CHEE 319 Process Dynamics and Control
Frequency Response of Linear Control Systems
■ 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 8 10 12 14 16 18 20
Recall: Sinusoidal input yields sinusoidal output characterized by and
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 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 Frequency Response
For a general transfer function
r(s) e−θ s(s − z )(s − z ) G(s) = = 1 m q(s) (s − p1)(s − pn)
Frequency Response summarized by
G( jω) = G( jω) e jϕ where G ( j ! ) is the modulus of G ( j ! ) and is the argument of G ( j!) | |
Note: Substitute for s = j ! 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 = j ! in the transfer function G(s)
ii) G ( j ! ) is a complex number. Its modulus is the amplitude ratio of the process and its argument is the phase shift.
iii) As ! , the frequency, is varied that G ( j ! ) gives a trace (or a curve) in the complex plane.
iv) The effect of the frequency, ! , 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 frequency- based design methods
17 Bode Plots
Low frequency asymptote ■ First order process
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 K 1 ⌧!j K K⌧!j G(j!)= = ⌧!j 1 1 ⌧!j 1+⌧ 2!2 ➤ Amplitude ratio and phase shift:
K AR = , = ⇡ + tan 1(⌧!) p1+⌧ 2!2 ➨ Same amplitude ratio but increase of 90 degrees in phase shift
21 Bode plot
■ First order unstable system
Bode Diagram
0
−10
−20
Magnitude (dB) −30
−40 0
−45 Stable
−90
Phase (deg) −135 Unstable
−180 −2 −1 0 1 2 10 10 10 10 10 Frequency (rad/sec)
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 -20 dB/cycle 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 in the right half plane ( a small number)
x
46 Nyquist plot
■ Pole at zero
xx
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 w in
2. Calculate AR
54 Bode Stability Criterion
To check for stability:
1. Compute open-loop transfer function 2. Solve for w in f=-p 3. Evaluate AR at w 4. If AR>1 then process is unstable
Find ultimate gain:
1. Compute open-loop transfer function without controller gain 2. Solve for w in f=-p 3. Evaluate AR at w 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 2 2 1+ ω 1+ 0.25ω 0.01ω −1 −1 −1⎛ 1 ⎞ φ = −0.1ω − tan (ω) − tan (0.5ω) − tan ⎜ ⎟ ⎝ ⎠ 0.1ω - At w=1.4128, f=-p, 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 wc,
➤ Ultimate gain, Ku, is 1/M where M is the amplitude of the open-loop system ➤ Ultimate Period 2π Pu = ωc
Ziegler-Nichols Tunings
P Ku/2 PI Ku/2.2 Pu/1.2 PID Ku/1.7 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 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 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 (the number of unstable poles of the closed-loop system)
➤ 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 corresponds to the point (-1,0) in the domain
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=jw in GOL(s)
3. Plot GOL(jw) 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 at the critical frequency
■ 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