CHEE 319
Process Dynamics and Control
Frequency Response of
Linear Control Systems
1
Frequency Response

Response of a linear control system
to a sinusoidal input signal as a function of frequency
2
First Order Process

Sinusoidal response
 Solving for

,
and
and taking the inverse Laplace transform
Sinusoidal response is the ultimate response of the 1st
order system
 rearranging
3
First order Process
Response to a sinusoidal input signal
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
2
4
6
Recall: Sinusoidal input
characterized by
and
8
10
12
14
16
18
20
yields sinusoidal output
4
First Order Process

Sinusoidal response of a 1st order process to a sinusoidal
input signal is:
 Sinusoidal with amplitude
and phase lag
 Important characteristics of the sinusoidal response of the process
are the amplitude ratio
and the phase lag, both functions of frequency
.
5
Second Order Process

Sinusoidal Response
where
6
Frequency Response
Q: Do we “have to” take the Laplace inverse to compute the
AR and phase shift of a 1st or 2nd order process?
No
Q: Does this generalize to all transfer function models?
Yes
Study of transfer function model response to sinusoidal inputs
is called “Frequency Domain Response” of linear
processes.
7
Frequency Response
Some facts for complex number theory:
i) For a complex number:
Im
Re
It follows that
where
such that
8
Frequency Response
Some facts:
ii) Let
and
then
iii) For a first order process
Let
Then
9
Frequency Response
Main Result:
The response of any linear process
input is a sinusoidal.
to a sinusoidal
The amplitude ratio of the resulting signal is given by the
Modulus of the transfer function model expressed in the
frequency domain,
.
The Phase Shift is given by the argument of the transfer
function model in the frequency domain.
10
Frequency Response
For a general transfer function
r ( s) e !" s ( s ! z1)L( s ! zm )
G ( s) =
=
q ( s)
( s ! p1)L( s ! pn )
Frequency Response summarized by
G ( j! ) = G ( j! ) e j"
where
G ( j! )
is the modulus of G(jω) and ϕ is the argument of G(jω)
Note: Substitute for s=jω in the transfer function.
11
Frequency Response
The facts:
For any linear process we can calculate the amplitude ratio
and phase shift by:
i) Letting s=jω in the transfer function G(s)
ii) G(jω) is a complex number. Its modulus is the amplitude ratio of
the process and its argument is the phase shift.
iii) As ω, the frequency, is varied that G(jω) gives a trace (or a curve)
in the complex plane.
iv) The effect of the frequency, ω, on the process is the frequency
response of the process.
12
Frequency Response
Examples:
1. Pure Capacitive Process
2. Dead Time
13
Frequency Response
Examples:
3. Lead unit
14
Frequency Response
Examples:
4. n process in series
Frequency response of
therefore
15
Frequency Response
Examples:
5. n first order processes in series
6. First order plus delay
16
Frequency Response

To study frequency response, we use two types of
graphical representations
1. The Bode Plot:
Plot of
(in dB)vs. on loglog scale
Plot of
(in degrees) vs.
on semilog scale
2. The Nyquist Plot:
Plot of the trace of

in the complex plane
Plots lead to effective stability criteria and frequencybased design methods
17
Bode Plots

First order process
Low frequency asymptote
High frequency
asymptote
Bandwidth
High frequency
Phase shift
18
Bode plots

Characteristic of the Bode plot
 low frequency asymptote
 High frequency asymptote (if
)
Yields a line on loglog scale,
The phase shift at high frequencies
19
Bode plot

Characteristics of the Bode plot:
 Cut-off frequency - this is the frequency
at which
 For first order systems
 The cut-off frequency is also the bandwidth of the first order
system
20
Bode Plot

First order unstable system
 Frequency response
 Amplitude ratio and phase shift:
 Same amplitude ratio but positive phase shift
21
Bode plot

First order unstable system
22
Bode plot

Lead unit:
23
Bode plot

Lead unit (right half plane zero)
24
Bode Plots

Filters
 Pass band is the range of frequencies where the signals pass
through the system at the same degree of amplification
 Low pass filter is a dynamical system with a pass band in the low
frequeny range
 High pass filter is a dynamical system with a pass band in the high
frequency range
 Band pass filter is a dynamical system with a pass band over a
certain range of frequencies
 Bandwidth is the width of the frequency interval over the pass
band of the filter
25
Bode plots

(Ideal) Low pass filter:
 With pass band gain
and bandwidth
26
Bode Plots

High-pass filter:
 Pass band gain
and cut-off frequency
27
Bode Plots

Pass band filter:
 Pass band gain
and cut-off frequencies
28
Bode plots

For second order processes
 Frequency response is given by
 Overdamped
 Critically damped
29
Bode Plots

Second order process
30
Bode plot

Characteristics:
 High frequency limit
 High frequency limit has slope -2 in Bode plot
 Underdamped process displays a maximum amplification greater
than the pass band gain
at the resonant frequency,
31
Bode plot

Overdamped system:
Slope=-1
Slope=-2
32
Bode plot

Characteristics:
 Amplitude and phase shift have two inflection points. One for each
pole of the system
 Overall asymptotic behavior is second order.
33
Bode Plot

Overdamped process with stable zeros
34
Bode plot

Overdamped process with stable zeros
35
Bode plot

Zeros reduce the relative order
 High frequency asymptote is a straight line (on loglog) with slope
equal to the realtive order
 LHP zeros increase the phase of the system
 RHP zeros decrease the phase of the system
 Also called nonminimum phase systems because they exhibit
more phase lag than any other system with the same AR
36
Bode plot

Delayed system:
 e.g. First order plus delay
 AR is unchanged but the phase shift has no high frequency
asymptote
37
Bode plot

First order plus delay:
38
Bode plot

System with pole at zero (integrator)
No finite dc gain
Slope = -2
39
Bode plot

Overall characteristics of Bode plots
 dc Gain is the value of AR at
 Slope of high frequency asymptote is the negative of the relative
order of the system
 Inflection points in AR and the phase shift correspond to poles and
zeros of the transfer function
 Each unstable poles and stable zeros lead to positive phase shift of
90 degrees as
 Each stable pole and Unstable zeros reduce phase shift by 90
degrees as
 Delay leads to a phase shift of
as
40
Nyquist Plot
Plot of
in the complex plane as
is varied on
Nyquist path
Relation to Bode plot
 AR is distance of
from the origin
 Phase angle, , is the angle from the Real positive axis
41
Nyquist Plot

First order system
42
Nyquist plot

Second order process
43
Nyquist plot

Third order process:
44
Nyquist Plot

Delayed system
45
Nyquist plot

System with pole at zero
 Pole at
is on the Nyquist path - no finite dc gain
 Strategy is to go around the singularity about
half plane (
a small number)
in the right
x
46
Nyquist plot

Pole at zero
x
47
Nyquist plot

System with integral action:
48
CHEE 319
Process Dynamics and Control
Frequency Domain
Analysis of Control Systems
49
PI Controller
50
PD Controller
51
PID Controller
52
Bode Stability Criterion
Consider open-loop control system
+
+
+
-
1. Introduce sinusoidal input in setpoint (
) and observe sinusoidal
output
2. Fix gain such
and input frequency such that
3. At same time, connect close the loop and set
Q: What happens if
?
53
Bode Stability Criterion
“A closed-loop system is unstable if the frequency of the
response of the open-loop GOL has an amplitude ratio
greater than one at the critical frequency. Otherwise it is
stable. “
Strategy:
1. Solve for ω in
2. Calculate AR
54
Bode Stability Criterion
To check for stability:
1. Compute open-loop transfer function
2. Solve for ω in φ=-π
3. Evaluate AR at ω
4. If AR>1 then process is unstable
Find ultimate gain:
1. Compute open-loop transfer function without controller gain
2. Solve for ω in φ=-π
3. Evaluate AR at ω
1
Kcu =
4. Let
AR
55
Bode Criterion
Consider the transfer function and controller
5e !0.1s
G ( s) =
( s + 1)(0.5s + 1)
- Open-loop transfer function
5e '0.1s
1 $
GOL ( s) =
0.4!# 1 +
&
( s + 1)(0.5s + 1) " 01
. s%
- Amplitude ratio and phase shift
AR =
5
1
1 + ( 2 1 + 0.25( 2
0.4 1 +
1
0.01( 2
1 %
) = !01
. ( ! tan !1 (( ) ! tan !1 (0.5( ) ! tan !1"$
'
# 01
. (&
- At ω=1.4128, φ=-π, AR=6.746
56
Bode Criterion
57
Ziegler-Nichols Tuning
Closed-loop tuning relation
 With P-only, vary controller gain until system (initially stable) starts to
oscillate.
 Frequency of oscillation is ωc,
 Ultimate gain, Ku, is 1/M where M is the amplitude of the open-loop
system
 Ultimate Period
2!
Pu =
"c
Ziegler-Nichols Tunings
P
PI
PID
Ku/2
Ku/2.2
Ku/1.7
Pu/1.2
Pu/2
Pu/8
58
Bode Stability Criterion

Using Bode plots, easy criterion to verify for closed-loop
stability
 More general than polynomial criterion such as Routh Array,
Direct Substitution, Root locus
 Applies to delay systems without approximation
 Does not require explicit computation of closed-loop poles
 Requires that a unique frequency yield phase shift of -180 degrees
 Requires monotonically decreasing phase shift
59
Nyquist Stability Criterion

Observation
 Consider the transfer function
 Travel on closed path containing
in the domain in the
clockwise direction
 in domain, travel on closed path encircling the origin in the
clockwise direction
 Every encirclement of
in s domain leads to one encirclement of
the origin in
domain
60
Nyquist Stability Criterion

Observation
 Consider the transfer function
 Travel on closed path containing
in the domain in the
clockwise direction
 in domain, travel on closed path encircling the origin in the
counter clockwise direction
 Every encirclement of
in s domain leads to one encirclement or
the origin in
domain
61
Nyquist Stability Criterion

For a general transfer function
 For every zero inside the closed path in the domain, every
clockwise encirclement around the path gives one clockwise
encirclement of the origin in the
domain
 For every pole inside the closed path in the domain, every
clockwise encirclement around the path gives one counter
clockwise encirclement of the origin in the
domain
 The number of clockwise encirclements of the origin in
domain
is equal to number of zeros
less the number of
poles
inside the closed path in the domain.
62
Nyquist Stability Criterion

To assess closed-loop stability:
 Need to identify unstable poles of the closed-loop system
 That is, the number of poles in the RHP

Consider the transfer function
 Poles of the closed-loop system are the zeros of
63
Nyquist Stability Criterion

Path of interest in the
RHP
domain must encircle the entire
• Travel around a half semi-circle or radius
that encircles the entire RHP
• For a proper transfer function
• Every point on the arc of radius are such that
collapse to a single point
domain
in
64
Nyquist Stability Criterion

Path of interest is the Nyquist path
 Consider the Nyquist plot of
65
Nyquist Stability Criterion

Compute the poles of
 Assume that it has
unstable poles
 Count the number of clockwise encirclements of the origin of the
Nyquist plot of
 Assume that it makes
clockwise encirclements
 The number of unstable zeros of
poles of the closed-loop system)
(the number of unstable
 Therefore the closed-loop system is unstable if
66
Nyquist Stability Criterion

Consider the open-loop transfer function

has the same poles as
 The origin in
domain
domain corresponds to the point (-1,0) in the
67
Nyquist Stability

In terms of the open-loop transfer function
 The number of poles of
gives
 The number of clockwise encirclements of (-1,0) of
gives the number of clockwise encirclements of the origin of
 The number of unstable poles of the closed-loop system is given
by
68
Nyquist Stability Criterion
“If N is the number of times that the Nyquist plot encircles the point (1,0) in the complex plane in the clockwise direction, and P is the
number of open-loop poles of GOL that lie in the right-half plane, then
Z=N+P is the number of unstable poles of the closed-loop
characteristic equation.”
Strategy
1. Compute the unstables poles of GOL(s)
2. Substitute s=jω in GOL(s)
3. Plot GOL(jω) in the complex plane
4. Count encirclements of (-1,0) in the clockwise direction
69
Nyquist Criterion
Consider the transfer function
and the PI controller
The open-loop transfer function is
 No unstable poles
 One pole on the Nyquist path (must go around small circle around
the origin in the right half plane)
70
Nyquist Stability

Nyquist plot
 There are 2 clockwise encirclements of (-1,0)
71
Nyquist Stability

Counting encirclements must account for removal of origin

the closed-loop system is unstable
72
Nyquist Criterion
Consider the transfer function
and the PI controller
The open-loop transfer function is
 No unstable poles
 One pole on the Nyquist path (must go around small circle around
the origin in the right half plane)
73
Nyquist Stability

Nyquist plot
 There are no clockwise encirclements of (-1,0)
74
Nyquist Criterion
Consider the transfer function
and the PI controller
The open-loop transfer function is
 No unstable poles
 One pole on the Nyquist path (must go around small circle around
the origin in the right half plane)
75
Nyquist Stability

Nyquist plot
 There are many clockwise encirclements of (-1,0)
76
Bode Stability

Stability margins for linear systems
77
Stability Considerations

Control is about stability

Considered exponential stability of controlled processes using:





Routh criterion
Polynomial
Direct Substitution
(no dead-time)
Root Locus
Bode Criterion (Restriction on phase angle)
Nyquist Criterion

Nyquist is most general but sometimes difficult to interpret

Roots, Bode and Nyquist all in MATLAB
78
Stability Margins

Stability margins for linear systems
79
Stability Margins

Gain margin:
 Let
be the amplitude ratio of
frequency

at the critical
Phase margin:
 Let
be the phase shift of
at the frequency where
80
Stability Margin
81
Surprise Quiz (20 minutes)

Consider the nonlinear system
 Using pole-placement, synthesize a control
such that
the characteristic polynomial of the closed-loop system has the
form
(choose
such that a unique solution exists)
7/10
 Compute the complementary sensitivity function and determine if
the closed-loop system has zero steady-state error
3/10
82