Root Locus Method

Transcription

Root Locus Method

Root Locus Method-introduction
• The relative stability and the transient performance of a closedloop system are directly related to the location of the closed-loop
roots of the characteristic equation in the s-plane.
• It is frequently necessary to adjust one or more system
parameters in order to obtain suitable root location.
• It is useful to determine the locus of roots in s-plane as a
parameter varied since the roots is a function of the system’s
parameter.
• The root locus technique is a graphical method for sketching
the locus of roots in the s-plane as a parameter is varied and
has been utilized extensively in control engineering practice.
• It provides the engineer with a measure of the sensitivity of
roots of the system a variation in parameter being considered.
• The root locus technique may be used to great advantage in
conjunction with the Routh-Hurwitz criterion.
• Who and when this method has been developed?
S-Plane
and Transient Response
Root Locus Method
• Developed by Evans while
he was a graduate student
at UCLA
• Uses the poles and zeros
of the open-loop system to
determine the closed-loop
poles when ONE parameter
is changing
Walter R. Evans, 1920-1999
Root Locus Concept
T =
KG ( s )
1 + KG ( s )
(Closed-loop Transfer function)
characteristic equation:
1+KG(s)=0
KG(s)=-1
=>
=>
π
-1
KG(s)
Vector or scale equation?
KG(s)=-1+j0
KG(s) ∠KG(s) = e− jπ
(Cartesian form)
(Polar form)
1
KG ( s ) = 1
∠KG(s) = ±k 3600 + π
The values of s that fulfill the angel and magnitude conditions are
The roots of the characteristic equation or the closed-loop poles.
Root Locus Concept
T =
characteristic equation:
KG ( s )
1 + KG ( s )
1+KG(s)=0
The root locus is the path of the roots of the characteristic
equation traced out in the s-plane as a system parameter
(K) is changed. (0<k<∞).
Root Locus Concept
K=0, this collapses to D(s )= 0.
Since the roots of D(s )= 0 are the poles of G(s),
those are the closed-loop poles for K=0.
When K is large,
When
thus the closed-loop poles tend to the roots of
i.e. the open-loop zeros. If
N(s )=0,
N(s)/D(s) is strictly proper,
the closed-loop poles tend to infinity.
If D(s)->∞
Identifying Points on the Locus
A location s* is on the locus if 1+KG(s*)=0 or,
equivalently, if
G(s*)=-1/K.
The phase condition:
arg[G(s*)]=
identifies a point on the locus,
and the magnitude condition:
|G(s*)|=1/K
identifies the gain at the point s*.
Geometric Interpretation
If
G(s) has been factored into the pole-zero form,
then the magnitude and phase of some G(s*) may be found
by drawing vectors from the singularities to the point s*:
*
∠G(s* ) = φ1 −θ1 −θ2 −θ3 −θ4
*
φ1 − θ1 − θ 2 − θ 3 − θ 4
∠G(s* ) = φ1 −θ1 −θ2 −θ3 −θ4
Second Order System
s 2 + 2ζωn s + ωn2 = 0
Assume K>0.
For K=0 the roots are 0,-2.
For K=1 system is critically damped with
poles at
s=-1.
For 0<K<1, system is overdamped with
roots at:
s=
For K>1, system is underdamped with
roots at:
s=
Resulting Root Locus
Root Locus Concept
Root Locus Concept
∠G(s*) =−(∠s1 +2+∠s1)
=-180˚
Example-root locus concept
Consider a second order system:
Find the root locus as a
function of parameter a.
Example-root locus concept
Root locus as
a function of
parameter a
Root locus construction procedure
Step 1: Write the characteristic equation
1+F(s)=0 as 1+KP(s)=0
Find the m zeros zi and n poles pj of P(s)
• Locate the poles and zeros on the s-plane
with selected symbols (o-zero, X-pole)
• The RL (root locus) starts at the n open-loop poles
• The RL ends at the open loop zeros, m of which are
finite, n-m of which are at infinity
The RL (root locus) starts at the n open-loop poles
and ends at zeros of P(s).
The RL ends at the open loop zeros, m of which are finite,
n-m of which are at infinity
Step 2: Locate the segments of the RL on the real axis
(they lie in sections of the real axis at the left of an odd
number of poles and zeros)
• Determine the number of separate branches (or loci).
• The number of branches is equal to the number of poles n
• The root locus is symmetrical with respect to the horizontal
real axis (because roots are either real or complex conjugate)
– Existence on the Real Axis
The root locus exists on the real axis to the left
of an odd number of poles and zeros.
Proof: Each point sr on the root locus must satisfy the
angle condition. Since the angular contribution along the
real axis due to complex-conjugate poles cancels, the
total angle of GH (s )
is due only to the contribution of
real poles (x) and zeros (0). Consider a point sr on the
real axis. For each pole and zero to the left of
Sr
X,o
∠( sr + zi ) = ∠( sr + p j ) = 1800
For each pole and zero to the right of
X,o
Sr
∠( s r + z i ) = ∠( s r + p j ) = 0
Hence
∠GH ( s) = 1800 × (n zr − n pr )
where n zr is the number of open-loop zeros to the right of sr
and n pr is the number of open-loop poles to the right of sr .
The angle
∠GH ( s ) = (2h + 1)π = (2h + 1)180 0 ,
n = 0,1,2,3...
only if nzr − npr = odd, which is equivalent to their sum
being odd.
The root locus exists on the real axis to the left of an odd
number of poles and zeros.
Example: A Second-Order System
?
?
Step 3: Asymptotes:
Branches of the root locus which diverge to ∞ (i.e. to
open-loop zeros at ∞ ) are asymptotic to the lines
with angle
( 2 q + 1)( 180 0 )
θ =
n p − nz
q = 0,1,2,...(n p − nz − 1)
Where n p is the number of open loop-poles,
n z is the number of open- loop zeros and
The asymptotes intersect the real axis at a point
called the pivot or centroid given by
np
σ=
nz
∑ p −∑z
j =1
j
i =1
n p − nz
i
np
σ=
nz
∑ p −∑z
j
j =1
i =1
i
n p − nz
Proof: Taking the limit of GH(s)
as approaches infinity :
⎡ nz
⎤
−
(
)
s
z
∏
i ⎥
⎢ i =1
k
=
(
)
GH
s
k
≅
= −1
⎢
⎥
lim
lim n p
n p − nz
(
−
)
s
σ
s →l arg e
s →l arg e ⎢
∏ (s − p j ) ⎥
⎢⎣ j =1
⎥⎦
Allowing the angle and magnitude conditions to be met
− k = (s − σ )
k = (s − σ )
n p −nz
n p −nx
∠ − k = ∠(s − σ )
If the angle of
asymptotes is
when s → ∞,
θ = ∠( s − σ )
:
( magnitude condition )
n p −nz
= (1 + 2 q )180 0
then the angle
condition can
be written as
( angle condition )
( n p − nz )∠( s − σ ) = ( n p − nz )θ = (1 + 2q)1800
(1 + 2q)1800
∴θ =
n p − nz
q = 0,1,..., (n p − nz − 1)
Root Locus Concept
(np − nz )∠(s −σ) = (np − nz )θ = (1+ 2q)1800
(n p − nz )∠( s − σ ) = (n p − nz )θ = (1 + 2q)1800
∴θ =
(1 + 2q)1800
n p − nz
(1+ 2q)1800
∴θ =
np − nz
np
σ=
q = 0,1,...,(np − nz −1)
nz
∑p − ∑z
j=1
j
i=1
np − nz
i
=
−2−0
= −1
2−0
Step 4: The actual point at which the root locus crosses
the imaginary axis can be evaluated using the
Routh-Hurwitz criterion
When the root locus crosses the imaginary axis, there is a
zero in the first column of the Routh-Hurwitz table, and other
elements of the row containing the zero are also zero.
A zero-entry appears in the first column, and
all other entries in that row are also zero
-Solution: Return to the previous row and form the
“Auxiliary Polynomial, qa(s)”, which will be a divisor
of the original q(s), divide out qa(s), and proceed.
The auxiliary polynomial is the polynomial immediately
precedes the zero entry in Routh array. The order of the
auxiliary polynomial is always even and indicates the
number of symmetrical roots pair.
Example:
q( s ) = s 3 + 2 s 2 + 4 s + K ,
s3
s2
s1
s0
1
2
8− K
2
K
0 < K < 8.
4
K
q(s) / qa (s) ⇒
2s2 + 8 s3 + 2s2 + 4s + 8
+ 4s
s3
qa (s )
2s2
2s2
0
0
When K=8
1/ 2s +1
+8
+8
Marginal stable
qa (s) = 2s 2 + Ks0 = 2s 2 + 8 = 2(s 2 + 4) = 2(s + j 2)(s − j 2).
q( s) = qa ( s)(1 / 2s + 1) = ( s + 2)(s + j 2)(s − j 2)
( when K = 8)
Step 5: Breakaway points:
• Breakaway points occur on the locus where two
or more loci converge or diverge. They often
occur on the real axis, but they may appear
anywhere in the s-plane.
• The loci that approach/diverge from a breakaway
point do so at angles spaced equally about the
breakaway point. The angles at which they
arrive/leave are a function of the number of loci
that approach/diverge from the break point.
Breakaway Points
If the root locus branches away (or break-in) the real axis
then K has an extremum there for real s .
Define
1
= p(s)
k (s) = −
GH ( s )
dk
This means to evaluate the points on the real axis where ds = 0
Why?
Let us assume that
p ( s ) = ( s − r1 ) n ( s − r2 )K( s − rm ) = K .
r1 is the repeated roots of p(s), therefore the breakaway point, n ≥ 2.
dK dp ( s )
=
= n( s − r1 ) n −1 ( s − r2 )...( s − rm )
Then
ds
ds
+ ( s − r1 ) n ( s − r3 )...( s − rm ) + ...
+ ( s − r1 ) n ( s − r2 )...( s − rm −1 ) = 0
Breakaway Points
dK dp ( s )
=
= n( s − r1 ) n −1 ( s − r2 )...( s − rm )
ds
ds
+ ( s − r1 ) n ( s − r3 )...( s − rm ) + ...
+ ( s − r1 ) n ( s − r2 )...( s − rm −1 ) = 0
The above equation has a factor of ( s − r1 ) in all the terms.
Therefore dK = dp ( s ) = ( s − r )[n ( s − r ) n − 2 ( s − r )...( s − r )
1
1
2
m
we have
ds
ds
+ ( s − r1 ) n −1 ( s − r3 )...( s − rm ) + ...
+ ( s − r1 ) n −1 ( s − r2 )...( s − rm −1 )] = 0
Hence,
( s − r1 ) is also the root of dp(s)/ds=0.
In other words, the breakaway point is the maximum of K=p(s).
Breakaway Points
Breakaway Points
Each break point is a point where a double (or higher
order) root exists for some value of K.
Breakaway Points
• Obtaining the breakaway points
Rewriting the characteristic equation to isolate :
The breakaway point occur when
Example:
Breakaway Points
Third-Order System
np
σ=
nz
∑ p −∑z
j =1
( 2 q + 1)( 180 0 )
θ =
n p − nz
j
i =1
n p − nz
i
q = 0,1,2,...(n p − nz − 1)
Third-Order System
Alternatively, one could evaluate p(s) between –2 and -3.
Third-Order System
np
σ=
nz
∑ p −∑z
j =1
( 2 q + 1)( 180 0 )
θ =
n p − nz
j
i =1
i
n p − nz
S=-2.45
q = 0,1,2,...(n p − nz − 1)
Step 6: Determine the angle of departure of the locus
from a pole and the angle of arrival of the
locus at a zero using the phase angle criterion
•The angle of departure (or arrival) is particularly of interest
for complex poles (and zeros) because the information is
helpful in completing the root locus.
• The angle of locus departure from a pole is the deference
between the net angle due to all other poles and zeros
and the criterion is ±180˚(2q+1), and similar for the locus
angle of arrival at a zero.
Angle departure and arrival
consider the third-order open-loop transfer function
K
.
2
2
( s + p 3 )( s + 2ζω n s + ω n )
q = 1,2,....
F ( s) = G( s) H ( s) =
∠p ( s ) = 1800 ± q3600 ,
The angles at a test point s1 ,
an infinitesimal distance from,
must meet the angle criterion.
Phase criterion
0
Therefore since θ 2 = 90, we have
θ1 + θ 2 + θ 3 = θ1 + 90 0 + θ 3 = +180 0 ,
√
or the angle of departure at pole p1 is
θ1 = 90 0 − θ 3 ,
√
The departure at pole p2 is the negative of that at p1
because p 1 and p2 are complex conjugates.
θ 1 = 90 0 − θ 3 ,
√
θ1 + θ 2 + θ 3 = −90 0 + θ 2 + θ 3 = +180 0 ,
or
θ 2 = 270 0 − θ 3 = −90 0 − θ 3
√
Another example of a departure angle is shown in
the departure angle is found form
θ − (θ + θ + 90 0 ) = 180 0. (phase criterion)
2
1
3
θ 2 − θ 3 = θ1 + 270 0
⇔ 90 0 + θ 2 − θ 3 = θ1 + 360 0.
Since (θ 2 − θ 3 ) = γ ,
we find that the departure angle is
θ 1 + 360 = θ 1 = 90 + γ .
0
0
√
√
Step 7: The final step of the root locus procedure are used
to determine a root location s1 , and the value K x
at the root location.
• Determine the root location that satisfy the
phase criterion at the root s x , x = 1,2,...n p .
• The phase criterion is
∠p ( s ) = 180 0 ± q360 0 ,
q = 1,2,....
• Determine the parameter value K x at a specific root s x
using the magnitude requirement. The magnitude
requirement s x at is
n
Kx =
∏ | (s + p j ) |
j =1
M
∏ | ( s + zi ) |
i =1
s = sx
Fourth-Order System
?
?
?
Fourth-Order System
n=4 and m= 0 implies that there are 4 infinite zeros.
N=4 implies that there are 4 separate loci
Fourth-Order System
Asymptotes:
Angles:
Centroid:
np
σ=
nz
∑ p −∑z
j =1
j
i =1
n p − nz
i
Fourth-Order System
Intersection with imaginary axis
=0
Fourth-Order System
Breakaway point:
s ≈ -1.6
Fourth-Order System
Angle of departure:
Angle of departure at pole
Because
Fourth-Order System
ζ=0.707
Fourth-Order System
Root Locus Summary
Root locus shows evolution of closed-loop poles as one
parameter changes
• A simple set of rules allow the loci to be sketched
• If details are needed, good computer packages are available to
plot root locus, Matlab: rlocus, rlocfind, rltool
• To develop insight into control design, a control engineer will
be able to determine the main features of root locus without a
computer
The 7 Steps to the Root Locus
Step 1:
Step 2:
Locus lie to the left of an odd number of poles and
zeros.
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Root locus examples
GH(s) =
s +1
s2 + 3s +1
1
0.8
0.6
0.4
Apply Steps 1-3
Imag Axis
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-4
-3
-2
-1
Real Axis
0
1
2
Root locus examples
s +1
GH(s) = 2
s + 3s + 3
1
0.8
0.6
0.4
0.2
Imag Axis
Apply steps 1-4 and
step 5 for breakaway
point
0
-0.2
-0.4
-0.6
-0.8
-1
-3
-2.5
-2
-1.5
-1
Real Axis
-0.5
0
0.5
1
Root locus examples
s −1
GH(s) = 2
s + 3s + 3
1
0.8
0.6
0.4
Step 4 for crossing and
0.2
Step 5 for breakaway point
…
Imag Axis
Apply steps 1-4,
0
-0.2
-0.4
-0.6
-0.8
-1
-3
-2.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
1.5
2
Root locus examples
s2 − s +1
GH ( s ) = 2
s + 3s + 3
1.5
1
0.5
step 4 for crossing
points
Imag Axis
Apply steps 1-3 and
0
-0.5
…
-1
-1.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
Root locus examples
2.5
GH(s) =
1
(s2 +3s +3)(s +2)
2
1.5
1
0.5
Imag Axis
Apply steps 1-4 and 6 :
0
-0.5
-1
-1.5
-2
-2.5
-3
-2.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
1.5
2
Root locus examples
GH(s) =
s −1
(s2 + 3s + 3)(s + 2)
8
6
4
2
Imag Axis
Apply steps 1-4, 6 and
determine the crossing
point by Ruth-Hurwitz
0
-2
-4
-6
-8
-3
-2.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
1.5
2
Self-Balancing Scale
For small deviations
Input voltage to motor:
Motor:
Lead screw:
Specifications
Type 1 system
Design by adjusting Km
• Note that the dc gain is lw/Wc and is independent of
the motor gain – this is the effect of the integrator.
(Type one system)
• The damping ratio requirement implies that the closed
loop poles are 60˚ (cosθ=ζ) up from the negative real
axis, and gives an overshoot of 16%.
• The settling time constraint requires that Ts<2=4/ζωn
or that ζωn>2 for the closed-loop dominant poles.
• Note also, for this third order system, that the above
conditions apply only if the complex closed loop pole
pair is dominant (i.e. the third pole is roughly 10 times
the distance away from the complex axis of the
dominant pair).
Selecting Km via Root Locus
Closed-loop transfer function:
The characteristic equation is:
To put it in root locus form:
3 Poles and two zeros
60˚
?
?
Type 1 system
double
poles
ζωn>2
60˚
Type 1 system
ζωn>2
PID Controller
• “Textbook”
PID controller:
which corresponds to
• In practice:
PID Controller
• PI controller: Used extensively in process
control on a broad range of applications due
to simplicity and relatively good performance
PI controller:
• PD controller: Used extensively in controlling
electromechanical systems
PD controller:
PID Controller
Consider the PID controller
The PID controller introduces a pole at the origin
and two zeros
PID Controller
Using a PID:
Z2
Z* 2
Improving Transient Response
Cannot be obtained
by gain adjustment
Laser Manipulator Control
• Find a suitable K so that ess <0.1 mm for a ramp
input r(t)=At where A=1 mm/s
For a ramp input
where
=
Hence
Design of a robot control system
• To achieve the rapid and accurate control a robot, it is is
important to keep arm stiff and yet lightweight.
• The specification for controlling the motion of a lightweight,
flexible arm are
1) a setting time < 2 second
2) a percent overshoot <10% for a step
3) a steady-state error of zero for a step
The transfer function
of the flexible arm
Complex zeros:
Complex poles:
First we consider K2=0,
Complex zeros:
Real poles: s=0; s=-10
Complex poles:
Root locus on real axis?
How the root locus look like?
double
poles
?
?
First we consider K2=0,
Complex zeros:
Real poles: s=0; s=-10
Complex poles:
Is this system stable for K1>0?
The system is unstable since two
roots of closed system appear in
the right-hand s-plane for K1>0.
double
poles
It is clear that we need to
introduce the use of velocity
feedback K2>0. Then we
have
and
We select
in order
to the adjustable zero near
the origin for canceling the
affect of the poles.
The system has 5 zeros and
7 poles.
Root locus on real axis
The system has 5
zeros and 7 poles.
?
Using steps 3-4
to check.
?
?
?
Since the system
has two net
Poles, it must be
stable for all
0<k1<∞
Double poles
Departure
angle
• When K1=0.8 and K2=5, we obtain a step response
with a percent overshoot of 12% and a settling time of
1.8 seconds. This is the optimum achievable response.
• If we use the following controller:
With z=1 and p=5,K2=5, when K1=5 we obtain a step
response with an overshoot of 8% and a settling time of
1.6 seconds.
The specification for controlling the motion of
a lightweight, flexible arm are
1. a setting time < 2 second
2. a percent overshoot <10% for a step
3. a steady-state error of zero for a step
Type two
System
ess=?
Disk Drive Read System
The system has
integral gain.
Goal: selecting K1 and K3, using
root locus method, to meet
the specifications.
Using
, we have the following responses:
• The system as designed meets all the specifications.
• The 20 ms setting time is the time it takes the system to
“practically reach the final value.
• In reality, the system drifts very slowly toward the final
value after quickly achieving 97% or 98% of the final value.
Root Locus Plots for Typical Transfer
Functions
Functions
Functions

Root Locus Method

Transcription

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