I know nothing about the engineering aspect, but you get a doppler shift, whatever the angle that the plane's velocity is to the line of sight.
If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.
The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).
See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect