CHEE 319 Process Dynamics and Control
Frequency Response of Linear Control Systems
1 FrequencyFrequency ResponseResponse
Response of a linear control system
to a sinusoidal input signal as a function of frequency
2 FirstFirst OrderOrder ProcessProcess
Sinusoidal response
Solving for , and and taking the inverse Laplace transform
Sinusoidal response is the ultimate response of the 1st order system
rearranging
3 FirstFirst orderorder ProcessProcess
Response to a sinusoidal input signal
2
1.5
1
0.5
0
-0.5
-1
-1.5 0 2 4 6 8 10 12 14 16 18 20
Recall: Sinusoidal input yields sinusoidal output characterized by and
4 FirstFirst OrderOrder ProcessProcess
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 SecondSecond OrderOrder ProcessProcess
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 to a sinusoidal input is 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 FrequencyFrequency ResponseResponse
For a general transfer function
r(s) e!" s(s ! z ) (s ! z ) G(s) = = 1 L m q(s) (s ! p1)L(s ! pn)
Frequency Response summarized by
j G( j!) = G( j!) e " 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 FrequencyFrequency ResponseResponse
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 frequency- based design methods
17 BodeBode PlotsPlots
Low frequency asymptote First order process
High frequency asymptote
Bandwidth
High frequency Phase shift
18 BodeBode plotsplots
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 BodeBode plotplot
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 BodeBode PlotPlot
First order unstable system
Frequency response
Amplitude ratio and phase shift:
Same amplitude ratio but positive phase shift
21 BodeBode plotplot
First order unstable system
22 BodeBode plotplot
Lead unit:
23 BodeBode plotplot
Lead unit (right half plane zero)
24 BodeBode PlotsPlots
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 BodeBode plotsplots
(Ideal) Low pass filter: With pass band gain and bandwidth
26 BodeBode PlotsPlots
High-pass filter: Pass band gain and cut-off frequency
27 BodeBode PlotsPlots
Pass band filter: Pass band gain and cut-off frequencies
28 BodeBode plotsplots
For second order processes
Frequency response is given by
Overdamped
Critically damped
29 BodeBode PlotsPlots
Second order process
30 BodeBode plotplot
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 BodeBode plotplot
Overdamped system:
Slope=-1
Slope=-2
32 BodeBode plotplot
Characteristics:
Amplitude and phase shift have two inflection points. One for each pole of the system
Overall asymptotic behavior is second order.
33 BodeBode PlotPlot
Overdamped process with stable zeros
34 BodeBode plotplot
Overdamped process with stable zeros
35 BodeBode plotplot
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 BodeBode plotplot
Delayed system:
e.g. First order plus delay
AR is unchanged but the phase shift has no high frequency asymptote
37 BodeBode plotplot
First order plus delay:
38 BodeBode plotplot
System with pole at zero (integrator)
No finite dc gain
Slope = -2
39 BodeBode plotplot
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 NyquistNyquist PlotPlot
First order system
42 NyquistNyquist plotplot
Second order process
43 NyquistNyquist plotplot
Third order process:
44 NyquistNyquist PlotPlot
Delayed system
45 NyquistNyquist plotplot
System with pole at zero Pole at is on the Nyquist path - no finite dc gain
Strategy is to go around the singularity about in the right half plane ( a small number)
x
46 NyquistNyquist plotplot
Pole at zero
xx
47 NyquistNyquist plotplot
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 4. Let K = cu AR
55 Bode Criterion
Consider the transfer function and controller
5e!0.1s G(s) = (s +1)(0.5s +1)
- Open-loop transfer function
'0.1s 5e ! 1 $ GOL (s) = 0.4#1+ & (s +1)(0.5s +1) " 0.1s% - Amplitude ratio and phase shift
5 1 1 AR 0.4 1 = + 2 1+ (2 1+ 0.25(2 0.01( !1 !1 !1" 1 % ) = !0.1( ! tan (() ! tan (0.5() ! tan $ ' # 0.1(& - At ω=1.4128, φ=-π, AR=6.746
56 BodeBode CriterionCriterion
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 Ku/2
PI Ku/2.2 Pu/1.2
PID Ku/1.7 Pu/2 Pu/8
58 BodeBode StabilityStability CriterionCriterion
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 NyquistNyquist StabilityStability CriterionCriterion
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 NyquistNyquist StabilityStability CriterionCriterion
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 NyquistNyquist StabilityStability CriterionCriterion
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 NyquistNyquist StabilityStability CriterionCriterion
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 NyquistNyquist StabilityStability CriterionCriterion
Path of interest in the domain must encircle the entire RHP
• 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 in domain
64 NyquistNyquist StabilityStability CriterionCriterion
Path of interest is the Nyquist path
Consider the Nyquist plot of
65 NyquistNyquist StabilityStability CriterionCriterion
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 (the number of unstable poles of the closed-loop system)
Therefore the closed-loop system is unstable if
66 NyquistNyquist StabilityStability CriterionCriterion
Consider the open-loop transfer function
has the same poles as
The origin in domain corresponds to the point (-1,0) in the domain
67 NyquistNyquist StabilityStability
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 NyquistNyquist StabilityStability
Nyquist plot
There are 2 clockwise encirclements of (-1,0) 71 NyquistNyquist StabilityStability
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 NyquistNyquist StabilityStability
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 NyquistNyquist StabilityStability
Nyquist plot
There are many clockwise encirclements of (-1,0) 76 BodeBode StabilityStability
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 StabilityStability MarginsMargins
Stability margins for linear systems
79 StabilityStability MarginsMargins
Gain margin: Let be the amplitude ratio of at the critical frequency
Phase margin: Let be the phase shift of at the frequency where
80 StabilityStability MarginMargin
81 SurpriseSurprise QuizQuiz (20(20 minutes)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