# The Differential Equations

The basic matrix differential equation of the airframe, with 0O = 0, and ДzE = —Ah is obtained from (4.9,18) and (7.6,4). Up to this point, we have neglected engine dy­namics and in effect regarded thrust as proportional to Sp. The matrix В of (7.6,4) is structured in that way. To accommodate the facts that Sp actually represents the throt­tle setting, not the thrust, and that the two are dynamically connected, we need to in­troduce two new symbols, y5 and c*. The quantity y5, when multiplied by XSp, etc. yields the aerodynamic force and moment increments AXC, etc., and c* is defined be­low. The differential equation of the airframe is then

x = Ax + Be* (8.5,4)

where x = [Дм w q в A h]r

c* = A y5f A = [аи]

В = [bu]

To obtain the differential equations of the three control elements, we begin with their transfer functions, which are specified for this example to be

Je(s) = (a0s~1 + a, + a2s){ 1 + re5)_l

= (a0 + a{s + a2s2)(s + Tes2)-1 (8.5,5)

Cp(s) = (Vі + bt + b2s)

= (b0 + bjS + b2s2)/s (8.5,6)

Jp(s) = 1/(1 + v) (8.5,7)

The first two of these contain proportional + rate + integral controls, all of which were found to be needed for good performance. The time domain equations that cor­respond to the elements of the controller are then as follows (verify this by taking their Laplace transforms):

в, = —к Ah (a)

теА8е + A 8e = a0e0 + ахёв + а2ёв (b) (8.5,8)

У 4 = V„ + V„ + b2eu (c)

y5 = A 8p – y5 (d)

After substituting the expressions for the two error signals, (8.5,8) yield three equa­tions for the controls

теА8е + A 8e = —ка2АЇг — a26 — kaxAh — axQ — ka0Ah — а0в (a)

y4 = – b2Au – bxAit – b0Au (b) (8.5,9)

Vs = “ Л (c)

For convenient integration we want a system of first-order equations and therefore have to do something about the second derivatives in (8.5,9). Since в = q we can re­place в with q. For the other second derivatives, we introduce three new variables, as follows:

Уі = дй О)

У2 = ДА (b) (8.5,10)

Уз = (с)

With these definitions, (8.5,9a and b) can now be rewritten in terms of first deriva­tives as

теУз + Уз = ~{a2q + ka^ + axq + а0в + ka0Ah + kaxy2) } j

У4 = ~ФіУ + b0Au + bxyx) ‘ ’

The state vector now consists of the original five variables from (8.5,4) plus the two control variables ASe and A8p, plus the five y, defined above, making 12 in all. We therefore require 12 independent equations. From the foregoing equations (8.5,4)

(8.5,10) , (8.5,11), and (8.5,9c) we can get 11 of the required differential equations. That fory, is obtained from (8.5,10a) by differentiating the first component of (8.5,4) and that for y2 by differentiating the fifth. The result of that operation is

y, = aX2w + axxyx + aX4q + bxxy3 + bX2y5 y2= – w + u0q

Finally, the 11 independent differential equations are assembled as follows:

Am = у і

w = a2XAu + a22w + a23q + b2XA8e q = a3]Au + a32w + a33q + b3XA8e d = q Ah = y2 A4 = Уз

У s = (~У s + Ь8р)/тр (8.5,13)

У = аххух + aX2w + aX4q + bxxy3 + bX2y5 y2 = u0q – w

y3 = —(a2q + ka2y2 + axq + aQ0 + ka0Ah + kaxy2 + у3)/те У 4 = “ОгУ і + b0Au + bxyx)