Systems of Proportional Control

What is A Proportional Control System ?

Stedy 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. 
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:
    • Pressure sensors for pressure and height of liquids,
    • Thermocouples for temperature,
    • 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:
    • 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.
    • 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.

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:
    • 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:
    • 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.
    • G(s) = Gdc/(st + 1)
  • In the simulator, the following items can be set.
    • Gdc - The DC gain for G(s)
    • t - The time constant for G(s)
    • K - The proportional gain in the controller
    • The Desired output, u, which corresponds to U(s) in the diagram above.
  • To operate the simulator,
    • You can start by just using the values that are pre-loaded into the simulator.
    • Click the Start button.  A plot will be generated.
    • 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.
    • G(s) = Gdc/(s2 + 2zwn + wn2)
  • In the simulator, the following items can be set.
    • Gdc - The DC gain for G(s)
    • z - The damping ratio for G(s)
    • wn - The undamped natural frequency for G(s)
    • K - The proportional gain in the controller
    • The Desired output, u.
  • To operate the simulator,
    • You can start by just using the values that are pre-loaded into the simulator.
    • Click the Start button.  A plot will be generated.
    • 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.
    • 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.   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.

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