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This example shows how to compute various measures of passivity for linear time-invariant systems.

A linear system G(s) is passive when all I/O trajectories $$(u(t),y(t))$$ satisfy

$${\int}_{0}^{T}{y}^{T}(t)u(t)dt>0,\phantom{\rule{1em}{0ex}}\forall T>0$$

where $${y}^{T}(t)$$ denotes the transpose of $$y(t)$$.

To measure "how passive" a system is, we use passivity indices.

The input passivity index is defined as the largest $$\nu $$ such that

$${\int}_{0}^{T}{y}^{T}(t)u(t)dt>\nu {\int}_{0}^{T}{u}^{T}(t)u(t)dt$$

The system G is "input strictly passive" (ISP) when $$\nu >0$$. $$\nu $$ is also called the "input feedforward passivity" (IFP) index and corresponds to the minimum feedforward action needed to make the system passive.

The output passivity index is defined as the largest $$\rho $$ such that

$${\int}_{0}^{T}{y}^{T}(t)u(t)dt>\rho {\int}_{0}^{T}{y}^{T}(t)y(t)dt$$

The system G is "output strictly passive" (OSP) when $$\rho >0$$. $$\rho $$ is also called the "output feedback passivity" (OFP) index and corresponds to the minimum feedback action needed to make the system passive.

The I/O passivity index is defined as the largest $$\tau $$ such that

$${\int}_{0}^{T}{y}^{T}(t)u(t)dt>\tau {\int}_{0}^{T}({u}^{T}(t)u(t)+{y}^{T}(t)y(t))dt$$

The system is "very strictly passive" (VSP) if $$\tau >0$$.

Consider the following example. We take the current $$I$$ as the input and the voltage $$V$$ as the output. Based on Kirchhoff's current and voltage law, we obtain the transfer function for $$G(s)$$,

$$G(s)=\frac{V(s)}{I(s)}=\frac{(Ls+R)(Rs+\frac{1}{C})}{L{s}^{2}+2Rs+\frac{1}{C}}.$$

Let $$R=2$$, $$L=1$$ and $$C=0.1$$.

```
R = 2; L = 1; C = 0.1;
s = tf('s');
G = (L*s+R)*(R*s+1/C)/(L*s^2 + 2*R*s+1/C);
```

Use `isPassive`

to check whether $$G(s)$$ is passive.

PF = isPassive(G)

`PF = `*logical*
1

Since PF = true, $$G(s)$$ is passive. Use `getPassiveIndex`

to compute the passivity indices of $$G(s)$$.

% Input passivity index nu = getPassiveIndex(G,'in')

nu = 2

% Output passivity index rho = getPassiveIndex(G,'out')

rho = 0.2857

% I/O passivity index tau = getPassiveIndex(G,'io')

tau = 0.2642

Since $$\tau >0$$, the system $$G(s)$$ is very strictly passive.

A linear system is passive if and only if it is "positive real":

$$G(j\omega )+{G}^{H}(j\omega )>0\phantom{\rule{1em}{0ex}}\forall \omega \in R.$$

The smallest eigenvalue of the left-hand-side is related to the input passivity index $$\nu $$:

$$\nu =\frac{1}{2}\underset{\omega}{\mathrm{min}}{\lambda}_{\mathrm{min}}(G(j\omega )+{G}^{H}(j\omega ))$$

where $${\lambda}_{\mathrm{min}}$$ denotes the smallest eigenvalue. Similarly, when $$G(s)$$ is minimum-phase, the output passivity index is given by:

$$\rho =\frac{1}{2}\underset{\omega}{\mathrm{min}}{\lambda}_{\mathrm{min}}({G}^{-1}(j\omega )+{G}^{-H}(j\omega )).$$

Verify this for the circuit example. Plot the Nyquist plot of the circuit transfer function.

nyquist(G)

The entire Nyquist plot lies in the right-half plane so $$G(s)$$ is positive real. The leftmost point on the Nyquist curve is $$(x,y)=(2,0)$$ so the input passivity index is $$\nu =2$$, the same value we obtained earlier. Similarly, the leftmost point on the Nyquist curve for $${G}^{-1}(s)$$ gives the output passivity index value $$\rho =0.286$$.

It can be shown that the "positive real" condition

$$G(j\omega )+{G}^{H}(j\omega )>0\phantom{\rule{1em}{0ex}}\forall \omega \in R$$

is equivalent to the small gain condition

$$||(I-G(j\omega ))(I+G(j\omega ){)}^{-1}||<1\phantom{\rule{1em}{0ex}}\forall \omega \in R.$$

The relative passivity index (R-index) is the peak gain over frequency of $$(I-G)(I+G{)}^{-1}$$ when $$I+G$$ is minimum phase, and $$+\infty $$ otherwise:

$$R={\Vert (I-G)(I+G{)}^{-1}\Vert}_{\infty}.$$

In the time domain, the R-index is the smallest $$r>0$$ such that

$${\int}_{0}^{T}||y-u|{|}^{2}dt<{r}^{2}{\int}_{0}^{T}||y+u|{|}^{2}dt$$

The system $$G(s)$$ is passive if and only if $$R<1$$, and the smaller $$R$$ is, the more passive the system is. Use `getPassiveIndex`

to compute the R-index for the circuit example.

R = getPassiveIndex(G)

R = 0.5556

The resulting $$R$$ value indicates that the circuit is a very passive system.