Lateral Frequency Response

The procedure for calculating the response of the airplane to sinusoidal movement of the rudder or aileron is similar to that used for longitudinal response in Sec. 7.6. The

Table 7.3

Control Derivatives—B747 Jet Transport (expressed in rad-1)





-1.368 X 10-2

-1.973 X 10-4



6.976 X 10-3






-3.818 X 10-6



aerodynamics associated with the two lateral controls are given by a set of control de­rivatives:


The Laplace transform of the system equation (4.9,19) is then

where В is


О о


For our numerical example we use the same jet transport and flight condition as in Sec. 7.6. A is given by (6.7,1) and the control derivatives are given in Table 7.3, from which, with the definitions of Table 7.1, the elements of В are calculated to be









The eight transfer functions are then as in (7.2,8), where f(s) is the characteristic polynomial of (6.7,2) (with s instead of A) and with the numerators as follows:

NvSa = 2.896s2 + 6.542s + 0.6220 (a)

NvSr = 5.642s3 + 379.4s2 + 167.9s – 5.934 (b)

NpSa = 0.1431s3 + 0.02730s2 + 0.1102s (c)

NpSr = 0.1144s3 – 0.1997s2 – 1.368s (d)

NrSa = -0.003741s3 – 0.002708s2 – 0.0001394s + 0.004539 (e) (7.9,5)

NrSr = -0.4859s3 – 0.2327s2 – 0.009018s – 0.05647 (f)

Ыф8а = ,1431s2 + 0.02730s + 0.1102 (g)

АГф8г = 0.1144s2 – 0.1997s – 1.368 (h)

From the transfer functions G^s) = Nij(s)/f(s), the frequency response functions Gj/iw) were calculated for both aileron and rudder inputs. The results for v. ф and r are shown on Figs. 7.26 and 7.27. The most significant feature in all of these re­sponses is the peak in the amplitude at the Dutch Roll frequency, and the associated sharp drop in phase angle.

At zero frequency we see from (7.9,5c and d) that the roll rate amplitude is zero for both inputs. All the other variables have finite values at ш — 0. Even for moderate

ш (rad/s) lb)

Figure 7.26 Frequency-response functions, rudder angle input. Jet transport cruising at high altitude, (a) Sideslip amplitude. (b) Sideslip phase, (c) Roll amplitude, (d) Roll phase, (e) Yaw-rate amplitude, if ) Yaw-rate phase.

control angles, however, the steady-state values of /3 = v/u0 and ф are very large (see Exercise 7.10). Hence the linearity assumption severely constrains these zero fre­quency solutions. If, however, we postulate that the control angles are so small that the linearity conditions are met, then there is a steady state with constant values of ф, A and r. This can only be a horizontal turn in which the angular velocity vector is

= [0 qss rJT qss = fl sin ф rss = П cos ф

Ю (rad/s) (e)

Figure 7.26 (Continued)

Leave a reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>