Passivity in State Space Systems

Question

Suppose I have the state space system $a\dot{x} = -x + \frac{1}{k}h(x) + u$, and output $y = h(x)$, where $a,k > 0$. The only information I'm given about $h(x)$ is that $h\in[0,k]$. I want to show that the system is passive using the storage function $V(x) = a\int_{0}^{x}h(\sigma)\ d\sigma$. The definition of a passive system is $u^\intercal y \geq \dot{V}$ for all $(x,u)$. Thus,

$$ \dot{V} = ah(x) \dot{x} = h(x) (-x + h(x)/k + u) = \frac{1}{k}h(x)(h(x) - kx) + h(x) u. $$

Rearranging this gives

$$ h(x)u = yu = \dot{V} + \frac{1}{k}h(x)[kx-h(x)]. $$

In order to prove passivity, I need to show that $\frac{1}{k}h(x)[kx - h(x)] \geq 0$, for all $x$. However, all I'm given is that $h\in[0,k]$ so how can this be true in general? Am I missing something really simple here?

Answer 1

In your problem, you define the supply rate (I will name it $s$) $s(y,u)=u^{\top}y$. And it is (strict) passive if $\dot{V} \leq s(y,u)$. I also consider that you are working taking the case where this consideration is valid $u^{\top}y \equiv uy$.

Saying that, I believe you can re-arrange your equation as follows: \begin{align*} \dot{V}(x) &= uy+\frac{1}{k}y^2-xy \\ &=y(u+\frac{1}{k}y) -xy \end{align*} then define as the new input $\nu := u+\frac{1}{k}y$, in that sense you will have the following equation: \begin{align*} \dot{V}(x) &= y\nu-xy\leq \nu y \end{align*}

The system is (strict) passive if $xh(x)> 0\quad\forall x\in\mathbb{R}$.

Answer 2

I assume that with the notation $h\in[k_1,k_1]$ it is meant that $h(x)$ is a possible nonlinear function that is sector bound, with $h(0)=0$, for positive $x$ it holds that $h(x)$ is upper bounded by $k_2\,x$ and lower bounded by $k_1\,x$ and for negative $x$ it holds that $h(x)$ is upper bounded by $k_1\,x$ and lower bounded by $k_2\,x$. These constraints can all also captured by

$$ [h(x) - k_1\,x]\,[h(x) - k_2\,x] \leq 0. $$

If with $h\in[0,k]$ it is indeed meant that $h(x)$ is sector bound, as described above, then it would follow that $h(x)\,[h(x) - k\,x] \leq 0$. From that inequality it also immediately follows that

$$ \frac{1}{k} h(x)\,[k\,x - h(x)] \geq 0. $$

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Also note that $-x+\dfrac1k h(x)$ is sector bound to $[-1,0]$.