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 K_{p} = 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:
We can also take a look at the open loop frequency response for this system.
As the gain is increased
- plotted here for K_{p} = 1 - the zero db crossing moves
up in frequency.
No matter how high the
gain gets, the phase margin is always at least 90^{o} .
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 -180^{o} .
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 -270^{o} .
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 -180^{o} ,
and moves toward -270^{o} .
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 K_{p} ) 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 = 2.2 t ,
in a first order system, where t is
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 5K_{p} . when K_{p} G(s) = 5K_{p} /(s
+ 5).
So, 5K_{p}
= 6, and K_{p} = 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 = K_{p} ,
we have K_{p} = 20.
When K_{p}
= 20, the Root Locus gain is 5K_{p} = 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 K_{p} = 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:
G_{L} = 1, t _{L}
= 2 sec.
For the controller:
G_{c} = K (variable), t _{c}
= 1 sec.
For the sensor:
G_{s} = 1, t _{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.
Call
the loop gain "K".
K = G_{L}
x G_{c} x G_{s}
Then, the expression for SSE (for a unit step
input) is:
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?
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:
Green - goes from pole
at s = -1.0 to the pole on the imaginary axis:
Purple: - goes from pole
at s = -5.0 to the pole on the imaginary axis:
From
that, we estimate the root locus gain as:
K_{RL}
~= 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:
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 w -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 -360^{o} as
frequency gets large.
The Bode' plot is shown below.
What can we glean from this plot?
The phase goes through
-180^{o} at a frequency of approximately f = 0.45.
At the crossing -180^{o} ,
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,
w _{n}
= 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,
w _{n}
= 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)
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:
K_{RL}
~= (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.
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.
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
Send
us your comments on these lessons .