# Roll Controller

This example is of another common component of an AFCS, a control loop that maintains the wings level when flying on autopilot, or that can be commanded to roll the airplane into a turn and hold it there. We shall see in this particular case that the resulting turn is virtually truly banked, even though no special provision has been made to control sideslip.

The block diagram of the system is shown in Fig. 8.26. It incorporates the yaw damper described in the previous section and adds two additional loops. The outer loop commands ф. The ф error is converted to a roll rate command by Jp, and it is the roll rate error that is then used to drive the aileron servo actuator. If the roll rate followed the command instantaneously, without lag, the bank angle response would be exponential (i. e., ф ф). In reality of course this ideal behavior is not achieved because of the airframe and servo dynamics.

For this example we use the state vector approach to system modeling in order to provide another illustration, one that differs in detail from that of Sec. 8.5.

As usual the starting point is the basic aircraft matrix equation,

x = Ax + Be (8.8,1)

in which x = v p г ф]т and c = [5a Sr]T.

The differential equations that correspond to the various control transfer functions in the figure are found as follows. For the yaw damper components, we have the same form of transfer functions as previously, that is,

(8.8,2)

s +

s H—

Krshr

s + Ays + ±

TvO / t.

From (8.8,4) we get the differential equation

For Jp we use the constant Kp, and for Ja we use a first order servoactuator

s +

The relation between Sa, p, and фс is seen from the diagram to be

Ф;(Рс P) ^а^р^Фс Ф) JaP

When we substitute for Ja and Jp the differential equation that results is

Ka 1 KaK„

ф-^р–8а+~^фс

T T T T

‘ (1 ‘ П 1 n • n

Equations (8.8,5) and (8.8,8) are the additional equations required to augment the basic system (8.8,1) to accommodate the addition of the two control angles as dependent variables. However, a little more manipulation is needed of (8.8,5). To put it in first-order form, we define the new variable

y = 8r (8-8,9)

and to put it in canonical form, we must eliminate r. This we do by using the third component equation in (8.8,1). When these steps have been aken the system can be assembled into the matrix equation

z = Pz + Q фс

where z = [u p г ф 8a Sr y]T

The matrices P and Q are:

Q = [0 0 0 0 (KaKpha) 0 0]T (8.8,12)

Equation (8.8,10) was solved by numerical integration for two cases, with the results shown on Figs. 8.27 and 8.28. The various gains and time constants used were se-

Figure 8.27 Response of roll controller to initial ф of 0.262 rad (15°). (а) ф, p, and 8a. (b) /3, г, ф, 8r. |

lected somewhat arbitrarily, as follows:

Kp = 1.5 Ku = -1.0 Kr = -1.6; r„ = .15 тг = .30 тм = 4.0

On the first of these figures, response to an initial bank error, we see that all the state variables experience a reasonably well damped oscillatory decay, and that the maxi-

0 5 10 15 20 25 30 Time, s (fe) |3, r, V|/, Sr Figure 8.28 Response of roll controller to roll command of 0.262 rad (15°). (а) ф, p, 8a. (b) Д r, ф, 8r. |

mum control angles required are not excessive—about 20° for the aileron and less than 1° for the rudder. The time taken for the motion to subside to negligible levels is equal to about two Dutch Roll periods. All the variables except ф subside to zero, whereas ф asymptotes to a new steady state. When level flight is reestablished, the airplane has changed its heading by about 1.8°.

The second figure shows the response to a 15° bank command. The new steady state is approached with a damped oscillation that takes about 15 s to decay. The steady state is clearly a turn to the right, in which r has a constant value and ф is increasingly linearly. All the other variables, including the two control angles, are very small. It is especially interesting that the sideslip angle is almost zero. Clearly this controller has the capability to provide the bank angle needed for a coordinated turn. (The angle of attack and lift would of course have to be increased.)

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