lecture abstract ee c128 / me c134 – feedback control...
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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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 2 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 3 / 61
4 Time response 4.2 Poles, zeros, & system response
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 4 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 5 / 61
4 Time response 4.2 Poles, zeros, & 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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 6 / 61
4 Time response 4.2 Poles, zeros, & 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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 7 / 61
4 Time response 4.2 Poles, zeros, & system 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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 8 / 61
4 Time response 4.3 First-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 9 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 10 / 61
4 Time response 4.3 First-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 11 / 61
4 Time response 4.3 First-order systems
Characteristics, [1, p. 166]
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.
Figure: 1st-order system response to aunit step
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 12 / 61
4 Time response 4.3 First-order systems
Characteristics, [1, p. 166]
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
Figure: 1st-order system response to aunit step
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 13 / 61
4 Time response 4.3 First-order systems
Characteristics, [1, p. 166]
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
Figure: 1st-order system response to aunit step
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 14 / 61
4 Time response 4.4 Second-order systems: intro
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 15 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 16 / 61
4 Time response 4.4 Second-order systems: intro
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 17 / 61
4 Time response 4.4 Second-order systems: intro
Underdamped response, [1, p. 169]
I 1 pole at origin from the unit step input
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 18 / 61
4 Time response 4.4 Second-order systems: intro
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 19 / 61
4 Time response 4.4 Second-order systems: intro
Undamped response, [1, p. 169]
I 1 pole at origin from the unit step input
I System poles: 2 imaginary at ±j!1
I Natural response: Undamped sinusoid
c(t) = A cos (!1t� �)
I Frequency: !1
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 20 / 61
4 Time response 4.4 Second-order systems: intro
Critically damped response, [1, p. 169]
I 1 pole at origin from the unit step input
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 21 / 61
4 Time response 4.4 Second-order systems: intro
Step response damping cases, [1, p. 172]
I Overdamped
I Underdamped
I Undamped
I Critically damped
Figure: Step responses for 2nd-order systemdamping cases
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 22 / 61
4 Time response 4.5 The general second-order system
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 23 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 24 / 61
4 Time response 4.5 The general second-order system
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 25 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 26 / 61
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)
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 27 / 61
4 Time response 4.6 Underdamped second-order systems
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)
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 28 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 29 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 30 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 31 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 32 / 61
4 Time response 4.6 Underdamped second-order systems
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. ⇣
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 33 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 34 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 35 / 61
4 Time response 4.6 Underdamped second-order systems
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 36 / 61
4 Time response 4.6 Underdamped second-order systems
Location of poles, [1, p. 182]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 37 / 61
4 Time response 4.6 Underdamped second-order systems
Location of poles, [1, p. 183]
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.
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 38 / 61
4 Time response 4.6 Underdamped second-order systems
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 ⇣
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 39 / 61
4 Time response 4.7 System response with additional poles
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 40 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 41 / 61
4 Time response 4.7 System response with additional poles
E↵ect on the 2nd-order system, [1, p. 187]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 42 / 61
4 Time response 4.7 System response with additional poles
E↵ect on the 2nd-order system, [1, p. 188]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 43 / 61
4 Time response 4.8 System response with zeros
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 44 / 61
4 Time response 4.8 System response with zeros
E↵ect on the 2nd-order system, [1, p. 191]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 45 / 61
4 Time response 4.8 System response with zeros
E↵ect on the 2nd-order system, [1, p. 191]
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.
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 46 / 61
4 Time response 4.8 System response with zeros
E↵ect on the 2nd-order system, [1, p. 191]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 47 / 61
4 Time response 4.8 System response with zeros
E↵ect on the 2nd-order system, [1, p. 191]
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 48 / 61
4 Time response 4.8 System response with zeros
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 49 / 61
4 Time response 4.9 E↵ects of nonlinearities upon time responses
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 50 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 51 / 61
4 Time response 4.9 E↵ects of nonlinearities upon time responses
Dead zone, [1, p. 197]
Figure: a. e↵ect of dead zone on load angular displacement response, b. Simulinkblock diagram
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 52 / 61
4 Time response 4.9 E↵ects of nonlinearities upon time responses
Backlash, [1, p. 198]
Figure: a. e↵ect of backlash on load angular displacement response, b. Simulinkblock diagram
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 53 / 61
4 Time response 4.10 Laplace transform solution of state equations
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 54 / 61
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
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
Y (s) = CX(s) +DU(s)
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 56 / 61
4 Time response 4.10 Laplace transform solution of state equations
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.
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 57 / 61
4 Time response 4.11 Time domain solution of state equations
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 58 / 61
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
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 59 / 61
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 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)
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 60 / 61
4 Time response 4.11 Time domain solution of state equations
Bibliography
Norman S. Nise. Control Systems Engineering, 2011.
Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 61 / 61