Spring 2003 16.61AC2-5 ● Now we have that △X X8 X E-1/42 26.26 △M for the longitudinal case B Typical values for the B747 X 16.54 X6n=0.3mg=849528 M6=-5.2·10 M≈0 Aircraft response y=G(su X=AX+ Bu -G(s)=C(sI-A-B y=CX We now have the means to modify the dynamics of the system, but first let's figure out what de and d, really doSpring 2003 16.61 AC 2–5 • Now we have that c = E−1         ∆Xc ∆Zc ∆Mc 0         = E−1         Xδe Xδp Zδe Zδp Mδe Mδp 0 0           δe δp   = Bu • For the longitudinal case B =                        Xδe m Xδp m Zδe m−Zw˙ Zδp m−Zw˙ I−1 yy [Mδe + ZδeΓ] I−1 yy Mδp + ZδpΓ 0 0                        • Typical values for the B747 Xδe = −16.54 Xδp = 0.3mg = 849528 Zδe = −1.58 · 106 Zδp ≈ 0 Mδe = −5.2 · 107 Mδp ≈ 0 • Aircraft response y = G(s)u X˙ = AX + Bu → G(s) = C(sI − A)−1B y = CX • We now have the means to modify the dynamics of the system, but first let’s figure out what δe and δp really do