me 433 – state space controleus204/teaching/me433/lectures/lecture02_handout.pdfthe system...
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ME 343 – Control Systems 1
ME 433 – State Space Control
Lecture 2
ME 343 – Control Systems – Summary Prof. Eugenio Schuster
ME 343 – Control Systems 2
Linearization
Dynamic System:
Equilibrium
Denote
Taylor Expansion
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ME 343 – Control Systems 3
Linearization
Dynamic System:
Equilibrium
Denote
Taylor Expansion
ME 343 – Control Systems 4
Linearization
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ME 343 – Control Systems 5
Laplace Transform
Function f(t) of time Piecewise continuous and exponential order
0- limit is used to capture transients and discontinuities at t=0 s is a complex variable (σ+jω)
There is a need to worry about regions of convergence of the integral
Units of s are sec-1=Hz A frequency
If f(t) is volts (amps) then F(s) is volt-seconds (amp-seconds)
∫ ∞
- = 0-
) ( ) ( dt e t f s F st ( ) [ ] ∫ ∞ +
∞ - - = =
j
j st ds e s F
j t f s F
α
α π ) ( 2 1 ) ( 1 L
ME 343 – Control Systems 6
Laplace Transform Examples Step function – unit Heavyside Function
After Oliver Heavyside (1850-1925)
Exponential function After Oliver Exponential (1176 BC- 1066 BC)
Delta (impulse) function δ(t)
⎩ ⎨ ⎧
≥ <
= 0 for , 1 0 for , 0
) ( t t
t u
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ME 343 – Control Systems 7
Laplace Transform Table Signal Waveform Transform
impulse
step
ramp
exponential
damped ramp
sine
cosine
damped sine
damped cosine
ME 343 – Control Systems 8
Laplace Transform Properties
Linearity: (absolutely critical property)
Integration property:
Differentiation property:
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ME 343 – Control Systems 9
Laplace Transform Properties
Translation properties:
s-domain translation:
t-domain translation:
Initial Value Property:
Final Value Property:
If all poles of F(s) are in the LHP
ME 343 – Control Systems 10
Laplace Transform Properties
Time Scaling:
Multiplication by time:
Convolution:
Time product:
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ME 343 – Control Systems 11
Laplace Transform
Exercise: Find the Laplace transform of the following waveform
Exercise: Find the Laplace transform of the following waveform
Exercise: Find the Laplace transform of the following waveform
ME 343 – Control Systems 12
Laplace Transform
The diagram commutes Same answer whichever way you go
Linear system
Differential equation
Classical techniques
Response signal
Laplace transform L
Inverse Laplace transform L-1
Algebraic equation
Algebraic techniques
Response transform
Tim
e do
mai
n (t
dom
ain)
Complex frequency domain (s domain)
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ME 343 – Control Systems 13
Solving LTI ODE’s via Laplace Transform
Initial Conditions:
Recall
For a given rational U(s) we get Y(s)=Q(s)/P(s)
ME 343 – Control Systems 14
Computing Transfer Functions via Laplace Transform
Assume all Initial Conditions Zero:
Input Output
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ME 343 – Control Systems 15
Laplace Transform
Exercise: Find the Laplace transform V(s)
Exercise: Find the Laplace transform V(s)
What about v(t)?
ME 343 – Control Systems 16
Rational Functions
We shall mostly be dealing with TFs which are rational functions – ratios of polynomials in s
pi are the poles and zi are the zeros of the function
K is the scale factor or (sometimes) gain
A proper rational function has n≥m A strictly proper rational function has n>m An improper rational function has n<m
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ME 343 – Control Systems 17
Partial Fraction Expansion - Residues at Simple Poles
Functions of a complex variable with isolated, finite order poles have residues at the poles
Residue at a simple pole:
ME 343 – Control Systems 18
Partial Fraction Expansion - Residues at multiple poles
Example:
Residue at a multiple pole:
2 ) 1 (
) 5 2 ( ) 1 ( lim ! 2 1
3 2 3
2 2
1 = ⎥ ⎥ ⎦ ⎤
⎢ ⎢ ⎣ ⎡
+ + +
- → s s s s
ds d
s 1 ) 1 (
) 5 2 ( ) 1 ( lim 3 2 3
1 = ⎥ ⎥ ⎦ ⎤
⎢ ⎢ ⎣ ⎡
+ + +
- → s s s s
ds d
s 3 ) 1 (
) 5 2 ( ) 1 ( lim 3 2 3
3 - =
+ + +
- → s s s s
s
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ME 343 – Control Systems 19
Partial Fraction Expansion - Residues at Complex Poles
Compute residues at the poles
Bundle complex conjugate pole pairs into second-order terms if you want … but you will need to be careful!
Inverse Laplace Transform is a sum of complex exponentials. But the answer will be real.
ME 343 – Control Systems 20
Inverting Laplace Transforms in Practice
We have a table of inverse LTs Write F(s) as a partial fraction expansion
Now appeal to linearity to invert via the table Surprise! Nastiness: computing the partial fraction expansion is best done by calculating the residues
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ME 343 – Control Systems 21
Inverse Laplace Transform
) 5 2 )( 1 ( ) 3 ( 20 ) ( 2 + + +
+ =
s s s s s F
Exercise: Find the Inverse Laplace transform of
ME 343 – Control Systems 22
Inverse Laplace Transform
Exercise: Find v(t)
Exercise: Find v(t)
What about v(t)?
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ME 343 – Control Systems 23
Not Strictly Proper Laplace Transforms
Find the inverse LT of
Convert to polynomial plus strictly proper rational function Use polynomial division
Invert as normal
3 4 8 12 6 ) ( 2
2 3
+ + + + +
= s s
s s s s F
ME 343 – Control Systems 24
Impulse Response
Dirac’s delta:
Integration is a limit of a sum ⇓
u(t) is represented as a sum of impulses
By superposition principle, we only need unit impulse response
System Response:
Response at t to an impulse applied at τ of amplitude u(τ )
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ME 343 – Control Systems 25
Impulse Response
Convolution:
s-domain:
t-domain:
The system response is obtained by convolving the input with the impulse response of the system.
The system response is obtained by multiplying the transfer function and the Laplace transform of the input.
Impulse response
Impulse response
ME 343 – Control Systems 26
Block Diagrams
Series:
+
+
Parallel:
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ME 343 – Control Systems 27
Block Diagrams Negative Feedback:
-
+ Error signal
Output
Feedback signal
Reference input
Rule: Transfer Function=Forward Gain/(1+Loop Gain)
ME 343 – Control Systems 28
Block Diagrams Positive Feedback:
+
+ Error signal
Output
Feedback signal
Reference input
Rule: Transfer Function=Forward Gain/(1-Loop Gain)
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ME 343 – Control Systems 29
Block Diagrams
Moving through a branching point:
Moving through a summing point:
+
+
+
+
ME 343 – Control Systems 30
Block Diagrams
-
+ + -
Example:
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ME 343 – Control Systems 31
Mason’s Rule
1 2 3 4
+
+ + +
+
+
+ +
nodes branches
Signal Flow Graph
ME 343 – Control Systems 32
Mason’s Rule Path: a sequence of connected branches in the direction of the signal flow without repetition Loop: a closed path that returns to its starting node Forward path: connects input and output
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ME 343 – Control Systems 33
Mason’s Rule
1 2 3 4
Example:
ME 343 – Control Systems 34
Impulse Response
Dirac’s delta:
Integration is a limit of a sum ⇓
u(t) is represented as a sum of impulses
By superposition principle, we only need unit impulse response
System Response:
Response at t to an impulse applied at τ of amplitude u(τ )
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ME 343 – Control Systems 35
Impulse Response
Convolution:
s-domain:
t-domain:
The system response is obtained by convolving the input with the impulse response of the system.
The system response is obtained by multiplying the transfer function and the Laplace transform of the input.
Impulse response
Impulse response
ME 343 – Control Systems 36
Time Response vs. Poles Real pole:
Time Constant
Impulse Response
Stable
Unstable
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ME 343 – Control Systems 37
Time Response vs. Poles
Real pole:
Time Constant
Step Response
Impulse Response
ME 343 – Control Systems 38
Time Response vs. Poles Complex poles: Impulse
Response
Undamped natural frequency
Damping ratio
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ME 343 – Control Systems 39
Time Response vs. Poles Complex poles:
Impulse Response
Stable
Unstable
ME 343 – Control Systems 40
Time Response vs. Poles Complex poles:
Impulse Response
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ME 343 – Control Systems 41
Time Response vs. Poles Complex poles:
Step Response
ME 343 – Control Systems 42
Time Response vs. Poles
Complex poles:
CASES:
Undamped
Underdamped
Critically damped
Overdamped
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ME 343 – Control Systems 43
Time Response vs. Poles
ME 343 – Control Systems 44
Time Domain Specifications
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ME 343 – Control Systems 45
Time Domain Specifications 1- The rise time tr is the time it takes the system to reach the vicinity of its new set point 2- The settling time ts is the time it takes the system transients to decay 3- The overshoot Mp is the maximum amount the system overshoot its final value divided by its final value 4- The peak time tp is the time it takes the system to reach the maximum overshoot point
ME 343 – Control Systems 46
Time Domain Specifications Design specification are given in terms of
These specifications give the position of the poles
Example: Find the pole positions that guarantee
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ME 343 – Control Systems 47
Time Domain Specifications Additional poles: 1- can be neglected if they are sufficiently to the left of the dominant ones. 2- can increase the rise time if the extra pole is within a factor of 4 of the real part of the complex poles.
Zeros: 1- a zero near a pole reduces the effect of that pole in the time response. 2- 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 (due to differentiation). 3- a zero in the RHP (nonminimum phase zero) will depress the overshoot and may cause the step response to start out in the wrong direction.
ME 343 – Control Systems 48
Stability
Impulse response:
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ME 343 – Control Systems 49
Stability
We want:
Definition: A system is asymptotically stable (a.s.) if
Characteristic polynomial:
Characteristic equation:
ME 343 – Control Systems 50
Stability
Necessary condition for asymptotical stability (a.s.):
Use this as the first test!
If any ai<0, the the system is UNSTABLE!
Example:
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ME 343 – Control Systems 51
Routh’s Criterion
Necessary and sufficient condition Do not have to find the roots pi!
Routh’s Array:
Depends on whether n is even or odd
ME 343 – Control Systems 52
Routh’s Criterion
How to remember this?
Routh’s Array:
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ME 343 – Control Systems 53
Routh’s Criterion
The criterion:
• The system is asymptotically stable if and only if all the elements in the first column of the Routh’s array are positive
• The number of roots with positive real parts is equal to the number of sign changes in the first column of the Routh array
ME 343 – Control Systems 54
Routh’s Criterion
Example 1:
Example 2:
Example 3:
Example 4:
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ME 343 – Control Systems 55
Routh’s Criterion
Example: Determine the range of K over which the system is stable
-
+
ME 343 – Control Systems 56
Routh’s Criterion
Special Case I: Zero in the first column We replace the zero with a small positive constant ε>0 and proceed as before. We then apply the stability criterion by taking the limit as ε→0
Example:
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ME 343 – Control Systems 57
Routh’s Criterion
Special Case II: Entire row is zero This indicates that there are complex conjugate pairs. If the ith row is zero, we form an auxiliary equation from the previous nonzero row:
Example:
Where βi are the coefficients of the (i+1)th row in the array. We then replace the ith row by the coefficients of the derivative of the auxiliary polynomial.
ME 343 – Control Systems 58
Properties of Feedback
Disturbance Rejection:
-
+ + +
+ +
Open loop
Closed loop
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ME 343 – Control Systems 59
Properties of Feedback
Disturbance Rejection:
Choose control s.t. for w=0,y≈r
Open loop:
Closed loop:
Feedback allows attenuation of disturbance without having access to it (without measuring it)!!!
IMPORTANT: High gain is dangerous for dynamic response!!!
ME 343 – Control Systems 60
Properties of Feedback
Sensitivity to Gain Plant Changes
-
+ + +
+ +
Open loop
Closed loop
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ME 343 – Control Systems 61
Properties of Feedback
Sensitivity to Gain Plant Changes
Let the plant gain be
Open loop:
Closed loop:
Sensitivity:
Example:
Feedback reduces sensitivity to plant variations!!!
ME 343 – Control Systems 62
Steady-state Tracking
The Unity Feedback Case
-
+
Test Inputs: k=0: step (position)
k=1: ramp (velocity)
k=2: parabola (acceleration)
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ME 343 – Control Systems 63
Steady-state Tracking
The Unity Feedback Case
-
+
Steady State Error:
Final Value Theorem
Type n System
ME 343 – Control Systems 64
Steady-state Tracking The Unity Feedback Case
Steady State Error:
-
+
Type n System
Type (n)
Type 0
Type 1
Type 2
Step (k=0) Ramp (k=1) Parabola (k=2)
Input (k)
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ME 343 – Control Systems 65
Steady-state Tracking Position Constant
Velocity Constant
Acceleration Constant
n: Degree of the poles of CG(s) at the origin (the number of integrators in the loop with unity gain feedback)
• Applying integral control to a plant with no zeros at the origin makes the system type ≥ I • All this is true ONLY for unity feedback systems • Since in Type I systems ess=0 for any CG(s), we say that the system type is a robust property.
ME 343 – Control Systems 66
Steady-state Disturbance Rejection
The Unity Feedback Case
-
+ +
+
Set r=0. Want Y(s)/W(s)=0.
Steady State Error: e=r-y=-y Final Value Theorem
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ME 343 – Control Systems 67
Steady-state Disturbance Rejection The Unity Feedback Case
Steady State Output:
Type (n)
Type 0
Type 1
Type 2
Step (k=0) Ramp (k=1) Parabola (k=2)
Disturbance (k)
-
+ +
+
ME 343 – Control Systems 68
Steady-state Disturbance Rejection Example:
-
+ +
+
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ME 343 – Control Systems 69
PID Controller PID: Proportional – Integral – Derivative
P Controller:
-
+
Step Reference:
• Proportional gain is high • Plant has a pole at the origin
True when:
High gain proportional feedback (needed for good tracking) results in underdamped (or even unstable) transients.
ME 343 – Control Systems 70
PID Controller P Controller: Example (P_controller.m)
-
+
Underdamped transient for large proportional gain Steady state error for small proportional gain
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ME 343 – Control Systems 71
PID Controller PI Controller:
-
+
Step Reference:
• It does not matter the value of the proportional gain • Plant does not need to have a pole at the origin. The controller has it!
Integral control achieves perfect steady state reference tracking!!! Note that this is valid even for Kp=0 as long as Ki≠0
ME 343 – Control Systems 72
PID Controller PI Controller: Example (PI_controller.m)
-
+
DANGER: for large Ki the characteristic equation has roots in the RHP
Analysis by Routh’s Criterion
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ME 343 – Control Systems 73
PID Controller PI Controller: Example (PI_controller.m)
Necessary Conditions:
This is satisfied because
Routh’s Conditions:
ME 343 – Control Systems 74
PID Controller PD Controller:
-
+
Step Reference:
PD controller fixes problems with stability and damping by adding “anticipative” action
• Proportional gain is high • Plant has a pole at the origin
True when:
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ME 343 – Control Systems 75
PID Controller PD Controller: Example (PD_controller.m)
-
+
The damping can be increased now independently of Kp The steady state error can be minimized by a large Kp
ME 343 – Control Systems 76
PID Controller PD Controller:
-
+
NOTE: cannot apply pure differentiation. In practice,
is implemented as
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ME 343 – Control Systems 77
PID Controller
PID: Proportional – Integral – Derivative
-
+
PID Controller: Example (PID_controller.m)
ME 343 – Control Systems 78
PID Controller: Ziegler-Nichols Tuning
• Empirical method (no proof that it works well but it works well for simple systems) • Only for stable plants • You do not need a model to apply the method
Class of plants:
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ME 343 – Control Systems 79
PID Controller: Ziegler-Nichols Tuning
METHOD 1: Based on step response, tuning to decay ratio of 0.25.
Tuning Table:
ME 343 – Control Systems 80
PID Controller: Ziegler-Nichols Tuning
METHOD 2: Based on limit of stability, ultimate sensitivity method.
-
+
• Increase the constant gain Ku until the response becomes purely oscillatory (no decay – marginally stable – pure imaginary poles) • Measure the period of oscillation Pu
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41
ME 343 – Control Systems 81
PID Controller: Ziegler-Nichols Tuning
METHOD 2: Based on limit of stability, ultimate sensitivity method.
Tuning Table:
The Tuning Tables are the same if you make:
ME 343 – Control Systems 82
PID Controller: Ziegler-Nichols Tuning
Actuator Saturates: - valve (fully open) - aircraft rudder (fully deflected)
(Output of the controller)
(Input of the plant)
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ME 343 – Control Systems 83
PID Controller: Ziegler-Nichols Tuning
What happens? - large step input in r - large e - large uc → u saturates - eventually e becomes small - uc still large because the integrator is “charged” - u still at maximum - y overshoots for a long time
-
+
ME 343 – Control Systems 84
PID Controller: Ziegler-Nichols Tuning
Plant with Anti-Windup:
Plant without Anti-Windup:
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ME 343 – Control Systems 85
PID Controller: Ziegler-Nichols Tuning In saturation, the plant behaves as:
For large Ka, this is a system with very low gain and very fast decay rate, i.e., the integration is turned off.
Saturation/Antiwindup: Example (Antiwindup_sim.mdl)
ME 343 – Control Systems 86
Root Locus Controller
-
+
Plant
Sensor
Writing the loop gain as KL(s) we are interested in tracking the closed-loop poles as “gain” K varies
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44
ME 343 – Control Systems
Root Locus Characteristic Equation:
The roots (zeros) of the characteristic equation are the closed-loop poles of the feedback system!!!
The closed-loop poles are a function of the “gain” K
Writing the loop gain as
The closed loop poles are given indistinctly by the solution of:
87
ME 343 – Control Systems
Root Locus
when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus)
Phase condition
Magnitude condition
Phase condition
Magnitude condition
88
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ME 343 – Control Systems
Root Locus by Characteristic Equation Solution
Example: -
+
Closed-loop poles:
89
ME 343 – Control Systems
We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS
Root Locus by Characteristic Equation Solution
90
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46
ME 343 – Control Systems
Phase and Magnitude of a Transfer Function
The factors K, (s-zj) and (s-pk) are complex numbers:
91
ME 343 – Control Systems
Phase and Magnitude of a Transfer Function
Now it is easy to give the phase and magnitude of the transfer function:
92
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ME 343 – Control Systems
Root Locus by Phase Condition
Example: -
+
belongs to the locus?
93
ME 343 – Control Systems
Root Locus by Phase Condition
belongs to the locus!
Note: Check code rlocus_phasecondition.m
94
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48
ME 343 – Control Systems
Root Locus by Phase Condition
We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS
95
ME 343 – Control Systems
Root Locus
Basic Properties:
• Number of branches = number of open-loop poles • RL begins at open-loop poles
• RL ends at open-loop zeros or asymptotes
• RL symmetrical about Re-axis
when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus)
96
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49
ME 343 – Control Systems
Root Locus
Rule 1: The n branches of the locus start at the poles of L(s) and m of these branches end on the zeros of L(s). n: order of the denominator of L(s) m: order of the numerator of L(s)
Rule 2: The locus is on the real axis to the left of and odd number of poles and zeros. In other words, an interval on the real axis belongs to the root locus if the total number of poles and zeros to the right is odd. This rule comes from the phase condition!!!
97
ME 343 – Control Systems
Root Locus
Rule 3: As K→∞, m of the closed-loop poles approach the open-loop zeros, and n-m of them approach n-m asymptotes with angles
and centered at
98
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50
ME 343 – Control Systems
Root Locus
Rule 4: The locus crosses the jω axis (looses stability) where the Routh criterion shows a transition from roots in the left half-plane to roots in the right-half plane.
Example:
99
ME 343 – Control Systems
Root Locus
Example:
100
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51
ME 343 – Control Systems
Root Locus
Design dangers revealed by the Root Locus:
• High relative degree: For n-m≥3 we have closed loop instability due to asymptotes.
• Nonminimum phase zeros: They attract closed loop poles into the RHP
Note: Check code lecture16_a.m
101
ME 343 – Control Systems
Root Locus
Viete’s formula:
When the relative degree n-m≥2, the sum of the closed loop poles is constant
102
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52
ME 343 – Control Systems
Phase and Magnitude of a Transfer Function
The factors K, (s-zj) and (s-pk) are complex numbers:
103
ME 343 – Control Systems
Phase and Magnitude of a Transfer Function
Now it is easy to give the phase and magnitude of the transfer function:
104
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53
ME 343 – Control Systems
Phase and Magnitude of a Transfer Function Example:
105
ME 343 – Control Systems
Root Locus- Magnitude and Phase Conditions
when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus)
106
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54
ME 343 – Control Systems
Root Locus
Selecting K for desired closed loop poles on Root Locus:
If so belongs to the root locus, it must satisfies the characteristic equation for some value of K
Then we can obtain K as
107
ME 343 – Control Systems
Root Locus
Example:
sys=tf(1,poly([-1 -5])) so=-3+4i [K,POLES]=rlocfind(sys,so)
Using MATLAB:
108
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55
ME 343 – Control Systems
Root Locus Example:
When we use the absolute value formula we are assuming that the point belongs to the Root Locus!
109
ME 343 – Control Systems
Root Locus - Compensators
Example:
Can we place the closed loop pole at so=-7+i5 only varying K? NO. We need a COMPENSATOR.
The zero attracts the locus!!! 110
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56
ME 343 – Control Systems
Root Locus – Phase lead compensator
Pure derivative control is not normally practical because of the amplification of the noise due to the differentiation and must be approximated:
How do we choose z and p to place the closed loop pole at so=-7+i5?
Phase lead COMPENSATOR
When we study frequency response we will understand why we call “Phase Lead” to this compensator.
ME 343 – Control Systems
Root Locus – Phase lead compensator Example:
Phase lead COMPENSATOR
112
Let us choose p=20
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57
ME 343 – Control Systems
Root Locus – Phase lead compensator Example:
Phase lead COMPENSATOR
113
ME 343 – Control Systems
Root Locus – Phase lead compensator
Selecting z and p is a trial an error procedure. In general:
• The zero is placed in the neighborhood of the closed-loop natural frequency, as determined by rise-time or settling time requirements. • The poles is placed at a distance 5 to 20 times the value of the zero location. The pole is fast enough to avoid modifying the dominant pole behavior.
The exact position of the pole p is a compromise between:
• Noise suppression (we want a small value for p) • Compensation effectiveness (we want large value for p)
114
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58
ME 343 – Control Systems
Root Locus – Phase lag compensator Example:
Phase lag COMPENSATOR
What can we do to increase Kp? Suppose we want Kp=10.
We choose:
115
ME 343 – Control Systems
Root Locus – Phase lag compensator Example:
116
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59
ME 343 – Control Systems
Root Locus – Phase lag compensator
Selecting z and p is a trial an error procedure. In general:
• The ratio zero/pole is chosen based on the error constant specification. • We pick z and p small to avoid affecting the dominant dynamic of the system (to avoid modifying the part of the locus representing the dominant dynamics) • Slow transient due to the small p is almost cancelled by an small z. The ratio zero/pole cannot be very big.
The exact position of z and p is a compromise between:
• Steady state error (we want a large value for z/p) • The transient response (we want the pole p placed far from the origin)
117
ME 343 – Control Systems
Root Locus - Compensators
Phase lead compensator:
Phase lag compensator:
We will see why we call “phase lead” and “phase lag” to these compensators when we study frequency response
118
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60
ME 343 – Control Systems
Frequency Response
• We now know how to analyze and design systems via s-domain methods which yield dynamical information
• The responses are described by the exponential modes – The modes are determined by the poles of the response Laplace
Transform
• We next will look at describing system performance via frequency response methods
• This guides us in specifying the system pole and zero positions
119
ME 343 – Control Systems
Sinusoidal Steady-State Response Consider a stable transfer function with a sinusoidal input:
• Where the natural system modes lie
– These are in the open left half plane Re(s)<0
• At the input modes s=+jω and s=-jω
Only the response due to the poles on the imaginary axis remains after a sufficiently long time
This is the sinusoidal steady-state response
The Laplace Transform of the response has poles
120
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ME 343 – Control Systems
Sinusoidal Steady-State Response
• Input
• Transform
• Response Transform
• Response Signal
• Sinusoidal Steady State Response
forced response natural response
121
ME 343 – Control Systems
Sinusoidal Steady-State Response
• Calculating the SSS response to • Residue calculation
• Signal calculation
) cos( ) ( φ ω + = t A t u
122
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62
ME 343 – Control Systems
Sinusoidal Steady-State Response
• Response to is
– Output frequency = input frequency – Output amplitude = input amplitude × |G(jω)| – Output phase = input phase + G(jω)
• The Frequency Response of the transfer function G(s) is given by its evaluation as a function of a complex variable at s=jω
• We speak of the amplitude response and of the phase response – They cannot independently be varied
» Bode’s relations of analytic function theory
123
ME 343 – Control Systems
Frequency Response
• Find the steady state output for v1(t)=Acos(ωt+φ)
• Compute the s-domain transfer function T(s) – Voltage divider
• Compute the frequency response
• Compute the steady state output
+ _ V1(s) sL
R V2(s)
+
-
124
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63
ME 343 – Control Systems
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω
Mag
nitu
de (d
B)
Phas
e (d
eg)
125
ME 343 – Control Systems
Bode Diagrams
126
The magnitude and phase of G(s) when s=jω is given by:
Linear in the phases
Nonlinear in the magnitudes
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ME 343 – Control Systems
Bode Diagrams
127
Why do we express in decibels?
Linear in the phases
Linear in the magnitudes (dB)
By properties of the logarithm we can write:
The magnitude and phase of G(s) when s=jω is given by:
ME 343 – Control Systems
Bode Diagrams
128
Why do we use a logarithmic scale? Let’s have a look at our example:
Expressing the magnitude in dB:
Asymptotic behavior:
LINEAR FUNCTION in logω!!! We plot as a function of logω.
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ME 343 – Control Systems
Bode Diagrams
129
Why do we use a logarithmic scale? Let’s have a look at our example:
Expressing the phase:
Asymptotic behavior:
LINEAR FUNCTION in logω!!! We plot as a function of logω.
ME 343 – Control Systems
General Transfer Function: Real poles/zeros
130
Magnitude and Phase:
Asymptotic behavior:
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66
ME 343 – Control Systems
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω
Mag
nitu
de (d
B)
Phas
e (d
eg)
131
-20dB/dec
ME 343 – Control Systems
Bode Diagrams
132
A cutoff frequency occurs when the gain is reduced from its maximum passband value by a factor :
Decade: Any frequency range whose end points have a 10:1 ratio
Bandwith: frequency range spanned by the gain passband Let’s have a look at our example:
This is a low-pass filter!!! Passband gain=1, Cutoff frequency=R/L The Bandwith is R/L!
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ME 343 – Control Systems
Frequency Response
133
Stable Transfer Function
• After a transient, the output settles to a sinusoid with an amplitude magnified by and phase shifted by .
• Since all signals can be represented by sinusoids (Fourier series and transform), the quantities and are extremely important.
• Bode developed methods for quickly finding and for a given and for using them in control design.
ME 343 – Control Systems
Bode Diagrams
134
The magnitude and phase of G(s) when s=jω is given by:
Linear in the phases
Nonlinear in the magnitudes
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68
ME 343 – Control Systems
Bode Diagrams
135
Why do we express in decibels?
Linear in the phases
Linear in the magnitudes (dB)
By properties of the logarithm we can write:
The magnitude and phase of G(s) when s=jω is given by:
ME 343 – Control Systems
General Transfer Function (Bode Diagrams)
136
Classes of terms:
1-
2-
3-
4-
The magnitude (dB) (phase) is the sum of the magnitudes (dB) (phases) of each one of the terms. We learn how to plot each term, we learn how to plot the whole magnitude and phase Bode Plot.
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ME 343 – Control Systems
General Transfer Function: DC gain
137
Magnitude and Phase:
ME 343 – Control Systems
General Transfer Function: Poles/zeros at origin
138
Magnitude and Phase:
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ME 343 – Control Systems
General Transfer Function: Real poles/zeros
139
Magnitude and Phase:
Asymptotic behavior:
ME 343 – Control Systems
General Transfer Function: Real poles/zeros
140
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71
ME 343 – Control Systems
General Transfer Function: Real poles/zeros
141
ME 343 – Control Systems
General Transfer Function: Complex poles/zeros
142
Magnitude and Phase:
Asymptotic behavior:
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ME 343 – Control Systems
General Transfer Function: Complex poles/zeros
143
ME 343 – Control Systems
General Transfer Function: Complex poles/zeros
144
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ME 343 – Control Systems
Frequency Response
145
Stable Transfer Function
• After a transient, the output settles to a sinusoid with an amplitude magnified by and phase shifted by .
• Since all signals can be represented by sinusoids (Fourier series and transform), the quantities and are extremely important.
• Bode developed methods for quickly finding and for a given and for using them in control design.
ME 343 – Control Systems
Bode Diagrams
146
The magnitude and phase of G(s) when s=jω is given by:
Linear in the phases
Nonlinear in the magnitudes
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ME 343 – Control Systems
Bode Diagrams
147
Why do we express in decibels?
Linear in the phases
Linear in the magnitudes (dB)
By properties of the logarithm we can write:
The magnitude and phase of G(s) when s=jω is given by:
ME 343 – Control Systems
General Transfer Function (Bode Diagrams)
148
Classes of terms:
1-
2-
3-
4-
The magnitude (dB) (phase) is the sum of the magnitudes (dB) (phases) of each one of the terms. We learn how to plot each term, we learn how to plot the whole magnitude and phase Bode Plot.
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ME 343 – Control Systems
Bode Diagrams
149
Example:
ME 343 – Control Systems
Frequency Response: Poles/Zeros in the RHP
150
Example:
• Same .
• The effect on is opposite than the stable case.
An unstable pole behaves like a stable zero An “unstable” zero behaves like a “stable” pole
This frequency response cannot be found experimentally but can be computed and used for control design.
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ME 343 – Control Systems
Bode Diagrams
151
A cutoff frequency occurs when the gain is reduced from its maximum passband value by a factor :
Decade: Any frequency range whose end points have a 10:1 ratio
Bandwith: frequency range spanned by the gain passband Let’s have a look at our example:
This is a low-pass filter!!! Passband gain=1, Cutoff frequency=R/L The Bandwith is R/L!
ME 343 – Control Systems
Frequency Response
152
Stable Transfer Function
BODE plots
NYQUIST plots
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ME 343 – Control Systems
Nyquist Diagrams
153
How are the Bode and Nyquist plots related?
They are two way to represent the same information
ME 343 – Control Systems
Nyquist Diagrams
154
Find the steady state output for v1(t)=Acos(ωt+φ)
Compute the s-domain transfer function T(s) Voltage divider
Compute the frequency response
Compute the steady state output
+ _ V1(s) sL
R V2(s)
+
-
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ME 343 – Control Systems
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω
Mag
nitu
de (d
B)
Phas
e (d
eg)
155
-20dB/dec
ME 343 – Control Systems
Nyquist Diagrams
156
1-
2-
3-
4-
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ME 343 – Control Systems
Nyquist Diagrams
157
ME 343 – Control Systems
Nyquist Diagrams
158
General procedure for sketching Nyquist Diagrams:
• Find G(j0)
• Find G(j∞)
• Find ω* such that Re{G(jω*)}=0; Im{G(jω*)} is the intersection with the imaginary axis.
• Find ω* such that Im{G(jω*)}=0; Re{G(jω*)} is the intersection with the real axis.
• Connect the points
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ME 343 – Control Systems
Nyquist Diagrams
159
Example:
1-
2-
3-
4-
ME 343 – Control Systems
Nyquist Diagrams
160
Example:
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ME 343 – Control Systems
Nyquist Diagrams from Bode Diagrams
161
ME 343 – Control Systems
Nyquist Stability Criterion
162
-
+
When is this transfer function Stable?
NYQUIST: The closed loop is asymptotically stable if the number of counterclockwise encirclements of the point (-1+j0) that the Nyquist curve of G(jω) is equal to the number of poles of G(s) with positive real parts (unstable poles)
Corollary: If the open-loop system G(s) is stable, then the closed-loop system is also stable provided G(s) makes no encirclement of the point (-1+j0).
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ME 343 – Control Systems
Nyquist Stability Criterion
163
ME 343 – Control Systems
Nyquist Stability Criterion
164
When is this transfer function Stable?
NYQUIST: The closed loop is asymptotically stable if the number of counterclockwise encirclements of the point (-1/K+j0) that the Nyquist curve of G(jω) is equal to the number of poles of G(s) with positive real parts (unstable poles)
-
+
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ME 343 – Control Systems
Neutral Stability
165
-
+
Root locus condition:
At points of neutral stability RL condition hold for s=jω
Stability: At If ↑K leads to instability If ↓K leads to instability
ME 343 – Control Systems
Stability Margins
166
The GAIN MARGIN (GM) is the factor by which the gain can be raised before instability results.
The PHASE MARGIN (PM) is the value by which the phase can be raised before instability results.
UNSTABLE SYSTEM
UNSTABLE SYSTEM
PM is the amount by which the phase of exceeds -180° when
GM is equal to at the frequency where .
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ME 343 – Control Systems
Stability Margins
167
PM
1/GM
ME 343 – Control Systems
Stability Margins
168
GM
PM
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ME 343 – Control Systems
Specifications in the Frequency Domain
169
1. The crossover frequency ωc, which determines bandwith ωBW, rise time tr and settling time ts.
2. The phase margin PM, which determines the damping coefficient ζ and the overshoot Mp.
3. The low-frequency gain, which determines the steady-state error characteristics.
ME 343 – Control Systems
Specifications in the Frequency Domain
170
The phase and the magnitude are NOT independent!
Bode’s Gain-Phase relationship:
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ME 343 – Control Systems
Specifications in the Frequency Domain
171
The crossover frequency:
ME 343 – Control Systems
Specifications in the Frequency Domain
172
The Phase Margin: PM vs. Mp
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ME 343 – Control Systems
Specifications in the Frequency Domain
173
The Phase Margin: PM vs. ζ
ME 343 – Control Systems
Frequency Response – Phase Lead Compensators
174
It is a high-pass filter and approximates PD control. It is used whenever substantial improvement in damping is needed. It tends to increase the speed of response of a system for a fixed low-frequency gain.
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ME 343 – Control Systems
Frequency Response – Phase Lead Compensators
175
1. Determine the open-loop gain K to satisfy error or bandwidth requirements: - To meet error requirement, pick K to satisfy error constants (Kp, Kv, Ka) so that ess specification is met. - To meet bandwidth requirement, pick K so that the open-loop crossover frequency is a factor of two below the desired closed-loop bandwidth.
2. Determine the needed phase lead → α based on the PM specification.
3. Pick ωMAX to be at the crossover frequency.
4. Determine the zero and pole of the compensator. z=1/T= ωMAX α1/2 p=1/ α T= ωMAX α-1/2
5. Draw the compensated frequency response and check PM.
6. Iterate on the design. Add additional compensator if needed.
ME 343 – Control Systems
Frequency Response – Phase Lag Compensators
176
It is a low-pass filter and approximates PI control. It is used to increase the low frequency gain of the system and improve steady state response for fixed bandwidth. For a fixed low-frequency gain, it will decrease the speed of response of the system.
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ME 343 – Control Systems
Frequency Response – Phase Lag Compensators
177
1. Determine the open-loop gain K that will meet the PM requirement without compensation.
2. Draw the Bode plot of the uncompensated system with crossover frequency from step 1 and evaluate the low-frequency gain.
3. Determine α to meet the low frequency gain error requirement.
4. Choose the corner frequency ω=1/T (the zero of the compensator) to be one decade below the new crossover frequency ωc.
5. The other corner frequency (the pole of the compensator) is then ω=1/ α T.
6. Iterate on the design
ME 343 – Control Systems
Frequency Response – Phase Lead Compensators
178
It is a high-pass filter and approximates PD control. It is used whenever substantial improvement in damping is needed. It tends to increase the speed of response of a system for a fixed low-frequency gain.
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ME 343 – Control Systems
Frequency Response – Phase Lead Compensators
179
1. Determine the open-loop gain K to satisfy error or bandwidth requirements: - To meet error requirement, pick K to satisfy error constants (Kp, Kv, Ka) so that ess specification is met. - To meet bandwidth requirement, pick K so that the open-loop crossover frequency is a factor of two below the desired closed-loop bandwidth.
2. Determine the needed phase lead → α based on the PM specification.
3. Pick ωMAX to be at the crossover frequency.
4. Determine the zero and pole of the compensator. z=1/T= ωMAX α1/2 p=1/ α T= ωMAX α-1/2
5. Draw the compensated frequency response and check PM.
6. Iterate on the design. Add additional compensator if needed.
ME 343 – Control Systems
Frequency Response – Phase Lag Compensators
180
It is a low-pass filter and approximates PI control. It is used to increase the low frequency gain of the system and improve steady state response for fixed bandwidth. For a fixed low-frequency gain, it will decrease the speed of response of the system.
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ME 343 – Control Systems
Frequency Response – Phase Lag Compensators
181
1. Determine the open-loop gain K that will meet the PM requirement without compensation.
2. Draw the Bode plot of the uncompensated system with crossover frequency from step 1 and evaluate the low-frequency gain.
3. Determine α to meet the low frequency gain error requirement.
4. Choose the corner frequency ω=1/T (the zero of the compensator) to be one decade below the new crossover frequency ωc.
5. The other corner frequency (the pole of the compensator) is then ω=1/ α T.
6. Iterate on the design