Lecture abstract EE C128 / ME C134 – Feedback Control Systemsee128/fa14/Lectures/EEC128-chap4.pdf · EE C128 / ME C134 – Feedback Control Systems Lecture – Chapter 4 – Time

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

Alexandre Bayen

Department of Electrical Engineering & Computer Science

University 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 E↵ect of additional poles

I E↵ect of zeros

I E↵ect 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 E↵ects 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: E↵ect 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�↵t,where �↵ 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: E↵ect of a real-axis pole upontransient response


4 Time response 4.3 First-order systems




Intro, [1, p. 166]

I1

st-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�at

Figure: 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. Thedi↵erence in time betweenc(t) = 0.9 and c(t) = 0.1.

Tr =2.2

a





I2% 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 �1, �2I Natural response: Summation of 2

exponentials

c(t) = K1e��1t

+K2e��2t

I Time constants: ��1, ��2



Underdamped response, [1, p. 169]


I System poles: 2 complex at �d ± j!d

I Natural response: Damped sinusoid withan exponential envelope

c(t) = K1e��dtcos(!dt� �)

I Time constant: �dI Frequency (rad/s): !d



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, !d: 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 ±j!1

I Natural response: Undamped sinusoid

c(t) = A cos (!1t� �)

I Frequency: !1



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��1t

+K2te��1t

I Time constant: �1I 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, !n

I The frequency of oscillation of the system without damping

I Damping ratio, ⇣

⇣ =

Exponential decay frequency

Natural frequency (rad/s)=

1

2⇡

Natural period (s)

Exponential time constant

I General TF

G(s) =b

s2 + as+ b=

!2n

s2 + 2⇣!ns+ !2n

wherea = 2⇣!n, b = !2

n, ⇣ =

a

2!n, !n =

pb



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

I Poles

s1,2 = �⇣!n ± !n

p

⇣2 � 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) =!2n

s(s2 + 2⇣!ns+ !2n)

...partial fraction expansion...

=

1

s+

(s+ ⇣!n) +⇣p1�⇣2

!n

p

1� ⇣2

(s+ ⇣!n)2+ !2

n(1� ⇣2)



Step response, [1, p. 177]

Time domain via inverse Laplace transform

c(t) = 1� e⇣!nt⇣

cos (!n

p

1� ⇣2)t+ ⇣p1�⇣2

sin (!n

p

1� ⇣2)t⌘

...trigonometry & exponential relations...

= 1� 1

p

1� ⇣2e�⇣!nt

cos(!n

p

1� ⇣2 � �)

where

� = tan

�1(

⇣p

1� ⇣2)



Responses for ⇣ values, [1, p. 178]

Response versus ⇣ plotted along atime axis normalized to !n

I Lower ⇣ produce a moreoscillatory response

I !n does not a↵ect 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 di↵erentiating 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) =!2n

s2 + 2⇣!ns+ !2n

...completing the squares in the denominator

...setting the derivative to zero

Tp =⇡

!n

p

1� ⇣2



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�⇣⇡p1�⇣2 ⇥ 100

⇣ given %OS

⇣ =

� ln(

%OS100 )

q

⇡2+ ln

2(

%OS100 )

Figure: %OS vs. ⇣



Evaluation of Ts, [1, p. 179]

Find the time for which c(t) reaches and stays within ±2% of the steadystate value, c

final

, i.e., the time it takes for the amplitude of the decayingsinusoid to reach 0.02

e�⇣!nt 1

p

1� ⇣2= 0.02

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

cos(!n

p

1� ⇣2t� �) = 1

Settling time

Ts =� ln(0.02

p

1� ⇣2)

⇣!n

Approximated by

Ts =4

⇣!n



Evaluation of Tr, [1, p. 181]

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

1. Designate !nt as thenormalized time variable

2. Select a value for ⇣

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

4. The normalized rise time !nTr

is the di↵erence between thosetwo values of !nt for thatvalue of ⇣

Figure: Normalized Tr vs. ⇣ for a2

nd-order underdamped response



Location of poles, [1, p. 182]

I Natural frequency, !n: Radialdistance from the origin to thepole

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

cos(✓) =�⇣!n

!n= ⇣

Figure: Pole plot for an underdamped2

nd-order system




I Damped frequency ofoscillation, !d: Imaginary partof the system poles

!d = !n

p

1� ⇣2

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

�d = ⇣!n

I Poles

s1,2 = ��d ± j!d

Figure: Pole plot for an underdamped2

nd-order system




I Tp / horizontal lines

Tp =⇡

!n

p

1� ⇣2=

⇡

!d

I Ts / vertical lines

Ts =4

⇣!n=

4

�d

I%OS / radial lines

%OS = e�⇣⇡p1�⇣2 ⇥ 100

⇣ = cos(✓)

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 =⇡

!n

p

1� ⇣2=

⇡

!d

I Ts / vertical lines

Ts =4

⇣!n=

4

�d

I%OS / radial lines

%OS = e�⇣⇡p1�⇣2 ⇥ 100

⇣ = cos(✓)

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


4 Time response 4.7 System response with additional poles




E↵ect 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 = ⇣!n ± j!n

p

1� ⇣2, s3 = �↵r




I Step response of the system in the frequency domain

C(s) =A

s+

B(s+ ⇣!n) + C!d

(s+ ⇣!n)2+ !2

d

+

D

s+ ↵r

I Step response of the system in the time domain

c(t) = Au(t) + e�⇣!nt(B cos(!dt) + C sin(!dt)) +De�↵rt




3 cases for the real pole, ↵r

I ↵r is not much greater than⇣!n

I ↵r � ⇣!n

I Assuming exponential decayis negligible after 5 timeconstants

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

I ↵r = 1

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 1


4 Time response 4.8 System response with zeros





I E↵ects 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: E↵ect 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) ⇡ an

1/(�b+c)s+b +

1/(�c+b)s+c

o

=

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 E↵ects of nonlinearities upon time responses




Saturation, [1, p. 196]

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



Dead zone, [1, p. 197]

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



Backlash, [1, p. 198]

Figure: a. e↵ect 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)⇤



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

h

adj(sI�A)det(sI�A)

i

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) +

Z t

0eA(t�⌧)Bu(⌧)d⌧

I Zero-input responseeAtx(0)

I Zero-state response / convolution integral

Z t

0eA(t�⌧)Bu(⌧)d⌧

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

h

adj (sI�A)det (sI�A)

i

= �(t)



Bibliography

Norman S. Nise. Control Systems Engineering, 2011.


Documents

Lecture abstract EE C128 / ME C134 – Feedback Control Systemsee128/fa14/Lectures/EEC128-chap4.pdf · EE C128 / ME C134 – Feedback Control Systems Lecture – Chapter 4 – Time