ZMP feedback in DCM feedback control law #79
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On a side note: feel free to open one separate post for each of your questions. This will make the overall discussion easier.
The post-print is actually out-of-phase with the code on this point: in #16 we switched from ZMP to DCM derivative error. As far as I remember the motivation for it was to be able to do pole placement #22, where the DCM derivative term is better decoupled from the proportional and integral ones than the previous ZMP term.
If the robot does not have F/T sensors a big question will be: how does it implement its force control? The DCM velocity error appears in the DCM feedback loop, the one that converts DCM tracking errors to a commanded ZMP. The point is then that, at the next stage, this commanded ZMP should be compared against a "measured ZMP". If the measured ZMP is observed by e.g. finite differentiation of the measured DCM, I would be concerned that it suffers from a lot of noise and delay. But I haven't tried this directly, so it's worth checking on a real robot 😉 |
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Hello Dr.Caron,
Thank you for maintaining this nice repo!
While reading your paper, I have the following question.
In the paper, you stated
We choose the latter in what follows (subtracting k_z(z^d - z^m))
instead of adding dcm velocity feedback to damp potential oscillations. Could you tell me the reason why you chose the zmp feedback instead of dcm velocity feedback to get the commanded ZMP law (equation before Eq.(9)) in the paper? In the case that robot does not have an F/T sensor to measure the zmp, do you think it is possible to use dcm velocity feedback law to get the commanded ZMP (eventually net contact wrench)?Another thing to ask is that in the CoM admittance control law (Eq.(21)), if the admittance gain A_{com} is a positive number, shouldn't it be written as
\ddot{c}_c = \ddot{c}_d + A_{com}(z^{qp} - z^m)
because we want the CoM to accelerate toward the z^{qp}?Thank you,
Seung Hyeon
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