I was wondering if it is possible to show that a variant of the inverse dynamics controller

$$\tau = \mathbf{M}(q)\ddot{q}^\mathrm{des} + \mathbf{K}_pe + \mathbf{K}_d\dot{e} + \mathbf{h}(q, \dot{q})$$

is able to stabilize the system

$$\mathbf{M}(q)\ddot{q} + \mathbf{h}(q, \dot{q}) = \tau$$


2 Answers 2


This answer is inspired by Ryszard Szwarc's answer in: https://math.stackexchange.com/questions/4509062/how-to-show-that-the-matrix-0-i-m-1k-p-m-1k-d-has-eigenvalues.

The dynamics of the system is in the form of \begin{equation} \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}} + \mathbf{h}(\mathbf{q}, \dot{\mathbf{q}}) = \tau \end{equation} and a variant of the inverse dynamics controller is in the form of \begin{equation} \tau = \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}}^\mathrm{des} + \mathbf{K}_p\Delta\mathbf{q} + \mathbf{K}_d\Delta\dot{\mathbf{q}} + \mathbf{h}(\mathbf{q}, \dot{\mathbf{q}}). \end{equation} Inputting the control into the dynamics gives us \begin{align*} \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}} + \mathbf{h}(\mathbf{q}, \dot{\mathbf{q}}) &= \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}}^\mathrm{des} + \mathbf{K}_p\Delta\mathbf{q} + \mathbf{K}_d\Delta\dot{\mathbf{q}} + \mathbf{h}(\mathbf{q}, \dot{\mathbf{q}})\\ \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}} &= \mathbf{M}(\mathbf{q})\ddot{\mathbf{q}}^\mathrm{des} + \mathbf{K}_p\Delta\mathbf{q} + \mathbf{K}_d\Delta\dot{\mathbf{q}}\\ -\mathbf{M}(\mathbf{q})\Delta{\ddot{\mathbf{q}}} &= \mathbf{K}_p\Delta\mathbf{q} + \mathbf{K}_d\Delta\dot{\mathbf{q}} \end{align*} which can be written as \begin{equation} \begin{bmatrix} \Delta\dot{\mathbf{q}}\\ \Delta\ddot{\mathbf{q}} \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & \mathbf{I}\\ -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p & -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_d \end{bmatrix}}_{\mathbf{A}(\mathbf{q})}\begin{bmatrix} \Delta\mathbf{q}\\ \Delta\dot{\mathbf{q}} \end{bmatrix}. \end{equation} First, we show that $\mathbf{A}(\mathbf{q})$ is full rank, which leads to having non-zero eigenvalues. Using the property that elementary operations do not change the rank of a matrix, instead of investigating the rank of $\mathbf{A}(\mathbf{q})$, we look at the rank of \begin{equation} \hat{\mathbf{A}}(\mathbf{q}) = \begin{bmatrix} \mathbf{I} & 0\\ -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_d & -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p \end{bmatrix} \end{equation} Then, by using Schur's complement, we have \begin{equation} \det\Bigg(\begin{bmatrix} \mathbf{I} & 0\\ -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_d & -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p \end{bmatrix}\Bigg) = \det(\mathbf{I})\det(-\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p) = \det(-\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p). \end{equation} Using another property \begin{equation} \mathrm{rank}(\mathbf{C}\mathbf{D}) \leq \min\{\mathrm{rank}(\mathbf{C}), \mathrm{rank}(\mathbf{D})\} \end{equation} if we see $\mathbf{D} = -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p$ and $\mathbf{C} = \mathbf{M}(\mathbf{q})$, then we have \begin{equation} \mathrm{rank}(-\mathbf{K}_p) \leq \min\{\mathrm{rank}(\mathbf{M}(\mathbf{q})), \mathrm{rank}(-\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p)\} \end{equation} which leads to the conclusion that $-\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p$ is full rank with a non-zero determinant. Then, we have $\hat{\mathbf{A}}(\mathbf{q})$ have a non-zero determinant and full rank, which finally gives us the conclusion that $\mathbf{A}(\mathbf{q})$ is full rank. We then have the eigenvalues and eigenvectors of $\mathbf{A}(\mathbf{q})$ satisfying \begin{equation} \begin{bmatrix} 0 & \mathbf{I}\\ -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_p & -\mathbf{M}(\mathbf{q})^{-1}\mathbf{K}_d \end{bmatrix}\begin{bmatrix} \mathbf{v}_1\\ \mathbf{v}_2 \end{bmatrix} = \lambda\begin{bmatrix} \mathbf{v}_1\\ \mathbf{v}_2 \end{bmatrix} \end{equation} with $\lambda \neq 0$, which gives us the relationship \begin{align} \mathbf{v}_2 &= \lambda\mathbf{v}_1\\ -\mathbf{M}(\mathbf{q})^{-1}(\mathbf{K}_p\mathbf{v}_1 + \mathbf{K}_d\mathbf{v}_2) &= \lambda\mathbf{v}_2 \end{align} This also tells us that $\mathbf{v}_1$ and $\mathbf{v}_2$ cannot be zero vectors. If one of them is the zero vector, then both of them will be zero, which gives a trivial solution to the equation above. Inputting the first equation into the second gives us \begin{equation} -\mathbf{M}(\mathbf{q})^{-1}(\mathbf{K}_p + \lambda\mathbf{K}_d)\mathbf{v}_1 = \lambda^2\mathbf{v}_1 \end{equation} then, by multiplying $\mathbf{M}(\mathbf{q})$ on both sides, we have \begin{equation} -\mathbf{K}_p\mathbf{v}_1 - \lambda\mathbf{K}_d\mathbf{v}_1 = \lambda^2\mathbf{M}(\mathbf{q})\mathbf{v}_1\quad\rightarrow\quad\lambda^2\mathbf{M}(\mathbf{q})\mathbf{v}_1 + \lambda\mathbf{K}_d\mathbf{v}_1 + \mathbf{K}_p\mathbf{v}_1 = 0 \end{equation} then, if we multiply $\mathbf{v}_1^T$ on the left, we have \begin{equation} \label{eq:final_eq} (\mathbf{v}_1^T\mathbf{M}(\mathbf{q})\mathbf{v}_1)\lambda^2 + (\mathbf{v}_1^T\mathbf{K}_d\mathbf{v}_1)\lambda + \mathbf{v}_1^T\mathbf{K}_p\mathbf{v}_1 = 0. \end{equation} We know that all of the coefficients are positive, then from the quadratic root formula \begin{equation} \frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{equation} we can determine all of the roots of the above equation have negative real parts.

Thus, we can show that $\mathbf{A}(\mathbf{q})$ is a Hurwitz matrix, which means the system of $\Delta{\mathbf{q}}$ and $\Delta{\dot{\mathbf{q}}}$ is stable. This tells us the controller $\tau$ can stabilize the system to zero error.

  • $\begingroup$ Nice work, is M(q) here the mass matrix in generalized coordinates? i.e. positive definite, symmetric? $\endgroup$ Commented Aug 12, 2022 at 18:23
  • $\begingroup$ Yes, the $\mathbf{M}(\mathbf{q})$ matrix is positive definite and symmetric, here $\mathbf{q}$ is the generalized coordinates. $\endgroup$ Commented Aug 12, 2022 at 21:45

This would be too long for a comment, so I will add some work as an answer:

The proof of $A(\mathbf{q})$ being invertible is not perfect. First, add some structure to the $K_p$ and $K_d$ matrices. Assuming they are gain matrices, they should probably be positive definite and diagonal (thus invertible). In this case $M(\mathbf{q})^{-1} K_p$ and $M(\mathbf{q})^{-1} K_d$ are very easily shown to be full rank. $\text{det}(AB) = \text{det}(A)\text{det}(B)$. We also know that $M(q)$, $K_p$, and $K_d$ are all invertible and thus have nonzero determinant. To finalize the argument you could still use Schur's complement to show that $A(\mathbf{q})$ is invertible as $\text{det}(I)\text{det}(-M(\mathbf{q})^{-1} K_p)$ is nonzero.

To touch on adding structure to your variables before making arguments, there is technically no reason $\mathbf{v}_{1}^{T}K_{d}\mathbf{v}_{1}$ and $\mathbf{v}_{1}^{T}K_{p}\mathbf{v}_{1}$ would be guaranteed to be positive - unless you state it beforehand that $K_d$ and $K_p$ are positive definite! This may be well known to those familiar with PD control, but it is still good to explicitly mention this. Likewise, we do not know that $-M(\mathbf{q})^{-1}K_{p}$ is invertible as $K_p$ is not necessarily invertible unless some underlying assumptions are made.

As far as I can tell everything else looks good!


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