Control Lectures (David Caughey)

Dynamics of Flight VehiclesM&AE 5070

David A. Caughey

Sibley School of Mechanical & Aerospace EngineeringCornell University

Ithaca, New York 14853-7501

May 2011

Longitudinal DynamicsMixed Response

0 10 20 30 40 50 60 70 80 90 100

−1.5

−1

−0.5

0

0.5

1

1.5

Time, sec

Sta

te V

aria

ble

Speed, u/u

0

Angle of attackPitch ratePitch angle

0 1 2 3 4 5 6 7 8 9 10

−1.5

−1

−0.5

0

0.5

1

1.5

Time, secS

tate

Var

iabl

e

Speed, u/u

0


Figure 1:Response of Boeing 747 in powered approach to perturbation in angle of attack. Level flight atM∞ = 0.25 at standard sea level conditions. (Heffley’s Condition 2.)

Longitudinal DynamicsModal Response

0 20 40 60 80 100 120 140 160 180 200

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time, sec

Sta

te V

aria

ble

Speed, u/u

0


0 1 2 3 4 5 6 7 8 9 10−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, secS

tate

Var

iabl

e

Speed, u/u

0


(a) Phugoid (b) Short Period

Figure 2:Response of Boeing 747 in powered approach to perturbation in angle of attack. Level flight atM∞ = 0.25 at standard sea level conditions. (Heffley’s Condition 2.)

Longitudinal DynamicsPhasor Plots

q

w/u0 Re

Im

u/u0

θ

Re

q

Im

w/u0

θ

(a) Phugoid (b) Short Period

Figure 3:Phasor plots for response of Boeing 747 in powered approach. Level flight atM∞ = 0.25 atstandard sea level conditions. (Heffley’s Condition 2.)

Longitudinal DynamicsEffect of c.g. Location

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

-0.8 -0.6 -0.4 -0.2 0

Imag

inar

y pa

rt o

f roo

t

Real part of root

Short-periodShort-period

Phugoid modePhugoid mode

Figure 4:Locus of roots of characteristic equation for response of Boeing 747 in powered approach. Levelflight at M∞ = 0.25 at standard sea level conditions. (Heffley’s Condition 2.) As static margin is reducedfrom 0.22 to -.05, short-period roots join on real axis at 0.0158, phugoid roots join on real axis at 0.0021, oneof the phugoid roots becomes neutrally stable at 0.0, and third oscillatory mode develops at -.0145.

Lateral/Directional DynamicsTypical Response

0 2 4 6 8 10 12 14 16 18 20−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Time, sec

Sta

te V

aria

ble

SideslipRoll rateRoll angleYaw rate

Figure 5:Response of Boeing 747 aircraft to unit perturbation in roll rate. Powered approach atM∞ = 0.25at standard sea level conditions (Heffley’s Condition 2).

Lateral/Directional DynamicsModal Response

0 1 2 3 4 5 6 7 8 9 10

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Time, sec

Sta

te V

aria

ble


0 10 20 30 40 50 60 70 80 90 100−0.2

0

0.2

0.4

0.6

0.8

1

Time, sec

Sta

te V

aria

ble


0 2 4 6 8 10 12 14 16 18 20−1.5

−1

−0.5

0

0.5

1

1.5

Time, sec

Sta

te V

aria

ble


(a) Rolling mode (b) Spiral mode (c) Dutch Roll mode

Figure 6:Response of Boeing 747 aircraft to unit perturbation in eigenvectors correspondingto the threelateral/directional modes. Powered approach atM∞ = 0.25 at standard sea level conditions (Heffley’sCondition 2).

Lateral/Directional DynamicsDutch Roll

φ

β

Im

Re

p

r

0 2 4 6 8 10 12 14 16 18 20−1.5

−1

−0.5

0

0.5

1

1.5

Time, sec

Sta

te V

aria

ble


(a) Phasor plot (b) Response

Figure 7:Dutch Roll response of Boeing 747 aircraft in powered approach atM∞ = 0.25 at standard sealevel conditions (Heffley’s Condition 2). (a) Phasor plot, and (b) Response to initial perturbation exciting onlythe Dutch Roll mode.

Lateral/Directional DynamicsDutch Roll – Dihedral Effect

-1

-0.5

0

0.5

1

-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0

Imag

inar

y pa

rt o

f roo

t

Real part of root

Rolling modeDutch Roll Dutch Roll

Spiral mode

Figure 8:Locus of roots of plant matrix for Boeing 747 aircraft in powered approach atM∞ = 0.25 understandard sea level conditions (Heffley’s Condition 2). Dihedral effect is varied from -.041 to -.561 in steps of-.04, while all other stability derivatives are held fixed at their nominal values. Rolling and spiral modesbecome increasingly stable as dihedral effect is increased; spiral mode becomes stable atCl β ≈ −.051.Dutch Roll mode becomes less stable as dihedral effect is increased and becomes unstableat Cl β ≈ −.532.

Lateral/Directional DynamicsDutch Roll – Weathercock Stability

-1

-0.5

0

0.5

1

-1.2 -1 -0.8 -0.6 -0.4 -0.2 0

Imag

inar

y pa

rt o

f roo

t

Real part of root

Rolling modeDutch Roll Dutch Roll

Spiral mode

Figure 9:Locus of roots of plant matrix for Boeing 747 aircraft in powered approach atM∞ = 0.25 understandard sea level conditions (Heffley’s Condition 2). Weathercock stabilityis varied Rom -.07 to 0.69 insteps of 0.04, while all other stability derivatives are held fixed at their nominal values. Rolling and spiralmodes become less stable as weathercock stability is increased; spiral mode becomes unstable atCnβ ≈ 0.657. Dutch Roll mode becomes increasingly stable as weathercock stability is increased, but isunstable for less thanCnβ ≈ −.032.

Longitudinal ControlImpulsive Elevator Response

0 2 4 6 8 10 12 14 16 18 20−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Time, sec

Sta

te V

aria

ble

Speed, uAngle of attackPitch ratePitch angle

0 50 100 150 200 250 300 350 400 450 500−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Time, sec

Sta

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aria

ble


Figure 10:Response of Boeing 747 for powered approach atM∞ = 0.25 and standard sea level conditionsto impulsive elevator input. Left plot is scaled to illustrate short-period response, and right plot is scaled toillustrate phugoid.

Longitudinal ControlResponse to Step Elevator Input

0 2 4 6 8 10 12 14 16 18 20−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


0 50 100 150 200 250 300 350 400 450 500−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


Figure 11:Response of Boeing 747 in powered approach atM∞ = 0.25 and standard sea level conditions toone-degree step elevator input. Left plot is scaled to illustrate short-period response, and right plot is scaled toillustrate phugoid.


0 50 100 150 200 250 300 350 400 450 500−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


0 2 4 6 8 10 12 14 16 18 20−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


Figure 12:Response of Boeing 747 in powered approach atM∞ = 0.25 and standard sea level conditions toone-degree step elevator input. Left plot illustrates phugoid resulting fromstep input; right plot addsperturbation to initial condition to cancel transient phugoid component of step input.

System response to step input is

x =“

eAt − I”

A−1Bη0 = eAt“

A−1Bη0

”

− A−1Bη0 (1)

So, system ultimately settles into the new equilibrium state given by

limt→∞

x(t) = −A−1Bη0 = [0.0459 − .0186 0.0 − .0160]T (2)

for the one-degree value ofη0. The new equilibrium state corresponds to an increase in flight speed at a reduced angle of attack,

and the aircraft has begun to descend.


0 2 4 6 8 10 12 14 16 18 20−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


0 10 20 30 40 50 60 70 80 90 100−0.1

−0.05

0

0.05

0.1

0.15

Time, sec

Sta

te V

aria

ble


Figure 13:Response of Boeing 747 in powered approach atM∞ = 0.25 and standard sea level conditions to one-degreestep elevator input. Left plot illustrates phugoid resulting from step input with perturbation added to the initial condition tocancel transient phugoid component of step input. Right plotillustrates degree to which the phugoid can be canceled byimpulsive control input.

For one-degree elevator step input, the system ultimately settles into the new equilibrium state given by

limt→∞

x(t) = −A−1Bη0 = [0.0459 − .0186 0.0 − .0160]T (3)

The phugoid excitation can be reduced by setting the initialcondition to the phugoid component of the final state

xPh = −Q−1A−1Bη0 (4)

or by including an impulsive control input designed to produce that initial perturbation (to the extent possible by the controlavailable) from the least-squares approximation to the solution of

Bηimp = xPh (5)

Longitudinal ControlStep Elevator Input – Comparison with Result of Static Longitudinal Analysis

This result is completely consistent with that of our study ofstatic longitudinal control, wherethe control sensitivity was found to be

dδe

dCL

«

trim=

Cmα

∆(6)

where∆ = −CLαCmδe + CmαCLδe (7)

Thus, from the static analysis we estimate for a step input of one degree in elevator

∆CL =δe

Cmα/∆=

π/180

(−1.26)/(7.212)= −.100 (8)

The asymptotic steady state of the dynamic analysis gives exactly the same result

∆CL = CLαα + CLδeδe = 5.70(−.0186) + 0.338(π/180) = −.100 (9)

State-Feedback ControlExample

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

Time, sec

Sta

te v

aria

ble

Original system, x

1

Original system, x2

With feedback, x1

With feedback, x2

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

Time, sec

Sta

te v

aria

ble

Original system, x

1

Original system, x2

With feedback, x1

With feedback, x2

(a) x(0) = [1.0 0.0] (b) x(0) = [1.0 0.1]

Figure 14:Response of linear, second-order system, showing effect of state variable feedback.Originalsystem hasλ1 = 0.0, λ2 = −3.0. Modified system hasωn = 25 sec−1 andζ = 0.707. (a)x(0) = [1.0 0.0]T ; (b) x(0) = [1.0 0.1]T .

Eigenvalues of original plant matrix areλ1 = 0 andλ2 = 3; corresponding eigenvectors are

u1 = [1 3/8]T and u2 = [1 0]T

so first initial condition has nou2 component. Final state of response of original system tosecond initial condition is parallel to first eigenvector.

Lateral/Directional Feedback ControlState Variable Feedback – Rudder Only

0 5 10 15 20 25 30−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Time, sec

Sta

te V

aria

ble


0 5 10 15 20 25 30−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Time, sec

Sta

te V

aria

ble


(a) (b)

Figure 15:Boeing 747 aircraft in powered approach at standard sea level conditions andM∞ = 0.25;response to 5 degree (0.08727 radian) perturbation in sideslip. (a) Original open-loop response; (b) Closedloop response with Dutch Roll damping ratio changed toζ = 0.30 using rudder state-variable feedback.

Lateral/Directional Feedback ControlState Variable Feedback – Effective Changes to Plant Matrix

The state and control vectors for lateral-directional motions are

x = [v p φ r]T and η = [δr δa]T (10)

and the plant and control matrices are given by

A =

0

B

B

B

B

@

Yv Yp g cosΘ0 Yr − u0Lv+ixNv1−ixiz

Lp+ixNp1−ixiz

0 Lr+ixNr1−ixiz

0 1 0 0Nv+izLv1−ixiz

Np+izLp1−ixiz

0Nr+izLr1−ixiz

1

C

C

C

C

A

(11)

and

B =

0

B

B

B

B

@

Yδr 0Lδr

+ixNδr1−ixiz

Lδa+ixNδa

1−ixiz0 0

Nδr+izLδr

1−ixiz

Nδa+izLδa

1−ixiz

1

C

C

C

C

A

(12)

respectively.The original plant matrix and thechangesto the plant matrix for improved Dutch-Roll dampingdue torudder-onlystate variable feedback control are

A =

0

B

B

@

−.0999 0.0000 0.1153 −1.000−1.604 −1.0932 0.0000 0.28500.0000 1.0000 0.0000 0.00000.4089 −.0395 0.0000 −.2454

1

C

C

A

and ∆A =

0

B

B

@

−.0025 −.0017 −.0023 0.0206−.0120 −.0082 −.0109 0.09840.0000 0.0000 0.0000 0.00000.0337 0.0230 0.0305 −.2766

1

C

C

A

(13)

respectively.

Lateral/Directional Feedback ControlState Variable Feedback – Aileron Only

0 5 10 15 20 25 30−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Time, sec

Sta

te V

aria

ble


0 5 10 15 20 25 30−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Time, sec

Sta

te V

aria

ble


(a) (b)

Figure 16:Boeing 747 aircraft in powered approach at standard sea level conditions andM∞ = 0.25;response to 5 degree (0.08727 radian) perturbation in sideslip. (a) Original open-loop response; (b) Closedloop response with Dutch Roll damping ratio changed toζ = 0.30 using aileron state-variable feedback.

Lateral/Directional Feedback ControlState Variable Feedback – Effective Changes to Plant Matrix




A =

0

B

B

B

B

@


Lp+ixNp1−ixiz

0 Lr+ixNr1−ixiz


Np+izLp1−ixiz

0Nr+izLr1−ixiz

1

C

C

C

C

A

(15)

and

B =

0

B

B

B

B

@

Yδr 0Lδr

+ixNδr1−ixiz

Lδa+ixNδa

1−ixiz0 0

Nδr+izLδr

1−ixiz

Nδa+izLδa

1−ixiz

1

C

C

C

C

A

(16)

respectively.

The original plant matrix and thechangesto the plant matrix for improved Dutch-Roll dampingdue toaileron-onlystate variable feedback control are

A =

0

B

B

@

−.0999 0.0000 0.1153 −1.000−1.604 −1.0932 0.0000 0.28500.0000 1.0000 0.0000 0.00000.4089 −.0395 0.0000 −.2454

1

C

C

A

and ∆A =

0

B

B

@

0.0000 0.0000 0.0000 0.00001.1387 −.2802 −.2169 1.30230.0000 0.0000 0.0000 0.0000−.0062 0.0015 0.0012 −.0071

1

C

C

A

(17)

respectively.

Optimal ControlMotivation

The Bass-Gura procedure often problematic for a variety of reasons:

• Best placement for the eigenvalues of the augmented matrix is not always obvious;

• Particular eigenvalue placement may require more control input than is available;

• For multiple input-output systems, we need strategies for howbest to allocate the gainsamong then× p elements, since we have onlyn eigenvalues to place;

• The process may not be controllable; i.e., if the rank of the controllability matrix V is lessthann, the method fails.

All these points argue for a control design strategy that, insome sense, optimizes the gain matrixfor stabilizing a given system. This is the goal of what has cometo be calledoptimal control.

Optimal ControlFormulation of Linear, Quadratic Control

Theoptimal controlof the linear system

x = Ax + Bη(t) (18)

is defined as the control vectorη(t) that drives the state from a specified initial statex(t) to adesired final statexd(tf ) such that a specified performance index

J =

Z tf

tg(x(τ), η(τ), τ) dτ (19)

is minimized. Forquadraticoptimal control, the performance index is specified in the form

g = xTQx + ηTRη (20)

whereQ andR are symmetric, positive-definite matrices, and the performanceindex becomes

J =

Z tf

t

`

xTQx + ηTRη´

dτ (21)

If the control law is assumed to be linear, i.e., of the form

η = −Kx + η′ (22)

then the determination of the gain matrixK that minimizesJ is called the linear quadraticregulator (LQR) problem.


For this control law the closed-loop response of the system to a perturbation is given by

x = [A − BK] x = A∗x (23)

whereA∗ = A − BK is theaugmentedplant matrix.We usually are interested in cases for which the matricesA, B, andK are independent of time,but the development here is easier if we allow the augmented matrix A∗ to vary with time. Inthis case, we need to express the solution to Eq. (23) in terms of the general state transitionmatrixΦ∗ as

x(τ) = Φ∗(τ, t)x(t) (24)

Equation (24) simply implies that the state of the system at anytimeτ depends linearly on thestate at any other timet. When the control law of Eq. (22) is substituted and Eq. (24) isused toexpress the evolution of the state variable, the performanceindex of Eq. (21) becomes

J =

Z tf

txT

(τ)h

Q + KTRKi

x(τ) dτ

=

Z tf

txT

(t)Φ∗T(τ, t)

h

Q + KTRKi

Φ∗(τ, t)x(t) dτ

= xT(t)

„Z tf

tΦ

∗T(τ, t)

h

Q + KTRKi

Φ∗(τ, t) dτ

«

x(t)

(25)

orJ = xT

(t)Sx(t) (26)

where

S(t, tf ) =

Z tf

tΦ

∗T(τ, t)

h

Q + KTRKi

Φ∗(τ, t) dτ (27)

Note that, by its construction, the matrixS is symmetric, since the weighting matricesQ andRare both symmetric.


The simple appearance of Eq. (26) belies the complexity of determiningS from Eq. (27) because it is almost impossible todevelop a useful expression for the state transition matrixin general. Instead, in order to find the gain matrixK that minimizesJ, it is convenient to find a differential equation that the matrix S satisfies. To this end, we note that since

J =

Z tf

txT

(τ)Lx(τ) dt (28)

whereL = Q + KTRK (29)

we can writedJ

dt= − xT

(τ)Lx(τ)˛

˛

˛

τ=t= −xT

(t)Lx(t) (30)

But, from differentiating Eq. (26), we have

dJ

dt= xT

(t)S(t, tf )x(t) + xT(t)S(t, tf )x(t) + xT

(t)S(t, tf )x(t) (31)

and, substituting the closed-loop differential equation,Eq. (23), forx gives

dJ

dt= xT

(t)h

A∗TS(t, tf ) + S(t, tf ) + S(t, tf )A∗(t)

i

x(t) (32)

Thus, we have two expressions for the derivative dJ/ dt: Eqs. (30) and (32). Both are quadratic forms in the initial statex(t),which must bearbitrary. The only way that two quadratic forms inx can be equal for any choice ofx is if the underlyingmatrices are equal; thus, we must have

−S = SA∗+ A∗TS + L (33)


Equation (33)

−S = SA∗+ A∗TS + L (34)

is a first-order differential equation for the matrixS, so it requires a single initial condition to completely specify its solution.We can use Eq. (27), evaluated att = tf to give the required condition

S(tf , tf ) = 0 (35)

Once a gain matrixK has been chosen to close the loop, the corresponding performance of the system is given by Eq. (26),whereS(t, tf ) is the solution of Eq. (33), which can be written in terms of the original plant and gain matrices as

−S = S (A − BK) +“

AT − KTBT”

S + Q + KTRK (36)

Our task, then, is to find the gain matrixK that makes the solution to Eq. (36) as small as possible – in the sense that thequadratic forms (Eq. (26)) associated with the matrixS are minimized. That is, we want to find the matrixS for which

J = xT Sx < xTSx (37)

for any arbitrary initial statex(t) and every matrixS 6= S.We will proceed by assuming that such an optimum exists, and use the calculus of variations to find it. The minimizing matrix Smust, of course, satisfy Eq. (36)

− ˙S = S“

A − BK”

+“

AT − KTBT”

S + Q + KTRK (38)

and anynon-optimum gain matrix, and its corresponding matrixS, can be expressed as

S = S + N and K = K + Z (39)

Substituting this form into Eq. (36) and subtracting Eq. (38) gives

−N = NA∗+ A∗

TN +

“

KTR − SB”

Z + ZT“

RK − BT S”

+ ZTRZ (40)

whereA∗

= A − BK = A − B“

K + Z”

(41)


Note that Eq. (40) has exactly the same form as Eq. (33) with

L =“

KTR − SB”

Z + ZT“

RK − BT S”

+ ZTRZ (42)

so its solution must be of the form of Eq. (27)

N(t, tf ) =

Z tf

tΦ

∗T(τ, t)LΦ

∗(τ, t) dτ (43)

Now, if J is a minimum, then we must have

xT Sx ≤ xT“

S + N”

x = xT Sx + xTNx (44)

and this equation requires that the quadratic formxTNx be positive definite (or, at least, positive semi-definite).But, if Z issufficiently small, the linear terms inZ (andZT ) in Eq. (42) will dominate the quadratic terms inZTRZ, and we could easilyfind values ofZ that would makeL, and henceN, negative definite. Thus,the linear terms in Eq. (42) must be absentaltogether. That is, for the gain matrixK to be optimum, we must have

KTR − SB = 0 = RK − BT S (45)

or, assuming that the weighting matrixR is not singular,

K = R−1BT S (46)

Equation (46) gives the optimum gain matrixK, once the matrixS has been determined. When this equation is substituted backinto Eq. (38) we have

− ˙S = SA + AT S − SBR−1BT S + Q (47)

This equation, one of the most famous in modern control theory, is called thematrix Riccati equation, consistent with the

mathematical nomenclature that identifies an equation witha quadratic non-linearity as a Riccati equation. The solution to this

equation gives the matrixS which, when substituted into Eq. (46), gives the optimum gain matrixK.


Due to the quadratic nonlinearity of the Riccati equation, itusually is necessary to solve it numerically. Since the matrix S issymmetric, Eq. (47) representsn(n + 1)/2 coupled, first-order equations. Since the “initial" condition is

S(tf , tf ) = 0 (48)

the equation must be integratedbackwardin time, since we are interested inS(t, tf ) for t < tf . When the control interval[t, tf ] is finite, the gain matrixK is generally time-dependent, even if the matricesA, B, Q, andR all are constant. But, suppose

the control interval isinfinite, so we want to find the gain matrixK that minimizes the performance index

J∞ =

Z

∞

t

“

xTQx + ηTRη

”

dτ (49)

In this case, integration of Eq. (47) backward in time will either grow without limit or converge to aconstantmatrix S. If it

converges to a limit, the derivative˙S must tend to zero, andS must satisfy thealgebraicequation

0 = SA + AT S − SBR−1BT S + Q (50)

and the optimum gain in the steady state is given by

K = R−1BT S (51)

The singlequadraticmatrix Eq. (50) representsn(n + 1)/2 coupled scalar, quadratic equations, so we expectn(n + 1)different (symmetric) solutions. The nature of these solutions is connected with issues of controllability and observability; forour purposes here, it is enough to know that

• If the system is asymptotically stable; or

• If the system defined by(A, B) is controllable, and the system defined by(A, C), where the weighting matrixQ = CTC, is observable,

then the algebraic Riccati equation has an unique positive definite solutionS that minimizesJ∞ when the control law

η = −Kx = −R−1BT Sx (52)

is used. (Note that there are stilln(n + 1) symmetric solutions; the assertion here is that, of these multiple solutions, one, and

only one, ispositive definite.)

Optimal ControlLinear, Quadratic Control: Example

We consider using optimal control to stabilize an inverted pendulum.The equation of motion for an inverted pendulum near its (unstable)equilibrium point, as illustrated in Fig. 17 is

mL2θ = mgLsinθ + T = mgLθ + T (53)

wherem is the mass of the pendulum,L is the pendulum length,g is theacceleration of gravity, andT is the externally-applied (control) torque;the second form of the right-hand side assumes the angleθ is small.

θ

T

g

L

m

Figure 17:Inverted pendulum affectedby gravityg and control torqueT.

If we introduce the angular velocityω = θ as a second state variable, Eq. (53) can be written in the standard state variable form

d

dt

„

θω

«

=

„

0 1γ 0

« „

θω

«

+

„

01

«

τ (54)

whereγ = g/L andτ = T/(mL2) are reduced gravity and input torque variables.Now, we seek the control law that minimizes the performance index

J∞ =

Z

∞

t

θ2+

τ2

c2

!

dt′ (55)

wherec is a parameter that determines the relative weighting of control input and angular deviation in the penalty function. Itisclear that this performance index corresponds to

Q =

„

1 00 0

«

and R =1

c2(56)


If we define the elements of the matrixS to be

S =

„

s1 s2s2 s3

«

(57)

then the optimum gain matrix is

K = R−1BT S = c2 ˆ

0 1˜

„

s1 s2s2 s3

«

=h

c2s2 c2s3

i

(58)

which is seen to be independent of the elements1. The terms needed for the algebraic Riccati equation

0 = SA + AT S − SBR−1BT S + Q (59)

are

SA =

„

s1 s2s2 s3

« „

0 1γ 0

«

=

„

s2γ s1s3γ s2

«

(60)

AT S =

„

0 γ1 0

« „

s1 s2s2 s3

«

=

„

s2γ s3γs1 s2

«

=`

SA´T (61)

and

SBR−1BT S =

„

s1 s2s2 s3

« „

01

«

c2 ˆ

0 1˜

„

s1 s2s2 s3

«

= c2

s22 s2s3

s2s3 s23

!

(62)

Thus, the Riccati equation is

0 =

„

s2γ s1s3γ s2

«

+

„

s2γ s3γs1 s2

«

− c2

s22 s2s3

s2s3 s23

!

+

„

1 00 0

«

(63)

which is equivalent to the three scalar equations

0 = 2s2γ − c2s22 + 1

0 = s1 + s3γ − c2s2s3

0 = 2s2 − c2s23

(64)


These can be solved in closed form. The first of Eqs. (64) gives

s2 =γ ±

p

γ2 + c2

c2(65)

and the third of Eqs. (64) gives

s3 = ±1

c

p

2s2 (66)

Since the elements ofS must be real,s2 must be positive (ors3 would be complex). Thus, wemust choose the positive root in Eq. (65). Further, the secondof Eqs. (64) gives

s1 = c2s2s3 − γs3 = s3

p

γ2 + c2 (67)

Thus, elementss1 ands3 have the same sign which, forS to be positive definite, must bepositive. Thus,

s2 =γ +

p

γ2 + c2

c2

s3 =1

c

p

2s2 =

√2

c2

h

γ +p

γ2 + c2i1/2

(68)

represents the unique solution for the corresponding elements for whichS is positive definite.


Thus, the (optimal) gain matrix is seen to be

K =ˆ

c2s2 c2s3˜

=

»

γ +p

γ2 + c2√

2h

γ +p

γ2 + c2i1/2

–

(69)

The augmented matrix is then given by

A∗ = A − BK =

0 1

−p

γ2 + c2 −√

2h

γ +p

γ2 + c2i1/2

!

(70)

and its characteristic equation is

λ2 +√

2h

γ +p

γ2 + c2i1/2

λ +p

γ2 + c2 = 0 (71)

which has roots

λ =

√2

2

ˆ

−√

γ + γ ± ı√

γ − γ˜

(72)

where we have introducedγ =

p

γ2 + c2 (73)


Note that asc/γ becomes large,γ becomes large relative toγ, so

limc/γ→∞

λ = −s

γ

2(1 ± ı) (74)

Thus, asc becomes large, the damping ratio of the systemapproaches a constant value of

ζ =1√

2

while the undamped natural frequency increases as

ωn =√

γ ≈√

c

Large values ofc correspond to a performance index in whichthe weighting of the control term is small compared to that ofthe deviations in state variables – i.e., to a situation in whichwe are willing to spend additional energy in control to main-tain very small perturbations of the state from its equilibriumposition.

−5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Real part

Imag

inar

y pa

rt

Figure 18:Locus of roots of characteristicequation of augmented plant matrix for invertedpendulum. Axes are scaled to give roots in unitsof γ. Open symbols represent roots atc/γ = 0, 1, 10, 100, 1000, with real rootcorresponding toc/γ = 0. Cyan lines representasymptotes in the limit of largec/γ.

On the other hand, asc becomes small, the weighting of the control term in the performance index becomes large compared tothat of the state variables. This is consistent with the factthat the gains in Eq. (69)

K1 = γ +

q

γ2 + c2 and K2 =√

2

»

γ +

q

γ2 + c2–1/2

decrease monotonically withc. In the limit c = 0, however, the gains remain finite, with

limc→0

K1 = 2γ and limc→0

K2 = 2√

γ

sincesomecontrol is necessary to stabilize this, otherwise unstable, system.

Optimal ControlLinear, Quadratic Control as a Stability Augmentation System

We here apply linear, quadratic optimal control to improve the stability of the Boeing 747 aircraft in powered approach atM = 0.25 at standard sea level conditions.We apply linear, quadratic, optimal control to minimize thesteady state performance index

J∞ =

Z

∞

t

„

xTQx +1

c2η

TRη

«

dτ (75)

wherec is again a parameter that determines the relative weights given to control action and perturbations in the state variable inthe penalty function. For lateral/directional motions at this flight condition, the plant matrix is the same as used earlier, whilethe control matrix is given by

B =

„

0.0182 0.0868 0.0000 −.24400.0000 0.3215 0.0000 −.0017

«T(76)

where the control vector isη = [δr δa]

T (77)

The weighting matrices in the performance index are taken tobe simply

Q = I and R = I (78)

whereQ is a 4× 4 matrix andR is a 2× 2 matrix.The MATLAB function

[S, L, G] = care(A,B,Q,R,T,E);

is used to solve the generalized matrix Riccati equation

ETSA + ATSE −“

ETSB + T”

R−1“

BTSE + TT”

+ Q = 0 (79)

which, with the additional input matrices are defined asT = zeros(size(B)) and E = eye(size(A))

reduces to Eq. (50). In addition to the solution matrixS, the MATLAB functioncarealso returns the gain matrix

G = R−1“

BTSE + TT”

(80)

and the vectorL = eig(A - BG, E)

containing the eigenvalues of the augmented matrix.

Lateral/Directional Feedback ControlLinear-Quadratic Control as a Stability Augmentation System

0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, sec

Sta

te V

aria

ble


0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, sec

Sta

te V

aria

ble

SideslipRoll rateRoll angleYaw rateAileronRudder

(a) No feedback (b)c = 0.001

Figure 19:Boeing 747 aircraft in powered approach at standard sea level conditions andM∞ = 0.25;response to unit perturbation in sideslip. (a) Original open-loop response; (b) Optimal closed loop responsewith performance parameterc = 0.001.


0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, sec

Sta

te V

aria

ble


0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, sec

Sta

te V

aria

ble


0 2 4 6 8 10 12 14 16 18 20−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time, sec

Sta

te V

aria

ble


(a)c = 0.50 (b)c = 1.0 (c)c = 2.0

Figure 20:Boeing 747 aircraft in powered approach at standard sea level conditions andM∞ = 0.25;response to unit perturbation in sideslip illustrating effect of varying weighting parameterc. Optimalclosed-loop responses with (a)c = 0.50; (b)c = 1.0; and (c)c = 2.0. Control deflections required tostabilize the motions are also shown.


-3

-2

-1

0

1

2

0 2 4 6 8 10 12 14

Pen

alty

func

tion

Time, sec.

c = 0.5c = 1.0c = 2.0

Figure 21:Penalty functions in performance index for optimal control solution; Boeing747 aircraft inpowered approach at standard sea level conditions andM∞ = 0.25. Upper curves arexTQx, and lowercurves are−ηTRη, as functions of time for response to unit perturbation in sideslip angleβ, so the areabetween the curves is equal to the performance indexJ∞.

Lateral/Directional Feedback ControlLinear-Quadratic Control – Effective Changes to Plant Matrix




A =

0

B

B

B

B

@


Lp+ixNp1−ixiz

0 Lr+ixNr1−ixiz


Np+izLp1−ixiz

0Nr+izLr1−ixiz

1

C

C

C

C

A

(82)

and

B =

0

B

B

B

B

@

Yδr 0Lδr

+ixNδr1−ixiz

Lδa+ixNδa

1−ixiz0 0

Nδr+izLδr

1−ixiz

Nδa+izLδa

1−ixiz

1

C

C

C

C

A

(83)

respectively.The original plant matrix and thechangesto the plant matrix for improved response usingLQCwith c = 2.0 are

A =

0

B

B

@

−.0999 0.0000 0.1153 −1.000−1.604 −1.0932 0.0000 0.28500.0000 1.0000 0.0000 0.00000.4089 −.0395 0.0000 −.2454

1

C

C

A

and ∆A =

0

B

B

@

−.0458 −.0114 0.0099 0.07730.3433 −.4532 −.4662 −.02530.0000 0.0000 0.0000 0.00000.6100 −.1498 −.1299 −1.0334

1

C

C

A

(84)

respectively.


-1

-0.5

0

0.5

1

-2.5 -2 -1.5 -1 -0.5 0

Imag

inar

y pa

rt o

f roo

t

Real part of root

Rolling Dutch rollSpiral

Figure 22:Boeing 747 aircraft in powered approach at standard sea level conditions andM∞ = 0.25; locusof roots of characteristic equation of augmented matrix as control weighting parameterc is increased.Symbols represent root locations forc = 0.001, 0.5, 1.0, 2.0, 5.0, 8.0, 9.0, 9.7, 9.76, 9.7727, 9.7728, 10.0;asc is increased, all roots move to the left (except for one of the Dutch Roll roots after that mode becomescritically damped between 9.7227< c < 9.7228).

Documents

Control Lectures (David Caughey)