root locus 2012
TRANSCRIPT
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Dr Ian R. Manchester
Dr Ian R. Manchester AMME 3500 : Root Locus
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Slide 7
Feedbackis important because of the ability tostabilize unstable systems, react to
disturbances, and reduce sensitivity to
changing system properties. But how does feedback affect system
properties? I.e. what happens to pole locations
under feedback?
Dr Ian R. Manchester AMME 3500 : Root Locus
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Slide 8Dr Ian R. Manchester AMME 3500 : Root Locus
For the moment let us examine aproportionalcontroller. (Other control structures such as integraland derivative terms may be lumped in to G(s)).
K
-
+
R(s) E(s) C(s)G(s)
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Slide 13Dr Ian R. Manchester AMME 3500 : Root Locus
We can also determine the closed loop poles as afunction of the gain for the system
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Slide 14Dr Ian R. Manchester AMME 3500 : Root Locus
The individual pole locations The root locus
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Slide 15Dr Ian R. Manchester AMME 3500 : Root Locus
n(decreasing Tr)
Remember our descriptionof system specs as a
function of pole location.
So by increasing gain wecan reduce settling timeup to a point, beyond that
we will induce large
overshoot.
The system alwaysremainsstable
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Slide 16Dr Ian R. Manchester AMME 3500 : Root Locus
We can easily derive theroot locus for a second
order system
What about for a general,possibly higher order,
control system?
Poles exist when thecharacteristic equation
(denominator) is zero
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Slide 18Dr Ian R. Manchester AMME 3500 : Root Locus
Rule 1 : Number of Branches the n branches of the rootlocus start at the poles
For K=0, this suggests that the denominator must be zero(equivalent to the poles of the OL TF)
The number of branches in the root locus therefore equalsthe number of open loop poles
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Slide 19Dr Ian R. Manchester AMME 3500 : Root Locus
Rule 2 : Symmetry - The root locus is symmetrical aboutthe real axis. This is a result of the fact that complex poles
will always occur in conjugate pairs.
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Slide 20Dr Ian R. Manchester AMME 3500 : Root Locus
Rule 3 Real Axis Segments According to the angle criteria,points on the root locus will yield an angle of (2k+1)180o. On the real axis, angles from complex poles and zeros are
cancelled.
Poles and zeros to the left have an angle of 0o. This implies that roots will lie to the left of an odd number of
real-axis, finite open-loop poles and/or finite open-loop zeros.
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Slide 22Dr Ian R. Manchester AMME 3500 : Root Locus
Consider the system atright
The closed loop transferfunction for this system
is given by
Difficult to evaluate theroot location as a
function of K
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Slide 23Dr Ian R. Manchester AMME 3500 : Root Locus
Open loop poles andzeros
First plot the OL polesand zeros in the s-plane
This provides us withthe likely starting(poles) and ending(zeros) points for the
root locus
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Slide 24Dr Ian R. Manchester AMME 3500 : Root Locus
Real axis segments Along the real axis, the
root locus is to the left
of an odd number of
poles and zeros
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Slide 25Dr Ian R. Manchester AMME 3500 : Root Locus
Starting and end points The root locus will start
from the OL poles and
approach the OL zeros as K
approaches infinity
Even with a rough sketch,we can determine what the
root locus will look like
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Slide 26Dr Ian R. Manchester AMME 3500 : Root Locus
Rule 5 Behaviour at infinity For larges and K, n-m of theloci are asymptotic to straight lines in thes-plane
The equations of the asymptotes are given by the real-axisintercept, a, and angle, a
Where k = 0, 1, 2, and the angle is given in radiansrelative to the positive real axis
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Slide 27Dr Ian R. Manchester AMME 3500 : Root Locus
Why does this hold? We can write the characteristic equation as
This can be approximated by
For large s, this is the equation for a system with n-mpoles clustered at s=
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Slide 28Dr Ian R. Manchester AMME 3500 : Root Locus
Here we have four OLpoles and one OL zero
We would thereforeexpect n-m = 3 distinct
asymptotes in the root
locus plot
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Slide 29Dr Ian R. Manchester AMME 3500 : Root Locus
We can calculate theequations of theasymptotes, yielding
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Slide 30Dr Ian R. Manchester AMME 3500 : Root Locus
For poles on the realaxis, the locus will
depart at 0o or 180o
For complex poles,the angle of departure
can be calculated by
considering the angle
criteria
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Slide 31Dr Ian R. Manchester AMME 3500 : Root Locus
A similar approach canbe used to calculate the
angle of arrival of the
zeros
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Slide 32Dr Ian R. Manchester AMME 3500 : Root Locus
We may also beinterested in the gain
at which the locus
crosses the imaginary
axis
This will determinethe gain with which
the system becomes
unstable
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Slide 33Dr Ian R. Manchester AMME 3500 : Root Locus
All of this probably seems somewhatcomplicated
Fortunately, Matlab provides us with tools forplotting the root locus
It is still important to be able to sketch the rootlocus by hand because
This gives us an understanding to be applied todesigning controllers
It will probably appear on the exam
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Slide 34Dr Ian R. Manchester AMME 3500 : Root Locus
As we saw previously, the specifications for asecond order system are often used indesigning a system
The resulting system performance must beevaluated in light of the true systemperformance
The root locus provides us with a tool withwhich we can design for a transient responseof interest
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Slide 35Dr Ian R. Manchester AMME 3500 : Root Locus
We would usually follow these stepsSketch the root locusAssume the system is second order and find the
gain to meet the transient response specifications
Justify the second-order assumptions by findingthe location of all higher-order poles
If the assumptions are not justified, systemresponse should be simulated to ensure that it
meets the specifications
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Slide 36Dr Ian R. Manchester AMME 3500 : Root Locus
Recall that for a second order system with nofinite zeros, the transient response parametersare approximated by
Rise time :Overshoot :Settling Time (2%) :
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Slide 37Dr Ian R. Manchester AMME 3500 : Root Locus
Recall the systempresented earlier
Determine a value of thegain K to yield a 5%
percent overshoot For a second order
system, we could find K
explicitly
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Slide 39Dr Ian R. Manchester AMME 3500 : Root Locus
Alternatively, we canexamine the Root Locus
x x
Im(s)
Re(s)
10 5 0
x
x
S=5+5.1j
=sin-1
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Slide 40Dr Ian R. Manchester AMME 3500 : Root Locus
We can use Matlabto generate the root
locus
!% define the OL system!sys=tf(1,[1 10 0])!% plot the root locus!rlocus(sys)!
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Slide 41Dr Ian R. Manchester AMME 3500 : Root Locus
We also need to verifythe resulting step
response
% set up the closed loop TF!
cl=51*sys/(1+51*sys)!% plot the step response!step(cl)!
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Slide 42Dr Ian R. Manchester AMME 3500 : Root Locus
Consider this system This is a third order
system with an
additional pole
Determine a value of thegain K to yield a 5%
percent overshoot
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Slide 43Dr Ian R. Manchester AMME 3500 : Root Locus
With the higher orderpoles, the 2nd order
assumptions are violated
However, we can usethe RL to guide our
design and iterate to
find a suitable solution
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Slide 44Dr Ian R. Manchester AMME 3500 : Root Locus
The gain found basedon the 2nd order
assumption yields a
higher overshoot
We could then reducethe gain to reduce the
overshoot
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Slide 45Dr Ian R. Manchester AMME 3500 : Root Locus
The preceding developments have beenpresented for a system in which the design
parameter is the forward path gain
In some instances, we may need to designsystems using other system parameters
In general, we can convert to a form in whichthe parameter of interest is in the required form
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Slide 46Dr Ian R. Manchester AMME 3500 : Root Locus
Consider a system of thisform
The open loop transferfunction is no longer of
the familiar form KG(s)H
(s)
Rearrange to isolate p1Now we can sketch theroot locus as a function of
p1
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Slide 47Dr Ian R. Manchester AMME 3500 : Root Locus
This results in thefollowing root locus as a
function of the
parameter p1
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Slide 48Dr Ian R. Manchester AMME 3500 : Root Locus
We have looked at a graphical approach torepresenting the root positions as a function ofvariations in system parameters
We have presented rules for sketching the rootlocus given the open loop transfer function
We have begun looking at methods for usingthe root locus as a design tool
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Slide 49Dr Ian R. Manchester AMME 3500 : Root Locus
NiseSections 8.1-8.6
Franklin & PowellSection 5.1-5.3