AERO 422: Active Controls for Aerospace Vehicles - Root ... Root Locus.pdf · Root Locus Plotting...

AERO 422: Active Controls for Aerospace VehiclesRoot Locus Design Method

Raktim Bhattacharya

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

Root Locus

Root Locus Plotting Guidelines Dynamic Compensator Design Example

Root LocusGeneralized Setting

C Pu+r e y

−y

Write1 + P (s)C(s) = 1 +KL(s) = 0

Roots depend on K generalized gain

1 +KL(s) = 0

1 +KN(s)

D(s)= 0

D(s) +KN(s) = 0

or L(s) = − 1

Kroot-locus form

AERO 422, Instructor: Raktim Bhattacharya 3 / 39


Simple ExampleC Pu+r e y

−y

P (s) =A

s(s+ c), C(s) = 1

Two rootsDepends on parameters A and c

r1, r2 =1

2±

√1− 4A

2r1, r2 =

c

2±

√c2 − 4

2

RL studies variation of r1, r2 with respect to A, c one at a time

MATLAB command rlocus(...) is used to generate theseplotshelp rlocus for more details



Simple ExampleVariation w.r.t A

C Pu+r e y

−y

Study variation w.r.t A, set c = 1

P (s) =A

s(s+ c), C(s) = 1

Root-locus form

1 + PC = 0 =⇒ 1 +A1

s(s+ 1)= 0

or 1

s(s+ 1)= − 1

A



Simple ExampleVariation w.r.t A (contd.)

1 +A1

s(s+ 1)= 0

−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Root Locus

Real Axis (seconds−1)

Imag

inar

y A

xis

(sec

onds

−1 )

MATLAB Code

s = tf(’s’);

sys = 1/s/(s+1);

rlocus(sys);

r1, r2 =1

2±

√1− 4A

2



Simple ExampleVariation w.r.t c

C Pu+r e y

−y

Study variation w.r.t c, set A = A∗ = 1

P (s) =1

s(s+ c), C(s) = 1

Root-locus form

1 + PC = 0 =⇒ 1 +1

s(s+ c)= 0

or s2 + cs+ 1 = 0

or (s2 + 1) + cs = 0

or L′(s) =s

s2 + 1= −1

c



Simple ExampleVariation w.r.t c (contd.)

s

s2 + 1= −1

c

−1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2−1.5

−1

−0.5

0

0.5

1

1.5

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

MATLAB Code

s = tf(’s’);

sys = s/(s^2+1);

rlocus(sys);

r1, r2 =c

2±

√c2 − 4

2


Guidelines for PlottingRoot Locus


Guidelines for Drawing Root LocusDefinition 1Root locus of L(s) is the set of values of s for which1 +KL(s) = 0 for values of 0 ≤ K <∞.

Definition 2Root locus is the set of values of s for which phase of L(s) is 180◦.Let the angle from a zero be ψi and angle from a pole be ϕi. Then∑

j

ψj −∑i

ϕi = 180◦ + 360◦(l − 1)

for integer l.



Step 1Draw poles and zeros of L(s)

C Pu+r e y

−y

Plot poles with ×0,−4± 4j for this examplePlot zeros with ONone for this example

Given P (s) = 1s[(s+4)2+16]

,C(s) = K.

Im

Re

x

x

x



Step 2Real axis portions of the locus

Im

Re

x

x

xs0

If we take s0 on the real-axiscontributions from complexpoles and zeros disappearAngle criterion :∑j

ψj −∑i

ϕi = 180◦ +360◦(l− 1)

ϕ1 = −ϕ2

−4 + 4j = − −4− 4j

s0 must lie to the left of oddnumber of real poles & zeros



Step 2Real axis portions of the locus (contd.)

Im

Re

x

x

xx

Let there bea pole at −2 anda zero at −4

How does the root locus change?



Step 3Asymptotes

Study behavior for large K,

L(s) = − 1

K

K → ∞ =⇒ L(s) = 0

For large values of K, roots will be close to zeros of L(s).But there are n poles and m zeros, with n > m.Where do n−m poles go?

They are asymptotic to lines with angles ϕr starting from s = α,where

ϕr =180◦ + 360◦(r − 1)

n−m, α =

∑pi −

∑zj

n−m.



Step 3Asymptotes (contd.)

For this examplen = 3,m = 0 =⇒ α = 60◦, 180◦, 300◦,

and α = −2.67.Im

Re

x

x

x



Step 4Departure Angles

Angle at which a branch of locus departs from one of the poles

rϕdep =∑

ψi −∑

ϕj − 180◦ − 360◦r,

where∑ϕj is over the other poles.

We assume there multiple poles of order q under consideration,and r = 1, · · · , q.

Summation∑ψj is over all zeros.



Step 4Arrival Angles

Angle at which a branch of locus arrives at one of the zeros

rψarr =∑

ϕj −∑

ψi + 180◦ + 360◦r,

where∑ψj is over the other zeros.

We assume there multiple zeros of order q under consideration,and r = 1, · · · , q.

Summation∑ϕj is over all poles.



Step 4Example

Im

Re

x

x

x



Step 5Imaginary axis crossing

Use Routh’s table to determine K for stability for

s3 + 8s2 + 32s+K = 0,

s3 1 32s2 8 Ks1 32−K/8 0s0 K 0

K > 0 and 32−K/8 > 0 =⇒ K > 256

Root locus crosses imaginary axis for K = 256.Substitute K = 256 and s = jω0 in characteristic equation,and solve for ω0.

(jω0)3 + 8(jω0)

2 + 32(jω0) + 256 = 0



Step 5Imaginary axis crossing (contd.)

Solve for ω0

(jω0)3 + 8(jω0)

2 + 32(jω0) + 256 = 0

=⇒ −8ω20 + 256 = 0, and − ω3

0 + 32ω0 = 0.

or ω0 = ±√32.

Im

Re

x

x

x



Step 6Arrival & departure angles at multiple root locations

Few examplesTwo segments come together at 180◦ and break away at ±90◦

Three locus segments approach at relative angles of 120◦ anddepart at angles rotated by 60◦

Read textbook for details



Final PlotIm

Re

x

x

x

−6 −4 −2 0 2 4 6

−6

−4

−2

0

2

4

6

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )


Dynamic Compensators


Lead CompensatorStabilizing effect

C Pu+r e y

−y

Re

Imx

Compensator form

C(s) = Ks/z + 1

s/p+ 1

p >> z > 0 p not too far to the left

Root locus:

s/z + 1

s/p+ 1P (s) = − 1

K

Moves the locus to the left



Lead CompensatorExample

C Pu+r e y

−y

Plant model P (s) = 1s(s+1)

−7 −6 −5 −4 −3 −2 −1 0 1−1.5

−1

−0.5

0

0.5

1

1.5

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

(a) C(s) = s/2 + 1

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

−16 −14 −12 −10 −8 −6 −4 −2 0 2−15

−10

−5

0

5

10

15

System: G2Gain: 12.4Pole: −6.74 + 5.34iDamping: 0.784Overshoot (%): 1.89Frequency (rad/s): 8.6

(b) C(s) = s/2+1s/15+1

−12 −10 −8 −6 −4 −2 0 2−15

−10

−5

0

5

10

15

System: G2Gain: 7.91Pole: −3.88 + 3.06iDamping: 0.785Overshoot (%): 1.87Frequency (rad/s): 4.94

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

(c) C(s) = s/2+1s/10+1

Root Locus: C(s)P (s) = − 1K

Location of z, p is based on trial and errorSelect desired closed-loop pole

▶ Arbitrarily pick z, then use angle criterion to select pAERO 422, Instructor: Raktim Bhattacharya 25 / 39


Lag CompensatorImproves steady state performance

C Pu+r e y

−y

Re

Imx

Compensator form

C(s) = Ks+ z

s+ p

z > p > 0 low frequency, near the origin

z is close to pRoot locus:

s+ z

s+ pP (s) = − 1

K

Boosts steady-state gain: z/p > 1.



Lag CompensatorExample

Plant : 1s(s+1) , Lead Compensator: K(s+2)

s+15

Kv := lims→0 sK(s+2)s+15

1s(s+1) = 90× 2/15 = 12.

Steady-state to ramp input = 1/Kv = 1/12 = 0.0833

How to increase Kv? reduce ess to ramp

Introduce a lag compensator: s+0.05s+0.01

Kv := lims→0 sK(s+0.05)s+0.01

s+2s+15

1s(s+1) = 5× 12 = 60

Steady-state to ramp input = 1/Kv = 1/60 = 0.0166

Lag compensators amplify gain at low frequencyHave no effect at high-frequency



Lag CompensatorCaution

−16 −14 −12 −10 −8 −6 −4 −2 0 2−20

−15

−10

−5

0

5

10

15

20

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

(a) Effect of Lag Compensator

−0.14 −0.12 −0.1 −0.08 −0.06 −0.04 −0.02 0 0.02 0.04−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

(b) Zoomed in

Closed-loop poles are near the zero at −0.05Very slow decay rate.Proximity of poles =⇒ low amplitudeMay affect settling time, especially for disturbance response

Put lag pole-zero at as high frequency possible, without affectingtransients AERO 422, Instructor: Raktim Bhattacharya 28 / 39

Design Example


Control of a Small AirplanePiper Dakota (from text book)

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 hastr < 1 and Mp < 10% =⇒ ωn > 1.8 rad/s and ζ > 0.6 2nd order



Control of a Small AirplaneOpen Loop Poles: −2.5± 5.81j,−0.015± 0.244j (stable)Open Loop Zeros: −2.5,−0.7 (no RHS zeros)

−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1−10

−8

−6

−4

−2

0

2

4

6

8

10

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

Figure: Root locus with proportional feedback



Control of a Small AirplaneProportional Controller

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (seconds)

Am

plitu

de

(a) Step response with proportional (b) Root locus with proportional

Not possible to satisfy ζ requirement with just proportionalcontroller



Control of a Small AirplaneLead Compensator

After trial and error, choose C(s) = K s+3s+25 , with K = 1.5

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (seconds)

Am

plitu

de

LeadProportional

(a) Step response with leadcompensator

−30 −25 −20 −15 −10 −5 0 5−60

−40

−20

0

20

40

60

0.10.160.230.320.44

0.62

0.84

10

20

30

40

50

60

10

20

30

40

50

60

0.050.10.160.230.320.44

0.62

0.84

0.05

System: G1Gain: 1.5Pole: −8.42 + 9.9iDamping: 0.648Overshoot (%): 6.92Frequency (rad/s): 13

Root Locus

Real Axis (seconds−1)Im

agin

ary

Axi

s (s

econ

ds−

1 )

(b) Root locus with leadcompensator

Has steady-state error ... have to fix this.AERO 422, Instructor: Raktim Bhattacharya 33 / 39


Control of a Small AirplaneLead Compensator + Integral Control

Fix Steady-State Errorintroduce integral control

C(s) = KDc(s)(1 +KI/s)

tune KI to get desired behaviourstudy root locus w.r.t KI

Characteristic Equation

1 +KDc(s)P (s) +KI

sKDc(s)P (s) = 0

Write this in L(s) = − 1KI

form



Control of a Small AirplaneLead Compensator + Integral Control (contd.)

Characteristic Equation

1 +KDc(s)P (s) +KI

sKDc(s)P (s) = 0

Write this in L(s) = − 1KI

form

L(s) =1

s

KDcP

1 +KDcP




−20 −15 −10 −5 0 5−15

−10

−5

0

5

10

15

Root Locus


Imag

inar

y A

xis

(sec

onds

−1 )

(a) Root locus with PI

−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5

Root Locus

Real Axis (seconds−1)Im

agin

ary

Axi

s (s

econ

ds−

1 )

(b) Root locus with PI (zoomed)

For KI > 0, ζ ↓ =⇒ Mp ↑




Choose small value of KI = 0.15

Higher overshoot at the cost of zero steady-state error

0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (seconds)

Am

plitu

de

PI+LeadLeadProportional

(a) Root locus with PI (zoomed)

20 22 24 26 28 30 32 34 36 38 400.94

0.96

0.98

1

1.02

1.04

1.06

Step Response

Time (seconds)

Am

plitu

de

PI+LeadLeadProportional

(b) Root locus with PI (zoomed)



Control of a Small Airplane – AnalysisControl u(t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Step Response

Time (seconds)

Ele

vato

r an

gle

(deg

)

High frequency in u(t) is undesirable rate limit & controller roll off

Large values for u(t) is undesirable saturation



Control of a Small Airplane – AnalysisHow good is this controller?

0 0.2 0.4 0.6−0.5

0

0.5

1

1.5

Ref to Control

Time (seconds)

Ele

vato

r an

gle

(deg

)

0 10 20−0.5

0

0.5

1

Ref to Error

Time (seconds)E

rror

(de

g)0 10 20

0

0.5

1

1.5

Ref to Output

Time (seconds)

Pitc

h A

ngle

(de

g)

0 20 40−5

−4

−3

−2

−1

0

Dist to Error

Time (seconds)

Err

or (

deg)

0 10 20−1.5

−1

−0.5

0

Dist to Control

Time (seconds)

Ele

vato

r an

gle

(deg

)

0 20 400

1

2

3

4

5

Dist to Output

Time (seconds)

Pitc

h an

gle

(deg

)


AERO 422: Active Controls for Aerospace Vehicles - Root ... Root Locus.pdf · Root Locus Plotting...

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