EE C128 / ME C134 { Feedback Control Systems · Lecture abstract Topics covered in this presentation I Poles & zeros I First-order systems I Second-order systems I E ect of additional

EE C128 / ME C134 – Feedback Control SystemsLecture – Chapter 4 – Time Response

Alexandre Bayen

Department of Electrical Engineering & Computer ScienceUniversity of California Berkeley

September 10, 2013

Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 1 / 61

Lecture abstract

Topics covered in this presentation

I Poles & zeros

I First-order systems

I Second-order systems

I Effect of additional poles

I Effect of zeros

I Effect of nonlinearities

I Laplace transform solution of state equations


Chapter outline

1 4 Time response4.1 Introduction4.2 Poles, zeros, and system response4.3 First-order systems4.4 Second-order systems: introduction4.5 The general second-order system4.6 Underdamped second-order systems4.7 System response with additional poles4.8 System response with zeros4.9 Effects of nonlinearities upon time responses4.10 Laplace transform solution of state equations4.11 Time domain solution of state equations


4 Time response 4.2 Poles, zeros, & system response




Definitions, [1, p. 163]

Poles of a TF

I Values of the Laplace transformvariable, s, that cause the TFto become infinite

I Any roots of the denominatorof the TF that are common tothe roots of the numerator

Figure: a. system showing input &output, b. pole-zero plot of the system;c. evolution of a system response



Definitions, [1, p. 163]

Zeros of a TF

I Values of the Laplacetransform variable, s, thatcause the TF to become zero

I Any roots of the numerator ofthe TF that are common to theroots of the denominator

Figure: a. system showing input &output, b. pole-zero plot of the system;c. evolution of a system response



System response characteristics, [1, p. 163]

I Poles of a TF: Generate theform of the natural response

I Poles of a input function:Generate the form of the forcedresponse

Figure: Effect of a real-axis pole upontransient response



System response characteristics, [1, p. 163]

I Pole on the real axis:Generates an exponentialresponse of the form e−αt,where −α is the pole locationon the real axis. The farther tothe left a pole is on thenegative real axis, the fasterthe exponential transientresponse will decay to zero.

I Zeros and poles: Generate theamplitudes for both the forcedand natural responses

Figure: Effect of a real-axis pole upontransient response


4 Time response 4.3 First-order systems




Intro, [1, p. 166]

I 1st-order system without zeros TF

G(s) =C(s)

R(s)=

a

s+ a

I Unit step input TF

R(s) = s−1

I System response in frequency domain

C(s) = R(s)G(s) =a

s(s+ a)

I System response in time domain

c(t) = cf (t) + cn(t) = 1− e−atFigure: 1st-order system;pole-plot



Characteristics, [1, p. 166]

I Time constant, 1a : The time

for e−at to decay to 37% of itsinitial value. Alternatively, thetime it takes for the stepresponse to rise to 63% of itsfinal value.

Figure: 1st-order system response to aunit step




I Exponential frequency, a: Thereciprocal of the time constant.The initial rate of change ofthe exponential at t = 0, sincethe derivative of e−at is −awhen t = 0. Since the pole ofthe TF is at −a, the fartherthe pole is from the imaginaryaxis, the faster the transientresponse.





I Rise time, Tr: The time for thewaveform to go from 0.1 to 0.9of its final value. Thedifference in time betweenc(t) = 0.9 and c(t) = 0.1.

Tr =2.2

a





I 2% Settling time, Ts: The timefor the response to reach, andstay within, 2% (arbitrary) ofits final value. The time whenc(t) = 0.98.

Ts =4

a



4 Time response 4.4 Second-order systems: intro




General form, [1, p. 168]

I 2 finite poles: Complex pole pairdetermined by the parameters a and b

I No zeros Figure: General 2nd-ordersystem



Overdamped response, [1, p. 169]

I 1 pole at origin from the unit step input

I System poles: 2 real at σ1, σ2I Natural response: Summation of 2

exponentials

c(t) = K1e−σ1t +K2e

−σ2t

I Time constants: −σ1, −σ2



Underdamped response, [1, p. 169]


I System poles: 2 complex at σd ± jωdI Natural response: Damped sinusoid with

an exponential envelope

c(t) = K1e−σdtcos(ωdt− φ)

I Time constant: σdI Frequency (rad/s): ωd



Underdamped response characteristics, [1, p. 170]

I Transient response: Exponentially decayingamplitude generated by the real part of thesystem pole times a sinusoidal waveformgenerated by the imaginary part of thesystem pole.

I Damped frequency of oscillation, ωd: Theimaginary part part of the system poles.

I Steady state response: Generated by theinput pole located at the origin.

I Underdamped response: Approaches asteady state value via a transient responsethat is a damped oscillation.

Figure: 2nd-order stepresponse componentsgenerated by complexpoles



Undamped response, [1, p. 169]


I System poles: 2 imaginary at ±jω1

I Natural response: Undamped sinusoid

c(t) = A cos (ω1t− φ)

I Frequency: ω1



Critically damped response, [1, p. 169]


I System poles: 2 multiple real

I Natural response: Summation of anexponential and a product of time and anexponential

c(t) = K1e−σ1t +K2te

−σ1t

I Time constant: σ1I Note: Fastest response without overshoot



Step response damping cases, [1, p. 172]

I Overdamped

I Underdamped

I Undamped

I Critically damped

Figure: Step responses for 2nd-order systemdamping cases


4 Time response 4.5 The general second-order system




Specification, [1, p. 173]

I Natural frequency, ωnI The frequency of oscillation of the system without damping

I Damping ratio, ζ

ζ =Exponential decay frequency

Natural frequency (rad/s)=

1

2π

Natural period (s)

Exponential time constant

I General TF

G(s) =b

s2 + as+ b=

ω2n

s2 + 2ζωns+ ω2n

wherea = 2ζωn, b = ω2

n, ζ =a

2ωn, ωn =

√b



Response as a function of ζ, [1, p. 175]

I Poles

s1,2 = −ζωn ± ωn√ζ2 − 1

Table: 2nd-order response as a functionof damping ratio


4 Time response 4.6 Underdamped second-order systems




Step response, [1, p. 177]

Transfer function

C(s) =ω2n

s(s2 + 2ζωns+ ω2n)

...partial fraction expansion...

=1

s+

(s+ ζωn) + ζ√1−ζ2

ωn√

1− ζ2

(s+ ζωn)2 + ω2n(1− ζ2)



Step response, [1, p. 177]

Time domain via inverse Laplace transform

c(t) = 1− eζωnt(

cos (ωn√

1− ζ2)t+ ζ√1−ζ2

sin (ωn√

1− ζ2)t)

...trigonometry & exponential relations...

= 1− 1√1− ζ2

e−ζωnt cos(ωn√

1− ζ2 − φ)

where

φ = tan−1(ζ√

1− ζ2)



Responses for ζ values, [1, p. 178]

Response versus ζ plotted along atime axis normalized to ωn

I Lower ζ produce a moreoscillatory response

I ωn does not affect the natureof the response other thanscaling it in time

Figure: 2nd-order underdampedresponses for damping ratio values



Response specifications, [1, p. 178]

I Rise time, Tr: Time requiredfor the waveform to go from0.1 of the final value to 0.9 ofthe final value

I Peak time, Tp: Time requiredto reach the first, or maximum,peak

Figure: 2nd-order underdampedresponse specifications



Response specifications, [1, p. 178]

I Overshoot, %OS: The amountthat the waveform overshootsthe steady state, or final, valueat the peak time, expressed asa percentage of the steadystate value

I Settling time, Ts: Timerequired for the transient’sdamped oscillations to reachand stay within ±2% of thesteady state value

Figure: 2nd-order underdampedresponse specifications



Evaluation of Tp, [1, p. 179]

Tp is found by differentiating c(t) and finding the zero crossing after t = 0,which is simplified by applying a derivative in the frequency domain andassuming zero initial conditions.

L[c(t)] = sC(s) =ω2n

s2 + 2ζωns+ ω2n

...completing the squares in the denominator

...setting the derivative to zero

Tp =π

ωn√

1− ζ2



Evaluation of %OS, [1, p. 180]

%OS is found by evaluating

%OS =cmax − cfinal

cfinal× 100

where

cmax = c(Tp), cfinal = 1

...substitution

%OS = e−ζπ√1−ζ2 × 100

ζ given %OS

ζ =− ln(%OS100 )√π2 + ln2(%OS100 )

Figure: %OS vs. ζ



Evaluation of Ts, [1, p. 179]

Find the time for which c(t) reaches and stays within ±2% of the steadystate value, cfinal, i.e., the time it takes for the amplitude of the decayingsinusoid to reach 0.02

e−ζωnt1√

1− ζ2= 0.02

This equation is a conservative estimate, since we are assuming that

cos(ωn√

1− ζ2t− φ) = 1

Settling time

Ts =− ln(0.02

√1− ζ2)

ζωn

Approximated by

Ts =4

ζωn



Evaluation of Tr, [1, p. 181]

A precise analytical relationshipbetween Tr and ζ cannot be found.However, using a computer, Tr canbe found

1. Designate ωnt as thenormalized time variable

2. Select a value for ζ

3. Solve for the values of ωnt thatyield c(t) = 0.9 and c(t) = 0.1

4. The normalized rise time ωnTris the difference between thosetwo values of ωnt for thatvalue of ζ

Figure: Normalized Tr vs. ζ for a2nd-order underdamped response



Location of poles, [1, p. 182]

I Natural frequency, ωn: Radialdistance from the origin to thepole

I Damping ratio, ζ: Ratio of themagnitude of the real part ofthe system poles over thenatural frequency

cos(θ) =−ζωnωn

= ζFigure: Pole plot for an underdamped2nd-order system




I Damped frequency ofoscillation, ωd: Imaginary partof the system poles

ωd = ωn√

1− ζ2

I Exponential dampingfrequency, σd: Magnitude ofthe real part of the systempoles

σd = ζωn

I Poles

s1,2 = −σd ± jωd

Figure: Pole plot for an underdamped2nd-order system




I Tp ∝ horizontal lines

Tp =π

ωn√

1− ζ2=

π

ωd

I Ts ∝ vertical lines

Ts =4

ζωn=

4

σd

I %OS ∝ radial lines

%OS = e−ζπ√1−ζ2 × 100

ζ = cos(θ)

Figure: Lines of constant Tp, Ts, and%OS. Note: Ts2 < Ts1 , Tp2

< Tp1,

%OS1 < %OS2.



Underdamped systems, [1, p. 184]

I Tp ∝ horizontal lines

Tp =π

ωn√

1− ζ2=

π

ωd

I Ts ∝ vertical lines

Ts =4

ζωn=

4

σd

I %OS ∝ radial lines

%OS = e−ζπ√1−ζ2 × 100

ζ = cos(θ)

Figure: Step responses of 2nd-ordersystems as poles move: a. withconstant real part, b. with constantimaginary part, c. with constant ζ


4 Time response 4.7 System response with additional poles




Effect on the 2nd-order system, [1, p. 187]

I Dominant poles: The two complex poles that are used to approximatea system with more than two poles as a second-order system

I Conditions: Three pole system with complex poles and a third poleon the real axis

s1,2 = ζωn ± jωn√

1− ζ2, s3 = −αr




I Step response of the system in the frequency domain

C(s) =A

s+B(s+ ζωn) + Cωd

(s+ ζωn)2 + ω2d

+D

s+ αr

I Step response of the system in the time domain

c(t) = Au(t) + e−ζωnt(B cos(ωdt) + C sin(ωdt)) +De−αrt




3 cases for the real pole, αrI αr is not much greater thanζωn

I αr � ζωnI Assuming exponential decay

is negligible after 5 timeconstants

I The real pole is 5× fartherto the left than thedominant poles

I αr =∞

Figure: Component responses of a3-pole system: a. pole plot, b.component responses: non-dominantpole is near dominant 2nd-order pair,far from the pair, and at ∞


4 Time response 4.8 System response with zeros





I Effects on the system responseI Residue, or amplitudeI Not the nature, e.g.,

exponential, dampedsinusoid, etc.

I Greater as the zeroapproaches the dominantpoles

I Conditions: Real axis zeroadded to a two-pole system

Figure: Effect of adding a zero to a2-pole system




Assume a group of poles and a zero far from the poles....partial-fraction expansion...

T (s) =s+ a

(s+ b)(s+ c)

=A

s+ b+

B

s+ c

=(−b+ a)/(−b+ c)

s+ b+

(−c+ a)/(−c+ b)

s+ c

If the zero is far from the poles, then a� b and a� c, and

T (s) ≈ a{

1/(−b+c)s+b + 1/(−c+b)

s+c

}=

a

(s+ b)(s+ c)

Zero looks like a simple gain factor and does not change the relativeamplitudes of the components of the response.




Another view...

I Response of the system, C(s)

I System TF, T (s)

I Add a zero to the system TF, yielding, (s+ a)T (s)

I Laplace transform of the response of the system

(s+ a)C(s) = sC(s) + aC(s)

I Response of the system consists of 2 partsI The derivative of the original responseI A scaled version of the original response




3 cases for aI a is very large

I Response → aC(s), a scaled version of the original response

I a is not very largeI Response has additional derivative component producing more

overshoot

I a is negative – right-half plane zeroI Response has additional derivative component with an opposite sign

from the scaled response term



Non-minimum-phase system, [1, p. 192]

Non-minimum-phase system:System that is causal and stablewhose inverses are causal andunstable.

I Characteristics: If thederivative term, sC(s), islarger than the scaled response,aC(s), the response willinitially follow the derivative inthe opposite direction from thescaled response.

Figure: Step response of anon-minimum-phase system


4 Time response 4.9 Effects of nonlinearities upon time responses




Saturation, [1, p. 196]

Figure: a. effect of amplifier saturation on load angular velocity response, b.Simulink block diagram



Dead zone, [1, p. 197]

Figure: a. effect of dead zone on load angular displacement response, b. Simulinkblock diagram



Backlash, [1, p. 198]

Figure: a. effect of backlash on load angular displacement response, b. Simulinkblock diagram


4 Time response 4.10 Laplace transform solution of state equations




Laplace transform solution of state equations, [1, p. 199]

State equationx = Ax+Bu

Output equationy = Cx+Du

Laplace transform of the state equation

zX(s)− x(0) = AX(s) +BU(s)

...combining all the X(s) terms

(sI −A)X(s) = x(0) +BU(s)

...solving for X(s)

X(s) = (sI −A)−1x(0) + (sI −A)−1BU(s)

=adj(sI −A)

det(sI −A)

[x(0) +BU(s)

]Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 55 / 61


Laplace transform solution of state equations, [1, p. 199]

State equationx = Ax+Bu

Output equationy = Cx+Du

Laplace transform of the state equation

Y (s) = CX(s) +DU(s)



Eigenvalues & TF poles, [1, p. 200]

Eigenvalues of the system matrix A are found by evaluating

det(sI −A) = 0

The eigenvalues are equal to the poles of the system TF....for simplicity, let the output, Y (s), and the input, U(s), be scalarquantities, and further, to conform to the definition of a TF, let x(0) = 0

Y (s)

U(s)= C

[adj(sI−A)det(sI−A)

]B +D

=C adj(sI −A)B +D det(sI −A)

det(sI −A)

The roots of the denominator are the poles of the system.


4 Time response 4.11 Time domain solution of state equations




Time domain solution of state equations, [1, p. 203]

Linear time-invariant (LTI) system

I Time domain solution

x(t) = eAtx(0) +

∫ t

0eA(t−τ)Bu(τ)dτ

I Zero-input responseeAtx(0)

I Zero-state response / convolution integral∫ t

0eA(t−τ)Bu(τ)dτ

I State-transition matrixeAt



Time domain solution of state equations, [1, p. 203]

Linear time-invariant (LTI) system

I Frequency domain unforced response

L[x(t)] = L[eAtx(0)] = (sI −A)−1x(0)

I Laplace transform of the state-transition matrix

eAtx(0)

I Characteristic equation

det(sI −A) = 0

I ??? equation

L−1[(sI −A)−1] = L−1[adj (sI−A)det (sI−A)

]= Φ(t)



Bibliography

Norman S. Nise. Control Systems Engineering, 2011.


EE C128 / ME C134 { Feedback Control Systems · Lecture abstract Topics covered in this presentation I Poles & zeros I First-order systems I Second-order systems I E ect of additional

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