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:

uJXHy

uGXFX

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)((2

1)(

0

dejCtf tj

)(2

1)(

dtetfjC tj

)(

2

1)(

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 Plot1. 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

magnitude

Bode Plot (a) magnitude

10-1

100

101

-180

-160

-140

-120

-100

-80

(rad/sec)

phase (

deg)

(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

magnitude

Bode Plot (a) magnitude

10-2

10-1

100

101

-150

-100

-50

(rad/sec)

phase (

deg)

(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

Magnitu

de (

dB

)

Plant

10-1

100

101

102

103

-270

-225

-180

-135

-90

Phase (

deg)

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

Magnitu

de (

dB

)

Plant

LEAD

PLant*LEAD

10-1

100

101

102

103

-270

-180

-90

0

90

Phase (

deg)

Bode Example of plant addition, Plant+ Lead

Frequency (rad/sec)

Bode Lead Design

-200

-150

-100

-50

0

50

Magnitu

de (

dB

)

Plant

LEAD

PLant*LEAD

Final with adjusted Gain

10-1

100

101

102

103

-270

-180

-90

0

90

Phase (

deg)

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

Magnitu

de (

dB

)

10-2

10-1

100

101

-270

-225

-180

-135

-90

Phase (

deg)

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*1

1*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

Magnitu

de (

dB

) Plant

LAG

PLant*LAG

10-2

10-1

100

101

-270

-225

-180

-135

-90

-45

0

Phase (

deg)

Bode Example of plant addition, Plant+ Lag

Frequency (rad/sec)

Bode Lag Design

-100

-80

-60

-40

-20

0

20

40

60

80

100

Magnitu

de (

dB

)

Plant

LAG

PLant*LAG

Final with adjusted Gain

10-2

10-1

100

101

-270

-225

-180

-135

-90

-45

0

Phase (

deg)

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

sssG

• (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

sssG

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

Pha

se (

deg)

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 (

dB)

Place Lead Zero near desired Gain Crossover Frequency

Phase Margin

Place Lag Zero at a decade belowGain Crossover Frequency