The formula for the gain of the
frequency response at w= w
n
is interesting because:
It depends only
upon the DC gain and the damping ratio, and, the smaller the damping
ratio, the higher the gain at the natural frequency.
Now, recall the other important
behavior at low frequencies and high frequencies.
For
small w,
the gain is just G
dc.For
large w,
the gain Gdc/w2.For
small w,
the gain is just Gdc,
assuming Gdc = 10 (or 20 db) on
the plot.
For
large w,
the gain is G
dc/w2,
- dropping off at -40 db/decade.
Here we assume that the natural
frequency is fn
= 20.
And, we can
insert the point at the resonant frequency, using our formula.
G(jwn)
=Gdc/j2z
For this example, we'll assume
z= 0.1. Remember:
Gdc=
10, and z
= 0.1,
so this works out to be a gain of 50 at
the resonant peak, the equivalent of 34 db. Do we have a problem here?
The peak is well above either
of the asymptotes at the natural frequency.
We should believe all of the
math we've done.
Is there really a problem here?
Should we look at the actual frequency response? Here it is. There's
the peak. It does exist.
Let's examine the parameters here
again to be sure that his all hangs together. The system parameters were:
Gdc=
10,
z
= 0.1,
wn
= 2 p
20, (since that natural frequency was 20 Hz.)
With these paramters, note the following in the
plot.
The DC gain is 20 db
which corresponds to a gain of 10.
The resonant peak is
pretty much right at 20 Hz as it should be.
The resonant peak is
about 13 or 14 db high.
A gain of 50 would be
14 db, do that also checks!
The high frequency slope
looks to be around -12 db/octave or -20 db/decade.
All of these observations confirm the
calculations, and they really point out that it can be important to
understand how the resonant peak depends upon the damping ratio.
To make that correspondence
between resonant peak and damping ratio as clear as possible, we have here
an example of a frequency response for another system. We'll let you control
the damping ratio, but we're going to set the DC gain and the natural
frequency. Hopefully, you'll see how this peak depends upon the system's
damping ratio. Use the right and left arrow controls to step the movie a
single step forward or backward.
Gdc
= 1.0.
Natural
frequency = 159 Hz.
Damping ratio -
variable and controllable by user.
What should we note about the
second order system response in the movie?
There is
a resonant peak
in the second order system response.
The
size of the resonant peak depends upon the
damping ratio.
For
damping ratios less than about 0.5 the peak is relatively insignificant.
Finally, we have to deal with
the phase. A Bode' plot isn't complete until you have the phase plot. Here's
a phase plot for a system with:
A damping ratio
of 0.1
An undamped
natural frequency of 159 Hz. (1000 rad/sec.)
Notice the following for this plot.
The phase
starts at zero degrees for low frequencies.
The phase
asymptotically approaches -180o
for high frequencies.
How the phase plot depends upon
damping ratio is something you should know. Next, we have a movie of phase
shift as a function of damping ratio.
For the system in the plot,
the parameters are:
Gdc
= 1.0.
Natural
frequency = 159 Hz.
Damping ratio -
variable.
Now, at this point you've seen
Bode' plots for second order system with complex poles. Second order
systems with real poles are really combinations of two first order systems,
and they will be covered in the next section.
At this point, one direction
to continue would be to continue to the next section. However, you might
want to go in the direction of looking at Nyquist plots for the systems
discussed above. In that case, use this link to go to the lesson on Nyquist
plots.
Sketching Bode' Plots For Larger Systems - Examples
There will be times when you
will need to have some sense of what a Bode' plot looks like for a larger
system. A useful skill is to be able to sketch what the plot should look
like so that you can anticipate what you'll get. That's particularly helpful
when you have a complex system and you enter a large transfer function.
It's not only helpful. You can often gain insight by playing "What if?"
games with a notepad and pencil.
In this section, we will
look at some larger systems and examine some overall properties of Bode'
plots for those systems.
We will start with a system
that is not all that large - a second order system with two real poles. Just
for discussion, we'll use the system with the transfer function shown below.
If we wanted to
sketch this Bode' plot we could start by looking at the DC gain.
Remember
that the DC gain is just G(jw) with w= 0.
Letting
w=
0 in G(jw),
we get:
G(0) = 10.
(20db)
At low
frequencies, the (.002s + 1) term in the denominator will still look
pretty much like 1.0.
However, as we
go up in frequency, the (.01s + 1) term will have an effect.
The (.01s
+ 1) term introduces a corner frequency
which
we discussed earlier in the section on Bode' plots for first order
systems.
The corner
frequency is at:
f =
100/2p= 15.9Hz.
At slightly higher frequencies, the
(.002s + 1) term will start to have an effect.
The (.002s + 1)
term will add another -20db/decade slope to the plot, for a total of -40
db.decade.
We get -40
db/decade because we now have two poles contributing to the roll-off,
and 2*(-20db/dec) = -40 db/dec.
The second
corner frequency is at f = 500/2p = 79.5Hz.
The straight
line approximation is high at the corners, but gives a pretty good idea
of where the actual Bode' plot lies.
Now, let us make this slightly
more complicated. Here's another transfer function.
Start by looking at the DC gain - as
before.
Remember that
the DC gain is just G(jw) with w = 0.
Letting w
= 0 in G(jw),
we get:
G(0) = 10.
(20db)
As we go up in
frequency from DC, the (.01s + 1) term will have an effect.
The (.01s + 1)
term introduces a corner frequency - as before.
The corner
frequency is at f = 100/2p = 15.9Hz.
Check the
slope. It should be -20 db/decade.
At slightly
higher frequencies, the (.002s + 1) term will start to have an effect.
The (.002s + 1)
term will add another -20db/decade - or wait a minute - is that +20
db/decade?
Because
it is a zero, it is +20db/dec,
and the corner frequency is at:
f =
500/2p= 79.5Hz.
For frequencies
above 79.5 Hz, the gain would be 10*.002/.01 = 2 or 6db.
And don't forget we still have one more
corner frequency. so let's add the last corner frequency.
We have another
corner frequency at:
f =
(1/.0001)/2p= 1590Hz. - Call that 1600 Hz.
Above 400 Hz,
we have another -20 db/decade added, but the total will now be -20
db/decade.
The straight
line approximation is off at the corners, but gives a pretty good idea
of where the actual Bode' plot lies.
This is a good point for some
reflection.
You should be able to sketch
even higher order systems. Real poles and zeroes give rise to straight
forward corner frequencies - either pole or zero factors - and you can
account for them. Problems could arise if the corner frequencies were more
"bunched" than they were in the examples. We deliberately chose two examples
which had some separation between corner frequencies and that allowed the
Bode' plots to straighten out between corners.
We have yet to talk about
two other topics - phase plots and second order factors with low damping -
i.e. with resonant peaks. We will first look at resonant peaks on Bode'
plots.
We'll work with
an example related to the one we've done with real poles. The transfer
function is shown below.
We start by
looking at the DC gain.
Letting
w
= 0 in G(jw),
we get: G(0) = 10. (20db)
At frequencies
below 100 Hz, the quadratic factor will stay essentially 1.
However, as we
go up in frequency, the (.01s + 1) term will have an effect,
We get a corner
frequency from the (.01s + 1) term.
The corner
frequency is at:
f =
100/2p
= 15.9Hz.
Now, we need to
consider the quadratic factor.
The
natural frequency, wn
= 1000.
The
damping ratio is z
= .05, since:
2zwn
= 100
The damping
ratio predicts a resonant peak of:
(1/2z)
= 10 (or 20 db)
at f
= 1000/2p
= 159Hz.
We also know
that the quadratic factor is going to add -40 db/decade above the
resonant peak at the natural frequency.
Finally, you can compare what
we have generated with what a detailed plot shows.
It's amazing how it all fits
together, isn't it?
Again, we are
off at the corner frequency, but overall, the straight line gives a good
idea of where the Bode' plot lies.
There are numerous other kinds
of examples we should examine. One simple example is a case in which the
resonant peak is below the corner frequency. You should be able to predict
what happens in that case. But just in case you can't we'll examine that
situation. Here's a transfer function of that type.
Here's what you need to do.
Calculate the
DC gain.
Check which
critical frequency is lowest - the corner frequency or the natural
frequency in the quadratic factor.
Think that
through now before moving on.
Now, we will show you
what we got.
The DC gain is
10 (20db). Here's a plot with a DC gain of 20 db. It shows 20 db at
all frequencies, and we know that needs to be fixed.
The natural
frequency is at:
f =
100/2p
= 15.9Hz.
The corner
frequency is at:
f =
1000/2p
= 159Hz.
The resonant
peak is a decade below the corner frequency.
Here is our
estimate of the frequency response.
We've shown the asymptotes, We
also can figure the damping ratio. We need to do that because we know
there is a resonant peak. We have the following:
wn2
= 10,000, so wn
= 100
2zwn
= 10, so
z = 10/200 = .05
Knowing the
damping ratio allows us to predict how high the resonant peak will be
above the straight line approximation.
Gain
above approximation = 1 /j2z
If
the damping ratio is .05, then this works out to a gain of 10, which
is equivalent to 20 db.
That allows
us to add one more point to our approximate Bode' plot. Here's the
plot.
Now, compare our estimate with the
actual response.
From the straight line approximation
you could generate a fairly accurate Bode' plot.
Summarizing this section, you should
note the following.
You actually
get a lot of insight about the shape of the plot if you can sketch it.
The visualization helps a great deal.
There are times
when a plot might surprise you, and you now have a technique that can
eliminate those surprises.
We haven't
covered all of the ground. There's more yet.
