You have a background in linear systems. Often when you are learning
linear systems you encounter problems like the following.
Given a system described
by one of the following
A transfer function,
A linear differential
equation with constant coefficients that relates the input and the output
of the system,
An impulse response,
A set of state equations,
And, given an input to
the system,
Determine the output (response)
of the system.
In
other words, if you are given a system and an input, you should be able
to find the output of the system. That's a problem that could be
difficult, but which has a straight forward approach that will get you
an answer. The approach might be different depending upon how the
system is described, but the approach is always straight forward.
It's not very likely that you will have that problem in your job.
If you do, there will probably be some sort of complication - like a nonlinearity
in the system - that makes your linear systems approach inapplicable.
However, there are many times when you have an inverse problem. You
may not know the system description, and you may need to figure out a description
for the system - transfer function, impulse response, differential equation,
state equations, whatever - knowing a sample of the input and the output.
That's a whole different kettle of fish.
If you think about linear systems, you probably got the idea that you could
always figure out a transfer function for a circuit. But, there are
many systems for which you can't get a good transfer function. Aircraft
may have transfer functions that vary widely with different conditions
(airspeed, altitude, fuel load, atmospheric conditions, etc.) and it may
not be easy to compute those transfer functions from physical data.
(The author of this page spent a few hours in an airplane with function
generators on the floor of the plane feeding low frequency sine waves to
the ailerons, etc. It's a good recipe for producing a queasy stomach.)
Chemical plants are another example of something that you have to control
but where you can't get a good handle on the transfer function of the system.
In situations like that you may need to have some tools that will let you
get a system description from a record of input and output signals.
Let's think about how you would go about that. We'll start with a
simple situation after we examine the goals for this lesson. The
method we will described is shown pictorially on an another page. Click
here for that.
Goals For This Lesson
It's often possible to measure the frequency response of a system using
a sinusoidal input. Those measurements can be used to produce a Bode'
plot of the frequency response. In a situation like that you can
use that measured data to calculate a transfer function for a system.
In this lesson we look at some simple systems with these goals in mind.
Given a Bode' plot for a system,
To
be able to determine the resonant frequency and damping
ratio for complex
poles in the system.
To be able to determine the DC gain for the system.
To be able to determine the resonant frequency and damping
ratio for complex
poles in the system.
We
assume that you are familiar with frequency response ideas and plots.
(Click here to review that material.)
At the end of this lesson you will be even more familiar with Bode' plots
for first order systems and second order systems. At that point we
will examine some higher order systems to see how what you have learned
about first and second order systems can be applied there.
Why
Use Frequency Techniques?
Frequency response methods are widely used.
Whenever you buy stereo
equipment specifications for the equipment is probably phrased in terms
of frequency response.
Historically, sinusoidal
signals were easiest to generate, so engineers started to use these methods,
and a lot of good methods have been devised over the years.
Frequency response plots
- particularly Bode' plots - have parts that are fairly good straight lines
and that makes it easy to measure system parameters like time constants,
damping ratios, etc.
Frequency
response methods are widely used. Often you can get a Bode' plot
for a system, and you need to determine what the system is. If you
can achieve these goals you're on your way to doing that.
One widely used way of getting a transfer function is the following.
Measure the frequency
response. Choose a set of frequencies - using experience, knowledge
of the system, whatever - and input sine waves at the chosen frequencies.
Measure amplitude gain
and phase shift.
Plot the data and from
the shape of the curve(s) determine the transfer function.
This is all more easily
said than done. There are several choices for plots, but you will
probably find that Bode' plots are the best choice because of their propensity
to assume straight lines over different regions of the plot.
In any event, that's the
essence of the method.
A
First Order Example System
Here's
an example plot. Look at the features that would help you identify
the system. Note that the complete plot includes both magnitude and
frequency plots.
The magnitude plot - the
red line - consists of two straight line segments joined by a curved section.
The two straight line
segments can be extended to meet at the corner frequency.
Click the button in, then
move the mouse outside the button and release to keep the line from disappearing
in the plot.
The low frequency asymptote
lets you compute the DC gain.
The high frequency asymptote
drops off at -20 db/decade.
The high frequency phase
asymptote is -90^{o}.
Both of these observations
support a conclusion that there is one more pole than zero in the system.
A simpler conclusion is
that the system is first order, but that conclusion is not one we can be
absolutely certain of.
Why
can't we assume that this system is first order? (Actually it is
first order, and the transfer function is 1/(s + 1) - a particularly simple
one that lets us make a few points we need to make.)
You don't know for sure
if you have a pole and zero close together. You could have something
like this. Here's the Bode' plot for a system with a transfer function
of:
G(s) = (.3 s + 1)/[s +
1)(.33 s + 10]
Note that there is a pole
at s = -1/.33 and a zero at s = -1/.3, and that the pole and zero are close
together. The pole is near f = 0.5.
The pole and zero are
close together so that there's only a slight bend in the magnitude plot
and a small blip in the phase plot. When the pole and zero are really
close you may see almost no effect. But they could be there.
Here's
another example. Let's examine this system the same way. This
system is a little more complex and presents a few other interesting features.
The magnitude plot - the
red line - consists of two straight line segments joined by a curved section
- and it is a very funny curve.
The two straight line
segments can be extended to meet at a frequency that turns out to be the
frequency of the resonant peak.
That frequency seems to
be about f_{n} = 0.6.
Where the lines meet is
considerably below the peak, and the lines do not generate a good approximation
to the actual curve because of the peak.
There are other things we can learn about this
example system.
The low frequency asymptote
lets you compute the DC gain. Here the asymptote is at -6
db, so the DC
gain is 0.5.
The high frequency asymptote
drops off at -40 db/decade.
The high frequency phase
asymptote is -180^{o}.
Both of these observations
support a conclusion that there are two more poles than zeros in the system.
A simpler conclusion is
that the system is second order, but that conclusion is not one we can
be absolutely certain of - for the same reasons we saw earlier for the
first order system.
Since
this system is a second order system, it not only has a natural frequency,
but there is a damping ratio to be determined. We can use the resonant
peak to get the damping ratio.
The resonant peak is 14
db high. That corresponds to a gain of 4.47
In an earlier lesson we
got a formula for the gain at the resonant peak.
4.47 = 1/(2z)
That implies z = 1/8.94
= .112
Is that right or wrong?
It's not a correct computation. The correct
formula is:
Height above DC gain =
1/(2z).
In the original lesson,
the expression is:
4.47 = G_{dc}/(2z).
The DC gain is more like
-6db. There is a 20 db difference from the DC gain to the peak -
a gain of 10 - so we have:
10 = 1/(2z)
So z = 1/20 = .05
And that's right!
Now, we have enough information to compute the transfer function.
The DC gain was found
to be 0.5, so G_{dc} = 0.5
We also found the natural
frequency, f_{n} = 0.6.
That corresponds to w_{n}
= 2p0.6
~= 3.77 r/sec.
Finally, z
= .05.
That gives us a transferfunction of:
G(s) = 7.11/(s^{2}
+ 3.77 s + 14.29)
A
More Complex Example
Here's
another example. This one looks fairly horrible. What can we
conclude? What evidence do we see?
The high frequency phase
asymptote is -270^{o}.
This is at least a third
order system.
If it's a third order
system, we still need a strategy to figure out what it is.
We can note the following.
There must be a second
order set of poles that are undamped. They produce the resonant peak.
That peak is at f = 0.159
Hz, so that's the natural frequency. Convert it to w_{n}.
and you get:
w_{n}
= 1.0
There may be two line
segments above and below the peak frequency. Check that by clicking
the button in the picture below. Looks good doesn't it?
Is there another corner
where the lower frequency line touches zero db?
Now, ponder the question
of the slope of the lower frequency line there.
Is it -20 db/dec?
We
need to answer the question about the magnitude slope.
It doesn't really look
like -20 db/dec. Check the lines again.
However, it may be that
there's not a large enough range of frequency for the slope to get to its'
asymptotic value - presumably -20 db/dec.
At high frequencies, the
slope looks more like -60 db/dec, and the phase looks like it is approaching
-270^{o}. It's a pretty good assumption that this
is a third order system.
It's a pretty good assumption,
but don't bet your life on it. The exact conclusion is that it
is at least a third order system. Check
the example above where we had a pole and zero close together where they
tended to mask each other even though they didn't exactly cancel out.
We could have that situation here too.
The
best bet is that there is a real pole that causes the phase to begin getting
negative around f = .01Hz, and the magnitude to dip there as well.
To find the system here,
you may need to try a few transfer functions and try to match this data.
An educated guess would
be to look for the point where the magnitude is 3 db down from the DC gain.
That's -9 db, and that's clearly near the midpoint between f = .01 and
f = .02 Hz. That's probably the corner frequency for the real pole.
Finish this. Find
G(s).
For
a final example, here's a deceptive system.
This system looks like
it is second order. But there's supporting and negating evidence
for that conclusion.
There's a peak that looks
like a resonant peak.
Phase goes to -180^{o}
at high frequencies.
High frequency rolloff
is -40 db/dec.
However, phase
is positive at f = .02Hz.
That can't happen
in a second order system. And, if you try to fit this data with a
second order system, the peak here is too broad.
This
system is really a third order system. Here's the transfer function.
G(s) = (5 s + 1)/[(s
+ 1)^{3}]
The zero is what causes
the rise in the frequency response.
The three poles at 1,
together with the zero produce the effects noted above.
Because the zero is
at a lower frequency than the poles, the phase gets slightly positive for
a while, but later tends to -180^{o} at high frequencies.
Can
we conclude anything from this? Here are some items to note.
If you have frequency
response data, you may be able to figure out what the transfer function
is.
Closely spaced poles and
zeroes may be missed.
There is the possibility
of confusing real poles and zeroes with a pair of complex poles.
There are surely other
things we have overlooked.
What
If?
There are some questions to ponder?
What if you only have
a magnitude plot - as sometimes happens?
You may still be able
to determine the transfer function. However, you will not have the
phase as a check, and there's always the possibility of a non- minimum
phase factor in the transfer function.
What if the data is not
taken at points that are dense enough?
It's possible to miss
highly resonant peaks. This is another reason you should have some
a priori information about the system, or you should know the form of the
transfer function from fundamental physical principles.