PID Control Systems - Proportional Control
Why Not Use A Proportional Controller?
Properties Of Proportional Controllers
Proportional Control Examples
Proportional Control Problems
Implementing Proportional Controllers
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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.


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:

        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.

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



        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:
    • 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.
    • There are special cases that can cause problems.
    • It's best to look at the situation in detail before drawing any conclusion.
    • 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.
    • 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.
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 = 2.2t, in a first order system, where tis the time constant.
  • That means that t is at most .2/2.2 = .09 sec.
    • 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.
    • 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,   tL = 2 sec.
  • For the controller:  Gc = K (variable), tc = 1 sec.
  • For the sensor:  Gs = 1,  ts = .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.

        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:
    • G(s)H(s) = (K/(2*.2))/((s + 0.5)(s + 1)(s + 5))
    • 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:
    • Length ~=3
  • Green - goes from pole at s = -1.0 to the pole on the imaginary axis:
    • Length ~=3.2
  • Purple: - goes from pole at s = -5.0 to the pole on the imaginary axis:
    • 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:
    • 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 jw-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.
    • If the gain changes a little bit (to 23),  you have a problem.
    • 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.
    • 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, wn = 2
  • The damping ratio, z = 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, wn = 2 and a damping ratio, z = 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?
    • 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-w 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:
    • 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:
    • 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:
    • 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.
    • 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.
    • Interesting!
     There is a question here that we have to answer.  First, we need to pose the question.
  • The question is:
    • "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:
  • 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.
    • 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.
  • 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.

Links To Related Lessons

Other Lessons On PID Controllers

Material That You Might Want To Review Moving Along - More Advanced Material
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