AERO 422: Active Controls for Aerospace Vehicles ...

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AERO 422: Active Controls for Aerospace Vehicles Frequency Response-Design Method Raktim Bhattacharya Intelligent Systems Research Laboratory Aerospace Engineering, Texas A&M University.

Transcript of AERO 422: Active Controls for Aerospace Vehicles ...

AERO 422: Active Controls for Aerospace VehiclesFrequency Response-Design Method

Raktim Bhattacharya

Intelligent Systems Research LaboratoryAerospace Engineering, Texas A&M University.

Frequency Response

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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 the system’s frequency response

AERO 422, Instructor: Raktim Bhattacharya 3 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

AERO 422, Instructor: Raktim Bhattacharya 4 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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.

AERO 422, Instructor: Raktim Bhattacharya 5 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Response to Sinusoidal InputGeneralization (contd.)

For a system P (s) and inputu(t) = Au sin(ω0t),

forced response isy(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.

AERO 422, Instructor: Raktim Bhattacharya 6 / 51

Fourier Analysis

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

AERO 422, Instructor: Raktim Bhattacharya 8 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

1.2

N=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

1.2

N=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

1.2

N=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

1.2

N=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

1.2

N=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

1.2

N=50

AERO 422, Instructor: Raktim Bhattacharya 9 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Fourier TransformStep function

Fourier transform reveals the frequency content of a signalAERO 422, Instructor: Raktim Bhattacharya 10 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

)

AERO 422, Instructor: Raktim Bhattacharya 11 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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)

AERO 422, Instructor: Raktim Bhattacharya 12 / 51

Bode Plot

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

First Order System

−40

−35

−30

−25

−20

−15

−10

−5

0

Magnitude (

dB

)

10−2

10−1

100

101

102

−90

−45

0

Phase (

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)

AERO 422, Instructor: Raktim Bhattacharya 14 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Second Order System

−40

−30

−20

−10

0

10

Ma

gn

itu

de

(d

B)

10−1

100

101

−180

−135

−90

−45

0

Ph

ase

(d

eg

)

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)

AERO 422, Instructor: Raktim Bhattacharya 15 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Lead Compensator

−15

−10

−5

0

5

10

Ma

gn

itu

de

(d

B)

10−2

10−1

100

101

102

103

0

30

60

Ph

ase

(d

eg

)

Lead Controller

Frequency (rad/s)

Phase leadlow gain at low frequencyhigh gain at high frequencyrelate it to derivative control

AERO 422, Instructor: Raktim Bhattacharya 16 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Lag Compensator

0

1

2

3

4

5

6

Ma

gn

itu

de

(d

B)

10−2

10−1

100

101

−20

−15

−10

−5

0

Ph

ase

(d

eg

)

Lag Controller

Frequency (rad/s)

Phase laghigh gain at low frequencylow gain at high frequencyrelate it to integral control

AERO 422, Instructor: Raktim Bhattacharya 17 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

Magnitude (

dB

)

Bode Diagram

Frequency (rad/s)

S

T

S+T

AERO 422, Instructor: Raktim Bhattacharya 18 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

All transfer functionsWith proportional controller

10−2

100

102

−30

−20

−10

0

10

Ma

gn

itu

de

(d

B)

Ger

Frequency (rad/s)

100

102

−80

−60

−40

−20

0

Ma

gn

itu

de

(d

B)

Ged

Frequency (rad/s)

10−2

100

102

−30

−20

−10

0

10

Ma

gn

itu

de

(d

B)

Gen

Frequency (rad/s)

100

102

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gyr

Frequency (rad/s)

100

102

−80

−60

−40

−20

0

Ma

gn

itu

de

(d

B)

Gyd

Frequency (rad/s)

100

102

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gyn

Frequency (rad/s)

AERO 422, Instructor: Raktim Bhattacharya 19 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Piper Dakota Control SystemDesigned with root locus method

SystemTransfer function from δe (elevator angle) to θ (pitch angle) is

P (s) =θ(s)

δe(s)=

160(s+ 2.5)(s+ 0.7)

(s2 + 5s+ 40)(s2 + 0.03s+ 0.06)

Control Objective 1Design an autopilot so that the step response to elevator input has tr < 1 andMp < 10% =⇒ ωn > 1.8 rad/s and ζ > 0.6 2nd order

Controller

C(s) = 1.5s+ 3

s+ 25(1 + 0.15/s)

AERO 422, Instructor: Raktim Bhattacharya 20 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Piper Dakota Control SystemTime Response

0 0.2 0.4 0.6−0.5

0

0.5

1

1.5

Ref to Control

Time (seconds)

Ele

va

tor

an

gle

(d

eg

)

0 10 20−0.5

0

0.5

1

Ref to Error

Time (seconds)

Err

or

(de

g)

0 10 200

0.5

1

1.5

Ref to Output

Time (seconds)

Pitch

An

gle

(d

eg

)

0 20 40−5

−4

−3

−2

−1

0

Dist to Error

Time (seconds)

Err

or

(de

g)

0 10 20−1.5

−1

−0.5

0

Dist to Control

Time (seconds)

Ele

va

tor

an

gle

(d

eg

)

0 20 400

1

2

3

4

5

Dist to Output

Time (seconds)

Pitch

an

gle

(d

eg

)AERO 422, Instructor: Raktim Bhattacharya 21 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Piper Dakota Control SystemFrequency Response

100

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Ger

Frequency (rad/s)

100

−40

−30

−20

−10

0

10

20

Ma

gn

itu

de

(d

B)

Ged

Frequency (rad/s)

100

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gen

Frequency (rad/s)

100

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gyr

Frequency (rad/s)

100

−40

−30

−20

−10

0

10

20

Ma

gn

itu

de

(d

B)

Gyd

Frequency (rad/s)

100

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gyn

Frequency (rad/s)

AERO 422, Instructor: Raktim Bhattacharya 22 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Piper Dakota Control SystemFrequency Response (contd.)

10−2

10−1

100

101

102

103

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gur

Frequency (rad/s)

10−1

100

101

102

103

−80

−60

−40

−20

0

20

Ma

gn

itu

de

(d

B)

Gud

Frequency (rad/s)

AERO 422, Instructor: Raktim Bhattacharya 23 / 51

Asymptotes

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Approximate Bode PlotUseful for Design & Analysis

Let open-loop transfer function be

KG(s) = K(s− z1)(s− z2) · · ·(s− p1)(s− p2) · · ·

Write in Bode form

KG(jω) = K0(jωτ1 + 1)(jωτ2 + 1) · · ·(jωτa + 1)(jωτb + 1) · · ·

K0 is the DC gain of the system.Example

G(s) =(s+ 1)

(s+ 2)(s+ 3)=⇒ G(jω) =

jω + 1

(jω + 2)(jω + 3)=

1

6

jω + 1

(jω/2 + 1)(jω/3 + 1)

AERO 422, Instructor: Raktim Bhattacharya 25 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Approximate Bode Plotcontd.

Transfer function in Bode Form

KG(jω) = K0(jωτ1 + 1)(jωτ2 + 1) · · ·(jωτa + 1)(jωτb + 1) · · ·

Three cases1. K0(jω)

n pole, zero at origin

2. (jω + 1)±1 real pole, zero

3.[(

jωωn

)2+ 2ζ jω

ωn+ 1

]±1

complex pole, zero

AERO 422, Instructor: Raktim Bhattacharya 26 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Case:1 K0(jω)n

pole, zero at origin

GainlogK0|(jω)|n = logK0 + n log |jw| = logK0 + n logw

PhaseK0(jω)|n = K0 + n jω = 0 + n× 90◦

10−1

100

101

102

103

10−4

10−3

10−2

10−1

100

101

102

Bode Plot

Frequency (rad/s)

Gain

(a) Gain

10−1

100

101

102

103

−80

−60

−40

−20

0

20

40

60

80

Bode Plot

Frequency (rad/s)

Phase

(b) Phase

AERO 422, Instructor: Raktim Bhattacharya 27 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Case:2 (jωτ + 1)±1real pole, zero

Gain(jωτ + 1) =

{≈ 1, ωτ << 1,≈ jωτ, ωτ >> 1.

Frequency ω = 1/τ is the break point

10−1

100

101

102

103

10−4

10−3

10−2

10−1

100

101

102

Bode Plot

Frequency (rad/s)

Gain

AERO 422, Instructor: Raktim Bhattacharya 28 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Case:2 (jωτ + 1)±1real pole, zero (contd.)

Phase

jωτ + 1 =

≈ 1, ωτ << 1, 1 = 0◦

≈ jωτ, ωτ >> 1, jωτ = 90◦

ωτ ≈ 1, jωτ + 1 = 45◦

10−1

100

101

102

103

−80

−60

−40

−20

0

20

40

60

80

Bode Plot

Frequency (rad/s)

Phase

AERO 422, Instructor: Raktim Bhattacharya 29 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Example

G(s) =200(s+ 0.5)

s(s+ 10)(s+ 50)

−80

−60

−40

−20

0

20

40

Ma

gn

itu

de

(d

B)

10−2

10−1

100

101

102

103

−180

−135

−90

−45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/s)

Bode Diagram

Frequency (rad/s)

−80

−60

−40

−20

0

20

40

Ma

gn

itu

de

(d

B)

0.01 0.1 1 10 100 1000−180

−150

−120

−90

−60

−30

0

Ph

ase

(d

eg

)

AERO 422, Instructor: Raktim Bhattacharya 30 / 51

Steady-State Errors

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Closed-loop systemC Pu+r e y

−y

Closed-loop transfer functionGer =

1

1 + PC= K0(jω)

n (jωτ1 + 1)(jωτ2 + 1) · · ·(jωτa + 1)(jωτb + 1) · · ·

Steady-state gain

lims→0

sGer(s)1

s⇔ lim

ω→0|Ger(jω)|

PC =200(s+ 0.5)

s(s+ 10)(s+ 50)

Typically analysis is done with open-loop system

10−3

10−2

10−1

100

101

102

−60

−50

−40

−30

−20

−10

0

10

Magnitude (

dB

)

Bode Diagram

Frequency (rad/s)

AERO 422, Instructor: Raktim Bhattacharya 32 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Open-loop systemC Pu+r e y

−y

Open-loop transfer functionPC =

200(s+ 0.5)

s(s+ 10)(s+ 50)= K0(jω)

n (jωτ1 + 1)(jωτ2 + 1) · · ·(jωτa + 1)(jωτb + 1) · · ·

Steady-state error step

ess =1

1 +Kp, Kp := K0.

Steady-state error ramp

ess =1

Kv

System type is the slope of the low frequencyasymptoteKv is the value of low frequency asymptote atω = 1 rad/s 10

−210

−110

010

110

210

3−80

−60

−40

−20

0

20

40

Magnitude (

dB

)

Bode Diagram

Frequency (rad/s)

AERO 422, Instructor: Raktim Bhattacharya 33 / 51

Stability Analysis

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

StabilityC Pu+r e y

−y

Given open-loop dataC(s) = K,P (s) = 1

s(s+1)2

−5 −4 −3 −2 −1 0 1 2−3

−2

−1

0

1

2

3

Root Locus

Real Axis (seconds−1

)

Imagin

ary

Axis

(seconds

−1)

Stable for K < 2

All points on root locus satisfy1 + P (s)C(s) = 0

P (s)C(s) = −1 =⇒ |P (s)C(s)| = 1and P (s)C(s) = 180◦

At neutral stability point s = jω,

|P (jω)C(jω)| = 1

P (jω)C(jω) = 180◦

AERO 422, Instructor: Raktim Bhattacharya 35 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Stability|P (jω)C(jω)| < 1 at P (jω)C(jω) = 180◦

−150

−100

−50

0

50

100

Magnitude (

dB

)

10−2

10−1

100

101

102

−270

−225

−180

−135

−90

−45

0

Phase (

deg)

Bode Diagram

Frequency (rad/s)

K=0.1

K=2

K=10

AERO 422, Instructor: Raktim Bhattacharya 36 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Gain MarginOpen loop Bode Plot

−150

−100

−50

0

50

100

Ma

gn

itu

de

(d

B)

10−2

10−1

100

101

102

−270

−225

−180

−135

−90

−45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/s)

K=0.1

K=2

K=10

Gain Margin (GM): factor by which gain can be increased at P (jω)C(jω) = −180◦

AERO 422, Instructor: Raktim Bhattacharya 37 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Phase MarginOpen loop Bode Plot

−150

−100

−50

0

50

100

Ma

gn

itu

de

(d

B)

10−2

10−1

100

101

102

−270

−225

−180

−135

−90

−45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/s)

K=0.1

K=2

K=10

Phase Margin (PM): amount by which phase exceeds −180◦ at |P (jω)C(jω)| = 1

AERO 422, Instructor: Raktim Bhattacharya 38 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Nyquist PlotRelates open-loop frequency response to number of unstable closed-loop polesResidue theorem in complex analysisPlot P (jω)C(jω) in the complex plainNumber of encirclements of −1 equals Z − P of 1 + P (s)C(s)

AERO 422, Instructor: Raktim Bhattacharya 39 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Nyquist Plotcontd.

Write P (s)C(s) = KG(s) = KN(s)D(s)

=⇒ 1 + P (s)C(s) =D(s) +KN(s)

D(s)

Poles of 1 + P (s)C(s) = Poles of G(s) none of them on RHP

Number of encirclements = number of zeros of 1 + P (s)C(s) on RHPnumber of poles of closed-loop system

AERO 422, Instructor: Raktim Bhattacharya 40 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Nyquist PlotExample: P (s)C(s) = K

s(s+1)2

−10 −5 0 5−3

−2

−1

0

1

2

3

Nyquist Diagram

Real Axis

Imagin

ary

Axis

K=1

K=2

K=10

AERO 422, Instructor: Raktim Bhattacharya 41 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Nyquist PlotDetermining Gain

Given P (s)C(s) = Ks(s+1)2

, what is K for stability?Encirclement of 1/K +G(s) = 0

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

−1.5

−1

−0.5

0

0.5

1

1.5

2

Nyquist Diagram

Real Axis

Imagin

ary

Axis

AERO 422, Instructor: Raktim Bhattacharya 42 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Nyquist PlotGain and Phase Margin

Nyquist plot of P (s)C(s)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Nyquist Diagram

Real Axis

Imagin

ary

Axis

AERO 422, Instructor: Raktim Bhattacharya 43 / 51

Frequency Domain Design

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

Develop conditions on the Bode plot of the open loop transfer functionSensitivity 1

1+PC

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

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

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

AERO 422, Instructor: Raktim Bhattacharya 45 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

Frquency 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 the constraintsSlope ≈ −1 at ωc ensures PM ≈ 90◦ stable if PM > 0 =⇒ PC > −180◦

AERO 422, Instructor: Raktim Bhattacharya 46 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

10−2

10−1

100

101

102

103

−250

−200

−150

−100

−50

0

50

Ma

gn

itu

de

(d

B)

Bode Diagram

Frequency (rad/s)

True

Model

Unc+

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

Steady-state error boundary

Sensor noise, disturbance Plant uncertainty

AERO 422, Instructor: Raktim Bhattacharya 47 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

Steady-state error boundary

Sensor noise, disturbance Plant uncertainty

Gyn = − PC

1 + PC

AERO 422, Instructor: Raktim Bhattacharya 48 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

Steady-state error boundary

Sensor noise, disturbance Plant uncertainty

Gyr =PC

1 + PC

Gyn = − PC1+PC

AERO 422, Instructor: Raktim Bhattacharya 49 / 51

Frequency Response Fourier Analysis Bode Plot Asymptotes Steady-State Stability Design

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

Steady-state error boundary

Sensor noise, disturbance Plant uncertainty

Gyd =P

1 + PC

Gyn = − PC1+PC

AERO 422, Instructor: Raktim Bhattacharya 50 / 51