automatic control
DESCRIPTION
Automatic Control. Review. Please Return Loan Clickers to the MEG office after Class! Today!. FINAL EXAM: Wednesday December 8 8:00 AM to 10:00 a.m. Feedback Terminology. - PowerPoint PPT PresentationTRANSCRIPT
Review
Please Return Loan Clickers to the MEG office after Class!
Today!
FINAL EXAM: Wednesday December 8 8:00 AM to 10:00 a.m.
Feedback Terminology
ControllerSum
Ksensor
Y(s)R(s) -Gplant
Disturbance W(s)+ or -
Actuator
Y is the 'ControlledOutput'
Sensor Output
Process(Brain) Power Unit
U(s)
Actuator OutputE(s) = R -Y = Deviationor ERROR, assuming
Ksensor = 1
In Block diagrams, we use not the time domain variables, but their Laplace Transforms. Always
denote Transforms by (s)!
ControllerSum
Ksensor
Y(s)R(s) -Gplant
Disturbance W(s)+ or -
Actuator
Y is the 'ControlledOutput'
Sensor Output
Process(Brain) Power Unit
U(s)
Actuator OutputE(s) = R -Y = Deviationor ERROR, assuming
Ksensor = 1
Deriving differential equations in state-variable form consists of writing them as a vector equation as follows:
uJXHyuGXFX
where is the output and u is the input
State-Variable Form
Transfer Function
Y(s)GplantR(s)
4 *s*Y(s) + Y(s) = R(s)Regroup:
Y(s)*(4s+1) = R(s)
Regroup:
Y(s) = 1__ R(s) 4s + 1
Fourier Transform:Let period T infinity
The interval betweenDiscrete frequencies 0
The Fourier series becomesthe Fourier Transform
dtBtAtf )sin)(cos)((21)(
0
dejCtf tj
)(21)(
dtetfjC tj
)(
21)(where
Compare with the definition of the Laplace Transform
G(s) = 2/(0.2s+1)
1. Note K and b
2. Draw |F| from low freq to b
3. Draw |F| from b , slope -1/decade
Bode Magnitude Plot
G(s) = 2/(0.2s+1)Bode Phase Plot
1. Phase = -450 at b
2. Draw from 0 to b/10 , slope =0
3. Draw from b/10 freq to 10*b
4. Min Phase is -900 from 10*b
Decibels• An alternate unit of
Magnitude or Gain
• Definition: xdB = 20* lg(x)
• dB Notation is widely used in Filter theory and Acoustics
x lg(x) X(db)
10 1 20
100 2 40
0.1 -1 -20
Bode Plot ConstructionG(s) = 2/(s)(s+1)
10-1
100
101
10-1
100
101
mag
nitu
de Bode Plot (a) magnitude
10-1
100
101
-180
-160
-140
-120
-100
-80
(rad/sec)
phas
e (d
eg)
(b) phase
1. Construct each Element plot
Integrator Slope = -1
Integrator Phase = -90 deg.
2. Graphical Summation
Gain = 2.
Slope = -2
Bode Plot of 1/(s(s+1)): Matlab Plot
10-1
100
101
10-2
100
102
mag
nitu
de Bode Plot (a) magnitude
10-1
100
101
-180
-160
-140
-120
-100
-80
(rad/sec)
phas
e (d
eg)
(b) phase
Bode Plot Construction
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
G(s) = 5*(s+1)/(10s+1)(100s+1)
1. Construct each Element plot
2. Graphical Summation: Complete plot. Note beginning and final values
K = 5 Slope = -1
Slope = -2
Slope = -1
Phase Plot Construction
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
G(s) = 5*(s+1)/(10s+1)(100s+1)
2. Graphical Summation of phase angles. Note beginning and final phase values. Here: = 0 at = 0, and = -90 final angle
K = 5
Initial Phase is zero to 0.001, follows the first Phase up to 0.01
- 90 deg./decade
0 deg./decade+45 deg./decade
Final phase:Constant - 90 deg
Bode Plot Construction: Matlab Plot
10-2
10-1
100
101
10-4
10-2
100
102
mag
nitu
de Bode Plot (a) magnitude
10-2
10-1
100
101
-150
-100
-50
(rad/sec)
phas
e (d
eg)
(b) phase
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
Break frequencies:Poles at 0.01, 0.1
Zero at 1
Given: An open-loop system
At = 0.1, the Magnitude is approximately
• (A) 1• (B) 0.1• (C) 0.01• (D) 0• (E) 1/(1000)
)110)(1100()1()(
ss
ssGopen
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
Break frequencies:Poles at 0.01, 0.1
Zero at 1
Given: An open-loop system
At = 0.1, the Magnitude is approximately
• (A) 1• (B) 0.1• (C) 0.01• (D) 0• (E) 1/(1000)
)110)(1100()1()(
ss
ssGopen
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
Break frequencies:Poles at 0.01, 0.1
Zero at 1
Given: An open-loop system
At =1, the phase angle is approximately
• (A) 0 degrees• (B) -45 degrees• (C) -135 degrees• (D) -180 degrees• (E) -90 degrees
)110)(1100()1()(
ss
ssGopen
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
Break frequencies:Poles at 0.01, 0.1
Zero at 1
Given: An open-loop system
At =1, the phase angle is approximately
• (A) 0 degrees• (B) -45 degrees• (C) -135 degrees• (D) -180 degrees• (E) -90 degrees
)110)(1100()1()(
ss
ssGopen
Bode Lead Design
1. Select Lead zero such that the phase margin increases while keeping the gain crossover frequency as low as reasonable.
2. Adjust Gain to the desired phase margin.
Bode Lead Design
-200
-150
-100
-50
0
Mag
nitu
de (d
B)
Plant
10-1
100
101
102
103
-270
-225
-180
-135
-90
Phas
e (d
eg)
Bode Example of plant addition, Plant = 2/[s(0.25s+1)(s/6+1)
Frequency (rad/sec)
Bode Lead Design
-200
-150
-100
-50
0
50
Mag
nitu
de (d
B)
Plant
LEAD
PLant*LEAD
10-1
100
101
102
103
-270
-180
-90
0
90
Phas
e (d
eg)
Bode Example of plant addition, Plant+ Lead
Frequency (rad/sec)
Bode Lead Design
-200
-150
-100
-50
0
50
Mag
nitu
de (d
B)
Plant
LEAD
PLant*LEAD
Final with adjusted Gain
10-1
100
101
102
103
-270
-180
-90
0
90
Phas
e (d
eg)
Bode Example of plant addition, gain adjusted, Plant* Lead
Frequency (rad/sec)
-margin = 51 deg. K = 100
Bode Lag Design
1. All other design should be complete. Gain K and phase margin are fixed
2. Select Lag zero such that the phase margin does not drop further. (Slow)
3. Steady State Gain should now be about 10 times larger than without Lag.
Bode Lag Design
-60
-40
-20
0
20
40
Mag
nitu
de (d
B)
10-2
10-1
100
101
-270
-225
-180
-135
-90
Phas
e (d
eg)
Bode plot of plant 1/[s(0.2s+1)(s+1)
Frequency (rad/sec)
Lag compensator |p| = 0.1*zG(s) =Construct each Element plot
Slope = 0 Gain = 0.1
1*11*1
spole
szero
Slope = 0
Phase = 0
Slope = 0
Slope = 0
Slope = -1
Slope = 0
Note Break Frequencies
Bode Lag Design
-80
-60
-40
-20
0
20
40
Mag
nitu
de (d
B) Plant
LAG
PLant*LAG
10-2
10-1
100
101
-270
-225
-180
-135
-90
-45
0
Phas
e (d
eg)
Bode Example of plant addition, Plant+ Lag
Frequency (rad/sec)
Bode Lag Design
-100
-80
-60
-40
-20
0
20
40
60
80
100
Mag
nitu
de (d
B) Plant
LAG
PLant*LAG
Final with adjusted Gain
10-2
10-1
100
101
-270
-225
-180
-135
-90
-45
0
Phas
e (d
eg)
Bode Example of plant addition, gain adjusted, Plant* Lag
Frequency (rad/sec)
-margin = 39 deg. K = 10
Lead Design Example• (a) P-control for phase margin of 45 degrees.
Controller gain K = 0.95
)11.0(*)1(2)( 2
ss
sG
• (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Initial design: Lead is too slow
)11.0(*)1(2)( 2
ss
sG
Lead is too slow. Lead Zero should be near the phase margin. Here: Place Lead zero around 3 rad/s.
• (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Improved design: Lead zero at 3, pole at 30 rad/s )11.0(*)1(
2)( 2
sssG
Lead zero at 3. Lead pole at 30.
New gain crossover at 5 rad/s
Final step: adjust gain K such that |F| = 0 dB at cr.
Result: The controller gain is now K = 3.4 (4 times better than P-
control)
Bode Lead and Lag Design:General placement rules
10-1
100
101
102
-270
-180
-90
0
Phas
e (d
eg)
Bode Example of plant addition, Plant = 2/[s(0.1s+1)(s+1)2
Frequency (rad/sec)
-100
-80
-60
-40
-20
0
20
Mag
nitu
de (d
B)
Place Lead Zero near desired Gain Crossover Frequency
Phase Margin
Place Lag Zero at a decade belowGain Crossover Frequency