Analysis of Control Systems by the Root

Transcription

Lectures 25-29
Analysis of Control Systems by the
Root-Locus Technique
Lecture objectives
In these lectures you will learn the following:
The definition of a root locus
How to sketch a root locus
How to refine the sketch of a root locus
How to use the root locus to find the poles of a closed-loop system
How to use the root locus to describe qualitatively the changes in transient
response and stability of a system as a system parameter is varied
How to use the root locus to design a parameter value to meet a transient
response specification for systems of order 2 and higher
Prof. K. Melhem (Qassim University)
Principles of Control Systems Academic year 2014-2015
368
Introduction to root locus
Root locus determines a graphical presentation of the closed-loop poles as a system parameter is
varied. Root locus is a powerful method of analysis and design for stability and transient response
of control systems. The real power of root locus lies in the ability of providing solutions for higher
order systems.
A typical feedback control system is shown in the figure below
G(s)
=
NG (s)
DG (s)
H(s)
=
NH (s)
DH (s)
where open-loop transfer function is KG(s)H(s) and closed-loop transfer function is
T (s) =
KG(s)
KNG (s)DH (s)
=
1 + KG(s)H(s)
DG (s)DH (s) + KNG (s)NH (s)
369
Introduction to root locus
Vector representation of complex numbers
Complex number in Cartesian coordinates: s = σ + jω. In polar form, s = M∠θ where magnitude
√
M = σ2 + ω2 and angle θ = tan−1 (ω/σ). By substituting s = σ + jω into a complex function
F(s) = s + a, we get another complex number F(s) = (σ + a) + jω.
Now, let us consider a complex function F(s) in a general form
(s + zi )
∏m
F(s) = ni=1
∏ j=1 (s + p j )
m=number of zeros n=number of poles
F(s), at any point s, is a complex number with magnitude M and angle θ such that
M
θ
=
=
|(s + zi )|
∏ zero lengths
∏m
= ni=1
∏ pole lengths
∏ j=1 |(s + p j )|
m
n
i=1
j=1
∑ zero angles − ∑ pole angles = ∑ ∠(s + zi ) − ∑ ∠(s + p j )
370
Defining the root locus through an example
A video camera similar to that shown below can automatically follow a subject. The tracking
system consists of a dual sensor and a transmitter, where one component is mounted on the camera
and the other worn by the subject. An imbalance between the outputs of the two sensors receiving
energy from the transmitter causes the system to rotate the camera to balance out the difference
and seek the source of energy.
The so-called root locus technique can be used to analyze and design the effect of loop gain upon
the video camera system’s transient response and stability. With the block diagram shown below,
we see that the closed-loop poles change location as the gain K is varied.
Applying the quadratic formula to the denominator of the closed-loop transfer function, we can
determine the variations of pole location for different values of K, as shown in the following table.
371
372
The data of the preceding table is now graphically displayed in the following figures, which show
each pole and its gain. We observe that as the gain K increases the closed-loop pole, which is at -10
for K = 0, moves toward the right, and the closed-loop pole, which is at 0 for K = 0, moves toward
the left. They meet at -5, break away from the real axis, and move into the complex plane. One
closed-loop pole moves upward and the other moves downward, without being able to tell which
pole moves up or which pole moves down. The representation of the paths of the poles as the gain is
varied is called a root locus.
373
The root locus which is a graphical representation of the pole paths as the gain varies:
• shows the changes in transient response and stability as the gain varies.
• relates the transient response characteristics of higher-order systems to the pole location.
• gives qualitative description of the control system’s performance with a less possible
computational burden.
374
Formal definition of root locus
Next we see how we make rapid sketch of the root locus for higher-order systems without having to
factor the denominator of the closed-loop transfer function. To this purpose, we will examine some
properties of the root locus.
Again, the closed-loop transfer function of a feedback control system is
T (s) =
KG(s)
1 + KG(s)H(s)
The root locus of this system determines the location of closed-loop poles as gain K varies. Thus, it
is determined by the following equation
KG(s)H(s) = −1 = 1∠(2k + 1)180◦
k = 0, ±1, ±2, · · ·
for which open-loop transfer function KG(s)H(s) is to be investigated.
Statement: A point on the root locus (a closed-loop pole for some particular value of K) should
satisfy both of the two conditions:
Magnitude condition: |KG(s)H(s)|
=
1
Angle condition: ∠KG(s)H(s)
=
∠G(s)H(s) = (2k + 1)180◦
375
How to check if an s plane point is on a root locus: Example
Suppose that we have a system with the following open-loop transfer function
KG(s)H(s) =
K(s + 3)(s + 4)
(s + 1)(s + 2)
A point in the s plane on the system root-locus (or a closed-loop system pole for some value of gain
K) must satisfy the angle condition.
θ1 + θ2 − θ3 − θ4
= 56.31◦ + 71.57◦ − 90◦ − 108.43◦ = −70.55◦
For example, the point −2 + j3 is not on the root locus (i.e., not a closed-loop pole for any gain).
√
Whereas, −2 + j( 2/2) is so and the respective gain is given by
K=
√
2
2 (1.22)
|s + 1||s + 2|
L4 L3
1
=
=
= 0.33
=
|G(s)H(s)|
|s + 3||s + 4|
L2 L1
(2.12)(1.22)
376
Basic rules for sketching the root locus
1. Number of branches: The number of branches of the root locus equals the number of closed-loop
poles.
2. Symmetry: The root locus is symmetrical about the real axis.
3. Real-axis segments: On the real axis, for K > 0, the root locus exists to the left of an odd number
of real-axis finite open-loop poles and/or finite open-loop zeros.
4. Starting and ending points: The root locus begins at the finite and infinite poles of G(s)H(s) and
ends at the finite and infinite zeros of G(s)H(s).
5. Behavior at infinity: The root locus approaches straight lines as asymptotes as the locus
approaches infinity. The equations of these asymptotes are given by the real-axis intercept and
angle in radians as follows:
σa =
(2k + 1)π
∑ finite poles − ∑ finite zeros
, θa =
♯finite poles − ♯finite zeros
where k = 0, ±1, ±2, ±3, · · · and the angle is given in radians with respect to the positive
extension of the real axis.
377
Basic rules for sketching the root locus: An example
Problem and its solution Sketch the root locus of the system shown below.
Let us calculate the asymptotes, as the
open-loop transfer function has zeros
at infinity. The real axis intercept is
evaluated as
σa =
(−1 − 2 − 4) − (−3)
4
=−
4−1
3
The angles of the lines that intercept
at -4/3 are
θa
=
=
(2k + 1)π
π/3 (k = 0), π (k = 1), 5π/3 (k = 2)
If the value for k continued to increase,
the angles would begin to repeat.
378
Additional rules for refining the sketch
Real-axis breakaway and break-in points
The root locus breaks away from the real axis at a point where the gain is maximum and breaks
into the real-axis at a point where the gain is minimum.
Breakaway and break-in points via differentiation: For the open-loop transfer function
KG(s)H(s) = K(s − 3)(s − 5)/(s + 1)(s + 2) differentiate K = −1/G(σ)H(σ) with respect to σ and set
derivative equal to zero yields
11σ2 − 26σ − 61
dK
=
=0
dσ
(σ2 − 8σ + 15)2
Solving for σ, we find σ = −1.45 and 3.82.
379
Calculation of jω-axis crossings
The jω-axis crossing determines the stability range of the closed-loop system as the gain K varies.
To find these points, Routh-Hurwitz criterion may be used.
The closed-loop transfer function for the system is
T (s) =
K(s + 3)
s4 + 7s3 + 14s2 + (8 + K)s + 3K
A complete row of zeros yields the possibility for imaginary axis roots. Thus, imaginary axis roots
are by −K 2 − 65K + 720 = 0 which gives K = 9.65. The even polynomial with K = 9.65 is
(90 − K)s2 + 21K = 80.35s2 + 202.7 = 0 ⇒ s = ± j1.59
Thus, the root locus crosses the jω-axis at ± j1.59 with a gain 9.65. Stability is for 0 < K < 9.65.
380
Angles of departure and arrival
The root locus departs from complex open-loop poles and arrives at complex open-loop zeros at
angles that can be calculated.
The open-loop transfer function for the system in the figure on the right is
KG(s)H(s) = K(s + 2)/(s + 3)(s2 + 2s + 2). The root locus departs from the pole −1 + j1 with angle θ1
such that −θ1 − θ2 + θ3 − θ4 = −θ1 − 90◦ + tan−1 (1/1) − tan−1 (1/2) = 180◦ from which θ1 = 108.4◦ .
381
Plotting and calibrating the root locus
All points on the root locus satisfy ∠G(s)H(s) = (2k + 1)180◦ . The gain K, at any point on the root
locus, is given by K = 1/|G(s)H(s)|.
Problem and its solution: Find the exact point at which the locus crosses the 0.45 damping ratio line
and the gain at that point.
If a few test points along the ζ = 0.45 line are selected, we found that the point at radius 0.747 is on
the root locus since the angles add up to −180◦ . The corresponding gain is K = |A||C||D||E|/|B| = 1.71.
382
Sketching a root locus and finding critical points: Example
Problem Sketch the root locus for the system shown below and find the following:
a. The exact point and gain where the locus crosses the 0.45 damping ratio line.
b. The exact point and gain where the locus crosses the jω-axis.
c. The breakaway point on the real axis.
d. The range of K within which the system is stable.
Solution First we sketch the root locus using the basic rules to yield the figure below.
a. The exact point where the root locus crosses the ζ = 0.45 line can be found either by searching
along the line β = − cos−1 0.45 + 180◦ = 116.7◦ for the point where the angles add up to an odd
multiple of 180◦ or by simply using a computer program. We find that the root locus crosses the
ζ = 0.45 line at 3.4∠116.7◦ with a gain K of 0.417.
383
Sketching a root locus and finding critical points: Example
b. The root locus crosses the jω-axis at ±3.9
with a gain K of 1.5.
c. The breakaway point can be found by first
settling
K(s2 − 4s + 20)
KG(s)H(s) =
(s + 2)(s + 4)
equal to -1, solving for K to yield K = −(s2 +
6s + 8)/(s2 − 4s + 20). and finally differentiating K with respect to s and then settling the
derivative equal to zero to yield
dK
10s2 − 24s − 152
=
=0
ds
(s2 − 4s + 20)2
Solving for s, we find s = −2.88 and s = 5.28.
The breakaway point is at s = −2.88 and the
other point at s = 5.88 does not correspond neither to a breakaway point nor to a break-in
point (K is now negative).
d. From the answer to b., the system is stable
for K between 0 and 1.5.
384
Transient response design via gain adjustment
Now that we know how to sketch a root locus, we show how to use it for the design of
transient response. The design process can be summarized by the following points:
1. Sketch the root locus for the given system.
2. Assume that the system is an underdamped second-order system without any zeros and
then find the gain (gain adjustment) to meet the transient response specification.
3. Justify the second-order approximation by finding the closed-loop zeros and all the
higher-order poles and prove that they have no appreciable effects on the transient
response by the dominant second-order pole pair (higher-order poles farther into the
left half-plane, pole-zero cancelation, zeros farther into the left half-plane).
4. If approximation cannot be justified, time response could still be simulated to make
sure the design is within tolerance.
385
An example
Problem: Design the value of gain K for the system shown in the figure below to yield 1.52%
overshoot. Furthermore, estimate the settling time, peak time, and steady state error.
Solution: We first sketch the root locus.
Breakaway points are found at
−0.62 with a gain of 2.511, and
at −4.4 with a gain of 28.89.
Break-in point is found to be
at −2.8 with a gain of 27.91.
Asymptotes to infinite zeros
are such that σa = −9.5/2 =
−4.75 and angles θa = π/2, 3π/2.
386
An example
Now, assume that the system can be approximated by an underdamped second-order system
without any zeros. Thus, a 1.52% overshoot corresponds to a damping ratio of ζ = 0.8. As shown in
the figure, the root locus intercepts with the 0.8 damping ratio line in three points: −0.87 ± j0.66,
−1.19 ± j0.90, and −4.6 ± j3.45 with respective gains of 7.36, 12.79, and 39.64.
For each of these points, the settling time and the peak time are evaluated using
Ts =
where −ζωn and ωn
p
4
,
ζωn
Tp =
π
p
ωn 1 − ζ2
1 − ζ2 are the real part and imaginary part of the closed-loop pole pair.
Here, the steady-state error produced in each case is given by the velocity constant
Kv = lim sKG(s) =
s→0
K1.5
(1)(10)
The results are depicted in the following table.
387
An example
388
An example
How valid are the second-order assumptions? From the preceding table, Case 1 and Case 2 yield third
closed-loop poles that are relatively far from the closed-loop zero. For these two cases there is no
pole-zero cancelation, and a second-order system approximation is not valid.
In the Case 3, the third closed-loop pole and the closed-loop zero are relatively close to each other,
and a second-order system approximation can be considered valid, provided that the residue of the
exponential decay after making a partial fraction expansion is much less than the residue of the
underdamped sinusoid. To check that, let us expand the step response as
C3 (s)
=
39.64(s + 1.5)
s(s + 1.8)(s + 4.6 + j3.45)(s + 4.6 − j3.45)
=
39.64(s + 1.5)
s(s + 1.8)(s2 + 9.2s + 33.06)
=
1
0.3
1.3(s + 4.6) + 1.6(3.45)
+
−
s s + 1.8
(s + 4.6)2 + 3.452
Thus, the residue of the exponential decay from the third pole is 0.3 while the residue of the
√
underdamped response from the dominant poles is 1.32 + 1.62 = 2.06. Hence, the dominant pole
response is 6.9 times large as the nondominant exponential response,, and we assume that a
second-order approximation is valid.
389
Root locus for positive-feedback systems
The properties of the root locus change dramatically if the feedback signal is added to the input
rather than subtracted.
We observe that a positive-feedback system can be thought of as a negative-feedback system with a
negative value of H(s). Hence, the closed-loop transfer function of a positive-feedback system is
T (s) =
KG(s)
1 − KG(s)H(s)
The root locus for positive-feedback systems consists of all points on the s plane where the following
condition is satisfied
KG(s)H(s) = 1 = 1∠k360◦ k = 0, ±1, ±2, ±3, · · ·
which is turn can be divided into
Magnitude condition: |KG(s)H(s)|
=
1
Angle condition: ∠KG(s)H(s)
=
∠G(s)H(s) = k360◦
390
The basic rules for sketching the root locus of positive-feedback systems are summarized as follows:
1. Number of branches: The number of branches of the root locus equals the number of closed-loop
poles.
2. Symmetry: The root locus is symmetrical about the real axis.
3. Real-axis segments: On the real axis, for K > 0, the root locus exists to the left of an even
number of real-axis finite open-loop poles and/or finite open-loop zeros.
4. Starting and ending points: The root locus begins at the finite and infinite poles of G(s)H(s) and
ends at the finite and infinite zeros of G(s)H(s).
5. Behavior at infinity: The root locus approaches straight lines as asymptotes as the locus
approaches infinity. The equations of these asymptotes are given by the real-axis intercept and
angle in radians as follows:
σa =
2kπ
∑ finite poles − ∑ finite zeros
, θa =
where k = 0, ±1, ±2, ±3, · · · and the angle is given in radians with respect to the positive
extension of the real axis.
What about the other calculations? In a search of crossing points (breakaway points, break-in points,
jω-axis crossing points) for positive-feedback systems, we look for the points where the angles to the
finite open-loop poles and zeros add up to a multiple of 360◦ instead of an odd multiple of 180◦
as for negative-feedback systems.
391
An example
Problem Sketch the root locus as a function of negative gain K for the negative-feedback system
below.
Solution The negative-feedback system with a negative gain can be converted to a positive-feedback
system with a positive gain. As shown in the figure on the right, we just push -1, associated with K
negative, to the right past the pickoff point. Therefore, as the gain of the equivalent system goes
through positive values of K, the root locus will be equivalent to that generated by the gain K of the
original system in the figure on the left as it goes through negative values.
By rule 3, the root locus exists on the entire positive extension of the real axis, between -1 and -2
and between -3 and -4. The σa intercept is found to be
σa =
(−1 − 2 − 4) − (−3)
4
=−
4−1
3
392
An example
The angles of the lines that intersect at -4/3 are given by
θa
=
=
2kπ
0(k = 0), 2π/3(k = 1), 4π/3(k = 2)
The final root locus sketch is shown in the figure above.
393
Comparison . . .
Table below shows various root-locus plots of negative- and positive-feedback systems. Heavy lines
and curves correspond to negative-feedback systems while dashed lines and curves correspond to
positive-feedback systems.
394
Suggested problems
Students are suggested to solve the following problems of the textbook:
E7.2, E7.3, E7.8, E7.10, E7.14, E7.15, E7.17, P7.6, P7.10, P7.38
Also, it is suggested to solve the following problems from the book [Control Systems
Engineering by N. Nise, Chapter 8]:
1, 11, 17, 19, 20, 21, 22, 24, 25, 29, 30
Students are encouraged to solve the assigned problems in hand before seeking help
from classmates or the teacher. Subsequently, the accompanying solutions can be
checked for confirmation.
395

Analysis of Control Systems by the Root

Transcription

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