# Why does an elevator deflection (step) result in a nonzero short-term-steady-state response in pitch rate?

I know from the simplified Short Period linearized equations (state space with Angle of Attack ($\alpha$) and Pitch Rate ($q$) as states) that the steady-state value of the pitch rate response to elevator deflection (step) is non-zero

I can also relate that to the spring-mass system and understand that the concise derivatives $M_{alpha}$ and $M_q$ work as spring and a damper, respectively.

I think (would appreciate confirmation) that the term short-term steady state should mean that the phugoid motion is going to take over eventually, being the dominant mode.

However, I would like to understand that physically, because $\alpha$ and $q$ seem to be both on steady state (impossible unless flight path changes). Also, I would expect $q$ to return to zero at some point. Does it mean this equation is oversimplified in this regard?

• I think it would help a lot if you would post the actual state space model used. Is it actually a state space model with two states (pitch and pitch rate) or are there also states for e.g. velocity? Sep 4 '18 at 11:19
• Actually, this applies to any general short period simplified representation. The velocity is assumed as constant for the mode and the system is represented in terms of α and q. To make it more clear, any system would have this same response, only varying according to the damping and frequency terms. What I need to understand is what is being left out so that this happens.
– Ben
Sep 4 '18 at 11:39
• I mostly asked to see which states are part of your state spate representation; indeed, the actual system matrix does not really matter. Sep 4 '18 at 11:50
• I see. I should've written: "...state space with Angle of Attack α and pitch rate q as the only states..."
– Ben
Sep 4 '18 at 11:54
• Feel free to use the 'edit' button to improve your question. And of course, welcome to Aviation.SE! Sep 4 '18 at 12:04

The equation is certainly oversimplified, hence the name 'short period': it only models short period pitch oscillations. In this model, there are only two states: pitch and pitch rate. Any other effect is thus neglected.

A step response in an elevator would in the short term definitely lead to a nonzero pitch rate, as the model predicts. In reality, the pitch rate will usually decay to zero, because the airspeed drops to zero (there may also be altitude density effects). Imagine what would happen if you would pull on the stick in an aircraft, and crank the throttle to make sure the speed does not decay. You would do a vertical loop, exactly like the $q$ term predicts.

Of course, then you run in to an extra complication: your model is linearized. The concept of large angles, let alone a loop, does not exist in the linearized world. Think of a linearized pendulum: that works fine for small angles, but as far as the model is concerned, a 360° angle means just a very large deflection, and not the reality of returning to zero deflection after a full loop. The model is then only valid for the small amplitude oscillations it is intended to predict.

A phugoid mode exchanges speed for altitude and vice versa. Neither of these states are represented in your short period linearized model. This behaviour is impossible to see in your model. You might as well ask why you don't see the effect of the Moon position in your model; it's simply not in the state space!

A long story short then: simplified models like these are great tools, but cannot and should not be used outside of what they are supposed to do. If you want to estimate instantaneous pitch rate from elevator deflection then this is the model for you. If you want to check for trim conditions, large attitude deviations, or even roll behaviour, then you need a better model.

• Excellent answer! The only minor note is that the name 'short period' is not because the model is just 'oversimplified'. It rather happens that when we linearise the full model, for typical airplanes the states ɑ and q (in pitch) change much faster than the others (altitude and speed), and thus it makes (some) sense to analyse them separately - always, as you note, keeping in mind the limitations. But for very small airplanes, and particularly models, the frequencies are much closer, and it may become incorrect to consider 'short-period' motion for them separately at all.
– Zeus
Sep 6 '18 at 7:06
• @Zeus I meant to say that the name implies that it only predicts short-term effects and is invalid in the long run. Of course, the model was created with the reason to get a simple estimate of the short-term behaviour, not the other way round. Thanks for your comment. Sep 6 '18 at 10:12
• Sorry, @Sanchises, my point was that the term short-period does not mean 'short term'! It means 'considering the states with quick response' (i.e. short period / high frequency). It is possible and valid to use a short-period-only linear model for quite a long stretch of time, as long as the conditions remain valid (V, H ~ const). For example, the airplane may exhibit pitch oscillations (being underdamped), while not departing the trajectory significantly. Or a simulator for the task of aiming a gunsight at a distant target might use a short-period model successfully.
– Zeus
Sep 7 '18 at 0:38
• I truly appreciate all the comments. I was afraid people would give replies on things I already knew and not address the real issue, but that second paragraph was exactly what I was looking for. Thanks!
– Ben
Sep 7 '18 at 7:04
• @Zeus True. I was reasoning from the idea that the error dynamics are unstable when using the short-term model as an observer precisely because the conditions (the system matrix) do not remain constant in general. I will edit the answer to be a little bit more precise. Sep 7 '18 at 12:44