automatic flight control - ch11

7/29/2019 Automatic Flight Control - Ch11

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Chapter 11

Automatic Flight Control

11.1 Simple Feedback Systems

11.1.1 First-Order Systems

x xu

a

b+

+

Figure 11.1: First-order system

The simple first-order system shown in figure 11.1 is represented by thedynamic equation of motion,

x = ax(t) + bu(t) (11.1)

The single eigenvalue of this system (with subscript OL to representOpen Loop) is

OL = a

The block diagram is simpler in transfer function form. The forced re-sponse becomes

231


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232 CHAPTER 11. AUTOMATIC FLIGHT CONTROL

x(s) = bs a u(s) (11.2)

The block diagram for the transfer function representation of this systemis shown in figure 11.2.

bsa

u(s) x(s)

Figure 11.2: Transfer function representation, first-order system

Now consider a feedback scheme in which the state x(t) is measured, themeasured value amplified by a factor k, and used to modify the input. Theinput u(t) is now the sum of kx(t) and a new signal r(t), or in terms of thecomplex variable s, u(s) = kx(s) + r(s) as shown in figure 11.3.

x(s)u(s)b

sa

r(s) +

+

k

Figure 11.3: Closed-loop system

The equation of motion of the closed-loop (CL) system becomes

x = ax(t) + b [kx(t) + r(t)] = (a + bk) x(t) + br(t) (11.3)

The closed-loop eigenvalue is then

CL = a + bk (11.4)

In other words, by proper choice of k, the system eigenvalue may beassigned arbitrarily.

The same result can be arrived at using the transfer function. We have


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11.1. SIMPLE FEEDBACK SYSTEMS 233

x(s) = bs au(s) =

bs a [kx(s) + r(s)] (11.5a)

1 bks a

x(s) =

b

s ar(s) (11.5b)s a bk

s a x(s) =b

s ar(s) (11.5c)

x(s) =b

s (a + bk)r(s) (11.5d)

11.1.2 Second-Order Systems

Open-loop Eigenvalues

A simple mass-spring-damper system (figure 11.4) is used to illustrate closed-loop control of second-order systems.

x2 x1x2u

a21

a22

b+

+ +

Figure 11.4: Second-order system

Here, x1 is the position and x2 is the velocity, with x1 = x2. The systemis represented generically by:

x1 = x2(t)

x2 = a21x1(t) + a22x2(t) + bu(t)(11.6)

The parameter a21 is related to the spring in the system (proportionalto displacement x1), and the parameter a22 to the damping (proportional tovelocity x2).


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We assume the system has complex eigenvalues, OL = j, so that

the response is oscillatory. The characteristic equation of this system is easilyverified to be s2 a22s a21 = 0. We may relate the parameters a21 and a22to the system natural frequency n and damping ratio as follows:

s2 a22s a21 = s2 + 2ns + 2n = 0 (11.7a)n =

a21 (11.7b) =

a222a21 (11.7c)

Using a little algebra, the real and imaginary parts of the system eigen-values, and , are related to the natural frequency, damping ratio, andsystem parameters:

OL = j = n jn

1 2 (11.8a) = n = a22

2(11.8b)

= n

1 2 = 12

a222 4a21 (11.8c)

In terms of the transfer functions, the state transition matrix is

[sI A]1 B =

s 1a21 s a22

1 0

b

=1

s2 a22s a21

s a22 1

a21 s

0

b

=1

s2 a22s a21

b

bs

The transfer functions are therefore

x1(s)u(s)

=b

s2 a22s a21 (11.9a)x2(s)

u(s)=

bs

s2 a22s a21 (11.9b)


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To construct a block diagram of this system, we can go from u(s) to x2(s),

then from x2(s) to x1(s). Since x1 = x2, a direct LaPlace transform showsthat x1(s) = x2(s)/s. Alternatively we may manipulate equations 11.9 toget the same result:

x1(s)

x2(s)=

x1(s)

u(s)

u(s)

x2(s)=

1

s(11.10)

The block diagram in the LaPlace diagram then becomes as shown infigure 11.5.

x1(s)x2(s)u(s)bss2a22sa21

1s

Figure 11.5: Transfer function representation, second-order system

A pole-zero map of the transfer function x1(s)/u(s) is shown in figure11.6.

Real

Imaginary

n

cos1

Figure 11.6: Pole-zero map, x1(s)/u(s)


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Position Feedback

Now consider feedback of the position variable, u(t) = k1x1(t) + r(t). Thisfeedback does not affect the kinematic equation x1 = x2, but changes theacceleration x2 as follows:

x2 = a21x1(t) + a22x2(t) + b [k1x1(t) + r(t)]

= (bk1 + a21) x1(t) + a22x2(t) + br(t)

Position feedback therefore affects only the spring parameter. The char-acteristic polynomial becomes s2 a22s (a21 + bk1).

In the LaPlace domain, whe have u(s) = k1x1(s) + r(s), with the blockdiagram shown in figure 11.9.

x1(s)x2(s)r(s) u(s)bs

s2a22sa21

1s

+

+

k1

Figure 11.7: Block diagram, position feedback

The closed-loop transfer function is easily determined,

x1(s) =b

s2 a22s a21 u(s)

=b

s2 a22s a21 [k1x1(s) + r(s)]

Simplifying,

x1(s)

r(s)=

b

s2

a22

s

(bk1

+ a21

)(11.11)

Since = a22/2, just as it was in the open-loop system, position feedbackdoes not change the damping term (real part of the eigenvalue) of the mass-spring-damper system. Therefore, as k1 is varied, the roots (eigenvalues) willmove vertically in the complex plane, as shown in figure 11.8.


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Real

Imaginary

Constant

Figure 11.8: Effect of position feedback

Rate Feedback

With rate feedback we have u(t) = k2x2(t) + r(t). The acceleration x2 thenbecomes:

x2 = a21x1(t) + a22x2(t) + b [k2x2(t) + r(t)]

= a21x1(t) + (bk2 + a22) x2(t) + br(t)

Rate feedback therefore affects only the damping parameter. The char-

acteristic polynomial becomes s2

(bk1 + a21) s a21. The block diagram ofthis system is shown in figure 11.9.

x1(s)

r(s)=

b

s2 (bk2 + a22) s a21 (11.12)


s2a22sa21

1s

+

+

k2

Figure 11.9: Block diagram, rate feedback


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Real

n

Constant n

...................................................................................................................................................................

.............................................................................................................

...............

Figure 11.10: Effect of rate feedback

The unchanged term now is n =a21. Therefore, as k2 is varied, the

roots will move in a circular arc about the origin, as shown in figure 11.10.

By combining position and rate feedback, as shown in figure 11.11, thetransfer function becomes

x1(s)

r(s)=

b

s2 (bk2 + a22) s (bk1 + a21) (11.13)


s2a22sa21

1s

+

++

k2

k1

Figure 11.11: Block diagram, position and rate feedback

Thus, the eigenvalues of the mass-spring-damper system may be placedin any arbitrary position.


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x(s)u(s)r(s)G(s)

+

+

K(s)

Figure 11.12: Generic multi-inputmulti-output system

11.1.3 A General Representation

A somewhat more general representation of the mass-spring-damper system

is shown in figure 11.12. In the figure, x(s), u(s), and r(s) are vectors, andG(s) and K(s) are matrices. With reference to figure 11.12,

u(s) = Kx(s) + r(s) (11.14a)

x(s) = GKx(s) + Gr(s) (11.14b)

x(s) = [I GK(s)]1 Gr(s) (11.15)

The following assignments relate figure 11.12 to the mass-spring-dampersystem:

x(s) =

x1

x2

(11.16a)

u(s) = {u} (11.16b)r(s) = {r} (11.16c)

G(s) = b

s2a22sa21

bss2a22sa21 (11.16d)

K(s) =

k1 k2

(11.16e)

With these assignments, we have


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x(s) =

bs2(bk2+a22)s(bk1+a21)

bss2(bk2+a22)s(bk1+a21)

r(s) (11.17)

11.2 Aircraft Control Applications

11.2.1 Roll Mode

The roll mode of an aircraft is approximated by

Ixx p = Lpp + L

In which

Lp < 0, L < 0

With substitutions

x = p, u = , a = Lp/Ixx, b = L/Ixx

x = ax + bu

The transfer function from u to x is

x(s)

u(s)=

b

s aThe roll mode time constant is

r = 1

a = Ixx

Lp

Consider the unaugmented aircraft with fixed gearing G between the lat-eral stick input s and ailerons . Assume for this example that the limitsof stick and aileron are 1 and that the gearing is 1:1:


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11.2. AIRCRAFT CONTROL APPLICATIONS 241

1 s 1, 1 1, G = 1

So that the unaugmented aircraft control law is:

= s, 1 s 1

For a step lateral stick input at t = 0 of magnitude 1 (for positive rollresponse),

s(t) = 1, t 0s(s) = 1/s

The steady-state roll rate is

pss =ba =

LLp

Now examine roll-rate feedback to improve (decrease) r:

= Kpp + su = Kpx + r

The closed-loop transfer functions from r to x is

x(s)

r(s)=

b

s (a bKp)The augmented roll-mode time constant raug is

raug = 1

a bKp = Ixx

Lp KpLSince we desire raug < r,


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0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

UnaugmentedAugmented


|Lp KpL| > |Lp|

so that Kp < 0 (the transfer function is negative).

For the same step lateral stick input at t = 0 of magnitude 1 the steady-state roll rate (augmented) pssaug is

pssaug =b

(a bKp) =L

Lp KpL

automatic flight control - ch11

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