Our teacher gave us an assignment where we have to study the longitudinal stability of an X-15. I have to study it at 353.9 m/s which is Mach 1.2. Another group has to study it at 235.9 m/s so Mach 0.8.

I made a Matlab code computing the eigenvalues of the A matrix because I have to plot the phugoid and the short period modes. With my speed I get :

Eigenvalues [at 353.9 m/s]:
  -0.6094 + 3.1782i
  -0.6094 - 3.1782i
  -0.0775 + 0.0000i
   0.0473 + 0.0000i

While with the other speed I get:

Eigenvalues [at 235.9 m/s]:
  -0.4093 + 2.1175i
  -0.4093 - 2.1175i
  -0.0070 + 0.0649i
  -0.0070 - 0.0649i

In the second case, it is clear that the phugoid is the -0.0070 ± 0.0649i and the short mode is -0.4093 ± 2.1175i. While in the first case, what can I deduce ?

Is there a dynamic mode which is cancelled by the supersonic speed?

EDIT: The A matrix is the coefficients of the longitudinal motion equations: enter image description here


1 Answer 1


At 353.9 m/s: your phugoid mode is unstable as well as degenerated (Poles lie on real axis and has damping ratio 1). while at 235.9 m/s: phugoid mode has become stable as well as oscillatory and has some damping and frequency.

so, nothing has been canceled its just phugoid mode poles changing positions. you can check it clearly by plotting root locus using rlocus command on MATLAB.

  • $\begingroup$ Thanks, it really helps! So what does my values -0.0775 and 0.0473 represents? I know that they are the roots on the root locus (one is stable, -0.0775 and one is unstable 0.0473) but physically, for the short period I have only one frequency while here my phugoid mode will have 2 frequencies? $\endgroup$ Apr 17, 2020 at 13:05
  • 1
    $\begingroup$ @Louis The phugoid has split into two real-valued eigenvalues. They are no longer "frequencies". $\endgroup$
    – JZYL
    Apr 17, 2020 at 13:52
  • $\begingroup$ @Louis The phugoid mode is now a first order mode which has time constant not frequencies $\endgroup$ Apr 17, 2020 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.