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In the literature, it is often seen that for a system:

\begin{align} \Lambda \ddot{e} + D_d \dot{e} + K_d e = F_{ext} \end{align}

A dynamic damping matrix is required for a robot to take into account the structure and changes of $\Lambda$ during movement. Hence the following method called the "double diagonalization".

Given a symmetric and positive definite matrix $\Lambda \in \mathcal{R}^{n\times n}$ and a symmetric matrix $K_d \in \mathcal{R}^{n\times n}$, one can find a non singular matrix $Q \in \mathcal{R}^{n\times n}$ and a diagonal matrix $B_0 \in \mathcal{R}^{n\times n}$ such that: \begin{align} \Lambda &= Q^\intercal Q\\ K_d &= Q^\intercal B_0 Q \end{align} Where the diagonal elements of $B_0$ are the generalized eigenvalues of $K_d$

The design of the damping matrix becomes: \begin{align} D_d = 2 Q^\intercal diag( \xi \sqrt(\lambda_{K,i}^\Lambda) Q) \end{align}

Such that \begin{align} Q^\intercal Q \ddot{x} + 2 Q^\intercal diag( \xi \sqrt(\lambda_{K,i}^\Lambda) Q) \dot{x} + Q^\intercal B_0 Q x = F_{ext} \end{align}

where $\xi_i$ is the damping factor in the range of $[0,1]$, $\lambda_{K,i}^\Lambda$ is the i'th diagonal element of $B_0$

- Yeah, I 'm pretty dumb (I'm guessing the solution is simple but I cannot see it). I cannot figure out how to solve for the matrices $Q$ and $B_0$, such that (in particular) $\Lambda = Q^\intercal Q$. What am I missing? Can elaborate for me?

How do I solve this?

Sources: (this article should be free and provide some context.) (Albu-Schaffer, Alin, et al., 2003)

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  • $\begingroup$ Great question! I've bookmarked this because I've had unsatisfactory results with a vanilla state feedback controller and I'm about to embark on my own impedance control project. I always find it particularly infuriating when papers skip major steps; they'll typically work through some derivation to a point, skip the important implementation steps, and then jump straight into some results figures. Anyways, good luck with the question, I'll be watching for answers! $\endgroup$ – Chuck May 26 at 12:46
  • $\begingroup$ @Chuck Yeah, I very much relate to that - that's precisely how I felt about this topic. And thanks for the encouragement :)! a bit late, now that someone has answered the question. but thanks. $\endgroup$ – Spaceman May 26 at 20:30
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A hint towards what the answer is given in the paper. Namely, one can use the generalized eigenvalue decomposition, which in this case can be formulated as finding eigenvalues $\lambda \in \mathbb{R}$ and eigenvectors $v \in \mathbb{R}^n$ such that

$$ (\lambda\,\Lambda - K_d)\,v = 0. \tag{1} $$

Consider two distinct solutions $(\lambda_i,v_i)$ and $(\lambda_j,v_j)$ of $(1)$, which can also be written as

\begin{align} \lambda_i\,\Lambda\,v_i &= K_d\,v_i, \tag{2a} \\ \lambda_j\,\Lambda\,v_j &= K_d\,v_j. \tag{2b} \end{align}

By pre-multiplying $(2a)$ by $v_j^\top$ and $(2b)$ by $v_i^\top$ one gets

\begin{align} \lambda_i\,v_j^\top \Lambda\,v_i &= v_j^\top K_d\,v_i, \tag{3a} \\ \lambda_j\,v_i^\top \Lambda\,v_j &= v_i^\top K_d\,v_j. \tag{3b} \end{align}

When using that $M = M^\top$ and $K_d = K_d^\top$ it follows that when subtracting the transpose of $(3b)$ from $(3a)$ one gets

$$ (\lambda_i - \lambda_j)\,v_j^\top \Lambda\,v_i = 0. \tag{4} $$

Thus when $\lambda_i \neq \lambda_j$ it follows that $v_j^\top \Lambda\,v_i = 0$, from which together with $(3a)$ it also follows that $v_j^\top K_d\,v_i = 0$. When $i \neq j$ but $\lambda_i = \lambda_j$ this orthogonality property between vectors $v_i$ and $v_j$, and matrices $\Lambda$ and $K_d$ does not immediately follow. It can be noted that $\lambda_i = \lambda_j = \lambda$ means that the kernel of $\lambda\,\Lambda - K_d$ has a dimension larger then one, such that for all $\alpha,\beta\in\mathbb{R}$ the vector $v = \alpha\,v_i + \beta\,v_j$ would satisfy $(1)$. However, most generalized eigenvalue problem solvers "choose" $\alpha$ and $\beta$ such that they do ensure that $v_j^\top \Lambda\,v_i = 0$ and $v_j^\top K_d\,v_i = 0$ whenever $\lambda_i = \lambda_j$.

When defining $V$ as the matrix whose columns are equal to $v_i\ \forall\,i=1,2,\cdots, n$ it follows from $v_j^\top \Lambda\,v_i = v_j^\top K_d\,v_i = 0$ that $V^\top \Lambda\,V$ and $V^\top K_d\,V$ are diagonal matrices. By also dividing each $i$th column of $V$ by $\sqrt{v_i^\top \Lambda\,v_i}$ one also obtains that $V^\top \Lambda\,V = I$ and $V^\top K_d\,V = \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)$. Solving this for $\Lambda$ and $K_d$ yields

\begin{align} \Lambda &= V^{-\top} V^{-1}, \tag{5a} \\ K_d &= V^{-\top} \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)\,V^{-1}, \tag{5b} \end{align}

which is equivalent to using $Q = V^{-1}$ and $B_0 = \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)$.

For more information of this derivation see section 3.3.2 from the book De Kraker, A. (2009). Mechanical vibrations. Shaker Publishing BV.

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    $\begingroup$ I love you, man. $\endgroup$ – Spaceman May 26 at 20:28

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