Course Outline
Week
1
2
3
4
5
6
7
8
Amme 3500 :
System Dynamics and Control
System Response
9
10
11
12
13
14
Dr. Stefan B. Williams
Date
1 Mar
8 Mar
15 Mar
22 Mar
29 Mar
5 Apr
12 Apr
19 Apr
26 Apr
3 May
10 May
17 May
24 May
31 May
Dr. Stefan B. Williams
Engineering as Design
Amme 3500 : System Response
Assignment Notes
Assign 1 Due
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No Tutorials
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Amme 3500 : Introduction
Slide 2
Problem Identification
•  Engineering Design is the systematic,
intelligent generation and evaluation of
specifications for products whose form and
function achieve stated objectives and
satisfy specified constraints
•  The purpose of design is to derive from a
set of specifications a description of a
product sufficient for its realization
Dr. Stefan B. Williams
Content
Introduction
Frequency Domain Modelling
Transient Performance and the s-plane
Block Diagrams
Feedback System Characteristics
Root Locus
Root Locus 2
Bode Plots
BREAK
Bode Plots 2
State Space Modeling
State Space Design Techniques
Advanced Control Topics
Review
Spare
Slide 3
•  Design constraints must be identified
–  Restrictions or limitations on a behaviour or a value or
some other aspect of a designed object s
performance
•  Requirement analysis
–  allows us to verify that we have reached a solution
–  a description of the requirements that characterise a
solution
–  the criteria to be used to verify that the requirements
have been met
–  produces an understanding of the nature and scope of
the remaining activities
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 4
1
Response Vs. Pole Location
•  As we saw previously,
a qualitative
understanding of the
effect of poles and
zeros on the system
response can help us
to quickly estimate
performance
Dr. Stefan B. Williams
Response Vs. Pole Location
Im{s}
stable
unstable
Re{s}
Slide 5
Amme 3500 : System Response
General First Order System
•  A first order system without
zeros can be described by:
•  The resulting impulse
response is
•  When
–  σ > 0, pole is located at σ < 0,
exponential decays = stable
–  σ < 0, pole is at σ > 0,
exponential grows = unstable
Dr. Stefan B. Williams
τ=
Slide 6
1
σ
•  This is the time taken for
the system to decay to 1/
e or 37% of its initial
value or rise to 63% of
step response
Step Response
Amme 3500 : System Response
Amme 3500 : System Response
•  We define the time
constant of the system
as
h(t ) = e−σ t
1
Dr. Stefan B. Williams
General First Order System
1
H (s) =
(s + σ )
y (t ) =
•  We would now like to gain a deeper insight
into the system performance as a function
of pole and zero locations
•  This will help us to describe the system
performance quantitatively
•  We will also see how this allows us to
design a system to meet certain design
constraints
(1 − e )
σ
−σ t
Slide 7
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 8
2
General First Order System
y (t ) = 1 − e −σ t
•  The Rise Time Tr is
defined as the time
for the waveform to
go from 0.1 to 0.9 of
its final value
•  The Settling Time Ts
is defined as the time
for the waveform to
reach and stay within
a certain percentage
of its final value
•  Values of 1%, 2% and
5% are often used
0.9 = 1 − e −σ t0.9
0.1 = 1 − e
−σ t0.1
TR = t0.9 − t0.1
=
=
Dr. Stefan B. Williams
General First Order System
2.31 0.11
−
σ
σ
2.2
Slide 9
Dr. Stefan B. Williams
Example
G(s) =
Step Response
0.9
0.8
0.7
–  1/σ = 0.02
Amplitude
•  The time constant is:
–  4/σ = 0.08
•  Rise Time is:
–  2.2/σ = 0.044
Dr. Stefan B. Williams
0.6
0.5
0.4
•  Settling time (2%) is:
0.3
0.2
0.1
0
0
0.02
4
(2%)
σ
4.6
TS =
(1%)
σ
3
TS =
(5%)
σ
Amme 3500 : System Response
Slide 10
General Second Order System
1
50
s + 50
TS =
σ
Amme 3500 : System Response
•  A system has a
transfer function
0.98 = 1 − e −σ Ts
0.04
0.06
0.08
0.1
Time (sec)
Amme 3500 : System Response
Slide 11
0.12
•  Many systems of interest are of higher
order
•  For a first order system we have a single,
real pole
•  Second order systems are quite common
and are generally written in the following
standard form
ωn2
G ( s) = C∞ 2
s + 2ςωn s + ωn2
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 12
3
Natural Frequency and Damping Ratio
•  The Natural Frequency, ωn, of a secondorder system is the frequency of oscillation
of the system without damping
•  The Damping Ratio, ! , of a second-order
is the ratio of the exponential decay
frequency and the natural frequency
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 13
General Second Order System
•  For a second order system, we can have
four distinct combinations of real and
imaginary poles
–  Two real poles -σ1, -σ2
–  Two identical real -σ1
–  Two complex poles -σ ± jωd
–  Two imaginary ± jωd
Dr. Stefan B. Williams
Slide 14
Amme 3500 : System Response
Natural Frequency and Damping Ratio
Natural Frequency and Damping Ratio
•  How does this relate to the pole location in
terms of their real and imaginary parts?
s = −σ ± jωd
•  Complex poles are always in complex
conjugate pairs so
•  By multiplying out and comparing the
denominator of the two forms we find that
(s + σ )2 + ωd = s 2 + 2ξωn s + ωn
2
σ = ζωn
d ( s) = ( s + σ − jωd )( s + σ + jωd )
= ( s + σ ) + ωd
2
Dr. Stefan B. Williams
ωd = ωn 1 − ζ 2
2
Amme 3500 : System Response
2
Slide 15
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 16
4
Natural Frequency and Damping Ratio
Natural Response of Second Order System
•  We d like to find the natural response of a general 2nd
order system
•  We can now relate
the natural frequency
and damping ratio to
the s-plane
y˙˙ + 2!" n y˙ + " n2 y = 0
s2Y (s) # sy 0 # y˙ 0 + 2!" n ( sY(s) # y 0 ) + " n2Y (s) = 0
q
σ = ζωn
Y (s) =
ωd = ωn 1 − ζ 2
C ( s) = y0
Slide 17
Amme 3500 : System Response
Dr. Stefan B. Williams
Natural Response of Second Order System
•  We d like to find the
natural response of a
general 2nd order system
•  After completing squares
ζ
( s + ζωn ) +
•  Take Inverse Laplace
and simplify
= y0
•  We can use this to
−ζω
estimate the system
⎛ 2π n ⎞
ω
y⎜
⎟ = y0 e
parameters
ω
⎝ d ⎠
2π n
•  By displacing the system
−ζω
ω
y n = y0 e
and measuring its
response we can identify
⎛y ⎞
2π n
ln ⎜ n ⎟ = −ζωn
the peaks and estimate
ωd
⎝ y0 ⎠
natural frequency and
⎛y ⎞
ζωn
1
damping ratio
=
ln ⎜ 0 ⎟
2
2
π
n
⎝ yn ⎠
•  This is known as the log ωn 1 − ζ
⎛y ⎞
1
decrement
ζ ;
ln ⎜ 0 ⎟
2π n ⎝ yn ⎠
n
ωn 1 − ζ
2
σ
ω
ωd d
( s + σ ) 2 + ωd2
⎛
⎞
σ
y (t ) = y0e−σ t ⎜ cos ωd t + sin ωd t ⎟
ω
d
⎝
⎠
Amme 3500 : System Response
Slide 18
Amme 3500 : System Response
⎛
⎞
σ
y (t ) = y0e−σ t ⎜ cos ωd t + sin ωd t ⎟
ωd
⎝
⎠
2π n
1− ζ 2
( s + ζωn ) + ωn2 (1 − ζ 2 )
C ( s ) = y0
s + 2ςωn
s 2 + 2ςωn s + ωn2
Natural Response of Second Order System
s + 2ςωn
C ( s) = y0 2
s + 2ςωn s + ωn2
(s + σ ) +
Dr. Stefan B. Williams
s2 + 2!" n s + " n2
•  For the case of zero initial velocity we have
θ = sin −1 (ζ )
Dr. Stefan B. Williams
(s + 2!" n ) y 0 + y˙ 0
Slide 19
2
n
Dr. Stefan B. Williams
Amme 3500 : System Response
d
d
Slide 20
5
A Familiar Mechanical Example
•  Earlier, we
considered this
mechanical system
•  Now we can consider
the natural and forced
response
M
y(t)
f(t)
My!!(t ) + K d y! (t ) + Ky (t ) = f (t )
(
)
M s 2Y ( s ) ! sy (0) ! y! (0) + K d ( sY ( s ) ! y (0)) + KY ( s ) = F ( s )
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 21
Example: Natural Response
•  Suppose we wish to find the natural
response of the system to an initial
displacement of 0.5m
•  Assume the mass of the system is 1kg
and the systems constant kd is 2 Nsec/m
and k is 20 N/m
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 23
Example: Natural Response
•  The natural response of the system can be
found by assuming no input force
( Ms
2
+ K d s + K )Y (s) = ( Ms + K d ) y(0) + M˙y (0)
Y (s) =
( Ms + K d ) y(0) + M˙y (0)
( Ms
2
+ Kd s + K)
•  Taking inverse Laplace transform will give
us the system response
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 22
Example: Natural Response
•  The natural response of the system can be
found by assuming no input force
Kd ⎞
⎛
⎜s+
⎟
s + 2ςωn
M ⎠
⎝
Y ( s ) = y (0)
= y0 2
K⎞
s + 2ςωn s + ωn2
⎛ 2 Kd
s+ ⎟
⎜s +
M
M⎠
⎝
Kd
K
2ζωn =
, ωn2 =
M
M
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 24
6
Example: Natural Response
•  Substituting values, we find
•  We can work backwards from the response to
estimate the parameters
Impulse Response
Impulse Response
0.5
System: h
Time (sec): 0
Amplitude: 0.5
0.4
1
," d = 19
20,
0.3
= 0.69 Hz = 4.36
0.2
Amplitude
%
(
$
y(t) = y 0e ' cos " d t +
sin " d t*
"d
&
)
#$t
1
%
(
= 0.5e #t ' cos 19t +
sin 19t *
&
)
19
⎛y ⎞
ζ ;
ln ⎜ 0 ⎟
2π n ⎝ yn ⎠
1 ⎛ 0.5 ⎞
=
ln ⎜
⎟
4π ⎝ 0.0279 ⎠
= 0.230
1
System: h
Time (sec): 2.88
Amplitude: 0.0279
0.1
0
-0.1
-0.2
-0.3
0.5
System: h
Time (sec): 0
Amplitude: 0.5
0.4
2 cycles
ωn ; ωd =
2.88s
0
1
2
3
4
5
rad
s
0.3
0.2
Amplitude
2!" n = 2," n2 = 20,
" n = 20,! =
Example: Natural Response
0
-0.1
-0.2
-0.3
6
System: h
Time (sec): 2.88
Amplitude: 0.0279
0.1
0
1
Time (sec)
Dr. Stefan B. Williams
Step Response of Second Order System
ω2
n
C (s) =
•  We d like to find the
s ( s 2 + 2ςωn s + ωn2 )
step response of the
ζ
standard form
( s + ζωn ) +
ωn 1 − ζ 2
2
1
−
ζ
1
•  After some partial C ( s) = −
s
( s + ζωn ) 2 + ωn2 (1 − ζ 2 )
fraction expansion
ζ
and completing
(s + σ ) +
ωd
squares
1− ζ 2
1
= −
s
2
3
4
5
6
Time (sec)
Slide 25
Amme 3500 : System Response
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 26
Step Response vs. Damping Ratio
•  Damping ratio
determines the
characteristics
of the system
response
( s + σ ) 2 + ωd2
•  Take Inverse
⎞
σ
−σ t ⎛
Laplace and simplify c(t ) = 1 − e ⎜ cos ωd t + sin ωd t ⎟
ω
d
⎝
⎠
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 27
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 28
7
Step Response vs. Damping Ratio
•  The damping ratio
describes the degree
of damping in the
system
•  For an underdamped
system, rising
damping will lower the
overshoot of the
system
c(t ) = A cos(ωnt − φ )
c(t ) = Ae −σ d t cos (ωd t − φ )
c(t ) = K1e −σ1t + K 2te −σ1t
c(t ) = K1e −σ1t + K 2e −σ 2t
Dr. Stefan B. Williams
Step Response vs. Damping Ratio
Amme 3500 : System Response
Slide 29
Time Domain Specifications
Dr. Stefan B. Williams
•  Rise time, settling
time and peak time
yield information
about the speed of
response of the
transient response
•  This can help a
designer determine if
the speed and nature
of the response is
appropriate
–  Peak Time, tp, is the time taken to reach maximum
overshoot
–  Overshoot, Mp, is the maximum amount the system
overshoots its final value
–  Settling time, ts, is the time for system transients to
decay
–  Rise time, tr, is the time it takes for the system to
reach the vicinity of its new set point
Amme 3500 : System Response
Slide 31
Slide 30
Time Domain Specifications
•  Specifications for a control system design often
involve requirements associated with the time
response of the system
Dr. Stefan B. Williams
Amme 3500 : System Response
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 32
8
Peak Time
•  Peak time is found by
differentiating and
finding the first zero
crossing after t=0
•  Substituting the peak
time back into the step
response yields
L[c!(t )] = sC ( s )
! n2
s 2 + 2!" n + ! n2
=
•  After some
manipulation we find
tp occurs when
sinwdt=0
Dr. Stefan B. Williams
Percent Overshoot
•  Notice this is only a
function of the damping
ratio. We can invert to
find the required
damping ratio for a
particular overshoot
π
tp =
ωn 1 − ζ 2
π
=
ωd
Slide 33
Amme 3500 : System Response
Dr. Stefan B. Williams
Percent Overshoot
Mp,%
•  We can plot the
relationship between
percent overshoot
and damping ratio
•  Two frequently used
specifications are
–  Mp=5% for ζ=0.7
–  Mp=16% for ζ=0.5
c(t p ) = 1 − e
= 1+ e
Mp =e
ζ =
−σ
π
ωd
−σ
π
ωd
⎛
⎞
σ
sin π ⎟
⎜ cos π +
ωd
⎝
⎠
πζ
−
1−ζ 2
,0 ≤ ζ <1
− ln( M p )
π 2 + ln 2 ( M p )
Slide 34
Amme 3500 : System Response
Settling Time
•  The settling time can
be derived in a similar
manner as for a first
order system
•  The settling time is
essentially decided by
the transient
exponential
Overshoot vs. Damping Ratio for second order system
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
16%
e −ζωnts = 0.02
4
so t s =
ζωn
=
4
σ
0.1
5%
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ζ
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 35
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 36
9
Rise Time
•  Rise time is not as straightforward
to compute for this case
•  A precise analytical relationship
does not exist
•  Instead we use an approximation
as suggested in Franklin (Section
3.4.1).
•  Nise suggests an alternative
approach based on a graph of
normalised rise time
Dr. Stefan B. Williams
Amme 3500 : System Response
Time Domain Specifications
tr ≅
•  Lines of constant
peak time,Tp ,
settling time, Ts ,
and percent
overshoot, %OS
and rise time
•  Note: Ts2 < Ts1 ;
Tp2 < Tp1;
%OS1 < %OS2
1.8
ωn
Slide 37
Step response and Pole Location
•  Step responses
of second-order
underdamped
systems
as poles move:
a. with constant
real part σ;
b. with constant
imaginary part jω;
c. with constant
damping ratio ζ
Dr. Stefan B. Williams
Amme 3500 : System Response
Dr. Stefan B. Williams
wn
Example : Transforming
Specifications
•  Find allowable region
in the s-plane for the
poles of a transfer
function to meet the
requirements
tr ≤ 0.6 sec
Slide 39
ωn ≥
Im(s)
sin-1z
wn
Re(s)
1.8
= 3.0rad / sec
tr
Mp ≤ 10%
ζ ≥ 0.6
ts ≤ 3 sec
σ≥
Dr. Stefan B. Williams
Slide 38
Amme 3500 : System Response
4
= 1.5
3
Amme 3500 : System Response
s
Slide 40
10
Example : Transient Response
through Component Design
Example : Transient Response
through Component Design
•  First the differential equation
J !!! + D!! + K ! = T
•  Find the transfer function
•  Given the system shown here, find J and
D to yield 20% overshoot and a settling
time of 2 seconds for a step input torque T
(t)
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 41
Example : Transient Response
through Component Design
•  From problem statement
Ts = 2 =
•  Hence
G( s) =
1/ J
D
K
s2 + s +
J
J
•  The second order properties are
ωn =
Dr. Stefan B. Williams
K
D
and 2ζωn =
J
J
Amme 3500 : System Response
Slide 42
Additional Poles and Zeros
•  The preceding developing holds for a
second order system with no zeros
•  What happens to the system response if
we add additional poles or zeros?
•  This depends on their location in the splane
4
ζωn
D
4
J
= 4 and ζ =
=2
J
2ωn
K
•  For a 20% overshoot, ζ = 0.456 so
J
= 0.052
K
•  From problem statement K = 5 so J = 0.26kgm2
and D = 1.04 Nms/rad
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 43
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 44
11
Additional Poles
•  Consider an
additional pole at a
•  By partial fraction
expansion and
completing squares
we have the step
response
H ( s) =
T ( s) =
(s + α )(s
Additional Poles
ωn2
2
+ 2ζωn + ωn2 )
A B( s + ζωn ) + Cωd
D
+
+
2
s
( s + ζωn ) + ωd2 ( s + α )
c(t ) = Au(t )
+ e−ζωnt ( B cos ωd t + C sin ωd t ) + De−α t
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 45
Dr. Stefan B. Williams
Additional Poles
Amme 3500 : System Response
Slide 46
Additional Poles
•  As the pole location is
moved to the left in
the LHP, the effect on
the response
becomes less and the
transient dies out
more quickly
Dr. Stefan B. Williams
Amme 3500 : System Response
•  How much further from the dominant poles does
the third pole have to be for its effect on the
second-order response to be negligible?
•  This depends on the accuracy required by the
design
•  We often assume the exponential decay is
negligible after five time constants
•  The accuracy of this assumption should always
be verified
Slide 47
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 48
12
Zeros
Zeros
( s + β )ω 2
n
•  Consider a zero at β
H z (s) = 2
s
+
2
ζω
+
( 2 n ωn2 )
•  For large β, the zero
sωn
βωn2
= 2
+ 2
2
acts as a scaling
s + 2ζωn + ωn s + 2ζωn + ωn2
factor
= sH ( s) + β H ( s )
•  For smaller β, the
derivative increases
the speed of
response and
overshoot
Step Response
0.7
Zero at -10
0.6
0.5
•  If we compare the
normalized step
response, we find that
the derivative term
increases the
responsiveness of the
system
Amplitude
0.4
Zero at -5
0.3
Zero at -3
0.2
No zero
0.1
0
Dr. Stefan B. Williams
0
0.5
Amme 3500 : System Response
1
1.5
Time (sec)
2
2.5
3
Slide 49
Zeros
Amme 3500 : System Response
Amme 3500 : System Response
Slide 50
Effects of Pole-Zero Patterns
•  We have seen that a pole
in the RHP makes a
system unstable
•  A zero in the RHP causes
the response to be
depressed
•  If the derivative term is
larger than the scaled
response, this can lead to
non-minimum phase
behaviour
Dr. Stefan B. Williams
Dr. Stefan B. Williams
•  For a second order system with no finite
zeros, the transient response parameters
are approximated by
–  Rise time :
tr ≅
1.8
ωn
⎧ 5%, ζ = 0.7
⎪
M p ≅ ⎨ 16%, ζ = 0.5
⎪20%, ζ = 0.45
⎩
–  Settling Time (2%) : t ≅ 4
s
–  Overshoot :
σ
Slide 51
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 52
13
Effects of Pole-Zero Patterns
•  An additional pole in the left half-plane (LHP) will
increase the rise time significantly if the extra
pole is within a factor of 5 of the real part of the
complex poles
•  A zero in the LHP will increase the overshoot if
the zero is within a factor of 4 of the real part of
the complex poles
•  A zero in the RHP will depress the overshoot
(and may cause the step response to be nonminimum phase)
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 53
Conclusions
•  We have looked at the characteristics of
first and second order systems
•  We have derived specifications that
describe the response of the system to a
step input
•  We have also looked at the effect of extra
poles and zeros on the system response
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 54
Further Reading
•  Nise
–  Sections 4.1-4.8
•  Franklin & Powell
–  Section 3.3-3.5
Dr. Stefan B. Williams
Amme 3500 : System Response
Slide 55
14