Steady State Error In Control Systems

(Step Inputs) Why Worry About Steady State Error? What Is Steady State Errror (SSE)? Systems With A Single Pole At The Origin ProblemsYou are at:  Analysis Techniques - Performance Measures - Steady State Error Click here to return to the Table of Contents

Why Worry About Steady State Error?

        Control systems are used to control some physical variable.  That variable may be a temperature somewhere, the attitude of an aircraft or a frequency in a communication system.  Whatever the variable, it is important to control the variable accurately.

        If you are designing a control system, how accurately the system performs is important.  If it is desired to have the variable under control take on a particular value, you will want the variable to get as close to the desired value as possible.  Certainly, you will want to measure how accurately you can control the variable.  Beyond that you will want to be able to predict how accurately you can control the variable.

        To be able to measure and predict accuracy in a control system, a standard measure of performance is widely used.  That measure of performance is steady state error - SSE - and steady state error is a concept that assumes the following:

The system under test is stimulated with some standard input.  Typically, the test input is a step function of time, but it can also be a ramp or other polynomial kinds of inputs.

The system comes to a steady state, and the difference between the input and the output is measured.

The difference between the input - the desired response - and the output - the actual response is referred to as the error.

Goals For This Lesson

        Given our statements above, it should be clear what you are about in this lesson.  Here are your goals.

  Given a linear feedback control system,   Be able to compute the SSE for standard inputs, particularly step input

signals.

  Be able to compute the gain that will produce a prescribed level of SSE in the system. Be able to specify the SSE in a system with integral control.

        In this lesson, we will examine steady state error - SSE - in closed loop control systems.  The closed loop system we will examine is shown below.  

The system to be controlled has a transfer function G(s). There is a sensor with a transfer function Ks.

There is a controller with a transfer function Kp(s) - which may be a constant gain.

What  Is SSE?

        We need a precise definition of SSE if we are going to be able to predict a value for SSE in a closed loop control system.  Next, we'll look at a closed loop system and determine precisely what is meant by SSE.

        In this lesson, we will examine steady state error - SSE - in closed loop control systems.  The closed loop system we will examine is shown below.

The system to be controlled has a transfer function G(s). There is a sensor with a transfer function Ks.

There is a controller with a transfer function Kp(s).

o A controller like this, where the control effort to the plant is proportional to the error, is called a proportional controller.

        In our system, we note the following: The input is often the desired output.  In other words, the input is what

we want the output to be.  If the input is a step, then we want the output to settle out to that value.

The output is measured with a sensor.  Often the gain of the sensor is one. That is especially true in computer controlled systems where the output value - an analog signal - is converted into a digital representation, and the processing - to generate the error, E(s) - takes place inside a computer.

    The signal, E(s), is referred to as the error signal. The error signal is the difference between the desired input and the

measured input. The error signal is a measure of how well the system is performing at

any instant.  When the error signal is large, the measured output does not match the desired output very well.

        A step input is often used as a test input for several reasons. A step input is really a request for the output to change to a new,

constant value. If the system is well behaved, the output will settle out to a constant,

steady state value.

The difference between the measured constant output and the input constitutes a steady state error, or SSE.

        It helps to get a feel for how things go.  So, below we'll examine a system that has a step input and a steady state error.  That system is the same block diagram we considered above.

For the example system, the controlled system - often referred to as the plant - is a first order system with a transfer function:

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

We will consider a system with the following parameters.

Gdc = 1 = 1

Ks = 1.

Kp  can be set to various values in the range of 0 to 10,

The input is always 1.

        Here is a simulation you can run to check how this works.  In this simulation, the system being controlled (the plant) and the sensor have the parameters shwon above.  You can adjust the gain up or down by 5% using the "arrow" buttons at bottom right.  You can also enter your own gain in the

text box, then click the red button to see the response for the gain you enter.

The actual open loop gain is shown in the text box above the red button.

        Vary the gain.  You can set the gain in the text box and click the red button, or you can increase or decrease the gain by 5% using the green buttons.  You should see that the system responds faster for higher gain, and

that it responds with better accuracy for higher gain. Try several gains and compare results.  The difference between the desired response (1.0 is the input = desired response) and the actual steady state response is the error.

Problem 1  For a proportional gain, Kp = 9, what is the value of the steady state output? 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:

Problem

P1  For a proportional gain, Kp = 9, what is the value of the steady state error?

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:

Problem

P2  For a proportional gain, Kp = 49, what is the value of the steady state output?

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:

Problem

P3  For a proportional gain, Kp = 49, what is the value of the steady state error?

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.

        When you do the problems above, you should see that the system responds with better accuracy for higher gain.  Try several gains and compare results using the simulation.  You need to understand how the SSE depends upon gain in a situation like this.  You need to be able to do that analytically.  That's where we are heading next.  Here is our system again.

       Now, we can get a precise definition of SSE in this system.  We have the following:

The input is assumed to be a unit step.  If the input is a step, but not a unit step, the system is linear and all results will be proportional.

We can calculate the output, Y(s), in terms of the input, U(s) and we can determine the error, E(s).  Later we will interpret relations in the frequency (s) domain in terms of time domain behavior.

We have:

o E(s) = U(s) - Ks Y(s) since the error is the difference between the desired response, U(s),

o The measured response, =  Ks Y(s).

o And we know:  Y(s) = Kp G(s) E(s).

Combine our two relations:

o E(s) = U(s) - Ks Y(s) and:

o Y(s) = Kp G(s) E(s), to get:

o E(s) = U(s) - Ks  Kp G(s) E(s)

Since E(s) = U(s) - Ks  Kp G(s) E(s) we can combine all the E(s) terms on the left side of this equation:

o E(s) [1 + Ks Kp G(s)] = U(s)

o or E(s) = U(s) / [1 + Ks Kp G(s)]

Now, we know that:

o E(s) = U(s) / [1 + Ks Kp G(s)]

This relation is in the frequency - s - domain, but has implications for time response behavior.

Assume a unit step input.  The transformed input, U(s), will then be given by:

o U(s) = 1/s

With U(s) = 1/s, the transform of the error signal is given by:

o E(s) = 1 / s [1 + Ks Kp G(s)]

The final value theorem can be applied to get the steady state error.

We get the Steady State Error (SSE) by finding the the transform of the error and applying the final value theorem.  To get the transform of the error, we use the expression found above.

        Since E(s) = 1 / s (1 + Ks Kp G(s)) applying the final value theorem

Multiply E(s) by s, and take the indicated limit to get:

Ess = 1/[(1 + Ks Kp G(0)]

We can draw a few conclusions from this expression.

The steady state error depends upon the loop gain - Ks Kp G(0). The term, G(0), in the loop gain is the DC gain of the plant.

To make SSE smaller, increase the loop gain.

o And, the only gain you can normally adjust is the gain of the proportional controller, Kp.

You should also note that we have done this for a unit step input.  If we have a step that has another size, we can still use this calculation to determine the error.  If the step has magnitude 2.0, then the error will be twice as large as it would have been for a unit step.  The system is linear, and everything scales.  We can take the error for a unit step as a measure of system accuracy, and we can express that accuracy as a percentage error.

If the response to a unit step is 0.9 and the error is 0.1, then the system is said to have a 10% SSE.

Problem 5  What loop gain - Ks Kp G(0) - will produce a system with 5% 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.

Problem

P4  What loop gain - Ks Kp G(0) - will produce a system with 1% 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.

Some Observations for Systems with Integrators

        This derivation has been fairly simple, but we may have overlooked a few items.

If the system has an integrator - as it would with an integral controller - then G(0) would be infinite.  That would imply that there would be zero SSE for a step input.

The pole at the origin can be either in the plant - the system being controlled - or it can also be in the controller - something we haven't considered until now.  However, it should be clear that the same analysis applies, and that it doesn't matter where the pole at the origin occurs physically, and all that matters is that there is a pole at the origin in the controller or the plant.

        Reflect on the conclusion above and consider what happens as you design a system.

You may have a requirement that the system exhibit very small SSE. You can get SSE of zero if there is a pole at the origin.

If there is no pole at the origin, then add one in the controller.  You will have reinvented integral control, but that's OK because there is no

patent on integral control.  Click here to learn more about integral control.

If you want to add an integrator, you may need to review op-amp integrators or learn something about digital integration.  Those are the two common ways of implementing integral control.  You can click here to see how to implement integral control.

 

 PID Control Systems - Proportional Control

Why Not Use A Proportional Controller? Properties Of Proportional ControllersProportional Control Examples Proportional Control Problems Implementing Proportional ControllersYou are at: Design Techniques - The PID Family of Controllers - Proportional Controllers Click here to return to the Table of Contents

Why Not Use A Proportional Controller?

        Of all the controllers you can choose to control a system, the proportional controller is the simplest of them all.

If you want to implement a proportional control system, it's usually the easiest to implement.

In an analog system, a proportional control system amplifies the error signal to generatethe control signal.  If the error signal is a voltage, and the control signal is also a voltage, then a proportional controller is just an amplifier.I

In a digital control system, a proportional control system computes the error from measured output and user input to a program, and multiplies the error by a proportional constant, then generates an output/control signal from that multiplication.

Goals For This Lesson

        Proportional control is a simple and widely used method of control for many kinds of systems.  When you are done with this lesson you will need to be able to use proportional control with some understanding.  Your goals are as follows:

    Given a system you want to control with a proportional controller,     Identify the system components and their function, including the

comparator, controller, plant and sensor.     Be able to predict how the system will respond using a proportional

controller - including speed of response, accuracy (SSE) and relative stability.

    Be able to use the root locus to make        those predictons.     Be able to use frequency response analysis to make those predictions.

Properties Of Proportional Controllers

        Proportional controllers have these properties:

The controller amplifies the error as shown in the block diagram below. So, the actuating signal (the input to G(s)) is proportional to the error.

        In the material that follows, we will examine some of the features of proportional control using a proportional controller.

       In a proportional controller, steady state error tends to depend inversely upon the proportional gain, so if the gain is made larger the error goes down.

        In this system, SSE is given by the expression

SSE = 1/(1 + KpG(0))

      As the proportional gain, Kp, is made larger, the SSE becomes smaller.  As the DC loop gain, KpG(0), becomes large, the error approaches becoming inversely proportional to the proportional gain, Kp.  That's true for most of the cases of interest, that is those with small SSE.

Proportional control has a tendency to make a system faster.

        If we think about the root locus for a system with proportional control we can note the following:

The proportional gain, Kp, is either the root locus gain, or the root locus gain is proportional to Kp.

The root locus for a first order system will have the form shown below.

There's one pole. The root locus moves to the left from that pole.

As the pole moves to the left, the time constant becomes smaller.

Higher order systems will not behave in the same way, but there may be tendencies to speed up the system, at least for some gain values.  Here is the root locus for a second order system.

        In this system, as the poles begin to move from the open loop poles, the "slow" pole (It doesn't move slower, but because it is closer to the imaginary axis the response due to that pole is slower.) at s = -1 initially moves to the left, indicating a faster response in that component of the response.  The "fast" pole starts from s = -2, and move to the right, indicating that that portion of the response slows down.  Still, the overall effect will be a speedup in the system, at least until the poles become complex.  You can see that in this clip that shows the unit step response of the system with the root locus above.

        Finally, we should note that proportional control does not change the order of the system, so a second order system stays a second order system when you use a proportional controller.  An integral controller, on the other hand, does increase the order of the system by one.

        You also need to know how to implement a proportional controller if you can achieve system performance specifications with a proportional controller. Click here to go to a short lesson on implementing proportional control.

Examples Of Proportional Control

Example

E1   Let's examine a few examples of systems with proportional control.  We'll use these same example systems when we discuss integral control and PID 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.

        The first example we will discuss is a system with a single pole.  For purposes of discussion, we'll take that pole to be as s = -1.  Here's the root locus for that system.

        The closed loop pole in the system moves to the left on the negative real axis as the gain increases, starting with Kp = 0.

As the closed loop pole moves to the left, the closed loop time response will be faster.

The system never goes unstable - not for any gain.

The transfer function here is:

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

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

As the gain is increased - plotted here for Kp = 1 - the zero db crossing moves up in frequency.

No matter how high the gain gets, the phase margin is always at least 90o.

The system never goes unstable - not for any gain.

Example

E2   The second example we will discuss is a system with two poles.  Here's the root locus.

Open loop poles are at -1 and -4. As the gain increases, the closed loop poles become imaginary, and

the response gets oscillatory.

The system never goes unstable, but damping gets smaller as gain increases.

        Now, examine the Bode' plot for this system.

At high frequencies, the phase approaches -180o. As the gain is increased, the phase margin gets smaller and smaller,

but is always positive - the system is stable.

A small phase margin implies oscillatory response.

Example

E3   Here's a third system with poles at -1, -3 and -5, and here is the root locus for this system.

At large gains, this system goes unstable. Initially, at low gains, the slowest pole moves to the left.

At higher gains, the two poles coalesce, then things start to get worse.  Poles become complex, and move towards the imaginary axis.

        Looking at the frequency response, we come to similar conclusions.

At large gains, this system goes unstable. At higher frequencies, the phase approaches -270o.

At higher gains, the phase margin becomes negative and the system becomes unstable.

Example

E4   Things get interesting with poles at the origin.  As our last example we'll consider a system with two poles at the origin, and one pole at -1.

There's a problem with this system.  Note that this plot has the origin shifted to the center of the plot because two branches of this locus move immediately into the RHP.

There are no gain values that make this system stable.

We have a problem!  Let's look at the frequency response.

        Looking at the frequency response, we note the following:

Phase starts at -180o, and moves toward -270o. Things don't look any better on the Bode' plot than they did on the root

locus.

Proportional control is not going to work here.  With this system, we need to look for some other control scheme.  With proportional control, you don't have a prayer.

Conclusions

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

Our original conclusions are generally correct.o There are special cases that can cause problems.

o It's best to look at the situation in detail before drawing any conclusion.

o Proportional control is great - because it's simple - if you can use it.  For some systems, it may not work at all, so it that's the case you'll have to use something more complex.

PID EXAMPLE PROBLEMS

 There are three example problems in this section.

The first problem is to control a system with just one real pole.  Click here to go to that problem.

A second problem considers a system with three real poles.  Click here to go to that problem.

A third problem considers a system with lightly damped poles.  Click here to go to that problem.

Example Problem 4

        This is a problem that asks you to design a controller for a very simple system.

In this system, there is one open loop pole.  G(s) = 5/(s + 5) and H(s) = 1. The system needs to have a settling time of .2 sec or less.

The system has to have a SSE of 5% for a step input. Determine gain values (for Kp) that will meet the specifications.

There are two pieces of information that you may need. The Bode' plot of 5/(s + 5). The Root Locus plot of the system.

Here are the Bode' plot and root locus.

The system needs to have a settling time of .2 sec or less. The system has to have a SSE of 5% for a step input.

o Determine gain values (for Kp) that will meet the specifications.

        Basically, you can choose to work in the frequency domain (using the Bode' plot), or the time domain (using the Root Locus).  Choose one now.

The frequency domain. The time domain.

Frequency Domain Approach - Example Problem 4

        OK, you chose the frequency domain.  Here is  the Bode' plot for the system to be controlled.

Determine the gain you need to meet the SSE spec of 5% for a step input.

Enter the gain you have determined.  The gain is not in db for this question, but if you get the gain right, then you'll have to give it in db also!

Remember, the plot has db on the left, and angle in degrees on the vertical axis on the right.

        Now, does this system meet the response speed specification of .2 sec or less?  Inquiring minds need to know that.  What is the response speed for a gain of 20 (26db)?

How do you figure response speed in the frequency domain?  That's a good question, and you can click the hotword for a brushup if you've forgotten.

Basically, you need to know the zero db crossing of the open loop frequency response.  So, add 26 db to the Bode' plot.

You should have figured that the zero db crossing is approximately at f = 16.

With an open loop zero db crossing at f = 16, that is the approximate closed loop 3-db bandwidth.

Rise time ~ .35/3db BW = .35/16 or .022 sec

That exceeds the .2 sec specification, so all is copasetic.

Click here to look at how this all works out in the time domain. Click here to skip the time domain section.

Time Domain Approach - Example Problem 4

        OK, you've chosen to work in the time domain.

Determine the gain you need to meet the SSE spec of 5% for a step input.

Note the gain you have determined.  The gain is not in db for this question!

        Now, does this system meet the response speed specification of .2 sec or less?  Inquiring minds need to know that.  What is the response speed for a gain of 20?

For a response speed of .2 sec, let's assume we mean a rise time of .2 sec.

Rise time = 2.2, in a first order system, where is the time constant.

That means that is at most .2/2.2 = .09 sec.

o That puts the closed loop pole at s = -1/.09 = -11.

If the closed loop pole is at -11, the root locus gain is 6 (pole distance!).

But the root locus gain is 5Kp. when KpG(s) = 5Kp/(s + 5).

So, 5Kp = 6, and Kp = 1.2 for a pole at -11.

o Larger gains move the pole further to the left.

        There is an unresolved question.  What is the actual rise time when SSE = 5%?

When SSE = 5%, the DC Gain is 20. Since DC Gain = Kp, we have Kp = 20.

When Kp = 20, the Root Locus gain is 5Kp = 100.

A root locus gain of 100 puts the closed loop pole at s = -105.

A pole at -105 has a time constant of .0095 sec.

A time constant of .0095 sec produces a rise time of 2.2 x .0095 = .021 sec.

Example Problem 4 - Conclusion

        We can finally conclude the following:

A gain of 20 will meet the SSE requirement, and the system will be more than fast enough to meet the speed of response specification.

        So, either way you got here - frequency response or time response, you came to the same conclusion.  The system can simultaneously meet the SSE and rise time sepecifications with Kp = 20.  (That's SSE less than or equal to 5% and rise time < .2 sec.)

If you did both the frequency and the time domain approach, you got a chance to compare how well the two ways work.  You should especially check the frequency response approach to see how well our estimates came to predicting the actual rise time.  You can only do that if you worked through both approaches and if you didn't do that, do it now.

Click here to go to the frequency section for the first problem.

More Proportional Control System Example Problems

        This section contains two more problems for you to work on to ensure that you understand proportional control and that you have met the goals for this lesson.

A Speed Control Example Problem

        This second 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 the 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.  Here is a block diagram of the system that describes visually what happens in the system.

        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.

        All of the models - for the Controller, Plant and Sensor - are shown in the block diagram above.

        That completes our model of the system, but now we need some numbers.  Here are the numbers we are going to use for this problem.

For the engine + load: GL = 1,   L = 2 sec. For the controller:  Gc = K (variable), c = 1 sec.

For the sensor:  Gs = 1,  s = .2 sec.

        Now that we have some numbers to work with, we are going to pose some questions about how well we can make this system work.  Here are a few.

Can the system operate with 5% SSE for a constant input? How quickly will the system respond when we have 5% SSE?

How much overshoot does the system exhibit when set for 5% SSE?

How sensitive is the system to changes in the gain at 5% SSE?

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

        We're going to look at the 5% SSE requirement first.  What loop gain do we need to get 5% SSE?  The system is shown below.  Determine the loop gain.  If you have trouble calculating the loop gain for a given SSE you may need to review material on SSE.

Click here to open the material on SSE. Click here to go to a problem on this topic.

        Call the loop gain "K". K = GL x Gc x Gs

Then, the expression for SSE (for a unit step input) is: SSE = 1/(1+K)

So, now the question is "What value must K be in order to have <5% SSE?"

Problem

P1  What loop gain - K - will produce a system with 5% 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:

        Now, we know that we need to have a gain, K, with a value of 19 or more.  (Why not just use 20?)  We need to examine the implications of a gain of 19 (or 20).

How will the system perform for a gain of 19? How can we answer the first question?  There are several options.  We

could try to answer the question either in the time domain or the frequency domain.

        In the time domain we can probably get some useful information from a root locus.  What do we need to remember?

The loop transfer function is: G(s)H(s) = K/[(2s + 1)(s + 1)(.2s + 1)] We need to rewrite the transfer function to put it into root locus form:

o G(s)H(s) = (K/(2*.2))/((s + 0.5)(s + 1)(s + 5))

o G(s)H(s) = (2.5K) / ((s + 0.5)(s + 1)(s + 5))

So we have 2.5K as the root locus gain with poles at -0.5, -1 and -5.0.

        Here's the root locus for the system.  We can see that there could be a problem.  The system becomes unstable for large enough gains.  How big is large enough?  Is this going to be a problem?

        We'll give you a chance to estimate the gain for which the system goes unstable.  Do that now, before you move on.  You need to estimate the gain at the point marked by the black dot where the root locus crosses into the right half of the s-plane.

        Now, we are showing the lines from the poles to the point where we want to estimate the gain.  There are three lines, black, green, and purple.  We can estimate the lengths as follows:

Black - goes from pole at s = -0.5 to the pole on the imaginary axis:o Length ~=3

Green - goes from pole at s = -1.0 to the pole on the imaginary axis:

o Length ~=3.2

Purple: - goes from pole at s = -5.0 to the pole on the imaginary axis:

o Length ~=6

        From that, we estimate the root locus gain as: KRL ~=  3 x 3.2 x 6 = 57.6

        At this point, we know the following: The maximum root locus gain is 57.6 The gain, K, and the root locus gain are related by:

o KRL =  2.5 K

Thus, the largest gain, K, is 57.6/2.5 = 23.04

There is virtually no margin for error here.  We need at least 19, and we can't go over 23.  Good grief!

        We can reasonably expect that a gain of 19 will produce a system that doesn't respond too well.

A gain of 19 has to give roots that are very lightly damped.  Look at the root locus again if you have to.

The system probably responds very slowly also.  For a gain of 19, the roots have to be very close to the j-axis.

        What conclusions can we draw?  Well, before you proceed, answer this question.

Would you build this system?

There's a pretty straight-forward answer to the question. You should not build this system! It's going to be on the edge.

o If the gain changes a little bit (to 23),  you have a problem.

o But you need a gain of 19 to meet the SSE specification.

It's going to be highly oscillatory - very low damping.

It's going to respond slowly.

Take your pick of the reason not to build.  What can you do?

        There are some other things to consider.

You could try integral control. You could try some sort of compensator.

You could try something in the PID family more complicated than integral control.

        There has to be some way you can make this system perform well.  Proportional control isn't going to do it.  At this point you can try any of the options above, or you could look at the analysis in the frequency domain, or just continue in this section.  Take your pick.

        In the frequency domain we can probably get some useful information by examining a Bode' plot for the loop transfer function and working with the Nyquist criterion.  What do we need to remember?

The loop transfer function is: G(s)H(s) = K/((2s + 1)(s + 1)(.2s + 1)) We will draw the Bode' plot for K = 1.

With three poles the phase will start at 0 and asymptotically approach -360o as frequency gets large.

The Bode' plot is shown below.

What can we glean from this plot?

The phase goes through -180o at a frequency of approximately f = 0.45.

At the crossing -180o, the magnitude is about -25db.

o That corresponds to a gain of 1/17.8.

We have a problem here.  We can't get a K of 19 because the system would be unstable at that gain value.  (Although it is right at the edge.)

        What conclusions can we draw from this?  Should we even try to build this system?

We should not try to build this system.  Even if our calculations are a little off and we can get the system to be stable, it will not be very stable.  It will have a very small phase margin.  We would have no margin for error.

        You may want to play some "What if?" questions with this system.  If you have access to software that will allow you to calculate the response of the system, calculate the step response for some gains less than the gain that puts you on the edge.

Example Problem 2

       You might be a little down on using proportional control after the first problem.  Here's one that looks even worse.  Here's the transfer function.

 The parameters in this transfer function are:

The undamped natural frequency, n = 2 The damping ratio, = 0.1

The zero is at s = -3, so the numerator factor is (s + 3).

        Let's consider this system a little further.  We can look at the behavior of the open loop system.  That's the system with no feedback.  Here is the step response of a system with an undamped natural frequency, n = 2 and a damping ratio, = 0.1.  This plot gives an idea of how slowly the oscillations die out for a damping ratio of 0.1.

It's a pretty good observation that this might be a particularly tough system to control because of the very light damping in the open loop system.

        It might help if we looked at a root locus for this system.  Here is a plot for a root locus of our system using proportional control.

This plot shows the open loop poles, and it is clear that they are very close to the j-w axis.

We know that two branches of the root locus start at those lightly damped poles.

Where do those branches go?

o Think about that question and try to answer it before you go on.

Here's the root locus.

There are some very interesting features to this plot.

Although the open-loop poles are very close to the j- axis, they move quickly into the left half of the s-plane.

For high enough gain the damping ratio of the closed loop poles begins to increase.

For even higher gains, the closed loop poles become real, and they even stay real for higher gains.

Estimate the gain that produces two real and equal poles before going on.

Problem

P2  What loop gain - K - will produce a closed loop system with poles at s = -6.4.  (Note, we estimate that the root locus shows double poles can exist at s = -6.4)

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:

        Let's check your calculation for that gain.  Here is the root locus plot with the different pole and zero distances to the possible pole on the negative real axis shown in different colors.  We are estimating that we could get a pole at -6.4

green: Length ~=6.5 orange: Length ~=6.5

purple: Length ~=3.4

From that, we estimate the root locus gain as:

o KRL ~=  (6.5x6.5)/3.4 = 12.4

        You may want to examine the step response of the system with the gain value we have chosen.  Here is a simulation of the system for a gain of 12.5

There is a small amount of overshoot, but, interestingly, there are no decaying oscillations in the response.  It does look as though the transient is composed of two decaying exponentials that add to give this response.

There is a steady state error, and it seems to be in the vicinity of 10%

If the root locus gain is 12.5, then the open-loop DC gain is:

o OLDCGain = (12.5x3)/4 = 9.375

        Now, we can compute the SSE we expect for the gain we have chosen. Since: OLDCGain = (12.5x3)/4 = 9.375 The SSE for a unit step input is:

o SSE = 1/(1+9.375) = .096 or 9.6%

We estimated 10%, and that's about what it's supposed to be.

Can we do better?

Let's look at the root locus again.  Here it is.

As the gain increases, the poles come together between -6 and -7. For further gain increases both poles stay real, although one of the

poles does move to the right.

The pole moving to the right might indicate a system with a slower response?

Should we try that or should we try integral control?

We won't know unless we check out higher gains.

Here is a plot of a computation done with a gain of 25.

The SSE is closer to 5%. If anything, this looks to be faster than the response for the lower gain.

And, there's less overshoot.

Even though this all might be counter- intuitive, we might be on to something here.

Let's try an even higher gain.

o Try 50?

Here is a plot of a computation done with a gain of 50.

The SSE is much smaller.  (We would expect only a few percent!) This seems even faster.

Again, there's even less overshoot.

o Interesting!

     There is a question here that we have to answer.  First, we need to pose the question.

The question is:o "Why does the system seem to improve - especially get faster -

as the gain increases when it is clear that one of the poles is moving to the right - implying a slower response?"

The answer has to do with the contribution that pole makes to the total response.

This system is simple enough that we can do a computation of the response for several gains.

        The closed loop transfer function is given by the expression below.

With a unit step input, the transform of the output is given by this expression:

o

The variable gain, K, here is the root locus gain we have been using in our computations.

Next, we will expand this expression with partial fractions using gain values of 25 and 50 - the gain values we computed the responses for earlier.

Here is the partial fraction expansion for a gain of 25.

o

o Notice the pole at s = -3.63 and the .239 value for its numerator.

Here is the partial fraction expansion for a gain of 50.

o

Notice the pole at s = -3.27 and the .093 value for its numerator.

        What do we see here? The pole does indeed move to the right! However, the contribution that the pole makes to the total response

gets smaller as the pole moves to the right.

Actually, that's something we should expect to happen as the pole gets closer to the zero.

So, it's not an accident that the performance actually improves.  There's a reason for it, even though it seems initially counterintuitive.

We probably don't get that behavior for free.  It probably takes more control effort to get the good response.

Summing Up

        To this point we have looked at two systems using proportional control.  What have we learned?

It's hard to make predictions about how well proportional control will work.  We examined a system with three real poles, and it proved to be a tough system to control.  For the specifications we had we couldn't get acceptable performance with proportional control.

On the other hand, we looked at a system with lightly damped poles and proportional control worked well there.  The lightly damped system would be a system that might be expected to be difficult to control.  It wasn't.  It would have been difficult without the fortunate location for the zero in the system.

        What can we conclude? Proportional control is arguably the easiest kind of control to

implement - and even to analyze.  A good strategy is to examine whether it can be used.  That should be the first thing to try in any design effort.

If proportional control works, then you're in good shape.  If not, there are many other alternatives.

PID Control Systems - Integral Control Why Not Use An Integral Controller? Properties Of Integral Controllers Implementing Integral Controllers Integral Control ExampleYou are at: Design Techniques - The PID Family of Controllers - Integral Controllers Click here to return to the Table of Contents

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.

An integral controller is not particularly difficult to implement. In an analog system, an integral control system integrates the error

signal to generate the control signal.  If the error signal is a voltage, and the control signal is also a voltage, then a proportional controller is just an analog integrator.

In a digital control system, an integral control system computes the error from measured output and user input to a program, and integrates the error using some standard integration algorithm, then generates an output/control signal from that integration.

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,

Be able to predict the effect of integral control on SSE. Be able to predict the effect of integral control on stability. Be able to predict the effect of integral control on speed  of

response.

Properties Of Integral Controllers

        Integral controllers have these properties.

The controller integrates the error as shown in the block diagram of an example system below.

o The integral controller has a transfer function of Ki/s

So, the actuating signal (the input to the system being controlled) is proportional to the integral of the error.

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

In an integral controller, steady state error should be zero.o The system would have to have a zero at the origin to make this

claim false.

o Of course, the closed loop system has to be stable.

Integral control has a tendency to make a system slower.  We'll talk about why that happens shortly.

        If we think about the root locus for a system with integral control we can note the following.

The integral gain, Ki, is either the root locus gain, or the root locus gain is proportional to Ki.

        Integral control has a tendency to make a system slower.  Consider a typical situation as shown below.

There's a pole at s = 0 from the integral controller. There may be a real pole from the system.

The root locus moves to the left from the pole at the origin.

o Click the green button on the plot to see that.

You may want to do a complete analysis of your system before you decide that integral control does give you the response you want.

        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.

There are now two closed loop poles. The closed loop poles move together and coalesce at s = -0.5, then

become complex, with a damping ratio that becomes smaller for larger gain.

The system never goes unstable - not for any gain.

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

As we increase the gain (The plot is for Ki = 1) the phase margin gets smaller because the phase for the entire system, including controller, starts at  -90o, and approaches -180o, for high frequencies.

No matter how high the gain gets, the phase margin is always positive.

The system never goes unstable - not for any gain.        The second example we will discuss is a system with two open loop poles.

Here's the root locus.  This is the root locus for the original system and is really the root locus you would have using proportional control.

Open loop poles are at -1 and -4.  But, we need to add the pole due to the integral controller at s = 0.  Here is the root locus of the system with integral control - adding a pole at s = 0.

As the gain increases, the closed loop poles become imaginary, and eventually move into the RHP.

        This second example illustrates something that happens often with integral control.

Here's the root locus again for the system with proportional control.  There's a root locus branch between the poles that splits for higher gains.

With integral control, a pole is added at s = 0, and there is now a branch of the root locus between the pole at s = 0 and the rightmost system pole.

There is now a pole constrained between those two poles, and when those poles do break off the axis, they tend to move toward the RHP.

o The pole constrained between 0 and -1 is slower than any pole you would get using proportional control for the original system.  Using proportional control all poles lie to the left of -1.  Using integral control there is always a pole to the right of -1, sometimes poles that are unstable.

        It is tempting to draw a conclusion, and that conclusion is often right.  Here it is.

Adding integral control to a system may sacrifice speed of response, and even stability, for the sake of zero SSE.

o Integral control produces a slower system because of the pole(s) constrained to the right of the rightmost pole in the system.

It is up to the designer to decide if that is an acceptable tradeoff.

It is not an acceptable tradeoff to trade zero SSE for instability.

        Now, examine the Bode' plot for this system.

At high frequencies, the phase approaches -270o, when the phase of the integral controller is included.

As the gain is increased, the phase margin gets smaller and eventually becomes negative as the system goes unstable.

        From the examples, there are some interesting things to note. Integral control is not a cure-all. It's best to look at the situation in detail before drawing any conclusion.

You will need to examine every other aspect of the system if you use integral control because stability, speed of response and other performance measures will all be affected by using integral control.

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).

Here's a pseudo-code segment that could implement an integral controller.

o MeasuredOutput = MeasureVolts(instrument); /Measure the output/

o Error = DesiredOutput - MeasuredOutput; /Compute Error/

o ErrorInt = ErrorInt + DT*(DesiredOutput - MeasuredOutput);

o VoltsOut = ProportionalGain*ErrorInt;  /Compute output voltage/

o OutputVoltage(VoltsOut);    /Output the control signal/

This code assumes that you have a function - MeasuredOutput - that will measure voltage using an instrument connected to the computer, and can output a voltage with a function - OutputVoltage - that uses another instrument connected to the computer.

        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.

The analog controller would have an integrator transfer function of 1/s. The digital controller would have an integrator transfer function of (z

DT)/(z - 1).

o There is a strong temptation to consider s, in the analog/continuous domain, to be the equivalent of (z - 1)/(z DT) in the digital/sampled domain.That conclusion would only be true

if the Euler integration algorithm were the integration algorithm.  We would get a different sampled transfer function if we used a different integration algorithm (like trapezoidal integration, for example), but we would probably still get one pole at z = 1.

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.

For the engine + load:o GL = 1,   L = 2 sec.

For the controller:

o GC = KI (variable), C = 1 sec.

For the sensor:

o GS = 1,   S = .2 sec.

        We have some questions. Can the system operate with 5% SSE for a constant input? How quickly will the system respond when we have 5% SSE?

How much overshoot does the system exhibit when set for 5% SSE?

How sensitive is the system to changes in the gain at 5% SSE?

        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.

Do the analysis in the time domain using root locus methods. Do the analysis in the frequency domain using Nyquist/Bode methods.

        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 two branches that start at 0 and -0.5 come together and head toward the right half plane.

However, those two branches stay in the left half plane for a while, giving pole locations that would produce very stable responses.

            The points noted above are the good points.  There could be one problem here.

The two branches that start at 0 and -0.5 come together and head toward the right half plane never go very far into the left half plane.

That might imply a slow response.  It looks as though all of the real parts are to the right of s = -0.2 roughly.

o That's a five second or longer time constant in the response.

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.  

It looks like there is a point on the negative real axis at s = -0.2 that gives poles that are furthest into the left half plane.

At that point, the root locus gain can be calculated from the product of pole distances, and is given by:

o KRL = .2x.3x.8x4.8 = .2304

        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.

G(s)H(s) = (2.5Ki) / (s(s + 0.5)(s + 1)(s + 5))o so, KRL = 2.5Ki

o for KRL =  0.23, 2.5Ki = 0.23, and Ki = 0.23/2.5 = .092

        Here is the unit step response for the system with an integral controller (Ki = .092).  The response is just about what you expect.

There's no overshoot. The response seems to be within 5% in about 20 sec.  With two poles

at -0.2 that seems about right.

There's no SSE.  For a unit step input, the output gets to exactly 1.

        Is there any thing wrong with this response? The only visible problem is that it is slow.  It would be better if it were

faster.o (On the other hand, maybe you can always say that!)

We haven't looked at how the response changes when the system changes,

Can we improve the system we have designed?

We ought to look at the system response for other gain values.  That could have an influence on system performance, and it will give use some inkling of how sensitive the system is to changes in parameters.

Here's the system response for a gain that is twice as large as the gain (in blue) we used previously, shown with the original gain.

This system has some overshoot, and has a somewhat shorter settling time, so this is possibly an improvement if the overshoot can be tolerated.

Here is the response of our system with a gain half of the original gain shown with the original system.

This system takes considerably longer to respond. Clearly this system is inferior to the original design.

        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.

The -180o crossing occurs at a frequency a little below f = 0.1 Hz. At f = 0.1 Hz, the magnitude is just a few db below zero.

It's a stable system, but just barely stable for this gain.

Earlier, in our root locus analysis we used a particular value for the gain of the integral controller:

o Ki = .092

        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.

The decibel scale is adjusted so that phase and magnitude can apprear on the same plot.  200 on the vertical scale is really 20 db!

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.

We will pick a phase margin arbitrarily, and we will choose 60o.  We can design to that phase margin and then check performance.

Now, select the gain to give the phase margin of 60o.

        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.

We have a clear idea of what can be done with integral control. We see the limits of what we can do.

We have some sense of what happens if the gain within the system changes - either the gain we add, or a gain in the system.

If we need this accuracy and we have to have faster response, we're going to have to do something else.

The End

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

Proportional Control Systems What Is A Proportional Control System? Steady State Analysis Calculating Steady State Error

What Is A Proportional Control System?

        Often control systems are designed using Proportional Control.  In this control method, the control system acts in a way that the control effort is proportional to the error.  You should not forget that phrase.  The control effort is proportional to the error in a proportional control system, and that's what makes it a proportional control system.  If it doesn't have that property, it isn't a proportional control systems.

        Here’s a block diagram of such a system.  In this lesson we will examine how a proportional control system works.

We assume that you understand where this block diagram comes from.  Click here to review the material in the introductory lesson where a typical block diagram is developed.

Here's what you need to get out of this lesson.

  Given a closed loop, proportional control system,   Determine the SSE for the closed loop system for a given proportional

gain. OR   Determine the proportional gain to produce a specified SSE in the system

Steady State Analysis

       To determine SSE, we will do a steady state analysis of a typical proportional control system.  Let's look at the characteristics of a proportional control system.

There is an input to the entire system.  In the block diagram above, the input is U(s).

There is an output, Y(s), and the output is measured with a sensor of some sort.  In the block diagram above, the sensor has a transfer function H(s).  Examples of sensors are:

o Pressure sensors for pressure and height of liquids,

o Thermocouples for temperature,

o Potentiometers for angular shaft position, and tachometers for shaft speed, etc.

        Continuing with our discussion of proportional control systems, the criticial properties of a proportional control system are how it computes the control effort.  The block diagram below shows how the computation is performed.

The measured output is subtracted from the input (the desired output) to form an error signal.

A controller exerts a control effort on the system being controlled

The control effort is proportional to the error giving this method its name of proportional control.

        We can do a steady state analysis of a proportional control system.  Let’s assume that the steady state output is proportional to the control effort.  Call the constant of proportionality DCGain.  The output is then given by:

Output = DC Gain x Control Effort

and

Control Effort = Kp * Error

Here, Kp is the gain of the proportional controller.

        Finally, we note that the error is:

Error = Input - Measured Output

Note that the measured output is just the output of the sensor.  Inserting the value for the output, we have:

Error = Input - Ks * Output

Here, Ks is the gain of the sensor.  (And note that the gain of the sensor might be unusual.  For example, it might have the units of volts/inch if the sensor is measuring the heigh of a liquid in a tank.)  And we can solve for the output in terms of the input.

Output = DC Gain x Control Effort

= DC Gain x Kp * Error

Output = DC Gain x Kp * (Input - Ks * Output)

Solving for the output, we get:

Output = DC Gain x Kp * Input/[1 + DC Gain * Ks * Kp ]

        Now, let us consider the output expression

When the controller gain, Kp, gets really large the output approaches:o Output = Input/Ks- for very large Kp and DCGain values.

        Now, let us consider the output expression again: If the sensor gain, Ks, is unity (1), then the output will be equal to the

input.o Output = Input for very large Kp and DCGain values.

        Finally, we can compute the steady state error for a unity feedback system.  Since the output is given by this expression:

Output = DC Gain x Kp * Input/[1 + DC Gain * Ks * Kp ]

Then, the error is given by this expression:

Error = Input/[1 + DC Gain * Ks * Kp ]

        The error expression tells us how much the output deviates from the input.

Problems

P1   In this system, you want the output to be close to the input.  Acceptable behavior is when the output is within 2% of the input.  Determine the gain, K, that will produce acceptable behavior when the DC gain of G(s) is 1.0.  Note that H(s) is 1.0 for this system since the output, Y(s), feeds directly back to the comparator to form the error.

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:

What Does It All Mean?

        There are many times when you want the output of a system to be equal to the input value.  If you can build a proportional control system with a high gain, then you can achieve that condition approximately.  You can't ever get there exactly because it will always take a finite error to give a finite output.  But, if the gain is large, then a small error can give the output you want with a small error.

If you want better error performance, you might want to consider using an integral controller, but that is covered in another lesson.

If you have an ON-OFF system (and many heating systems are like that) you might want to consider pulse width modulation (PWM) for your system.  PWM can be used to give a proportional type of action in a system that is really ON-OFF.

If you want to know details of how the system reaches steady state, you'll need to learn more about the dynamics of control systems.  There's more in the more advanced lesson on proportional control.

Calculating SSE

        Earlier we showed that the error in the system is given by:

Error = Input/[1 + DC Gain * Ks * Kp ]

Since the error is the difference between the input and the measured output it is a measure of how well the system performs.  Steady state (DC) error is the error value that the system reaches after any transients die down.  If the input is constant, then this expression gives the steady state error (SSE) for the system with this input.  SSE is frequently use as one of several measures of how well a system performs.

        Let's look at an example system.  The block diagram of the system is shown below.  Let's calculate the SSE.

The proportional controller multiplies the error by Kp. The system being controlled has a DC gain of 2 (That's 10/5.)

We will examine how to get an SSE that is less than 2%.

        Now, the expression for the error will let us calculate the error.  Let's turn it around and ask what proportional gain, Kp, will give a SSE of 2% - a fractional error of .02.

We have an expression for the error:o Error = Input/[1 + DC Gain * Ks * Kp ]

The error is proportional to the input, and is less than the input.

The ratio of Error to Input - the fractional error - is given by:

o Fractional Error  = 1/[1 + DC Gain * Ks * Kp ]

We will need to have a denominator of 50 to get SSE  = 2%.

A denominator of 50 implies DC Gain x Kp = 49, or Kp = 49/2 = 24.5

        If you want to calculate the SSE for a unit step input given a value for Kp, you only need to use the expression for SSE, given below.  Here's that result for our example

Assume the proportional control gain is given by: Kp = 50. The DC gain of the controlled system is 2.

The error formula will evaluate to: 1/(1 + 50x2) ~= .01

        Now, next you can experiment with a few simulators that let you see the performance of some simple systems.  These demos are duplicated from the introductory lesson on control systems.

Example/Experiment

E1    In this simulator, the system is the one shown in the block diagram below.  To simplify things we have used a sensor with a gain of 1, and shown the feedback path as a gain of one.

In the simulator, we assume that G(s) is a first order system.o G(s) = Gdc/(s + 1)

In the simulator, the following items can be set.

o Gdc - The DC gain for G(s)

o The time constant for G(s)

o K - The proportional gain in the controller

o The Desired output, u, which corresponds to U(s) in the diagram above.

To operate the simulator,

o You can start by just using the values that are pre-loaded into the simulator.

o Click the Start button.  A plot will be generated.

o Observe the final value that the system achieves, and compare that to the desired final value.

Now, double the gain - from 5 to 10 - by entering a new value in the gain text box, and run the simulation again (You will have to clear the previous plot to do that.) and observe the final value again.

Compare the results and determine if the claims above about getting a small error with a large gain are true.

Does the system perform more accurately with the higher gain?

        Now you should have seen that the system performs better with a higher gain.  It is more accurate, and - if you didn't notice - it is also faster for the higher gain.  It's tempting to conclude that you always want higher gain because you will get better performance.  Let's check that on a second order system.

Example/Experiment

E2           In this simulator, the system is the one shown in the block diagram below.  It's the same configuration that we had before.

In the simulator, we assume that G(s) is a first order system.

o G(s) = Gdc/(s2 + 2nn2)

In the simulator, the following items can be set.

o Gdc - The DC gain for G(s)

o The damping ratio for G(s)

o nThe undamped natural frequency for G(s)

o K - The proportional gain in the controller

o The Desired output, u.

To operate the simulator,

o You can start by just using the values that are pre-loaded into the simulator.

o Click the Start button.  A plot will be generated.

o If you want to change anything, enter the new data, then click the Reset button which appears when the plot is complete.  That clears the plot and brings back the start button.

o The output is indicated as the simulation runs.

Now, double the gain - from 5 to 10 - by entering a new value in the gain text box, and run the simulation again (You will have to clear the previous plot to do that.) and observe the final value again.

Compare the results and determine if the claims above about getting a small error with a large gain are true.

Does the system perform more accurately with the higher gain?

Does the system perform better with the higher gain?

        Now, higher order systems are important, but they can exhibit behavior that can make you pull your hair out.  Below we have a simulator for a third order system.  This simulator will let you enter values for the gains of all the blocks in a system that has three poles.  You can also change any of the time constants.

Example/Experiment

E3  Here is the simulator.  Using the simulator, investigate how the system performs when you change the gain in the first block.  Keep the time constants at the pre-loaded values.

Summary

        In this lesson you should have learned that the open loop gain determines how accurate a proportional control system is.  The simulations should have driven that point home.  If not, you should look at the simulation again and try several gains to appreciate that relationship.

        However, in more complex systems the dynamics will be different.  Changing the proportional gain will not necessarily make the system faster.  In fact, increasing the proportional gain might produce disastrous effects in a system.  In later lessons you'll have to come to grips with that.  That's it for this lesson.  The next lesson should be the lesson on integral control.  Click here to go on to that lesson.  Or, you may want to go on to the advanced lesson on proportional control.  In that advanced lesson you will start to work on the consequences of controlling more complex systems.  You may want to prepare yourself for that lesson by looking at the lessons on root locus or the Nyquist stability criterion.

Implementing Proportional Controllers

Analog Implementation:

       Proportional control in an analog system is often implemented with operational amplifier circuits.  The circuit below produces an output voltage given by:

Vout = (Rf /R1)V1 - (Rf /R2)V2

        Clearly, this circuit can generate an output voltage that is proportional to an error signal, V1 - V2, and can be used as a proportional controller when resistance values are chosen so that:

R1 = R2

Digital Implementation:

        Proportional control in a digital system is often implemented in code in some programming language like C.

       Here's a pseudo-code segment that could implement a proportional controller.  

/Measure the output/ MeasuredOutput = MeasureVolts(instrument);

/Compute Error/ Error = DesiredOutput - MeasuredOutput;

/Compute output voltage/ VoltsOut = ProportionalGain*Error;

/Output the control signal/ OutputVoltage(VoltsOut);

        This code assumes that you have a function - MeasureOutput - that will measure voltage using an instrument connected to the comuter, and can output voltage with a function - OutputVoltage - that uses another instrument connected to the computer.  It also assumes the somewhere earlier in the program the desired output and the proportional gain have been assigned values.  @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

Proportional Control Systems What Is A Proportional Control System? Steady State Analysis Calculating Steady State Error

What Is A Proportional Control System?

        Often control systems are designed using Proportional Control.  In this control method, the control system acts in a way that the control effort is proportional to the error.  You should not forget that phrase.  The control effort is proportional to the error in a proportional control system, and that's what makes it a proportional control system.  If it doesn't have that property, it isn't a proportional control systems.

        Here’s a block diagram of such a system.  In this lesson we will examine how a proportional control system works.

We assume that you understand where this block diagram comes from.  Click here to review the material in the introductory lesson where a typical block diagram is developed.

Here's what you need to get out of this lesson.   Given a closed loop, proportional control system,   Determine the SSE for the closed loop system for a given proportional

gain.

OR   Determine the proportional gain to produce a specified SSE in the system

Steady State Analysis

       To determine SSE, we will do a steady state analysis of a typical proportional control system.  Let's look at the characteristics of a proportional control system.

There is an input to the entire system.  In the block diagram above, the input is U(s).

There is an output, Y(s), and the output is measured with a sensor of some sort.  In the block diagram above, the sensor has a transfer function H(s).  Examples of sensors are:

o Pressure sensors for pressure and height of liquids,

o Thermocouples for temperature,

o Potentiometers for angular shaft position, and tachometers for shaft speed, etc.

        Continuing with our discussion of proportional control systems, the criticial properties of a proportional control system are how it computes the control effort.  The block diagram below shows how the computation is performed.

The measured output is subtracted from the input (the desired output) to form an error signal.

A controller exerts a control effort on the system being controlled

The control effort is proportional to the error giving this method its name of proportional control.

        We can do a steady state analysis of a proportional control system.  Let’s assume that the steady state output is proportional to the control effort.  Call the constant of proportionality DCGain.  The output is then given by:

Output = DC Gain x Control Effort

and

Control Effort = Kp * Error

Here, Kp is the gain of the proportional controller.

        Finally, we note that the error is:

Error = Input - Measured Output

Note that the measured output is just the output of the sensor.  Inserting the value for the output, we have:

Error = Input - Ks * Output

Here, Ks is the gain of the sensor.  (And note that the gain of the sensor might be unusual.  For example, it might have the units of volts/inch if the sensor is measuring the heigh of a liquid in a tank.)  And we can solve for the output in terms of the input.

Output = DC Gain x Control Effort

= DC Gain x Kp * Error

Output = DC Gain x Kp * (Input - Ks * Output)

Solving for the output, we get:

Output = DC Gain x Kp * Input/[1 + DC Gain * Ks * Kp ]

        Now, let us consider the output expression

When the controller gain, Kp, gets really large the output approaches:o Output = Input/Ks- for very large Kp and DCGain values.

        Now, let us consider the output expression again: If the sensor gain, Ks, is unity (1), then the output will be equal to the

input.o Output = Input for very large Kp and DCGain values.

        Finally, we can compute the steady state error for a unity feedback system.  Since the output is given by this expression:

Output = DC Gain x Kp * Input/[1 + DC Gain * Ks * Kp ]

Then, the error is given by this expression:

Error = Input/[1 + DC Gain * Ks * Kp ]

        The error expression tells us how much the output deviates from the input.

Problems

P1   In this system, you want the output to be close to the input.  Acceptable behavior is when the output is within 2% of the input.  Determine the gain, K, that will produce acceptable behavior when the DC gain of G(s) is 1.0.  Note that H(s) is 1.0 for this system since the output, Y(s), feeds directly back to the comparator to form the error.

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:

What Does It All Mean?

        There are many times when you want the output of a system to be equal to the input value.  If you can build a proportional control system with a high gain, then you can achieve that condition approximately.  You can't ever get there exactly because it will always take a finite error to give a finite output.  But, if the gain is large, then a small error can give the output you want with a small error.

If you want better error performance, you might want to consider using an integral controller, but that is covered in another lesson.

If you have an ON-OFF system (and many heating systems are like that) you might want to consider pulse width modulation (PWM) for your system.  PWM can be used to give a proportional type of action in a system that is really ON-OFF.

If you want to know details of how the system reaches steady state, you'll need to learn more about the dynamics of control systems.  There's more in the more advanced lesson on proportional control.

Calculating SSE

        Earlier we showed that the error in the system is given by:

Error = Input/[1 + DC Gain * Ks * Kp ]

Since the error is the difference between the input and the measured output it is a measure of how well the system performs.  Steady state (DC) error is the error value that the system reaches after any transients die down.  If the input is constant, then this expression gives the steady state error (SSE) for the system with this input.  SSE is frequently use as one of several measures of how well a system performs.

        Let's look at an example system.  The block diagram of the system is shown below.  Let's calculate the SSE.

The proportional controller multiplies the error by Kp. The system being controlled has a DC gain of 2 (That's 10/5.)

We will examine how to get an SSE that is less than 2%.

        Now, the expression for the error will let us calculate the error.  Let's turn it around and ask what proportional gain, Kp, will give a SSE of 2% - a fractional error of .02.

We have an expression for the error:o Error = Input/[1 + DC Gain * Ks * Kp ]

The error is proportional to the input, and is less than the input.

The ratio of Error to Input - the fractional error - is given by:

o Fractional Error  = 1/[1 + DC Gain * Ks * Kp ]

We will need to have a denominator of 50 to get SSE  = 2%.

A denominator of 50 implies DC Gain x Kp = 49, or Kp = 49/2 = 24.5

        If you want to calculate the SSE for a unit step input given a value for Kp, you only need to use the expression for SSE, given below.  Here's that result for our example

Assume the proportional control gain is given by: Kp = 50. The DC gain of the controlled system is 2.

The error formula will evaluate to: 1/(1 + 50x2) ~= .01

        Now, next you can experiment with a few simulators that let you see the performance of some simple systems.  These demos are duplicated from the introductory lesson on control systems.

Example/Experiment

E1    In this simulator, the system is the one shown in the block diagram below.  To simplify things we have used a sensor with a gain of 1, and shown the feedback path as a gain of one.

In the simulator, we assume that G(s) is a first order system.o G(s) = Gdc/(s + 1)

In the simulator, the following items can be set.

o Gdc - The DC gain for G(s)

o The time constant for G(s)

o K - The proportional gain in the controller

o The Desired output, u, which corresponds to U(s) in the diagram above.

To operate the simulator,

o You can start by just using the values that are pre-loaded into the simulator.

o Click the Start button.  A plot will be generated.

o Observe the final value that the system achieves, and compare that to the desired final value.

Now, double the gain - from 5 to 10 - by entering a new value in the gain text box, and run the simulation again (You will have to clear the previous plot to do that.) and observe the final value again.

Compare the results and determine if the claims above about getting a small error with a large gain are true.

Does the system perform more accurately with the higher gain?

        Now you should have seen that the system performs better with a higher gain.  It is more accurate, and - if you didn't notice - it is also faster for the higher gain.  It's tempting to conclude that you always want higher gain because you will get better performance.  Let's check that on a second order system.

Example/Experiment

E2           In this simulator, the system is the one shown in the block diagram below.  It's the same configuration that we had before.

In the simulator, we assume that G(s) is a first order system.

o G(s) = Gdc/(s2 + 2nn2)

In the simulator, the following items can be set.

o Gdc - The DC gain for G(s)

o The damping ratio for G(s)

o nThe undamped natural frequency for G(s)

o K - The proportional gain in the controller

o The Desired output, u.

To operate the simulator,

o You can start by just using the values that are pre-loaded into the simulator.

o Click the Start button.  A plot will be generated.

o If you want to change anything, enter the new data, then click the Reset button which appears when the plot is complete.  That clears the plot and brings back the start button.

o The output is indicated as the simulation runs.

Now, double the gain - from 5 to 10 - by entering a new value in the gain text box, and run the simulation again (You will have to clear the previous plot to do that.) and observe the final value again.

Compare the results and determine if the claims above about getting a small error with a large gain are true.

Does the system perform more accurately with the higher gain?

Does the system perform better with the higher gain?

        Now, higher order systems are important, but they can exhibit behavior that can make you pull your hair out.  Below we have a simulator for a third order system.  This simulator will let you enter values for the gains of all the blocks in a system that has three poles.  You can also change any of the time constants.

Example/Experiment

E3  Here is the simulator.  Using the simulator, investigate how the system performs when you change the gain in the first block.  Keep the time constants at the pre-loaded values.

Summary

        In this lesson you should have learned that the open loop gain determines how accurate a proportional control system is.  The simulations should have driven that point home.  If not, you should look at the simulation again and try several gains to appreciate that relationship.

        However, in more complex systems the dynamics will be different.  Changing the proportional gain will not necessarily make the system faster.  In fact, increasing the proportional gain might produce disastrous effects in a system.  In later lessons you'll have to come to grips with that.  That's it for this lesson.  The next lesson should be the lesson on integral control.  Click here to go on to that lesson.  Or, you may want to go on to the advanced lesson on proportional control.  In that advanced lesson you will start to work on the consequences of controlling more complex systems.  You may want to prepare yourself for that lesson by looking at the lessons on root locus or the Nyquist stability criterion.