PID Control Systems - Integral Control
Why Not Use An Integral Controller?
Properties Of Integral Controllers
Implementing Integral Controllers
Integral Control Example
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Why Not Use An Integral Controller?

        An integral controller has one very good quality.  An integral controller will normally ensure zero SSE in a control system - for step (constant) inputs.


Goals For This Lesson

        Here's what you need to work toward in this lesson.

  Given a system in which you want to use integral control,

Properties Of Integral Controllers

        Integral controllers have these properties.

        We can examine some of the features of integral control using an integral controller.

        If we think about the root locus for a system with integral control we can note the following.         Integral control has a tendency to make a system slower.  Consider a typical situation as shown below.
        Finally, we should note that integral control changes the order of the system, so a second order system becomes a third order system when you use an integral controller.  An proportional controller, on the other hand, does not increase the order of the system.

        You also need to know how to implement an integral controller if you can achieve system performance specifications with an integral controller.

        Let's examine a few examples of systems with integral control.  We'll use these same example systems we used when we discussed proportional control.

        We're going to make some standard assumptions about our systems.  First of all, in the prototype system configuration, we'll assume H(s) = 1.  In other words, we'll always have unity feedback.  Here is the system we will use.

The first example is a system with a single pole.  We assume that pole to be at s = -1.  Here's the root locus for that system after adding an integral controller.

We can also take a look at the frequency response for this system.

        The second example we will discuss is a system with two open loop poles.
        This second example illustrates something that happens often with integral control.
        It is tempting to draw a conclusion, and that conclusion is often right.  Here it is.         Now, examine the Bode' plot for this system.


        From the examples, there are some interesting things to note.
Implementing Integral Control

        Integral control in an analog system is often implemented with operational amplifier circuits.  The circuit below is an operational amplilfier integrator that can be used to integrate a signal.

The integrating circuit produces an output voltage given by this expression.

Clearly, this circuit can generate an output voltage that is proportional to an error signal, V1 - V2, and used as a proportional controller, so that gives us a way to implement integral control analog-fashion.

        Integral control in a digital system is often implemented in code in some programming language like C.  That's the digital version of integral control.  To implement integral control you use an approximation to the integral.  An approximation that is widely used and easy to implement approximates the integral using rectangular areas as shown in the figure.

The integral computation is updated by adding an area equal to the latest measurement multiplied by the sampling period between measurements (the width of the rectangle in the plot above).         The pseudo-code segment implements a particular integration algorithm - the Euler Integration Algorithm - and other integration algorithms would have different code and different Z-domain transfer functions.

        If you want to predict the effect of an integrator on system performance, you will need to have the transfer function for the intgrator.  For that, consider the difference equation that the integrator satisfies.

ErrorInt = ErrorInt + DT*(DesiredOutput - MeasuredOutput);
In other words:
ErrorIntk = ErrorIntk-1  + DT*(Errork)
        This equation assumes that the error is measured/computed, and that the computation for the integral of the error takes place quickly so that the D/A in the system can produce the output quickly.

        Take the transform of this difference equation to get:

ErrorInt[z] = [1/z] ErrorInt[z] + DT*Error[z]
so:
ErrorInt[z]/Error[z] = DT/(1 - 1/z) = (z DT)/(z - 1)
and, this system has a pole added at z = 1, and a zero added at the origin of the z plane.

        Let's reflect on the analog and digital implementations of integral controllers.


Example Problem

       This problem concerns the control of an internal combustion engine that is used to rotate a load.  That load is shown in the diagram below as the plant.  It's really a rotating load that is connected to the engine.  That engine is controlled by a controller, and the speed of rotation is measured with a sensor.

        When the motor starts to turn the load, it gradually increases in speed and asymptotically approaches a steady state speed.  We're going to model the plant as a first order system to show that.  The transfer function model for the load is shown above.

        The controller also is modelled with a single time constant.  When an error is applied to the controller, the engine torque does not change instantaneously.  We're going to model the controller with a similar transfer function.  That's next.  However, the controller for the motor needs a control signal computed by a PID unit.  That's shown in the block diagram above.

        The transfer function model for the controller is shown below.  Now we have to model the sensor.  We're going to use the same kind of model again.  When the speed changes, the sensor measurement doesn't change instantaneously but rather approaches the steady state measurement asymptotically with time-constant behavior.  The transfer function model for the sensor is incorporated in the block diagram below.  It is the same kind of model again.  When the speed changes, the sensor measurement doesn't change instantaneously but rather approaches the steady state measurement asymptotically with time-constant behavior.

 That completes our model of the system, but now we need some numbers.

        Here's the numbers we are going to use for this problem.

        We have some questions.         We'll try to go through these questions one-by-one to get some feel for what we can make the system do.

        You should find that there are difficulties controlling this system and meeting the SSE specifications using proportional control.  (Click here to go to that solution.)  Integral control should be able to solve that problem by ensuring 0% SSE.  However, to design the system - particularly to choose a gain value for the integral control - will take some analysis.

 You have two options.

        We'll start by getting a root locus for the system with integral control.  That's shown below.  There are a few points to note in the locus.

            The points noted above are the good points.  There could be one problem here. We need to examine the response of this system in the time domain.  What we can do is pick a gain and compute the response for the system with the given gain.  Here's a blow-up of the root locus that shows some of the detail near the origin.
          In order to do a simulation to calculate time response for this system, we need to relate the root locus gain to the gain in the system.         Here is the unit step response for the system with an integral controller (Ki = .092).  The response is just about what you expect.
        Is there any thing wrong with this response? Can we improve the system we have designed?
Here is the response of our system with a gain half of the original gain shown with the original system.

        Now, let's examine the system from a frequency domain viewpoint.  We'll start by getting a Bode' plot for the system with integral control.  That's shown below, drawn for Ki = 1.  There are a few points to note on this plot.

        We can adjust the gain of the system to get a better phase margin.  Here is a video that permits you to vary the gain and see the resulting phase margin as well as the magnitude and phase plots.

Note the following in that plot.

Play the video to see how changing the gain changes the phase margin.  Now, let us adjust the gain so that we get a reasonable system.         Now, select the gain to give the phase margin of 60o.  Before you continue, mark the gain.

Here is the computed step response for Ki = 0.18

 This response was with a sixty degree phase margin.  We could relax that somewhat.  A higher gain would give a lower phase margin, but we might expect a little faster response.  Still this compares reasonably favorably with the response obtained using root locus methods.


        We aren't going to pursue integral control beyond this point for this particular system.
The End

        That's it for this lesson.   You're ready to apply what you know in an integral control system.


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Other Lessons On PID Controllers

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