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:
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, Kpis 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. 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.
Given a closed loop, proportional control system,
Determine the SSE for the closed loop system for a given proportional gain.
