# Longitudinal Response

To treat this case we need the matrices A and В of (7.1,4). A is given in (4.9,18), but we have not yet given В explicitly. On the right side of (4.9,18) we have the product ВДс given by

A Z,

(m – ZJ

AM, M* AZ,

Iy Iy (m – Zw)

0

We now have to specify the control vector c and the corresponding aerodynamic forces and moment. For longitudinal control, we assume here that the available con­trols are well enough represented by

c = [8e 8p]T (7.6,2)

and that the incremental aerodynamic forces and moment that result from their actua­tion are given by a set of control derivatives XSe and so on, in the form

 AX, *5 Xs ‘ °e Op as; AZ, = Zs Zs Oe Op AS AM, MSe Mgp P

Additional elements can be added to c and to (7.6,3) if the situation requires it.

The use of constant derivatives, as in (7.6,3), to describe the force output of the propulsion system in response to throttle input does not allow for any time lag in the buildup of engine thrust since it implies that the thrust is instantaneously proportional to the throttle position. This is not unreasonable for propeller airplanes, but it is not a good model for jets in situations when the short-term response is important, as for ex­ample in a balked landing. To allow for this effect when the system is modeled in the Laplace domain, one can use control transfer functions instead of control derivatives. That is, one can replace, for example, XSp by Gxs (5). If the system model is in the time domain, the same result can be obtained by adding an additional differential equation and an additional variable. This latter method is illustrated in the example of Sec. 8.5.

By substituting (7.6,3) into (7.6,1) we derive the matrix В to be

Ms, M„ZSe MSp M^ZSp

Iy ly(m – ZJ Iy Iy(m – ZJ
0 0

With A and В known, we can compute the desired transfer functions and responses. (In this example the elevator angle is in radians, and English units are used for all other quantities). We calculate responses for the same jet transport as was used previ­ously in Sec. 6.2, with A given by (6.2,1). The nondimensional elevator derivatives are:

CXSc = -3.818 X КГ6 CJ6 = -0.3648 Cmit=~ 1.444

from which the dimensional derivatives are calculated as

= CxspulS = -3.717 Z5<, = CzspulS = -3.551 X 105 MSr = CmSr hpu20Sc = -3.839 X 107

For the throttle, we arbitrarily choose a value of XsJm = 0.3 g when Sp = 1, and ZSp and MSp = 0. With these values we get for the matrix B:

(7.6,5)