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