Why do I need I-gain in my outer-loop?

Question

I'm implementing a set of loops to control pitch-and-roll angular positions.

In an inner-loop, motor speeds are adjusted to achieve desired angular rates of rotation (the "inner-loop setpoints").

An outer-loop decides these desired angular rates (the "inner-loop setpoints") based on the aircraft's angular positions.

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Outer-loop

• Frequency = ~400Hz
• Outer PV = input angular position (in degrees)
• Outer SP = desired angular position - input angular position (in degrees)

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Inner-loop

• Frequency = ~760Hz
• Inner PV = input angular rotation (in degrees-per-second)
• Inner SP = constant1 * Outer MV (in degrees-per-second)
• PWM = Inner MV / constant2 (as percentile)

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I understand what I-gain does and why this is important, but I'm not able to see any practical reason for also having I-gain specified in the outer-loop. Surely the inner-loop would compensate for any accumulated error, leaving no error to compensate for in the outer-loop, or is my thinking flawed?

Any example gain values to elaborate would be greatly appreciated.

Answer 1

In my own experience, it is not necessary Integral term in any of these loops. I got a fine PD tuning of both inner and outer loop. See for instance this paper, where I firstly tuned PD only with PSO.

Wikipedia shows a simple animation of each of the effects of P-I-D in any tuning https://en.wikipedia.org/wiki/PID_controller#Manual_tuning.

Answer 2

Integral gain is how you eliminate steady state error. An integral gain in the inner loop means you'll (eventually) have no steady state error on the angular velocity.

What ensures no steady state error on angular position, which is arguably the more important of the two? Outer loop integral error.

Answer 3

Let's assume that the inner loop, whose input is desired angular rate and output is motor input, can track the desired angular rate. Some internet fellow has taken care of that for us. I'm going to be considering a step change in desired position:

Let $e = x_{des} - x$ and $ \dot e = -\dot x$ this will be our outer loop controller

$$\dot x = k_p e + k_d \dot e$$

substituting in $\dot e$ gives

$$ -\dot e = k_p e + k_d \dot e$$

which can be rearranged as

$$\dot e = - \frac{k_p}{k_d + 1} e$$

which is a first order system. I've ignored inner loop dynamics but that should be fine, inner loop dynamics should be at least an order of magnitude faster than the outer loop dynamics.

I am not saying that integral control is not useful, just that it is not strictly necessary in the outer loop. It is easy to imagine a situation where the inner loop is not behaving correctly that results in a steady state error.

The px4 flight stack for multicopters approaches attitude control with pd control for angular position and pid control for body rate control.

https://pixhawk.org/users/multirotor_pid_tuning (link broken)

Answer 4

Each control system has its own characteristics (overshoot, damping, settling time, response time, etc).

The outer loop is just another control system, so it may have a steady state (residual) error. And the I term is necessary to get rid of it.

If you had a third level controller (e.g lateral acceleration) you would still need another I term. And if you had a higher level outer loop (e.g. Speed), you would still need yet another I term in that loop. If you had yet another higher level (outer) loop (e.g. Position), you would again need an I term. Otherwise you run the risk of not reaching your target location.

Of course if the intermediate control variables are not important (e.g acceleration), you might just cancel that I term of that loop. However, any time you want to get rid of the steady state error of any of your variables (in any of the loops), you use an integral term.