cl i l t lclassical control - upc universitat politècnica...

Control & Guidance 2011

Enginyeria Tècnica d'AeronàuticaA ióesp. en Aeronavegació

Escola d'Enginyeria de Telecomunicació i Aeroespacial de Castelldefels

Adeline de Villardi de Montlaur

Cl i l t lCl i l t lClassical controlClassical control

Control and guidance

Slide 1

Cl i l t lCl i l t lClassical control1 P t i ti ti

Classical control1 P t i ti ti1. Parametric estimation

2 Steady state error

1. Parametric estimation

2 Steady state error2. Steady state error

3. Root locus

2. Steady state error

3. Root locus

4. Controllers4. Controllers

5. Frequency response5. Frequency response

6. Bode diagrams6. Bode diagrams


Slide 2

Study properties of the response of the system:

desired angle of attack αrefd

actual angle of attack αdisplacement

Study properties of the response of the system:

LONGITUDINAL CONTROL LONGITUDINAL

DYNAMICS

speed urefg

speed udisplacement of elevator δe

CONTROL SYSTEM DYNAMICS

t l l f tt k

SENSORS:

actual angle of attack αspeed u

SENSORS: INS,

Anemometer

1- Parametric estimation1- Parametric estimationControl and guidance

Slide 3

1- Parametric estimation1- Parametric estimation

Temporal methods:Temporal methods:

a. Firsta. First--order systemsorder systems

b. Secondb. Second--order systemsorder systems

c. Higherc. Higher--order systemsorder systems


Slide 4

Fi tFi t d td t


A first-order system is defined by a first-order


A first order system is defined by a first order

differential equation:

s1K

)s(R)s(Y)s(G)t(Kr)t(y)t(y L

s1)s(R

τ: system time constantτ: system time constant

K: gain

Electrical/mechanical examples


Slide 5

examples

Fi tFi t d td t


Impulse response


Impulse response

1)s(R)]t([L 1K)s(Y )()]([ s1

using the inverse Laplace transform, the impulse

response is:

0teK)t(yt

0te)t(y


Slide 6

a Firsta First order systemsorder systems


Impulse response 1


0teK)t(yt

0.8

0.9

y(t))(y

0.6

0.7

)Tangent slope in 0:

y( )

0 3

0.4

0.5y(t)K)t(dy

g p

0.1

0.2

0.320tdt

r(t)

0 1 2 3 4 5 6 70

t


Slide 7

Fi tFi t d td t


Step response: response to a unit step function


p p p p

1KK

)1(K)s(Y

1)s(R)]t(u[L s1ss)s1(

)(s

)()]([

i th i L l t f th tusing the inverse Laplace transform, the step response

or indicial response is:

0te1K)t(yt

0te1K)t(y


Slide 8

Fi tFi t d td t


St 9

10


y(t)Step response:

t

8

9y( )

te1K)t(y

6

7

)Tangent slope in 0: 4

5y(t)

2

3

K

dt)t(dy r(t)

0 1 2 3 4 5 6 70

1

t

dt 0t

Example


Slide 9

t

b S db S d d td t


A second-order system is defined by a second-order


A second order system is defined by a second order

differential equation:

)t(ra)t(yb)t(yb)t(yb 0012

0a)s(Y)(G01

22

0

bsbsb)s(R)()s(G

012)(

Electrical/Mechanical examples


Slide 10

ect ca / ec a ca e a p es

b Secondb Second order systemsorder systems


It can be factorized to emphasize particular


parameters:2KK)s(Y

2nn

2n

2 s2sK

1s2s

K)s(R)s(Y)s(G

nn

12

with K: system gain (corresponds to final value for a unit step function)

ωn: undamped natural frequency

ζ: damping factor (ζ>0)Control and guidance

Slide 11

ζ: damping factor (ζ 0)



Step response:2nK)s(Y


p p

Response depends on the poles of the transfer function

2nn

2 s2s)s(R Response depends on the poles of the transfer function

0s2s 2nn

2

2442

s2n

2n

2n

1444let2

22n

2n

2n

2 di i i t’ i d d ζ l→ discriminant’s sign depends on ζ value

→ poles and response's properties depend on ζ valueControl and guidance

Slide 12

poles and response s properties depend on ζ value



ζ>1 Over-damped movement (non-oscillatory modes)


Real and negative poles: 1s 2n21

2n

2n K1K1)s(Y

n2,1

212nn

2 ssssss2ss)s(Y

111

Development in simple fractions:

22112n

sss1

sss1

12s1K)s(Y


Slide 13



ζ>1 Over-damped movement (non-oscillatory modes)1

Step Response


tsts

Inverse Laplace transform:0.8

1

y(t)y(t)

2

ts

1

ts

2n

se

se

121K)t(y

21

0.6

e

y(t)

0.4

Am

plit

ude

Tangent slope in 0:

)t(dy0.2 =4

=1.5

0dt)t(dy

0t

0 5 10 15 20 25 300

Time (sec )


Slide 14



ζ=1 Critically damped movement (non-oscillatory modes)


Double real negative poles: n2,1s

22n

sK

s1)s(Y

nss


n 11K)(Y


n

2n

n

sssK)s(Y


Slide 15



ζ=1 Critically damped movement (non-oscillatory modes)


1Step Response

Inverse Laplace transform:y(t)

tnnet11K)t(y

0 6

0.8y( )

Tangent slope in 0:

)t(dy 0.4

0.6

Am

plit

ude

0dt)t(dy

0t

0.2

0 2 4 6 80

Time (sec )


Slide 16

Time (sec )



ζ<1 Under-damped movement (oscillatory modes)


Conjugated complex poles: 21js Conjugated complex poles:

2K1

n2,1 1js

22n

2n

n

1ssK

s1)s(Y

Development in simple fractions…p p


Slide 17



ξ<1 Under-damped movement: Inverse Laplace transform:


t1sin1

t1cose1K)t(y 2n2

2n

tn

S R 1

1 4

1.6

1.8Step Response

=0.7=0.1

Tangent slope in 0:1

1.2

1.4

litud

e

Tangent slope in 0:

0)t(dy 0 4

0.6

0.8Am

p

0dt)(y

0t

0 10 20 30 400

0.2

0.4


Slide 18

Time (sec )



Characteristic parametersvaluefinaloutput


K: gain

M: maximum overshoot : represents the value of thevaluefinalinputvaluefinaloutput

M: maximum overshoot : represents the value of the highest peak of the system response measured with

t t th f l (fi l l )respect to the reference value (final value)

tp: peak time: time needed for the response to arrive at its p p pfirst peak

T: period

t : settling timeControl and guidance

Slide 19

ts: settling time



Characteristic parameters: for second-order systems


K 2

21tan eeM

natural(dampedζ1ωω 2

nd

2p1

t

frequency)

( p

2n

p1

22 2

nd 122T


Slide 20

n

b Secondb Second--order systemsorder systems1- Parametric estimation1- Parametric estimation

Obtain:b. Secondb. Second order systemsorder systems3,5

Step ResponseObtain:KM

2,5

3 r(t)Mtpt 5%

2

itude

ts5%

T

1

1,5Am

pl

Deduce:ζ

0,5

1

y(t)

ζωn

0 1 2 3 4 5 6 7 8 9 10 11 120

Time (sec)

y( )Calculate: G(s)=Y(s)/R(s)


Slide 21

Time (sec)( ) ( ) ( )

Hi hHi h d td t


c. Higherc. Higher--order systemsorder systems→ characterize the transitory state of any-order systems

generally y(t)= linear combination of elementary time

functions defined by the nature (real or complex) of the

characteristic equation roots: system modes:characteristic equation roots: system modes:

• real poles: non- oscillatory modes, exponential term in

the response

• complex poles: oscillatory modes, exponential term

multiplied by sine or cosineControl and guidance

Slide 22

p y

Hi hHi h d td t


High order systems can be simplified using:

c. Higherc. Higher--order systemsorder systems

High order systems can be simplified using:

dominant poles

poles further from the imaginary axis have a

weaker contribution

1 pole near 1 zero

if there is a zero near a pole this pole contributionif there is a zero near a pole, this pole contribution

will be weak


Slide 23

Study error of the response of the system:

desired angle of attack αrefspeed u actual angle of attack αdisplacement

Study error of the response of the system:


DYNAMICS

speed urefg



t l l f tt k

SENSORS:


SENSORS: INS,

Anemometer

2- Steady state error2- Steady state errorControl and guidance

Slide 24

2- Steady state error2- Steady state error

G(s)Y(s)+

-R(s) Ess

Steady State error:

ess= difference between the entry signal and the exit signal

e = “what we want minus what we get”ess = what we want minus what we get


Slide 25


System’s type:y yp

Given the transfer function:2

)ss1) (s1)(s1(s)...dscs1)...(bs1)(as1(K)s(G 2N

2

)...ss1)...(s1)(s1(s

with K: system gain,

and N: number of poles in the origin

→ N = system’s typeControl and guidance

Slide 26


Definition of Steady State error:

)t(y)t(rlimetss

t

e > 0 : exit signal has not reached the entry referenceess > 0 : exit signal has not reached the entry reference

ess < 0 : exit signal is higher than the entryss g g y


Slide 27


)t(y)t(rlime )t(y)t(rlimetss

Moving to the Laplace space: Final value theorem:

)s(R)s(G1)s(G)s(Rslim)s(Y)s(Rslime

0s0sss )s(G1

)s(R Depends on the entry

)s(G1

)s(Rslime0sss

p y

+ on the system’s type


Slide 28

)...dscs1)...(bs1)(as1(K)(G2


G(s)Y(s)+R(s)

)...ss1)...(s1)(s1(s)) ()((K)s(G 2N

G(s)-

1. Position error: error for a step function entry: r(t)=u(t)

1111

)s(Glim1

1)s(G1

1lims1

)s(G11slime

0s0s0sp

0type10s

Itype0

ypK1


Slide 29

1 Position error: error for a step function entry: r(t)=u(t)


1

Step Response

1. Position error: error for a step function entry: r(t)=u(t)

0 8

1

Error

0.6

0.8

ude

0.4

Am

plitu

0.2

0 2 4 6 8 10 120

Ti ( )


Slide 30

Time (sec )

2- Steady state error2- Steady state error)...dscs1)...(bs1)(as1(K)(G2

G(s)Y(s)+R(s)

)...ss1)...(s1)(s1(s)) ()((K)s(G 2N

2 Speed error: error for a ramp function entry: r(t)= t

( )-

2. Speed error: error for a ramp function entry: r(t)= t

1lim1lim11slime

0type

)s(sGlim

)s(sGslim

s)s(G1slime

0s0s20sv

Itype10type

IItype0

ItypeK


Slide 31

IItype0

2- Steady state error2- Steady state error)...dscs1)...(bs1)(as1(K)(G2

G(s)Y(s)+R(s)

)...ss1)...(s1)(s1(s)) ()((K)s(G 2N

G(s)-

3. Acceleration error: error for a parabolic entry: r(t)= t2

1li1li11li

I0

)s(Gslim

)s(Gsslim

s)s(G1slime 20s220s30sv

IItype1Ior0type

IIItype0

IItypeK


Slide 32

IIItype0


Error based on type + entryyp y

Input:Type:

step ramp parabolic

0 constant ∞ ∞

I 0 constant ∞

II 0 0 constant


Slide 33


Example: G(s) θ(s)+θref(s) δe(s)K

C t th i t d t t f it t f ti

-

Compute the error in steady state for a unit step function entry and for a system with the following open loop transfer function: 10s2)s( function:

4s1.0s1.0s2

)s()s(

2e

• for K=1, K=10, K=100,

)(e

0 12sθ(s) • for K=1 and

40.1sss0.12s

(s)δθ(s)

2e


Slide 34

Design a simple proportional controller in order to satisfy some



constraints on the response of the system


DYNAMICS

speed urefg



t l l f tt k

SENSORS:


SENSORS: INS,

Anemometer

3- Root locus3- Root locusControl and guidance

Slide 35

3- Root locus3- Root locus

• Root locus technique• Root locus technique

• Gain setting• Gain setting

• Effect of zeros and poles• Effect of zeros and poles


Slide 36

Root locus techniqueRoot locus technique



• Introduced by W. R. Evans in 1949: developed a series of rules

that allow the control system engineer to quickly draw the

root locus diagram = locus of all possible roots of the

characteristic equation: 1+K G(s)= 0

= locus of all possible poles in closed loop= locus of all possible poles in closed loop

as K varies from 0 to infinity

• The resulting plot helps us in selecting the best value of K

G f f f• Gives information for the transitory part of the response

(stability, damping factor, natural frequency)


Slide 37




Let

m21 zszszs)s(G

A d b tit t it i th h t i ti ti

n21 pspsps)(

And substitute it in the characteristic equation

zszszsk

Kkwhere0pspspszszszsk1n21

m21


Slide 38




The characteristic equation is complex and can be written in terms

of magnitude and angle as follows

zszszsk 21 1

pspspszszszsk

n21

m21

180)1q2(pszsn

1ii

m

1ii

)1mn(...,2,1,0qfor1i1i


Slide 39

Root locus technique: RulesRoot locus technique: Rules



If we rearrange the magnitude criteria as

k1

pspspszszszs m21

Rule 1: The number of separate branches of the root locus plot is

kpspsps n21

equal to the number of poles of the transfer function (n)

Branches of the root locus originate at the poles of G(s) for k=0Branches of the root locus originate at the poles of G(s) for k=0

and terminate at either the open-loop zeroes or at infinity for k=+∞

n separate branches, n-m infinite branches, m finite branches


Slide 40




Rule 2: Because the complex poles are always “conjugated”Rule 2: Because the complex poles are always conjugated ,

the root locus branches are symmetric with respect to the real axis


Slide 41




Rule 3:Rule 3:

Segments of the real axis that are part of the root locus:

points on the real axis that have an odd number of poles and

zeroes to their right


Slide 42



Root locus technique: RulesRoot locus technique: RulesRule 4: Asymptotes

The root locus branches that approach the open-loop zeroes atThe root locus branches that approach the open loop zeroes at

infinity do so along straight-line asymptotes that intersect the real

axis at the center of gravity of the finite poles and zeroesaxis at the center of gravity of the finite poles and zeroes

zpm

i

n

i

mn

p1i

i1i

i

The angle that the asymptotes make with the real axis is given by

)1(210f1q2º180

)1mn(...,2,1,0qformnq

a


Slide 43




Rule 5: breakaway points

If a portion of the real axis is part of the root locus and a branch is

between two poles the branch must break away from the real axis p y

so that the locus ends on a zero as k approaches infinity. The

breakaway points on the real axis are determined by solvingbreakaway points on the real axis are determined by solving

kfor0zszszsk1 m21

and then finding the roots of the equation dk/ds=0

kfor0pspsps

1n21

and then finding the roots of the equation dk/ds=0

Only roots that lie on a branch of the locus are of interest


Slide 44




Rule 6: Intersection with the imaginary axisRule 6: Intersection with the imaginary axis

Solve the characteristic equation for s=jω (equation of the

imaginary axis)

zjzjzjk 0

pjpjpjzjzjzjk1n21

m21


Slide 45




Rule 7: for complex poles and zeroes only:Rule 7: for complex poles and zeroes only:

The angle of departure of the root locus from a pole of G(s) or

arrival angle at a zero of G(s) can be found by the following

expression

If you consider a test point t:nm

180ptztn

1ii

m

1ii


Slide 46

Root locus technique: examplesRoot locus technique: examples



Example 1:

Root Locus2

Example 1:

1

1.5

ary

Axi

s0

0.5

3s2s2s)s(G 2

Imag

in

1

-0.5

03s2s

-1.5

-1

Real Axis-4 -3 -2 -1 0

-2


Slide 47




Example 2: 4Root Locus

Example 2:

2

3

2s1ss1)s(G

0

1

ary

Axi

s

2s1ss

2

-1

0Im

agin

a

-3

-2

-6 -4 -2 0 2-4

Real Axis


Slide 48

Setting of the gain and natural frequencySetting of the gain and natural frequency


Setting of the gain and natural frequencySetting of the gain and natural frequencyBasic operation: to adjust the gain K to obtain a damping factor

given by the poles in closed loop and fixed by the damping factor ζgiven by the poles in closed-loop and fixed by the damping factor ζ

)R (Cf second-order systems: i

s)sRe(

straight line doing an angle φ with the real axis (cos φ= ζ) sets an

isg g g φ ( φ ζ)

intersection point with the poles position, and k (and then K) is

obtained solving the characteristic equationobtained solving the characteristic equation

Natural frequency for a second-order system: ωn=|si|


Slide 49

Gain settingGain setting



0)s(KG1 0

pspspszszszsK1 m21

The system total gain is computed thanks to the module condition

pspsps n21

szszszspspspKk m21

szszsz n21

“total gain” = product of the distances from the poles of G(s) to the

intersection point (= target pole) divided by the product of the

distances from zeros of G(s)


Slide 50




0)s(KG1

0pspsps

K1

If there are no zeros:

pspsps n21

spspspK m21 “total gain” =product of the distances between the poles of G(s)

( )and the intersection point (= target pole)

Examples


Slide 51

p




40.5

Root Locus

1

2

3

0.5

2s1ss1)s(G

0

1

ary

Axi

sDesigner requirement:we want ζ=0.5

2

-1

0

Imag

ina

ζ=0.5=cos φ → φ=60º

-3

-2

0.5

ζ φ φ

Examples

-6 -5 -4 -3 -2 -1 0 1 2-4

Real Axis


Slide 52

p




1

Root Locus3 0.5

S t

2s1ss1)s(G

1

2System: sys

Gain: 1.04Pole: -0.332 + 0.577i

Damping: 0.499Overshoot (%): 16.4

Designer requirement:we want ζ=0.5

gin

ary

Axi

s

0

1 Frequency (rad/sec): 0.666

ζ=0.5=cos φ → φ=60º

Imag

2

-1

ζ φ φ

3 5 3 2 5 2 1 5 1 0 5 0 0 5 1 1 5 2

-3

-2

0.5

Examples

Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2


Slide 53

p




1

Root Locus3 0.5

2s1ss1)s(G

1

2System: sysGain: 1.04Pole: -0.332 + 0.577iDamping: 0.499Overshoot (%): 16.4

Sys tem: sysGain: 1.04Pole: -2.33

Designer requirement:we want ζ=0.5 in

ary

Axi

s

0

1( )

Frequency (rad/sec): 0.666Damping: 1Overshoot (%): 0

Frequency (rad/sec): 2.33

This corresponds to Im

ag-1 System: sys

Gain: 1.04Pole: -0.332 - 0.578iDamping: 0.498p

k=1.04

-3

-2

0.5

Damping: 0.498Overshoot (%): 16.5Frequency (rad/sec): 0.667

Examples

Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2


Slide 54

p




2

Root Locus

1 5

20.7

3s2s2s)s(G 2

0 5

1

1.5


agin

ary

Axi

s

0

0.5

Ima

-1

-0.5

-4 -3 -2 -1 0-2

-1.50.7

Examples

Real Axis-4 -3 -2 -1 0


Slide 55

p



Gain settingGain settingRoot Locus

20.7

2

1

1.5 System: sysGain: 1.32Pole: -1.66 + 1.69iDamping: 0.7O h t (%) 4 6

3s2s2s)s(G 2


nary

Axi

s

0

0.5Overshoot (%): 4.6Frequency (rad/sec): 2.37

This corresponds to

Imag

in

-1

-0.5 System: sysGain: 1.32Pole: -1.66 - 1.69ip

k=1.32

2

-1.5

1

0.7

Damping: 0.7Overshoot (%): 4.6Frequency (rad/sec): 2.37

ExamplesReal Axis

-4 -3 -2 -1 0-2


Slide 56

p

R l ti t bilit i iR l ti t bilit i i


Relative stability: gain marginRelative stability: gain margin

“Re(s)<0” criterion informs about the absolute stability of a system ( ) y y

but it says nothing about its relative stability

h f i i f h i bili h= how far it is from the instability → system strength

Gain margin: maximum proportional factor that can be introduced

into the control loop until the system becomes critically stable.

CrG

kM actual

G kM

Examples


Slide 57

p




1

Root Locus3 0.5

2s1ss1)s(G

1

2

System: sys

Designer requirement:we want ζ=0.5 in

ary

Axi

s

0

1 System: sysGain: 5.97Pole: 0.000761 + 1.41iDamping: -0.00054Overshoot (%): 100Frequency (rad/sec): 1 41

This corresponds to k=1

Imag

-1

Frequency (rad/sec): 1.41

kcr=6

MG=6 -3

-2

0.5

Examples

Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2


Slide 58

p

Roots locus exerciseRoots locus exercise



2

1

1

1.5

2s2s2ss1)s(G 2

0.5

-0.5

0

1 5

-1

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

-1.5

Real Axis


Slide 59

Real Axis




1 5

2

0.5

1

1.5

0

0.5

2ary

Axi

s

-0 .5

0

Imag

in

-1 5

-1

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

1.5

0.5

Real Axis


Slide 60

ea s




1 5

2

0.5

1

1.5

ary

Axi

s

0

0 .5

2

Imag

in

-0 .5

0

S

-1 5

-1

System: sysGain: 1.62Pole: -0.373 - 0.627iDamping: 0.511Overshoot (%): 15.4

Real Axis

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

1.5

0.5

( )Frequency (rad/sec): 0.73


Slide 61

ea s




1 5

2

0.5

Root Locus

1

1.5

0

0.5

2ry A

xis

-0.5

0

Imag

inar

-1

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

-1.5

0.5


Slide 62

Real Axis




Note that even though the closed-loop poles have this value of

damping factor, the transitory response is not exactly sub-damped

with that characteristic, because the ζ formula has been used as if

it was a 2nd order system.

However the approximation is valid to obtain a good ζ magnitudeHowever, the approximation is valid to obtain a good ζ magnitude

order, the influence of poles and zeros on the response is seen in

the following studythe following study


Slide 63

Additional poleAdditional pole



1. A second-order system is consideredy

2. A pole is added in s=-p

• system reference signal first affected by a first-order system and then by a 2nd order one

• for a step function, signal attenuated by an exponential, which isfor a step function, signal attenuated by an exponential, which is the 2nd order system entry

→ exit has less overshoot and it takes more time to reach its→ exit has less overshoot and it takes more time to reach its final value


Slide 64




)4.52.3)(15.0(4.2

21

sssf

45234.2

22

ssf

4.52.3 ss


Slide 65

Additional zeroAdditional zero


Additional zeroAdditional zero

4523)2(2.1

21

ss

sf4.52.3 ss

4.2f

Zeros (negative)

4.52.322

ssf

( g )

• increase the initial slope,

• make the system faster so it reaches its final value earlier,

• can produce overshoot


Slide 66

Effect of an additional pole in the roots locusEffect of an additional pole in the roots locus

3- Root locus3- Root locuspp

Transfer function of a 8Root Locus

vehicle cruise-control system: 4

6

4820

2

)33.3s)(1s)(06.0s(48.2)s(G

-4

-2

8

-6

4

-10 -8 -6 -4 -2 0 2 4-8

Real Axis


Slide 67

Effect of an additional pole in the roots locusEffect of an additional pole in the roots locus3- Root locus3- Root locus

A pole is added on 0 (i t t )

Root Locus

4

6

(integrator)

48.2)s(G is

2

4

)33.3s)(1s)(06.0s(s)(

Roots locus moved Imag

inar

y A

x

0

0 15Root Locus

to the right

+ unstableI

-4

-2

0 05

0.1

0.15

+ unstable

R l A i-8 -6 -4 -2 0 2 4 6

-6-0.05

0

0.05

Real Axis

0 2 0 15 0 1 0 05 0 0 05 0 1 0 15-0.15

-0.1


Slide 68

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15

Effect of an additional zero in the roots locusEffect of an additional zero in the roots locus


A zero is added i 0 12

Root Locus3

)333)(1)(060()12.0s(48.2)s(G

in -0.12

1

2

)33.3s)(1s)(06.0s(s

gina

ry A

xis

0

1

Imag

-1

5 4 3 2 1 0 1-3

-2

Root locus moved to the left + stable, and fasterReal Axis

-5 -4 -3 -2 -1 0 1


Slide 69

Design a controller/compensator in order to satisfy some



constraints on the response of the system


DYNAMICS

speed urefg



t l l f tt k

SENSORS:


SENSORS: INS,

Anemometer

4- Controllers4- ControllersControl and guidance

Slide 70

4- Controllers4- Controllers

• Proportional controller: P: KP

I t l t ll I K• Integral controller : I:

• Derivative controller: D: sKI

sKD

tp M ts steady-state error

P decreases increases small changes decreases

I decreases increases increases eliminates (=0)

D small changes decreases decreases small changes

• these correlations may not be exactly accurate because K K and K• these correlations may not be exactly accurate, because KP, KI, and KDare dependent of each other

• changing 1 of these variables can change the effect of the other 2Control and guidance

Slide 71

changing 1 of these variables can change the effect of the other 2


• 2 kinds of controllers improve the transitory response:

Lead Compensator:00

0C zpwith

pszs)s(G

adds 1 zero and 1 pole, but zero is more important: it moves the t l t th l ft i t bilit ( t i f t d h

0ps

root locus to the left: improves stability (system is faster and has less overshoot)

Proportional Derivative Compensator:

dd 1 i t bilit

sKK(s)G DPC

adds 1 zero: improves stability


Slide 72


• 2 kinds of controllers improve the steady state response:

Lag Compensator:00

0C pzwith

pszs)s(G

add 1 zero and 1 pole, but pole is more important: it moves the t l t th i ht d t bilit ( t i l d

0ps

root locus to the right: decreases stability (system is slower and has more overshoot), but decreases the steady state error

Proportional Integral Compensator: KK(s)G I

PC gs

( ) PC


Slide 73


• 2 kinds of controllers improve both transitory and steady state response:

Lead - Lag Compensator:

110010

C zpandpzwithzszs)s(G

110010

C pppsps

)(

Proportional Integral Derivative (PID) Compensator:

K sDIPC K

sKK(s)G


Slide 74

5- Frequency response5- Frequency response

11 FourierFourier transformstransforms andand propertiesproperties

22 FrequencyFrequency responseresponse

33 ExamplesExamplespp


Slide 75

1 Fourier transforms and properties1 Fourier transforms and properties


1 Fourier transforms and properties1 Fourier transforms and properties

The Fourier transform of a function x(t) is a function of theThe Fourier transform of a function x(t) is a function of the pulsation ω:

dt)t(xe)(X)]t(x[F tj

dt)t(e)()]t([

→ It transforms a signal from the time domain to the frequency domain


Slide 76

1 Transforms and properties1 Transforms and properties


The inverse Fourier transform recovers the original function


The inverse Fourier transform recovers the original function x(t):

d)(Xe

21)](X[F)t(x tj1 2

This is true for an absolutely integrable signal:

dt)t(x 2 dt)t(x


Slide 77



Linearity:


Linearity:

)(Y)(X)]t(y)t(x[F )()()]t(y)t([

)t(dx Derivation:

)(Xjdt)t(dxF

)(Xj)t(xdF nn

)(Xjdt)(F n


Slide 78



Additional properties


Additional properties

X1)at(xF

a

Xa

)at(xF

Duality Xtx F

x2tXXtx

F x2tX


Slide 79



C l i h


Convolution theorems

FFtff F

FFtff

FFtffF

21F

21

FFtff 2121

dsstfsftffwhere 2121


Slide 80




Time delay

XeTtxF Tj


Slide 81



Important pairs of transforms


δ(t-t0), unit impulse )t(f )(F

0tje

2π δ(ω-ω0)tj 0e e

u(t), unit step1

δ(ω)

)t(ue at

a)t(ue t2cos 0 002

1 0

t2sin 02

00j21


Slide 82

j2

2 Frequency response2 Frequency response


In previous examples we examined the free response of an airplane with step changes in control input


step changes in control input

Other useful input function is the sinusoidal signal. Why?

1. Input to many physical systems takes the form or either a step change or sinusoidal signal

2. An arbitrary function can be represented by a series of step changes or a periodic function can be decomposed by means of Fourier analysis into a series of sinusoidal waves

→ if we know the response of a linear system to either a step or sinusoidal input then we can construct the system's response to an arbitrary input by the principle of superposition


Slide 83




Example:

Examine the response of an airplane subjected to an external disturbance such as a wind gust

Wind gust can be a sharp edged profile or a sinusoidal profile (these 2 types of gust inputs occur quite often in nature) + arbitrary gust profile can be constructed by step and sinusoidal functions


Slide 84

1.5



0.5

1


Arbitrary wind gust profiles:0

0.9

1

-1

-0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8 9 10-1.5

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2


Slide 85




Definition of “ frequency response”:

Response in steady state to a sinusoidal input

We will demonstrate that the steady state response is another

sinusoidal with the same frequency


Slide 86




Similarity with Laplace functions with regard to the

operational properties (ex: differentiation)

→ the transfer function models can be transformed from one

method to the other replacing jω with s (or s with jω). (for

causal signals: signals defined for positive time)


Slide 87




Given any system:

R(s) G(s) C(s)

Hypothesis: stable system

Sinusoidal input

22stsinLsRtsintr


Slide 88




It can be demonstrated that the steady state response is:

tsinjG)t(c

jGImjGRejGwith 22

jGRejGImarctanjGargand jGRe


Slide 89

Parametric estimationParametric estimation


Parametric estimationParametric estimation

a. Firsta. First--orderorder

a. Firsta. First orderorder

Frequency response: 21j1K

j1KjG

1j1

Gain: KjGGain: 21

jG

Delay: arctany arctan


Slide 90


b. Secondb. Second--orderorder

KjG

2n j2

jGn

22n

Gain:

222

21

KjG

nn

21

D l

nωω2ξ

arctanDelay:

2

nωω1

arctan


Slide 91

n


For a system composed by series of blocks:c. c. HigherHigher--orderorder

=

C(s)G1(s)

R(s)G2(s) G3(s)

=

G(s)C(s)R(s)

GGGGi h

jGjGjGjG

sGsGsGsGwith 321

jGjGjGjG 321

321 jGjGjGjG 321

321 jGjGjGjG


Slide 92

6- Bode diagram6- Bode diagram

1 Introduction1 Introduction1. Introduction1. Introduction

2. Construction rules2. Construction rules

3. Stability3. Stability


Slide 93

G l f th B d di

6- Bode diagrams6- Bode diagrams

Goals of the Bode diagrams:

To show the frequency response characteristics in a graphical form

2 graphics for the frequency using a logarithmic scale:

• one for the logarithm of a function magnitude (in decibels):

• one for the phase angle (in degrees):

dB)j(G

)j(Garg one for the phase angle (in degrees):

The decibel is a unit measure used to compare a certain value with a

)j(Garg

reference one. It is basically used to measure a signal power, and it is

defined as:

)j(GP2 )j(Glog20

1)j(G

log10PPlog10)j(G

ref

medidadB


Slide 94


Semi-logarithmic axes: with lineal scale for the magnitude or the phase, and logarithmic for the frequencyp , g q y

Represents the complex transfer function adding each pole or zero effect, which compose this function (adding property of the log)


Slide 95

effect, which compose this function (adding property of the log)

GainGain

6- Bode diagrams6- Bode diagramsGainGain

The gain is a factor that only modifies the magnitude and its angular value is 0º; that is the gain value remains constant for any frequencyvalue is 0 ; that is, the gain value remains constant for any frequency value, because it does not depend on it.35

33 5

34

34.5

agn

itud

e (d

B)

Bode diagram for a K 50 i

32.5

33

33.5

Ma

1 K=50 gain

0

0.5

1

e (d

eg)

-1

-0.5

0

Pha

se


Slide 96

100 10 11

Integral and derivative factorsIntegral and derivative factors6- Bode diagrams6- Bode diagrams

gg

An integral factor or a pole centered in zero, has a transfer function of:

j11

Its magnitude is therefore:

j

j1)j(G

s1)s(G

Its magnitude is, therefore:

)log(201log20)j(GdB

For a logarithmic frequency axis: it corresponds to a straight negative

line of -20 dB per decade

line of -20 dB per decade

The phase is:1j1

900

1arctanjarg

j1arg)j(Garg


Slide 97

I t l d d i ti f tI t l d d i ti f t


Integral and derivative factorsIntegral and derivative factors

For a derivative factor or a zero centered in zero, the results are

deduced using a similar development:

)log(20)j(GdB

90)j(Garg


Slide 98

Integral and derivative factorsIntegral and derivative factors


Integral and derivative factors Integral and derivative factors 20

B ode Diagram

Bode diagrams of the0

10

tude

(dB

)

Bode diagrams of the derivative and the

integrator-10

Mag

ni

g-20

90

0

Pha

se (

deg)

100 101

-90


Slide 99

Frequency (rad/sec)

FirstFirst--order factors: firstorder factors: first--order poleorder pole


FirstFirst order factors: firstorder factors: first order poleorder pole

j11)j(G

2211log20Its magnitude is:

j1 1

01log201log20)j(G11for 22 It seems more complicated, but approximations are made:

01log201log20)j(G,1fordB

log201log20)j(G,11for 22dB

gg)j(,dB

• substitute the curve by its two asymptotes

• magnitude is 0 dB until it reaches the point where both asymptotesg p y pmeet: ωτ=1, this point is called cut frequency

• from there: other asymptote with a 20 dB per decade slope• from there: other asymptote, with a -20 dB per decade slope.

• point where approximation error is maximum corresponds to the cutfrequency and the error is 3 dB


Slide 100

frequency and the error is 3 dB.



similar for the phase, the phase real value is:


1arctan

1j1arg

j11arg)j(Garg 22

However the approximation in this case is:

0)j(Garg11for

•

90)j(Garg11for

)j(g

90)j(Garg1for


Slide 101



1

FirstFirst order factors: firstorder factors: first order poleorder pole0

Bode Diagram

s11)s(G

-20

-10

nitu

de (

dB)

Bode diagram of a -40

-30Mag

n

gfirst-order system

-400

deg)

-45

Pha

se (

d

10-2

10-1

100

101

102

-90

Frequency (rad/sec)


Slide 102

Frequency (rad/sec)



For the study of the first-order zeros, similar development, there is only a sign change: j1)j(G


a sign change: j1)j(G

2222 1l101l20M it d 2222 1log101log20 Magnitude:

Angular contribution:

1

arctanj1arg)j(Garg


Slide 103



Example:


20Bode Diagram

Example:

10)s(G 10

nitu

de (

dB)

1s2)s(G

-10

0

Mag

n010

45

0(d

eg)

90

-45

Pha

se

10-2

10-1

100

101

-90

Frequency (rad/sec)


Slide 104

SecondSecond order factorsorder factors


SecondSecond--order factors order factors

22 1

22nn

2

2n

2nn

2

2n

2j1

12j

)j(Gs2s

)s(G

nn

The magnitude is:

222

21log201log20)j(G

nn

2

n

22

n

21log20

21

log20)j(G

nn


Slide 105

SecondSecond--order factorsorder factors 222


SecondSecond order factors order factors

it can be approximated this way:

nn

21log20)j(G

2n

dB0)1log(20)j(G,1for

2

nn

log20)j(G,1for

F l f i t i ht li t 0dBFor low frequencies: straight line at 0dB

For high frequencies: straight line with a –40 dB per decade slope.

Both asymptotes cross on ω=ωn.

However, in the second-order poles a resonance effect can appear.However, in the second order poles a resonance effect can appear.

In the frequency domain the resonance is shown as a peak close to the cut frequency; the resonance peak value is conditioned to the ζ valueControl and guidance

Slide 106

cut frequency; the resonance peak value is conditioned to the ζ value.

SecondSecond--order factorsorder factors


SecondSecond order factors order factors

F h h n

2

For the phase: 2n

1

tan

n

0)0tan(Arc,2.0generally,1For

180)0tan(Arc,5generally,1For

nn

90)tan(Arc,1For

nn

n

The phase graphic form depends also on ζ


Slide 107

SecondSecond--order factors order factors 6- Bode diagrams6- Bode diagrams

Bode diagram of a second-order pole for different ζ values


Slide 108

SecondSecond--order factors order factors 6- Bode diagrams6- Bode diagrams

Example: -20

Bode Diagram

25s3s5.2)s(G 2

-40

gnitu

de (

dB)

25s3s

-80

-60Mag

0

90

-45

0

(deg

)

-135

-90

Pha

se

10-1

100

101

102

-180

Frequency (rad/sec)


Slide 109

Stability conditionStability condition


Stability condition Stability condition

)º180G()0G( forº180andfor0G Notation: )180G()0G(

The Bode diagram in open loop is studied

Stability condition:

0GforAndº180,forIf )0G(

0GforOrº180,forIf )0G(

0G,forAnd )º180( 0G,forOr )º180(

THEN the system is STABLE THEN the system is UNSTABLETHEN the system is STABLE THEN the system is UNSTABLE

in closed loop in closed loop


Slide 110

Stability conditionStability condition


Stability condition Stability condition

STABLE


Slide 111

Stability condition Stability condition 6- Bode diagrams6- Bode diagrams

yy

UNSTABLE


Slide 112

Stability marginsStability margins


Stability margins Stability margins

Common values:Common values:

Minimum gain

margin: 10 a 12 dB,

Minimum phaseMinimum phase

margin: 45 a 50º


Slide 113

Stability marginsStability margins


Stability margins Stability margins

50Bode Diagram

-50

0

nitu

de

(dB

)

2)1)((1G(s) -150

-100Mag

90 2)1)(ss(s( )

-180

-135

-90

e (d

eg)

-270

-225

180

Pha

se

10-2

10-1

100

101

102

270

Frequency (rad/sec)


Slide 114

Stability margins Stability margins 6- Bode diagrams6- Bode diagrams

2)1)(ss(s1G(s)

MG=14.5dB

with roots locus:

MG=6, G ,

and note that:

20log(6)=15.5dB


Slide 115


2)1)(ss(s1G(s)

MΦ=51º


Slide 116


50Bode Diagram

)33.3s)(1s)(06.0s(48.2)s(G

Transfer function of a vehicle cruise-control system

0

50

e (d

B)

-100

-50

Mag

nitu

de

-150

90-45

0

g)

225-180-135

-90

Pha

se (d

eg

10-2

100

102

-270-225

Frequency (rad/sec)


Slide 117

Frequency (rad/sec)


A pole is added on 0 (integrator): Bode diagram shifted downward : + unstable


Slide 118


A zero is added in -0.12: Bode diagram shifted upward : + stable

Bode Diagram

0

50

100(d

B)

Bode Diagram

-100

-50

0

Mag

nitu

de

-150

100

135

-90

)

-225

-180

-135

Pha

se (d

eg)

10-2

100

102

-270

-225

F ( d/ )


Slide 119

Frequency (rad/sec)

REFERENCESREFERENCESREFERENCESREFERENCES

G.F Franklin, J.D. Powell, A. Emani-Naeini, Feedback Control of DynamicSystems, 4a Edición, Prentice-Hall, 2002

P. Lewis, Sistemas de Control en Ingeniería, Prentice Hall, 1999

W B lt C t l E i i 2ª Edi ió L 1998W. Bolton, Control Engineering, 2ª Edición, Longman, 1998

D. Arzelier, D. Peaucelle, Représentation et analyse des systèmes linéaires, Tomes 1 et 2, Version 1, ENSICA, 1999


Slide 120

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