systems analysis and controlcontrol.asu.edu/classes/mae318/318lecture08.pdf · 2019. 2. 3. ·...
TRANSCRIPT
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Systems Analysis and Control
Matthew M. PeetArizona State University
Lecture 8: Response Characteristics
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Overview
In this Lecture, you will learn:
The Effects of Feedback on Dynamic Response
• Changes in Transfer Function• Stability• Impact of Poles on Dynamic Response
Real Poles
• Steady-State Error• Rise Time• Settling Time
Complex Poles
• Complex Pole Locations• Damped/Natural Frequency• Damping and Damping Ratio
M. Peet Lecture 8: Control Systems 2 / 26
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The Effect of Feedback ControlRecall the Upper Feedback Interconnection
G(s)K(s)+
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y(s)u(s)
Feedback:
• Controller: û(s) = K(s)(û(s)− ŷ(s))• Plant: ŷ(s) = Ĝ(s)û(s)
The output signal is ŷ(s),
ŷ(s) =Ĝ(s)K̂(s)
1 + Ĝ(s)K̂(s)û(s)
M. Peet Lecture 8: Control Systems 3 / 26
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Controlling the Inverted Pendulum Model
Open Loop Transfer Function
Ĝ(s) =1
Js2 − Mgl2
Controller: Static Gain: K̂(s) = KInput: Impulse: û(s) = 1.
Closed Loop: Lower Feedback Interconnection
ŷ(s) =Ĝ(s)K̂(s)
1 + Ĝ(s)K̂(s)û(s) =
KJs2−Mgl2
1 + KJs2−Mgl2
=K
Js2 − Mgl2 +K
M. Peet Lecture 8: Control Systems 4 / 26
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Effect of Feedback on the Inverted Pendulum Model
Closed Loop Impulse Response:Lower Feedback Interconnection
ŷ(s) =KJ
s2 − Mgl2J +KJ
Traits:
• Infinite Oscillations• Oscillates about 0.
0 2 4 6 8 10 12−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Impulse Response
Time (sec)
Am
plitu
de
Open Loop Impulse Response:
ŷ(s)
=1
J
√2J
Mgl
1s−
√Mgl2J
− 1
s+√
Mgl2J
Unstable! 0 5 10 15
0
2
4
6
8
10
12
14
16
18x 10
5 Impulse Response
Time (sec)
Am
plitu
de
Figure: Impulse Response withg = l = J = 1, M = 2
M. Peet Lecture 8: Control Systems 5 / 26
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Controlling the Suspension System
Open Loop Transfer Function:Set mc = mw = g = c = K1 = K2 = 1.
Ĝ(s) =s2 + s+ 1
s4 + 2s3 + 3s2 + s+ 1
Controller: Static Gain: K̂(s) = k
x1
x2
mc
mw
u
Closed Loop: Lower Feedback Interconnection:
ŷ(s) =Ĝ(s)K̂(s)
1 + Ĝ(s)K̂(s)û(s)
=k(s2 + s+ 1)
s4 + 2s3 + (3+k)s2 + (1+k)s+ (1+k)
M. Peet Lecture 8: Control Systems 6 / 26
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Controlling the Suspension ProblemEffect of changing the Feedback, k
Closed Loop Step Response:
ŷ(s) =
k(s2 + s+ 1)
s4 + 2s3 + (3 + k)s2 + (1 + k)s+ (1 + k)
1
s
High k:
• Overshot the target• Quick Response• Closer to desired value of f
Low k:
• Slow Response• No overshoot• Final value is farther from 1.
Questions:
• Which Traits are important?• How to predict the behaviour? Figure: Step Response for different k
M. Peet Lecture 8: Control Systems 7 / 26
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Stability
The most basic property is Stability:
Definition 1.
A system, G is Stable if there exists a K > 0 such that
‖Gu‖L2 ≤ K‖u‖L2
Note: Although this is the true definition for systems defined by transferfunctions, it is rarely used.
• Bounded input means bounded output.• Stable means y(t)→ 0 when u(t)→ 0.
M. Peet Lecture 8: Control Systems 8 / 26
F4_PIO.mp4Media File (video/mp4)
f22_crash.mp4Media File (video/mp4)
boeing777_PIO_Roll.mp4Media File (video/mp4)
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Stability Depends on Pole Locations
Definition 2.
The Closed Right Half-Plane,CRHP is the set of complex numberswith non-negative real part.
{s ∈ C : Real(s) ≥ 0}
Im(s)
Re(s)
CRHP
Im(s)
Re(s)
Figure: Unstable
Theorem 3.
A system G is stable if and only if it’stransfer function Ĝ has no poles in theClosed Right Half Plane.
• Check stability by checking poles.• x is a pole• o is a zero
M. Peet Lecture 8: Control Systems 9 / 26
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Predicting Steady-State Error
Definition 4.
Steady-State Error for a stable system is the final difference between inputand output.
ess = limt→∞
u(t)− y(t)
5 10 15 20 25 30t
0.4
0.5
0.6
0.7
yHtL
Figure: Suspension Response for k = 1
• Usually measured using the stepresponse.
I Since u(t) = 1,ess = 1− limt→∞ y(t)
M. Peet Lecture 8: Control Systems 10 / 26
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Predicting Steady-State Value Using the Residue
5 10 15 20 25 30t
0.4
0.5
0.6
0.7
yHtL
Recall: For any system G, by partial fraction expansion:
ŷ(s) = Ĝ(s)1
s=r0s
+r1
s− p1+ . . .+
rns− pn
Soy(t) = r01(t) + r1e
p1t + . . .+ rnepnt
which meanslimt→∞
y(t) = r0
and henceess = 1− r0
M. Peet Lecture 8: Control Systems 11 / 26
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The Final Value Theorem
The steady-state error is given by r0.
ess = 1− r0Recall: The residue at s = 0 is r0 and is found as
r0 = Ĝ(s)|s=0 = lims→0
Ĝ(s)
Thus the steady-state error is
ess = 1− lims→0
Ĝ(s)
This can be generalized to find the limit of any signal:
Theorem 5 (Final Value Theorem).
limt→∞
y(t) = lims→0
sŷ(s)
• Assumes the limit exists (Stability)• Can be used to find response to other inputs
I Ramp, impulse, etc.
M. Peet Lecture 8: Control Systems 12 / 26
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The Final Value Theorem for Systems in Lower Feedback
Lower Feedback Interconnection:
ŷ(s) =G(s)K
1 +G(s)Kû(s)
Error Response for Lower Feedback Interconnection:
ê(s) = û(s)− ŷ(s)
=1 +G(s)K
1 +G(s)Kû(s)− G(s)K
1 +G(s)Kû(s) =
1 +G(s)K −G(s)K1 +G(s)K
û(s)
=1
1 +G(s)Kû(s)
So the steady-state error is
limt→∞
e(t) = lims→0
sê(s) = lims→0
s
1 +G(s)Kû(s)
Error in Step Response: If û(s) = 1s , then
limt→∞
e(t) = lims→0
1
1 +G(s)K
Conclusion: Increasing K always reduces steady-state error to step!M. Peet Lecture 8: Control Systems 13 / 26
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The Final Value Theorem for Systems in Upper Feedback
Upper Feedback Interconnection:
ŷ(s) =G(s)
1 +G(s)Kû(s)
Error Response for Upper Feedback Interconnection:
ê(s) = û(s)− ŷ(s)
=1 +G(s)K
1 +G(s)Kû(s)− G(s)
1 +G(s)Kû(s) =
1 +G(s)K −G(s)1 +G(s)K
û(s)
=1 +G(s)(K − 1)
1 +G(s)Kû(s)
Error in Step Response: If û(s) = 1s , then
limt→∞
e(t) = lims→0
1 +G(s)(K − 1)1 +G(s)K
Conclusion: Increasing K doesn’t help with step response!
M. Peet Lecture 8: Control Systems 14 / 26
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Steady-State ErrorNumerical Example
Ĝ(s) =k(s2 + s+ 1)
s4 + 2s3 + (3 + k)s2 + (1 + k)s+ (1 + k)
The steady-state response is
yss = lims→0
sŷ(s) = lims→0
Ĝ(s)
= lims→0
k(s2 + s+ 1)
s4 + 2s3 + (3 + k)s2 + (1 + k)s+ (1 + k)
=k
1 + k
The steady-state error is
ess = 1− yss = 1−k
1 + k
=1
1 + k• When k = 0, ess = 1• As k →∞, ess = 0
M. Peet Lecture 8: Control Systems 15 / 26
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Ramp Response for Systems in Lower Feedback
Lower Feedback Interconnection: Recall the steady-state error is
limt→∞
e(t) = lims→0
sê(s) = lims→0
s
1 +G(s)Kû(s)
Error in Ramp Response: If û(s) = 1s2 , then
limt→∞
e(t) = lims→0
1
s(1 +G(s)K)
Conclusion: limt→∞ e(t) =∞!
Preview of Integral Feedback: However, if we choose K → Ks
limt→∞
e(t) = lims→0
1
s(1 +G(s)Ks )= lims→0
1
s+G(s)K
Conclusion: Increasing K improves the ramp response!
• More on this Later....
M. Peet Lecture 8: Control Systems 16 / 26
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Dynamic Response CharacteristicsTwo Types of Response
From Partial Fractions Expansion, you know that motion is determined by thepoles of the Transfer Function!
n(s)
(s2 + as+ b)(s− p1) · · · (s− pn)û(s) =
k1s+ k2s2 + as+ b
û(s)+r1
s− p1û(s)+· · ·+ rn
s− pnû(s)
• Simplify the response by considering response of each pole.
Figure: Real Pole
5 10 15 20 25 30t
0.4
0.5
0.6
0.7
yHtL
Figure: Complex Pair of Poles
We start with Real Poles
M. Peet Lecture 8: Control Systems 17 / 26
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Step Response CharacteristicsReal Poles
The Solution: Step response of a real pole.
ŷ(s) =r
s− pû(s) =
r
s− p1
s=
rp
s− p−
rp
sy(t) =
r
p
(ept − 1
)• Is it stable? (p < 0?)
Cases: Stability or Instability?
• If p > 0, then limt→∞ y(t)→∞• If p < 0, then limt→∞ y(t)→ − rp
Final Value:
yss = −r
p
Question: How quickly do we reachthe final value?
M. Peet Lecture 8: Control Systems 18 / 26
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Step Response CharacteristicsRise Time
y(t) =r
p
(ept − 1
), yss = −
r
p
Definition 6.
The rise time, Tr, is the time it takes togo from .1 to .9 of the final value.
If y(t1) = .1yss = −.1 rp , then t1 is:
−.1 = ept1 − 1ln(1− .1) = pt1
t1 =ln .9
p=.11
−p
Likewise if y(t2) = −.9 rp , then
t2 =ln .1
p=
2.31
−p
Rise time (Tr) for a Single Pole is:
Tr = t2 − t1 =2.31
−p− .11−p
=2.2
−p
M. Peet Lecture 8: Control Systems 19 / 26
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Step Response CharacteristicsSettling Time
Definition 7.
The Settling Time, Ts, is the time it takes to reach and stay within .99 of thefinal value.
5 10 15 20 25 30t
0.4
0.5
0.6
0.7
yHtL
Figure: Complex Pair of Poles
LINK: Bouncing BallsLINK: More Bouncing Balls!LINK: Newton’s Cradle
M. Peet Lecture 8: Control Systems 20 / 26
http://www.youtube.com/watch?v=7mXh-LeQb3Ahttps://www.youtube.com/watch?v=pZlYl0l2lFshttp://www.youtube.com/watch?v=mFNe_pFZrsA
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Step Response CharacteristicsSettling Time
y(t) =r
p
(ept − 1
), yss = −
r
p
Definition 8.
The Settling Time, Ts, is the time ittakes to reach and stay within .99 ofthe final value.
Solve y(Ts) =rp
(epTs − 1
)= −.99 rp
for Ts:
−.99 = epTs − 1ln(.01) = pTs
Ts =ln .01
p= −4.6
p
The settling time for a Single Pole is:
Ts =4.6
−p
M. Peet Lecture 8: Control Systems 21 / 26
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Solution for Complex Poles
ŷ(s) =ω2d + σ
2
s2 + 2σs+ ω2d + σ2
1
s=
σ2 + ω2d(s+ σ)2 + ω2d
1
s= − s+ 2σ
(s+ σ)2 + ω2d+
1
s
The poles are at s = −σ ± ωdı. The solution is:
y(t) = 1− eσt(cos(ωdt) +
σ
ωdsin(ωdt)
)
The result is oscillation with anExponential Envelope.
• Envelope decays at rate σ• Speed of oscillation is ωd, the
Damped Frequency
M. Peet Lecture 8: Control Systems 22 / 26
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Step Response CharacteristicsDamping Ratio
Besides ωd, there is another way to measureoscillation
Definition 9.
The Natural Frequency of a pole atp = σ + ıωd is ωn =
√σ2 + ω2d.
• for ŷ(s) = 1s2+as+b1s , ωn =
√b.
• Radius of the pole in complex plane.
Resonant Frequency.
• Also known as resonant frequency
Im(s)
Re(s)
ωn
M. Peet Lecture 8: Control Systems 23 / 26
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Step Response CharacteristicsDamping Ratio
Besides σ, there are other ways tomeasure damping
Definition 10.
The Damping Ratio of a pole at
p = σ + ıω is ζ = |σ|ωn .
• for ŷ(s) = 1s2+as+b1s , ζ =
a2√b.
• Gives the ratio by which theamplitude decreases per oscillation(almost...).
M. Peet Lecture 8: Control Systems 24 / 26
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Damping
We use several adjectives to describeexponential decay:
• UndampedI Oscillation continues forever,σ = ζ = 0
• UnderdampedI Oscillation continues for many
cycles. ζ < 1
• Critically DampedI No oscillation or overshoot.ζ = 1, ωd = 0
• OverdampedI When ζ > 1, poles are real (not
complex)
M. Peet Lecture 8: Control Systems 25 / 26
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Summary
What have we learned today?
Characteristics of the Response
Real Poles
• Steady-State Error• Rise Time• Settling Time
Complex Poles
• Complex Pole Locations• Damped/Natural Frequency• Damping and Damping Ratio
Continued in Next Lecture
M. Peet Lecture 8: Control Systems 26 / 26
Control Systems