VelocityImpedance is actually a PassiveController

[hubernikus]

The VelocityImpedance controller as used by the CartesianTwist is in reality a passive controller, this can be check by looking at the underlying math (des == desired; fb == feedback)
q_des = w_des * dt ; q_fb = 0

tau_ctrl = K(q_des - q_fb) + D (w_des - w_fb) = (K *dt + D) w_des - D
using new dynamics which are given by: hat{w}_des = (K *dt + D) / D w_des we can rewrite the control law as a simple passive one:
tau_ctrl = D (hat{w}_des - w)

I think using the simplified control law, the controller can be simplified (in the number of operations), hence made simpler for a user to understand as well as lower computational cost.
Additionally, reducing the redundant parameter K will allow easier tuning.

[domire8]

Fair point. We will add this issue to our backlog and see if we can address it. Thanks!

[eeberhard]

I'm not sure if that's quite right. I will rewrite your steps with math formatting to make sure I understand them.

Starting with the definition $q_{des} = w_{des} * dt$; $q_{fb} = 0$, I agree this is the formulation in the VelocityImpedance controller. And, we have the impedance control law:

$$\tau = K(q_{des} - q_{fb}) + D (w_{des} - w_{fb})$$

So you substitute and expand (I think you were missing the feedback term w_{fb} in your first comment):

$$\tau = K(w_{des} * dt - 0) + D * w_{des} - D *w_{fb}$$

$$\tau = (K * dt + D) * w_{des} - D * w_{fb}$$

Then, if you define a modified desired velocity:

$$\hat{w}{des} = (K *dt + D) * w{des} / D $$

you can rewrite the final control law as:

$$\tau = (K * dt + D) * w_{des} - D * w_{fb} = D * (\hat{w}{des} - w{fb})$$

I have two points. First, where do you see a computational saving? You still have to calculate $\hat{w}$. Or maybe you suggest that we calculate the matrix $K * dt + D$ in advance and store it as some $\hat{D}$, so that we can have $\tau = \hat{D} * w_{des} - D * w_{fb}$.

But that leads to the second point, which is that $q_{des} = w_{des} * dt$ represents an integration, and not a normal product. The angular velocity is integrated into angular displacement, and the K matrix then scales the imaginary vector part of the quaternion. So I don't think you can just factor out dt and have the same behavior.

Let me know if I misunderstood anything.

[hubernikus]

Yes, good point. Did not know nice formulas were possible.

The computational saving is minor, I agree. But in the end the step of integrating from $q_{des} = w_{des} *dt$ and then taking the angular difference $\Delta q = q_{des} - q_0$ (since the desired velocity is set to a zero state), is the same as just multiplying the $\Delta_q = w_{des} * \hat k$ a factor $\hat k$ (which is 1 if you take the actual orientation vector, and an factor which depends on the rotation-angle otherwise).

I agree, that it is not uniquely a multiplication, but also requires a slight modification of the damping matrix $D$. However, I feel it is a bit misleading letting a user define a proportional $K$ matrix, which actually only acts on the desired velocity, but not the orientation.