Chapter 5 Time Response - Prince of Songkla Universityfivedots.coe.psu.ac.th/.../240-209/Slides/controlch5.pdf · 2006. 11. 17. · Chapter 5 Time Response. 240-209: Time Response

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Chapter 5

Time Response


Outline

1 Poles, Zeros, and System Response2 First order systems3 Second order systems4 Time domain Solution of State equations


Poles of a Transfer Function

The values of the Laplace transform variable,s, that cause the transfer function to infinite.Any roots of the denominator of transfer function that are common to roots of the numerator.


Zeros of a Transfer Function

The values of the Laplace transform variable,s, that cause the transfer function to zero.Any roots of the numerator of transfer function that are common to roots of the denominator.


C s =s2s s5

Unit step response of systems


Pole–Zero plot of the system on s-plane


C s =s2s s5

=As

Bs5

A=[ s2s5]s 0

= 25

By using Partial Fraction Expansion

B=[ s2s ]

s 5= 3

5


c t = 25

35 e−5t

By Inverse Laplace transform

C s = 2 /5s 3/5

s5


System response and response component generated by the pole or zero.


Effect of a real-axis pole upon transient response


First-Order Systems


Pole plot of first order system


Unit-step response of First-Order System

C s =R s G s = as sa

c t =c f t c n t =1−e−at

By Inverse Laplace Transform


At t = 1/a , 2/a and 3/a

Unit-step response of First-Order System

c(t) = 1 – e-at

c(t) = 0.63, 0.86 and 0.95


Time response definition

Rise time is define as the time for the waveform to go from 0.1 to 0.9 of its final value.

Tr = 2.31/a – 0.11/a

Settling time is the time for the response to reach, and stay within 2% of its final value.

Ts = 4/a


First-order system response to a unit step


Laboratory results of a system unit step response test


Second-Order Systems


Overdamped Response


Poles plot of Overdamped System


Overdamped Response

C s = 9s s29s9

= 9s s7.854s1.146

c t =K 1K 2 e−7.854tK 3 e

−1.146t



Overdamped Response


Underdamped Response


Poles plot of Underdamped Response



C s = 9s s22s9

s=−1± j 8

C s =K 1

s K 2

s−1 j 8

K 3

s−1− j 8

c t =1−e−t cos8 t88 sin8 t

c t =1−1.06 e−t cos8 t−19.470





Second-order unit step response components generated by complex poles


Undamped Response


Poles plot of Undamped Response


Undamped Response

C s = 9s s29

s=± j3

c t =K 1K 4 cos 3t−


Undamped Response


Critically Damped Response





C s = 9s s26s9

s=−3

c t =K 1K 2e−3tK 3 te

−3t

K 1=1



Two real poles at same point

c t =K 1e−1 tK 2 t e

−2 t

c t =1−3te−3t−e−3t

Then




Natural responses of second order system

1. Overdamped response

c t =K 1e−p 1 tK 2e

−p 2 t

Poles : two real at - 1 ,

2

Natural response :


2. Underdamped response

c t =K 1e−d t cosd t−

Poles : two real at - d ± j

d

Natural response :


3. Undamped response

c t =A cos1 t−

Poles : two real at ± jn

Natural response :


4. Critically damped responses

c t =K 1e−p 1 tK 2 t e

−p 1 t

Poles : two real at - 1

Natural response :


Step responses for second-order system damping cases


The General Second-Order Systems

G s =n

2

s22n sn2

Transfer function

- Natural Frequency- Damping Ratio

n


Poles of the Transfer function

s1,2=−n±n 2−1


The General Second-Order Systems Second-order response as a function of damping ratio.

=0

Poles Response


Poles Response

01


Poles Response

=1


Poles Response

1


Step response for the general second-order system

C s =n

2

s s22n sn2

C s = 1s−

sn

1−2n1−2

sn2n

2 1−2


Inverse Laplace Transform

c t =1−e−n tcosn1−2 t

1−2sin n1−2 t

=1− 1

1−2e−n t cosn1−2 t−

=tan−1/1−2


Underdamped Second-Order Systems

Second-orderunderdampedresponses fordamping ratio values


Second-order underdamped response specifications


Natural FrequencyNatural Frequency of a second order system is the frequency of oscillation of the system without damping.

Damping ratioDamping ratio is the proportion between exponential decay frequency and natural frequency (rad/second)

= Exponentialdecay frequencyNatural frequency rad / second

= 12

natural period seconds Exponential time constant


Peak timePeak time is the time require to reach the first, or maximum, peak.

T P=

n1−2

Settling timeSettling time is the time it takes for the amplitude of the output c(t) to reach 0,98 or 98%

T s=−ln 0.021−2

n


Percent overshootPercent overshoot is a ratio between the different between system time response and final value.

OS=c max−c final

c final×100

OS=e−/1−2×100

%

%

= −ln OS /100

2ln2 OS /100

%

%


Percent overshoot vs. damping ratio


Time constantTime constant has the units (1 seconds), or frequency, thus we call the parameter a the exponential frequency.

c t =1−e−at

t= 1a

Time constantTime constant is the time require for step response to rise to 63% of its final value.


Delay timeDelay time is the time require for the system time response y(t) to reach 50% of final value.


Normalized risetime vs. damping ratio for a second-order underdamped response


Pole plot for an underdamped second-order system


Note: Ts2 < Ts

1 ; Tp

2 < Tp

1; %OS1 < %OS2

Lines of constant peak time,Tp , settlingtime,Ts , and percent overshoot, %OS


Constant real part

Step responses of second-order underdamped systems as poles move


Constant imaginary part


Constant damping ratio


System Response with Additional Poles

Component responses of a three-pole system : Case I : non-dominant pole is near dominant second-order poles pair. Case II : non-dominant pole is far from the pair. Case III : non-dominant pole is at infinity.



Example 4.8 (Nise)

T 1s =24.542

s24s24.542

T 2 s=24.542

s10s 24s24.542

T 3s =24.542

s3s24s24.542


c 1t =1−1.09 e−2t cos4.532t−23.8

c 2 t =1−0.29e−10t−1.189e−2t cos4.532t−53.34

c 3t =1−1.14 e−3t−0.707 e−2t cos4.532t78.63


Step responsesof system T1(s),system T2(s), andsystem T3(s)


T s =sa

sb sc =A

sbB

sc

T s =−ba/−bc

sb −ca /−cb

sc

If zero is far from poles, then a is large compared to b and c.

T s = asb sc

System Response with Zeros


Effect of addinga zero to a two-pole system

System Response with Zeros

p 1,2=−1± j2.282


Step response of a nonminimum-phase system


a. Effect of amplifiersaturation on loadangular velocityresponse;b. Simulink blockdiagram

Effects of Non-linearities Upon Time Response


a. Effect ofdeadzone onload angulardisplacementresponse;b. Simulink blockdiagram


a. Effect of backlashon load angulardisplacementresponse;b. Simulink blockdiagram


Time domain Solution of State equations

e at=1at 12! a

2 t 2... 1k ! a

k t k...

e A t=IA t 12! A

2 t 2... 1k ! A

k t k...

x.=ax x t =e at x 0

Scalar differential equation

x.=Ax x t =eAt x 0

Vector-matrix differential equation


x.=Ax t Bu t

e−A t [x.t −Ax t ]=e−A t Bu t

Rearrange and multiply both sides by e−A t

ddt [e

−A t x t ]=e−A t x.t −Ae−A t x t

Derivative of the product e−A t x t


Integrating both sides yields

[e−A t x t ]0t=e−A t x t −x 0=∫0

te−A Bu d

We obtain

x t =e A t x 0∫0

te A t−Bu d

t =e A t - state-transition matrix

=t x 0∫0

tt−Bu d


s I−A−1 is the Laplace transform of the state-transtion

matrix t

L−1[s I−A−1]=t

det s I−A=0 - characteristic equation

Documents

Chapter 5 Time Response - Prince of Songkla Universityfivedots.coe.psu.ac.th/.../240-209/Slides/controlch5.pdf · 2006. 11. 17. · Chapter 5 Time Response. 240-209: Time Response