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.

• First, a lag compensator is a device that provides phase lag in its' frequency response.
• If the compensator has phase lag - and never a phase lead - then there are implications about where the corner frequencies are in the Bode' plot.
• Other implications are that the phase lag compensator will have only certain types of pole-zero patterns in the s plane.
A lag compensator will have a transfer function of the form:

• Since a lead compensator has only positive phase angle, we must have:
• wz > wp
• A Bode' plot will make this clearer.
Here's a Bode' plot for a transfer function, G(s), with:

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

• Notice:
• wz > wp
• since:
• wp = .1 rad/sec, and
This Bode' plot shows the essential characteristic of a phase lag compensator.
• There is one pole and one zero.  Both are real.
• Phase is always negative!
• wz > wp
• If, for example, wz < wp , we would have a lead network, not a lag network.
Here note the following in the plot of the example we encountered earlier.
• Phase angle is always negative.
• The pole at wz = .1, or f = .0159 Hz, causes the magnitude plot to bend down, and the phase to become negative near f = .159 Hz.
• Later, the zero at 1.59 Hz. brings the Bode' plot's slope back to zero for high frequencies, and the phase back to zero.
Below, we have a video that shows how lag networks behave as the ratio of the pole to the zero changes.

• In the video, the ratio of pole to zero is Alpha.  The pole is at -1, so the corner frequency - in hertz - is .159.
• The zero starts out equal to the pole, so the zero corner is at .159 and then, as Alpha decreases, the zero moves higher in frequency.
• Notice the phase behavior especially on the plot.  Phase is always negative, but you can get larger phase angles when the pole-to-zero ratio is larger.
You can see the features of a lag network.
• A lag network has one pole and one zero - although you can add multiple lag networks in a system.
• The pole corner frequency (Bode' plot) is lower than the zero corner frequency.
• The lag network adds phase lag to a system.
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.

• In this system, we assume that you have a lag network, and an adjustable gain in the lag network.
• In that case, the compensator transfer function would be:
• Gc(s) = Kc(s + a)/(s + b)
• And we must have a > b, for a lag compensator.
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.

• SSE < 2%
• Phase Margin > 50o
The Bode' plot below has a line indicating where the phase reaches -130o.

Examining the Bode' plot we can conclude the following:

• The zero db crossing should occur at around f = 0.8 if we want to get a phase margin of 50o.
• If the zero db crossing is set at f = 0.8, the db gain at that point is-30 db.
• If the gain is -30 db, we can add 30 db gain to the system.
• If we add 30 db of gain, the DC gain will move from -12 db to +18 db.  A gain of 30 db is a gain of about 32.
• A DC gain of 18 db is a gain of 8.
• A DC gain of 8 will produce a SSE of .11 or 11%.
• Here's the Bode' plot after adjusting the gain.

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

• 11% SSE.
• About 24% overshoot (up to 1.1 when it settles out to 0.89).
• A rise time (10-90%) that looks to be a little less than 0.5 sec, and about what we predicted.

Note the following.
• If the SSE if 11%, then we cannot simultaneously achieve both specifications above.
• If we raise the gain to get the SSE smaller, we will also bring the phase margin below what we need to have.
• If we wanted to get the SSE to <2%, we would need to increase the DC gain to 50.  Since it is now at 8, we need a further increase of (50/8) or a factor of 6.25.
• Also note that since the zero db crossing is at f = 0.8, we expect the closed loop bandwidth to be near that frequency, and the rise time to be given by:
• Rise time ~= .35/0.8 = .44 sec.
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.
• It's possible to meet the stability requirements (rise time and phase margin), and they depend mostly on the behavior of the Bode' plot near the zero db crossing.
• The accuracy requirement is really a requirement on the DC gain - i.e. the behavior at low frequencies.
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.

• We use the lag compensator to allow us to increase the DC gain of the system.  We focus on the low frequency behavior.
• We want to increase the DC gain without disturbing the stability properties.  In other words, we are willing to accept the system's behavior around the zero-db crossing for the open-loop Bode' plot.
• In the example system the zero db crossing of the uncompensated system was aat f = 0.8.  If we want to use a lag compensator in that system we will do that by trying to leave the behavior around 0.8 Hz undisturbed.  That means that the pole and zero in the compensator will have to produce their phase and magnitude changes at frequencies lower than f = 0.8 Hz.
• The implications of these considerations are:
• We need a zero to pole frequency ratio of 6.25 - the gain increase we need.
• The zero will have to be around f = 0.08Hz.  By putting the zero there, there will be little effect on the phase at f = 0.8Hz.
• If the zero is a 0.08Hz, then the pole has to be at .08/6.25, or at .013Hz
• The actual compensator will be (s + .08*2*p)/(s + .013*2*p) = (s + .503)/(s + .082).
• And remember that we want a DC gain of 50, and we have:
• Total transfer function = K (s + .503)/[(s + .082)(s + 1)(s + 4)
• To get a DC gain of 50, we need K = 32 or thereabouts.
• Check that out with the simulator.
• And, since these are very low frequencies, the compensator may be best implemented digitally.
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:
• 2.5% SSE.
• About 15% overshoot if you count from the final value.
• A rise time (10-90%) that looks to be a little less than 0.5 sec, and about what we predicted.
• BUT!
• The rise time doesn't tell the whole story because there is a very slow "exponential tail" in this response, and we don't get near the steady state in anything like 0.5 sec.
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

• When the closed-loop poles move away from the open-loop poles they come together at s = -2.5
• After coming together at s = -2.5, the closed-loop poles have a real part of -2.5.
• A real part of -2.5 implies a settling time of approximately 1.1 sec.  (Click here for the relationship between the real part of the pole and the settling time.)
• Earlier, we found this response for a DC gain of 8.
• The DC gain of 8 corresponds to a Root Locus gain of 32.  In other words, the forward transfer function, GcGp(s), is given by:
• Gc(s)Gp(s) = 32/[(s + 1)(s + 4)]
• A root locus gain of 32 is off the scale in the root locus plot above, and the roots are at s = -2.5 + j5.45 and -2.5 - j5.45.
• Here's the step response of the closed loop system for this gain.  This plot was shown above and is just reproduced here.

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.

• The centroid (center of gravity) shifts to the right when the lag compensator is added because the zero is larger than the pole.
• The shift isn't really very large, but there is definitely a shift.
• Putting the pole and zero at higher frequencies - keeping the pole-to-zero ratio the same - would cause the centroid to shift even further because the pole and zero would both be further into the left half plane.
• The asymptotes for the branches going to infinity will be shifted to the right because of the centroid's shift to the right.
• If there is root locus activity near the zero, then the locus will have a tendency to bend toward the zero.

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.

• K1 = 2.0, t1 = 0.5s
• K2 = 1.0, t2 = 4.0s
• K3 = 0.1, t3 = 2.0s
We want to control this system to meet several specifications.
• The first specification is that we want to have a SSE of 2%.

Problem

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

You should have found that the value of the DC gain needs to be around 50.  Now, we have:
• K1 = 2.0,
• K2 = 1.0,
• K3 = 0.1,
So we have:
• K1 x K2 x K1 = 0.2
• That means the proportional gain, Kp, needs to be 250, to get:
• Kp x K1 x K2 x K1 = 50
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.

• The system looks like it is unstable!
• We should check that conclusion before committing to it completely.
• The  complete transfer function is:
• GCL(s) = (250x0.2)/[(.1s + 1)(s + 1)(2s + 1) + (250x0.2)]
Note that the root locus gain for this system is 50/(10x.5) = 10.  Let's examine the root locus for this system.
• Note that the poles are the negative reciprocals of the time constants, i.e.:
• -1/0.5 = -2.
• -1/4 = -0.25
• -1/2 = -0.5
Here's the root locus, and the points at a root locus gain of 10 are emphasized.
• The system is unstable!
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.
• We cannot meet the SSE specification.  The system is unstable for the gain that would produce the SSE.
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.