Cleveland State University MCE441: Intr. Linear Control ...⊲ Second-Order Systems Time-Domain...

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Cleveland State UniversityMCE441: Intr. Linear Control Systems

Lecture 7:Time ResponsePole-Zero MapsInfluence of Poles and ZerosHigher Order Systems and Pole Dominance Criterion

Prof. Richter

Test Inputs

⊲ Test Inputs

Pole-Zero Maps

First-Order Systems

First-Order StepResponse

First-Order RampResponse

Second-OrderSystems

Time-DomainTransientSpecifications

Transient Specs:Step Response

Design Aids:2nd-Order Systems

Design Chart:2nd-Order Systems

Pole Locations andResponse

Effect of ZerosHigher-OrderSystems: PoleDominance

Effect of a Zero

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� Real inputs contain randomness (for example, noise):unpredictable.

� Test inputs help to predict response quality under otherinputs.

– step inputs are useful to simulate sudden changes(startup, jump loading, etc)

– sinusoidal inputs are used to test frequency response

(more later)– impulses are used to simulate shock conditions– ramp inputs are used to simulate transitions between

setpoints

Pole-Zero Maps

Test Inputs

⊲ Pole-Zero Maps

First-Order Systems



Second-OrderSystems







Effect of a Zero

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Re

Im

p1p2

p4

p5

z1

z2

z1 p4

p5

G(s) = (s−z1)(s−z1)(s−z2)(s−p5)(s−p5)(s−p2)2(s−p1)(s−p4)(s−p4)(s−p5)(s−p5)

First-Order Systems

Test Inputs

Pole-Zero Maps

⊲First-OrderSystems



Second-OrderSystems







Effect of a Zero

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First-order systems are described by the I/O differential equation

τ c+ c = r(t)

or in TF formC(s)

R(s)=

1

τs+ 1

The number τ is called the time constant of the system. The TFhas one real pole at s = −

1τ.

Step Response

Applying the unit step input R(s) = 1

sgives the response

c(t) = 1− e−tτ

First-Order Step Response

Test Inputs

Pole-Zero Maps

First-Order Systems

⊲First-Order StepResponse


Second-OrderSystems







Effect of a Zero

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(Stable) first-order systems always track the step command (the error approaches zero asymptotically).The time it takes for the output to reach approx. 98% of its steady value is called settling time, also referred toas 2 % settling time. It equals 4 time constants.

First-Order Ramp Response

Test Inputs

Pole-Zero Maps

First-Order Systems


⊲First-Order RampResponse

Second-OrderSystems







Effect of a Zero

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First-order systems cannot track a ramp command. The tracking error is always equal to the time constant.

Second-Order Systems

Test Inputs

Pole-Zero Maps

First-Order Systems



⊲Second-OrderSystems







Effect of a Zero

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The standard form of a second-order system is:

C(s)

R(s)=

w2n

s2 + 2ζwns+ w2n

where wn and ζ are called the natural frequency (in rad/s) anddamping ratio (dimensionless), respectively. The response willdepend on the nature of the poles:

� Underdamped: 0 < ζ < 1� Marginally Stable: ζ = 0� Critically Damped: ζ = 1� Overdamped: ζ > 1

Case 1: Underdamped (0 < ζ < 1)

Test Inputs

Pole-Zero Maps

First-Order Systems










Effect of a Zero

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The poles are complex conjugates. The response to a unit step input isgiven by

c(t) = 1−e−ζwnt

√

1− ζ2sin

(

wdt+ tan−1

√

1− ζ2

ζ

)

where wd is the damped natural frequency:

wd = wn

√

1− ζ2

The damped natural frequency is the one obtained by counting thecycles per unit time in an experimental (and underdamped) responseand converting the result to radians per second.

Underdamped Case

Test Inputs

Pole-Zero Maps

First-Order Systems










Effect of a Zero

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wn

wd = wn

√

1 − ζ2

Re

Im

−ζwn

β

0

−wd

cos β = ζ

Poles: −ζwn ± wdj

Marginally Stable and Overdamped Cases

Test Inputs

Pole-Zero Maps

First-Order Systems










Effect of a Zero

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Marginally Stable Case (ζ = 0)When ζ = 0 the poles lie on the imaginary axis and the response to astep input is

c(t) = 1− coswnt

that is, the oscillation does not die out. The system oscillates with itsnatural frequency wn.

Critically Damped Case (ζ = 1)When ζ = 1 we get a real pole of multiplicity two at s = −wn. Theresponse is given by c(t) = 1− e−wnt(1 + wnt). No oscillations occurfor a step input.

Overdamped Case (ζ > 1)

Test Inputs

Pole-Zero Maps

First-Order Systems










Effect of a Zero

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When ζ > 1 the poles lie are real and unequal. The responsefeatures two decaying exponential terms (no oscillations):

c(t) = 1 +wn

2√

ζ2 − 1

(

e−s1t

s1−

e−s2t

s2

)

where the poles are

s1,2 = (ζ ±√

ζ2 − 1)wn

Normalized Second-Order Response

Test Inputs

Pole-Zero Maps

First-Order Systems










Effect of a Zero

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Time-Domain Transient Specifications

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems

⊲







Effect of a Zero

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� Many practical design cases call for time-domainspecifications

� Step response specs are often demanded (worst-case, drasticsituation)

� Common specifications

1. Rise time (Tr)2. Peak time (Tp)3. Percent overshoot (P.O.)4. Settling time (Ts)

Transient Specs: Step Response

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems


⊲Transient Specs:Step Response





Effect of a Zero

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0 Tr Tp Tset 2%0

Setpoint

FV

Second−Order Step Response Characteristics

Time (sec)

Setpoint

0.98*F.V

FinalValue

Overshoot

1.02*F.V

SteadyError

Design Aids: 2nd-Order Systems

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems



⊲

Design Aids:2nd-OrderSystems




Effect of a Zero

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� 2 % settling time: ts =4

ζwn

� Rise time: tr =π−βwd

� Peak time: tp =πwd

� Peak value: Mpt = (e−( ζ

√

1−ζ2)π

+ 1)y(∞)

� Percent overshoot: P.O.= 100e−( ζ

√

1−ζ2)π

Design Chart: 2nd-Order Systems

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems




⊲

Design Chart:2nd-OrderSystems



Effect of a Zero

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Pole Locations and Response

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems





⊲Pole Locationsand Response


Effect of a Zero

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Effect of Zeros

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems






⊲ Effect of ZerosHigher-OrderSystems: PoleDominance

Effect of a Zero

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� The above design formulas are valid when no zeros arepresent.

� In general, zeros spoil the response by increasing theovershoot.

� If zeros are found on the r.h.p., the system is said to benonminimum phase, and the response is worsened.

� Zeros may be neglected in some cases (more about this laterin this lecture)

Higher-Order Systems: Pole Dominance

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems






Effect of Zeros

⊲

Higher-OrderSystems: PoleDominance

Effect of a Zero

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� When more than two poles are present, or when two distinctreal poles are present, it can be difficult to predict theresponse without computer simulation

� Frequently, one or two poles are dominant. Remember thateach single real pole gives rise to a term of the form a

s(s+a)when a step input is applied. If we invert the term, thecontribution of the pole is 1− e−at. So if the time constantof the pole (1/a) is very small, the transient is very short anddoes not contribute much to the overall response. On theother hand, poles with large time constants are the onesdictating the response. We say that such poles are dominant.The same applies to complex conjugate pairs, where the timeconstant taken into consideration is 1

ζwn.

Pole Dominance Criterion

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems






Effect of Zeros

⊲


Effect of a Zero

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� Begin by finding all poles of the TF and finding their timeconstants. Remember: For a real pole at s = −a, the timeconstant is 1/a, and for a complex conjugate pair with0 < ζ < 1, it is 1

ζwn.

� Sort out the time constants: (high) τ1, τ2..., τn (low).� Attempt to find a gap of a factor of eight or more between

two consecutive time constants, starting with the highest. Ifsuch a gap is found, neglect all poles on the fast side of thegap (small time constants).

� Careful! when reassembling the TF. You may beinadvertently changing the TF gain (a.k.a. DC gain): Makesure that the responses to a step input have the same finalvalue. If not, correct the reduced TF with an appropriateconstant.

Pole Dominance: Example

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems






Effect of Zeros

⊲


Effect of a Zero

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G(s) = 3s4+6.4s3+14.4s2+27.2s+32

Poles: s1 = −2, s2 = −4, s3,4 = −0.2 ± 1.989i(ζ = 0.1, wn = 2). Time constants for real poles:τ1 = 0.5 and τ2 = 0.25. For the complex poles : τ = 5.The factored TF is: G(s) = 3

(s+2)(s+4)(s2+0.4s+4).

Time constants: 5, 0.5, 0.25. A gap of a factor of 10 is found, therefore we neglect the fast poles ats2 = −4 and s1 = −2.If we simply eliminate the factors (s + 2) and (s + 4), we have a gain mismatch. In order to have the samefinal value, we correct the gain to obtain:

Gr(s) =3/8

s2 + 0.4s + 4

Check with the final value theorem: lims→0 sGr(s) = lims→0 sG(s).

Effect of a Zero

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems







⊲ Effect of a Zero

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For second-order systems, we can predict the effect of a single real zero on the overshoot. Consider

G(s) =(w2

n/a)(s + a)

s2 + 2ζwns + w2n

If a/ζwn > 8, neglect the zero due to dominance: Gr(s) =w2

ns2+2ζwns+w2

n. If not, use the chart:

Example

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems








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Sketch the response of the following TF when the input is a stepof size 2:

G(s) =3(s+ 5)

(s+ 23)(s2 + 5s+ 16)

Solution

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems








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Solution

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems








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Matlab Simulation

Test Inputs

Pole-Zero Maps

First-Order Systems



Second-OrderSystems








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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

c(t)

t

c(tp)=0.0919

ts=1.3 s

c(∞)=0.0815 Overshoot=12.76%

Cleveland State University MCE441: Intr. Linear Control ...⊲ Second-Order Systems Time-Domain...

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Transcript of Cleveland State University MCE441: Intr. Linear Control ...⊲ Second-Order Systems Time-Domain...