Lag Compensators
Why Use Lag Compensators?
Goals For This Lesson
What Is A Lag Network?
Bode' Plot Analysis
Root Locus Analysis
Design Example


     Implementing Digital Lags

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Why Use Lag Compensators?

        Lag compensators are sometimes the best controller to use to get a system to do what you want it to do.  Like other compensators, lag compensators can be used to adjust frequency response by adding equal numbers of poles and zeroes to a systems.  Those added singularities may possibly be manipulated to give better stability, better performance and general improvement.  Lag compensators are one tool that should be in the toolkit of every control system designer.


Goals Of This Lesson

 Before we get very far, we need to establish what our goals are for lag compensators.

  Given a compensator transfer function,
  Know which compensators are leads, and which are lags.
  Given a lag compensator,
  Determine the effects of the lag compensator on closed loop system behavior using Bode' plot methods.
  Determine the effects of the lag compensator on closed loop system behavior using Root Locus methods.
Other goals of this lesson include:
  Given a system to be controlled.
  Determine if a lag compensator can be used to satisfy closed loop system specifications.
  Given a system in which a lag compensator can be used,
  Determine the parameters of a lag compensator to produce a closed loop system that meets specifications.

What Is A Lag Compensator?

        A lag compensator can be thought of in several different ways.

        A lag compensator will have a transfer function of the form:

        Here's a Bode' plot for a transfer function, G(s), with:

G(s) = (s + 1)/(10s + 1)

        This Bode' plot shows the essential characteristic of a phase lag compensator.         Here note the following in the plot of the example we encountered earlier. Below, we have a video that shows how lag networks behave as the ratio of the pole to the zero changes.

        You can see the features of a lag network.         In the next section, we'll look at how you can use a lag network in a system.
Using Lag Compensators

        Usually, a lag network is used as part of the controller in a feedback system.  Here's where the lag network would be used.

        Now, let's look at how the lag network affects how the system behaves.  We'll look at root locus effects and then Bode' plot effects.
Bode' Plot Effects

        Let's first examine how adding a lag compensator to a system affects the Bode' plot - and therefor the performance - of the system.    Consider a system with this transfer function.

Gp(s) = 1/[(s + 1)(s + 4)]

Here is the Bode' plot for this transfer function.

This is the Bode' plot for the system as it stands - with no compensator or controller.  Let us first see how well we can control the system as it stands.  Here are some specifications to try to achieve.

The Bode' plot below has a line indicating where the phase reaches -130o.

Examining the Bode' plot we can conclude the following:

Here is the response for the gain we have just computed.  This response shows the following:


Note the following. If you want to check things on this system, click here to open a simulated system + compensator in a separate window.  From the comments above, there are some general conclusions to draw for this example.         Now, we can examine what happens when a lag compensator is added.  We will try to go through this slowly - step-by-step.

        When we use a lag compensator we are probably defining our problem as one where the DC gain cannot be made high enough - as in the example above.  The important points in our thought processes are these.

Here is a Bode' plot of the compensated system.  In this plot, the solid lines are for the original (uncompensated) system while the dotted lines are for the compensated system.  (Red for db, blue for phase)  Note how the magnitude and phase plots of both systems are pretty much the same around the zero db crossing.

        Here is the unit step response for a system with that gain calculated above.

There seems to be something wrong here.  It looks like this system has about 5% SSE.  However, that's not true!  If you examine what happens over a longer time frame, you get something like the response shown below.  You can see the same effect in the simulator where you can see how the exponential tail approaches the steady state.

This response shows the following:         There is clearly something here that we need to understand.  The time response of this system effectively contains two parts - a fast response that overshoots a little, and a slow response that seems to take forever to crawl up to the final value.  The time response is not easily evaluated because of the general shape it seems to have.  We are going to examine this through the lens of the Root Locus.  Since the root locus is a time-domain based approach we may be able to get more insight into what is happening here because the root locus is the tool that addresses time response more directly.

Root Locus Analysis

        Let's review the situation.  We have a system with this transfer function.

Gp(s) = 1/[(s + 1)(s + 4)]

This system is embedded in a control loop like this:

Now, examine the root locus for this system

Thinking about this root locus, we can conclude the following.

        Now, what happens when we add the compensator?

To see the root locus for the compensated system click the button on the right.  You should be able to see that the root locus doesn't change much except for the addition of a branch between the compensator pole and the compensator zero.  That branch is critical.  Here is the root locus for the compensated system.

 Note the following points that you can see in this plot.


A System Design Example

        Here's a model of a chemical process.  This process has three time constants and three gains.

Parameters for this system are as follows.

        We want to control this system to meet several specifications.
Problem

P1  What loop gain - Ks Kp G(0) - will produce a system with 2% SSE?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Your grade is:



        You should have found that the value of the DC gain needs to be around 50.  Now, we have: So we have: We said we'd like to show you a simulation for a gain, Kp, of 250.  This is what we have.  It's not pretty.

Note that the root locus gain for this system is 50/(10x.5) = 10.  Let's examine the root locus for this system.
Here's the root locus, and the points at a root locus gain of 10 are emphasized. Here is the root locus again.  Here poles for a root locus gain of 4.22 are marked with black dots.  The system has poles on the imaginary axis for a root locus gain of 4.22.

We can draw one conclusion from this analysis.         Clearly, we need to do something if we want to control this system.  Let us consider the possibility of adding a lag compensator.  We will examine how we can change the system's behavior with this compensator.  Note, that we are still assuming an adjustable proportional gain as well as this compensator transfer function.

(s + 1)/(20s + 1)

        We will start by considering the effect on a root locus.  Here is the root locus for the compensated system.  It's not much different from the root locus for the system with no compensation.

Compute the root locus gain for instability just to see if we can figure out what is possible.  Here is the root locus with points marked where the plot enters the right half plane.  At those points, the root locus gain is 13.7.

That means that the transfer function is:

KpGcG(s) = 13.7(s + 1)/[(.1s + 1)(s + 1)(2s + 1)(.05s + 1)]

That means that the DC gain is 13.7 which will not give 2% SSE.

        The question is what to do to meet the SSE requirement.  We need to look at this problem a little differently.  Let's try looking at this from a Bode' plot viewpoint.