Question 25.5: Consider a plant having a nominal model given by G(s) = 1/(s...

This is a stable and strictly proper plant. However, it has a NMP zero located at s = $z_{o}$ = 1. We can then use z-interactors to synthesize Q(s) (see subsections §25.5.1 and §25.5.2).

We first need to find the left z-interactor $ψ_{L}(s)$ for the matrix $G_{o}(s)$ and the matrix $H_{o}(s)$.

To compute $ψ_{L}(s)$ we note that this plant is similar to that in example 25.4, where α = 1 was chosen. It is then straightforward to prove that

$\psi _{L}(s)=\begin{bmatrix} 1 & 0 \\ \frac{s+1}{s-1} & -\frac{s+1}{s-1} \end{bmatrix}$   and  $[\psi _{L}(s)]^{-1}=\begin{bmatrix} 1 & 0 \\ 1 & -\frac{s+1}{s-1} \end{bmatrix}$    (25.5.30)

We then compute $H_{o}(s)$, which is given by

$H_{o}(s)=\psi _{L}(s)G_{o}(s)=\frac{1}{(s+1)(s+3)}\begin{bmatrix} s+1 & 1 \\ s+1 & 0 \end{bmatrix}$    (25.5.31)

We also need to compute the exact model inverse $[G_{o}(s)]^{-1}$. This is given by

$[G_{o}(s)]^{-1}=\frac{(s+1)(s+3)}{(s-1)}\begin{bmatrix} 1 & -1 \\ -2 & s+1 \end{bmatrix}$      (25.5.32)

From (25.5.28) and the above expressions we have that

$T_{o}(s)=G_{o}(s)Q(s)=[\psi _{L}(s)]^{-1}H_{o}(s)Q(s)=[\psi _{l}(s)]^{-1}D_{Q}(s)$    (25.5.28)

$T_{o}(s)=[\psi _{L}(s)]^{-1}D_{Q}(s)=\begin{bmatrix} 1 & 0 \\ 1 & -\frac{s-1}{s+1} \end{bmatrix} D_{Q}(s)$    (25.5.33)

We first consider a choice of $D_{Q}(s)$ to make $T_{o}(s)$ diagonal. This can be achieved with a lower triangular $D_{Q}(s)$, i.e.

$D_{Q}(s)=\begin{bmatrix} D_{11}(s) & 0 \\ D_{21}(s) & D_{22}(s) \end{bmatrix}$    (25.5.34)

Then

$T_{o}(s)=\begin{bmatrix} 1 & 0 \\ 1 & -\frac{s-1}{s+1} \end{bmatrix}\begin{bmatrix} D_{11}(s) & 0 \\ D_{21}(s) & D_{22}(s) \end{bmatrix}=\begin{bmatrix} D_{11}(s) & 0 \\ D_{11}-\frac{s-1}{s+1}D_{21}(s) & -\frac{s-1}{s+1} D_{22}(s) \end{bmatrix}$    (25.5.35)

and

We immediately observe from (25.5.35) that, to make $T_{o}(s)$ diagonal, it is necessary that $(s+1)D_{11}(s)=(s−1)D_{21}(s)$. This means that the NMP zero will appear in channel 1, as well as in channel 2 of the closed loop. As we will see in Chapter 26, the phenomenon that decoupling forces NMP zeros into multiple channels is notpeculiar to this example, but more generally a trade off associated with full dynamicdecoupling.

If, instead, we decide to achieve only triangular decoupling, we can avoid havingthe NMP zero in channel 1. Before we choose $D_{Q}(s)$, we need to determine, from (25.5.36), the necessary constraints on the degrees of $D_{11}(s)$, $D_{21}(s)$ and $D_{22}(s)$, so as to achieve a biproper Q(s).

From (25.5.36) we see that Q(s) is biproper if the following conditions are simultaneously satisfied

(c1) Relative degree of $D_{22}(s)$ equal to 2.

(c2) Relative degree of $D_{21}(s)$ equal to 1.

(c3) Relative degree of $D_{11}(s) − D_{21}(s)$ equal to 2.

Conditions c2 and c3 are simultaneously satisfied if $D_{11}(s)$ has relative degree 1, and at the same time $D_{11}(s)$ and $D_{21}(s)$ have the same high frequency gain.

Furthermore, we assume that it is required that the MIMO control loop be decoupled at low frequencies (otherwise, steady state errors appear for constant references and disturbances). From equation (25.5.35), we see that this goal is attained if $D_{11}(s)$ and $D_{21}(s)$ have low frequency gains of equal magnitudes but opposite signs.

A suitable choice which simultaneously achieves both biproperness of Q(s) and decoupling at low frequencies is

$D_{21}(s) = \frac{s-\beta }{s+\beta}D_{11}(s)$    (25.5.37)

where $β ∈ \mathbb{R}^{+}$ is much larger than the desired bandwidth, say β = 5.

We can now proceed to choose $D_{Q}(s)$ in such a way that (25.5.37) is satisfied, together with the bandwidth constraints.
A possible choice is

$D_{Q}(s)=\begin{bmatrix} \frac{0.5(s+0.5)}{s^{2}+0.75s+0.25} & 0 \\ \\ \frac{0.5(s+0.5)(s-5)}{(s^{2}+0.75s+0.25)(s+5)} & \frac{0.25}{s^{2}+0.75s+0.25} \end{bmatrix}$      (25.5.38)

The performance of this design can be evaluated via simulation with the SIMULINK file mimo3.mdl. Note that you must first run the MATLAB program in file pmimo3.m