aero 632: design of advance flight control system ... preliminaries.pdf · raktim bhattacharya...

AERO 632: Design of Advance Flight Control SystemPreliminaries

Raktim Bhattacharya

Laboratory For Uncertainty QuantificationAerospace Engineering, Texas A&M University.

Signals & Systems Laplace Transforms Transfer Functions System Interconnection Frequency Domain Analysis Design

PreliminariesSignals & SystemsLaplace transformsTransfer functions – from ordinary linear differential equationsSystem interconnectionsBlock diagram algebra – simplification of interconnectionsGeneral feedback control system interconnection.

C P+u+r +

d

+

n

e +y ym−ym

AERO 632, Instructor: Raktim Bhattacharya 2 / 46

Signals & Systems


Signals & Systems

Pu(t) y(t)

Actuator applies u(t)

Sensor provides y(t)

Feedback controller takes y(t) and determines u(t) to achievedesired behaviorThe controller is typically implemented as software, running ina micro controller

Imperfections exist in real world▶ sensors have noise▶ actuators have irregularities▶ plant P is not fully known



System Response to u(t)

Pu(t) y(t)

Given plant P and input u(t), what is y(t)?P is defined in terms of ordinary differential equationsy(t) is the forced + initial condition response.

Linear Dynamics

mx+ cx+ kx = u(t) dynamics

y(t) = x(t) measurement

Nonlinear Dynamics

x− µ(1− x2)x+ x = u(t) dynamics

y(t) = x(t) measurement

In this class we focus on linear systems



Linear Systems

Pu(t) y(t)

Dynamics is defined by linear ordinary differential equationSuper position principle applies

u1(t) 7→ y1(t)u2(t) 7→ y2(t)

=⇒ (u1(t) + u2(t)) 7→ (y1(t) + y2(t))


Laplace Transforms


Laplace TransformsGiven signal u(t), Laplace transform is defined as

L{u(t)} :=

∫ ∞

0u(t)e−stdt

Exists when

limt→∞

|u(t)e−σt| = 0, for some σ > 0

Very useful in studying linear dynamical systems and designingcontrollers



Properties Laplace TransformsLinear operator

Additive

L{u1(t) + u2(t)} =

∫ ∞

0(u1(t) + u2(t)) e

−stdt

=

∫ ∞

0u1(t)e

−stdt+

∫ ∞

0u2(t)e

−stdt

= L{u1(t)}+ L{u2(t)}

Superposition

L{au(t)} = aL{u(t)} , a is a constant



Properties (contd.)1. U(s) := L{u(t)}2. L{au1(t) + bu2(t)} = aL{u1(t)}+ bL{u2(t)} = aU1(s) + bU2(s)

3. 1

sU(s) ⇐⇒

∫ t

0u(τ)dτ

4. U1(s)U2(s) ⇐⇒ u1(t) ∗ u2(t) Convolution

5. lims→0

sU(s) ⇐⇒ limt→∞

u(t) Final value theorem

6. lims→∞

sU(s) ⇐⇒ u(0+) Initial value theorem

7. − dU(s)

ds⇐⇒ tu(t)

8. L{du

dt

}⇐⇒ sU(s)− su(0)

9. L{u} ⇐⇒ s2U(s)− su(0)− u(0)



Important Signals

1. L{δ(t)} = 1 δ(t) is impulse function

2. L{1(t)} =1

s1(t) is unit step function at t = 0

3. L{t} =1

s2

4. L{sin(ωt} =ω

s2 + ω2

5. L{cos(ωt} =s

s2 + ω2


Transfer Functions


Spring Mass Damper System

m u(t)

Equation of Motionmx+ cx+ kx = u(t)

Take L{·} on both sidesL{mx+ cx+ kx} = L{u(t)}mL{x}+ cL{x}+ kL{x} = L{u(t)}m

(s2X(s)− sx(0)− x(0)

)+ c (sX(s)− x(0)) + kX(s) = U(s)

(ms2 + cs+ k)X(s) = U(s) x(0) and x(0) are assumed to be zero



Transfer Function

m u(t)

Pu(t) y(t)

(ms2 + cs+ k)X(s) = U(s) =⇒ X(s)

U(s)=

1

ms2 + cs+ k

Choose output y(t) = x(t) =⇒ Y (s) = X(s).Therefore

P (s) :=Y (s)

U(s)=

1

ms2 + cs+ kTransfer function



Transfer Function (contd.)In general

P (s) =N(s)

D(s)

where N(s) and D(s) are polynomials in s

Roots of N(s) are the zerosRoots of D(s) are the poles – determine stability



Response to u(t)Given

– input signal u(t) and transfer function P (s).

Determine– output response y(t)

1. Laplace transformU(s) := L{u(t)}

2. Determine Y (s) := P (s)U(s)

3. Laplace inverse

y(t) := L−1 {Y (s)} = L−1 {P (s)U(s)}

Pu(t) y(t)


System Interconnection


Block DiagramRepresentation of System Interconnections

SeriesParallelFeedbackA simple exampleA complex example



Series ConnectionG1 G2

u y



Parallel Connection

G1

G2

+u y



Feedback ConnectionG

r + y

−



Simple ExampleG1

r +

−G2 G3

G4

−

+

y



Complex Example

G1r +

−G2 G3

G4

−

+

y+


Frequency Response


Response to Sinusoidal InputP

u(t) y(t)

Let u(t) = Au sin(ωt)Vary ω from 0 to ∞

A linear system’s response to sinusoidal inputs is called thesystem’s frequency response



Response to Sinusoidal InputExample

Let P (s) = 1s+1 , u(t) = sin(t)

y(t) =1

2e−t − 1

2cos(t) + 1

2sin(t)

=1

2e−t︸︷︷︸

natural response

+1√2

sin(t− π

4)︸︷︷︸

forced response

Forced response has form Ay sin(ωt+ ϕ)

Ay and ϕ are functions of ω



Response to Sinusoidal InputGeneralization

In general

Y (s) = G(s)ω0

s2 + ω20

=α1

s− p1+ · · · αn

s− pn+

α0

s+ jω0+

α∗0

s− jω0

=⇒ y(t) = α1ep1t + · · ·+ αne

pnt︸︷︷︸natural

+Ay sin(ω0 + ϕ)︸︷︷︸forced

Forced response has same frequency, different amplitude and phase.



Response to Sinusoidal InputGeneralization (contd.)

For a system P (s) and input

u(t) = Au sin(ω0t),

forced response is

y(t) = AuM sin(ω0t+ ϕ),

where

M(ω0) = |P (s)|s=jω0 = |P (jω0)|,magnitude

ϕ(ω0) = P (jω0) phase

In polar formP (jω0) = Mejϕ.


Fourier Analysis


Fourier Series ExpansionGiven a signal y(t) with periodicity T ,

y(t) =a02

+∑

n=1,2,···an cos

(2πnt

T

)+ bn sin

(2πnt

T

)

a0 =2

T

∫ T

0y(t)dt

an =2

T

∫ T

0y(t) cos

(2πnt

T

)dt

bn =2

T

∫ T

0y(t) sin

(2πnt

T

)dt



Fourier Series ExpansionApproximation of step function

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=6

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=8

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=10

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=20

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2N=50



Fourier TransformStep function

Fourier transform reveals the frequency content of a signalAERO 632, Instructor: Raktim Bhattacharya 32 / 46


Fourier TransformStep function – frequency content

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

y(t)

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

ω

y(ω

)



Signals & SystemsInput Output

Pu(t) y(t)

Fourier Series Expansionsuperposition principle

P∑

i ui(t)∑

i yi(t)

Fourier Transform

PU(jω) Y (jω)

ui(t) = ai sin(ωit)

yiforced(t) = aiM sin(ωit+ ϕ)

Y (jω) = P (jω)U(jω)

Suffices to study P (jω) |P (jω)|, P (jω)


Bode Plot


First Order System

−40

−35

−30

−25

−20

−15

−10

−5

0

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/s)

P (s) = 1/(s+ 1)

loglog scaledB = 10 log10(·)20dB = 10 log10(100/1)

0 10 20 30 40 50 60 70 80 90 100−1

0

1

y(t)

Fre q = 0.10 rad/s

0 1 2 3 4 5 6 7 8 9 10−1

0

1

y(t)

Fre q = 1.00 rad/s

0 1 2 3 4 5 6 7 8 9 10−1

0

1

y(t)

Fre q = 5.00 rad/s

0 1 2 3 4 5 6 7 8 9 10−1

0

1

y(t)

Fre q = 10.00 rad/s

u(t) = A sin(ω0t)

yforced(t) = AM sin(ω0t+ϕ)



Second Order System

−40

−30

−20

−10

0

10

Mag

nitu

de (

dB)

10−1

100

101

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/s)

P (s) = 1/(s2 + 0.5s+ 1)

ωn = 1 rad/s

0 10 20 30 40 50 60 70 80 90 100−2

0

2

y(t)

Fre q = 0.10 rad/s

0 10 20 30 40 50 60 70 80 90 100−2

0

2

y(t)

Fre q = 1.00 rad/s

0 1 2 3 4 5 6 7 8 9 10−1

0

1

y(t)

Fre q = 5.00 rad/s

0 1 2 3 4 5 6 7 8 9 10−1

0

1

y(t)

Fre q = 10.00 rad/s

u(t) = A sin(ω0t)

yforced(t) = AM sin(ω0t+ϕ)



S(jω) + T (jω) = 1

10−2

10−1

100

101

102

10−2

10−1

100

101

ω rad/s

Magnitude|S

(jω)|

10−1

100

101

102

10−3

10−2

10−1

100

101

ω rad/s

Magnitude|T

(jω)|

C P+u+r +

d

+

n

e +y ym−ym

P (s) = 1(s+1)(s/2+1)

C(s) = 10

S = Ger = 11+PC

= 11+10P

T = Gyr = PC1+PC

= 10P1+10P

10−2

10−1

100

101

102

−60

−50

−40

−30

−20

−10

0

10

Mag

nitu

de (

dB)

Bode Diagram

Frequency (rad/s)

STS+T



All transfer functionsWith proportional controller

10−2

100

102

−30

−20

−10

0

10M

agni

tude

(dB

)

Ger

Frequency (rad/s)10

010

2−80

−60

−40

−20

0

Mag

nitu

de (

dB)

Ged

Frequency (rad/s)10

−210

010

2−30

−20

−10

0

10

Mag

nitu

de (

dB)

Gen

Frequency (rad/s)

100

102

−60

−40

−20

0

20

Mag

nitu

de (

dB)

Gyr

Frequency (rad/s)10

010

2−80

−60

−40

−20

0

Mag

nitu

de (

dB)

Gyd

Frequency (rad/s)10

010

2−60

−40

−20

0

20

Mag

nitu

de (

dB)

Gyn

Frequency (rad/s)


Controller DesignConsiderations


Design Using Bode Plot of P (jω)C(jω)Loop Shaping

Develop conditions on the Bode plot of the open loop transferfunction

Sensitivity 11+PC

Steady-state errors: slope and magnitude at limω → 0

Robust to sensor noiseDisturbance rejectionController roll off =⇒ not excite high-frequency modes ofplantRobust to plant uncertainty

Look at Bode plot of L(jω) := P (jω)C(jw)



Frequency Domain SpecificationsConstraints on the shape of L(jω)

Stea

dy-s

tate

erro

r bou

ndar

y

Sens

or n

oise

, pla

nt

unce

rtain

ty

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

Steady-state error boundary

Sensor noise, disturbance Plant uncertainty

Choose C(jω) to ensure |L(jω)| does not violate theconstraintsSlope ≈ −1 at ωc ensures PM ≈ 90◦

stable if PM > 0 =⇒ PC > −180◦



Plant UncertaintyP (jω) = P0(jω)(1 + ∆P (jω))

10−2

10−1

100

101

102

103

−250

−200

−150

−100

−50

0

50

Mag

nitu

de (

dB)

Bode Diagram

Frequency (rad/s)

TrueModelUnc+Unc−

Stea

dy-s

tate

erro

r bou

ndar

y

Sens

or n

oise

, pla

nt

unce

rtain

ty

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1





Sensor CharacteristicsNoise spectrum

Stea

dy-s

tate

erro

r bou

ndar

y

Sens

or n

oise

, pla

nt

unce

rtain

ty

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1



Gyn = − PC

1 + PC



Reference TrackingBandlimited else conflicts with noise rejection

0 5 10 15 20 25

0.01

0.02

0.03

0.04

0.05

Spectrum of r(t)

Frequency (Hz)

|X(f

)|

0 5 10 15 20 25

0.01

0.02

0.03

0.04

0.05

Spectrum of n(t)

Frequency (Hz)

|X(f

)|

Stea

dy-s

tate

erro

r bou

ndar

y

Sens

or n

oise

, pla

nt

unce

rtain

ty

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1



Gyr =PC

1 + PC

Gyn = − PC1+PC



Disturbance RejectonBandlimited else conflicts with noise rejection

0 5 10 15 20 25

0.01

0.02

0.03

0.04

0.05

Spectrum of n(t)

Frequency (Hz)

|X(f

)|

0 5 10 15 20 25

0.01

0.02

0.03

0.04

0.05

Spectrum of d(t)

Frequency (Hz)

|X(f

)|

Stea

dy-s

tate

erro

r bou

ndar

y

Sens

or n

oise

, pla

nt

unce

rtain

ty

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1

!!c

1

|P(j!)C

(j!)|

slope ⇡ 1



Gyd =P

1 + PC

Gyn = − PC1+PC


aero 632: design of advance flight control system ... preliminaries.pdf · raktim bhattacharya...

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