial controllers don’t use the independent form of the equation shown above.Instead, they use the dependent form of the equation:
The K is typically the same as the proportional gain, Kp. The factor τi determines how much of the error is going to be applied tothe accumulated reset on each scan. So in the big mathy equation, Ki can essentially be replaced by the faceplate parameters: K∕τi.
What’s important to understand from this is that the gain that affects the proportional action of a controller also affects the integralaction. But, the integral time constant τi only affects the integral action.
In pseudo code this would look like:
Error := Setpoint ProcessValue;
Reset := Reset + K/tau_i * Error;
Output := K * Error + Reset;
The unit’s minutes per repeat for the integral time constant τi comes from the fact that if the error stays constant, that is how longwould it take for the integral accumulator to repeat the proportional change in output.
Note: Another way of specifying the integral tuning parameter is in seconds, and then it is the reciprocal of seconds per repeat. Ifthe integral time constant is in seconds, the bigger the number, then the slower the response. If the integral is in seconds perrepeat, the opposite is true.
Derivative
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Now let’s look at derivative:
Again, this is another mathy looking equation with a simple explanation. The mathy definitionfirst, the output will be changed by the derivative (or rate of change) of the error function. Whatthis means is that the output will be affected by the change in error from one scan to another.
Adding this to our pseudocode gives us:
Error := Setpoint ProcessValue;
Reset := Reset + K/tau_i * Error;
Output := K * Error + Reset + ((PreError – Error) * K/tau_d));
PreError := Error; //Save the error for the next scan
What is intended is for the output to change as soon as the process variable begins to move either toward or away from thesetpoint. What results can be a very quick response to a change in error from one scan to another.
The intention of derivative action is to respond to changes as they begin to occur. For example, if a temperature is starting to rise,the valve will begin to open as soon as it sees the change instead of waiting for it to cross a setpoint. This can result in a veryrapid response to a small change. This rapid response can become unstable if there is noise in the process variable or on asetpoint change. So, the derivative action is often filtered separately and is sometimes calculated on PV only to ignore setpointchanges.
Summary
So now we have reviewed the three components of the PID algorithm. One way they have been described is in terms of the flowof time. P depends on the present error, I on the accumulation of past errors, and D on the prediction of future errors based oncurrent rate of change.
This post was written by Scott Hayes. Scott is a senior engineer at MAVERICK Technologies, a leading system integratorproviding industrial automation, operational support, and control systems engineering services in the manufacturing and processindustries. MAVERICK delivers expertise and consulting in a wide variety of areas including industrial automation controls,distributed control systems, manufacturing execution systems, operational strategy, and business process optimization. Thecompany provides a full range of automation and controls services – ranging from PID controller tuning and HMI programming toserving as a main automation contractor. Additionally MAVERICK offers industrial and technical staffing services, placing onsiteautomation, instrumentation and controls engineers.
Related News: PID math demystified, part 4 10.06.2014 08:00 PID math demystified, part 3: More on derivative control 21.10.2013 15:00
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Real World Engineering
This is a blog from the trenches—written by engineers at Maverick Technologies who are implementing andupgrading control systems every day across every industry. This isn’t what they teach you in engineeringschool. These are lessons learned from years on the job, encountering the obstacles and issues that arepart of the real world of control and process engineering.
PID math demystified, part 3: More on derivative controlConsider how a PD controller would work, without an integral function. Would you ever want to use that approach?Scott Hayes
To investigate how derivative action works, let’s look at a proportional derivative or PD controller. PID controllers are far morecommon than PD alone, but we already have an understanding of the integral component’s effects from the first two parts of thisseries, so we don’t need to review it again.
For a PD controller, we add the derivative of the error into the equation. Similar to what we discussed in the previous posts, we’renot really interested in deriving a derivative of the error function. Conceptually, the derivative in this case refers to how fast theerror is changing. So, if we take the change in error divided by the change in time we get the slope.
To explain how this works, let’s look at the pseudo code for this controller. The calculation of error is the same as before: setpointminus process value. Since the derivative reduces down to the change in error, the output is now the same proportionalcomponent as before: gain times error. The current error minus last error is multiplied by the gain and divided by the derivativetime constant. The current error is stored in last error for use in the next scan.
Error : = Setpoint – ProcessValue;
Output : = K * Error + K/tau_i * (Error – LastError);
LastError : = Error; // save for next scan
So what does this mean?
The proportional component is affected by the error at that time, in the present. The integral component is affected by anaccumulation of the error, or the past. The derivative component is a measure of how fast the error is changing, or a prediction ofthe future error.
How is this prediction of the future used? At first glance you might think that this term would be used to get you to your setpointthat much faster. But, that is not really the case. In practice, the derivative component is used to detect when the processvariable is changing too fast, and it puts the brakes on to prevent overshooting the setpoint.
So if the derivative component acts as brakes on the momentum, how does it get you to the setpoint faster? It does this by
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So if the derivative component acts as brakes on the momentum, how does it get you to the setpoint faster? It does this byallowing you to use a higher proportional gain to get you there quickly, but dampening the overshoot that would normally makethat level of gain unstable.
The tuning constants for derivative control are typically the same units as the time constant for reset. A couple of otherconsiderations though are important. On many controllers the derivative term is filtered independently. This prevents signal noiseor spurious disturbances from being interpreted as a change in momentum, which causes the derivative action to overreact. Also,on some controllers the derivative does not actually derive from the error, but instead on the process variable signal alone. Thisprevents a change is set point from being seen as a change in momentum.
This video summarizes our PID discussions with a graphical explanation.
I hope this helps to demystify PID, at least a little.
Read Demystifying PID part 1.
Read Demystifying PID part 2.
Also read Understanding Derivative in PID Control.
This post was written by Scott Hayes. Scott is a senior engineer at MAVERICK Technologies, a leading automation solutionsprovider offering industrial automation, strategic manufacturing, and enterprise integration services for the process industries.MAVERICK delivers expertise and consulting in a wide variety of areas including industrial automation controls, distributedcontrol systems, manufacturing execution systems, operational strategy, business process optimization and more.
Related News: PID math demystified, part 4 10.06.2014 08:00 PID math demystified, part 2 13.05.2013 14:24 PID math demystified, part 1 06.05.2013 14:02
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PID math demystified, part 4Posted by Scott Hayes on June 11, 2014 @ 12:56 pm
You’ve see the equations, but have you thoughtabout how those elements work together? Part4: Feed forwardIn the first three blog posts (see Part 1, Part 2, and Part 3) we covered the basics of PID math. Westarted with the basic proportionalonly controller and worked in the integral and derivativecomponents. Well before get too far, let’s review the proportionalonly controller:
u(t) = Kpe(t)
u(t) is the output of the controller at the end of the scan. If the output of the controller is a valve, thenthe output is the valve position that the controller is requesting after it has seen the inputs.
Kpe(t) is the proportional component, the P in PID. e(t), usually called “error,” is simply thedifference between the setpoint and the process variable. Kp gain is a factor that is multiplied by theerror to give you the new output—the new valve position. It’s that simple. The error at that instant ofthe scan is calculated and the new output is calculated.
The pseudo code that we discussed was:
Error = Setpoint – ProcessValueOutput = K * Error
With this code the output changes immediately after the error changes. I like to use a cruise control ona car as an example. As soon as the car starts to slow down as it starts to climb a hill, the change in
3/17/2015 PID math demystified, part 4 | MAVERICK Rethinking Automation
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error (speed) immediately causes the output to change.
But what if you know that the speed is about to slow down? What if you can see the hill coming?That’s where feed forward comes in. The first step is detecting a condition that will affect the processvalue. This can be an upstream flow rate that affects a level controller, a pressure change that affects aflow controller or any number of variables. The important thing is that the feed forward signal has tohave a predictable affect on the controlled variable.
The math behind it is conceptually pretty simple. The feed forward signal is multiplied by a gain andthen added directly to the output:
Error = Setpoint – ProcessValueOutput = K * Error + FeedForward * Kf
That way as long as the feed forward signal doesn’t change, the controller is not affected. But, achange in the feed forward signal immediately makes a proportional type shift in the output.
The math does get more complicated when the time that it takes for the signal to affect the processvalue has to be considered. In that case a lead or lag calculation must be done to slow down or speedup the feed forward signal. But that’s a topic for the next post.
This post was written by Scott Hayes. Scott is a senior engineer at MAVERICK Technologies, aleading automation solutions provider offering industrial automation, strategic manufacturing, andenterprise integration services for the process industries. MAVERICK delivers expertise andconsulting in a wide variety of areas including industrial automation controls, distributed controlsystems, manufacturing execution systems, operational strategy, business process optimization andmore.
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