systems analysis and controlcontrol.asu.edu/classes/mae318/318lecture08.pdf · 2019. 2. 3. ·...

Systems Analysis and Control

Matthew M. PeetArizona State University

Lecture 8: Response Characteristics

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

The Effect of Feedback ControlRecall the Upper Feedback Interconnection

G(s)K(s)+

-

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)


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


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


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)


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


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.


F4_PIO.mp4Media File (video/mp4)

f22_crash.mp4Media File (video/mp4)

boeing777_PIO_Roll.mp4Media File (video/mp4)

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


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)


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


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.


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

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!


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


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....


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


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?


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


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


http://www.youtube.com/watch?v=7mXh-LeQb3Ahttps://www.youtube.com/watch?v=pZlYl0l2lFshttp://www.youtube.com/watch?v=mFNe_pFZrsA

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


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


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


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...).


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)


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


Control Systems

systems analysis and controlcontrol.asu.edu/classes/mae318/318lecture08.pdf · 2019. 2. 3. ·...

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