Input-Output Parameterization

Using the Final Value Theorem, we can try to design a controller that meets some steady-state specs. However, it’s still difficult to design a controller that satisfies desired transient specs. Some examples of other control design methods:

• PID control
• Pole placement

Limitations of these methods:

1. Requires lots of manual tuning of parameters, often based on trial and error to satisfy transient specs.
2. May not be feasible to satisfy desired transient plants

With these in mind, we come up with a new method to design controllers. The idea is to:

1. Design closed-loop transfer functions to obtain closed-loop stability
   • We are designing the closed-loop transfer functions directly instead of designing the controller
2. Recover a controller $D[z]$ that results in those closed-loop transfer functions.

To do so, we will use a clever change of variables called input-output parameterization (IOP).

Define the IOP equations (feasibility constraints) for variables $X[z],W[z],V[z]$:

$$\begin{array}{rl}& X[z]+G[z]W[z]=1\text{(i)}\\ & V[z]+G[z]X[z]=0\text{(ii)}\\ & X[z],W[z],V[z]\text{ are real, rational, and proper}\text{(iii)}\end{array}$$

Since the plant $G[z]$ is known (constant/not variable), these equations are linear in the variables $X[z],W[z],V[z]$.

Recall that for closed-loop stability, we just need to check:

$$\left[\begin{array}{c}E\\ U\end{array}\right]=\left[\begin{array}{cc}\frac{1}{1+GD}& \frac{-G}{1+GD}\\ \frac{D}{1+GD}& \frac{1}{1+GD}\end{array}\right]\left[\begin{array}{c}R\\ D\end{array}\right]$$

Theorem and Proof

IOP Theorem

• a. If $D[z]$ results in closed-loop stability, then $X[z]=T_{re}[z]=T_{du}[z]$, $W[z]=T_{ru}[z]$, and $V[z]=T_{de}[z]$, satisfy the IOP equations (i)-(iii).
• b. If $X[z],W[z],V[z]$ satisfy the IOP equations (i)-(iii), and if we choose our controller $D[z]=\frac{W[z]}{X[z]}$, then $T_{re}[z]=X[z]$, $T_{du}[z]=X[z]$, $T_{ru}[z]=W[z]$, and $T_{de}[z]=V[z]$.

Proof.

Part (a):

• Given: $D[z]$ results in closed-loop stability
• WTS: $X[z]=T_{re}[z]=T_{du}[z],W[z]=T_{ru}[z]$ satisfy the IOP equations

Proof of part (a):

• $D[z]$ results in closed-loop stability
• $⟹T_{re},T_{du},T_{ru},T_{de}$ are BIBO stable [definition of closed-loop stability]
• $⟹T_{re},T_{du},T_{ru},T_{de}$ are stable [theorem from class]
• $⟹X=T_{re}=T_{du},W=T_{ru},V=T_{de}$ satisfy equation (iii)
• We have:

$$X[z]+G[z]W[z]=T_{re}[z]+G[z]T_{ru}[z]=\frac{1}{1+GD}+G\frac{D}{1+GD}=\frac{1+GD}{1+GD}=1$$

which satisfies (i)

• We have $$V[z]+G[z]X[z]=T_{de}[z]+G[z]T_{re}[z]=\frac{-G}{1+GD}+G\frac{1}{1+GD}=0$$ which satisfies (ii)

Part (b):

• Given: $X[z],W[z],V[z]$ satisfy the IOP equations (i)-(iii)
• WTS: $D[z]=\frac{W[z]}{X[z]}⟹T_{re}=T_{du}=X,T_{ru}=W,T_{de}=V$

Proof of part (b):

• $X,W,V$ satisfy the IOP equations (i)-(iii)
• $T_{re}=T_{du}=\frac{1}{1+GD}=\frac{1}{1+G\frac{W}{X}}=\frac{1}{1+\frac{1-X}{X}}=\frac{1}{\frac{1-X+X}{X}}=\frac{1}{\frac{1}{X}}=X$
  • Because $D=\frac{W}{X}$, $GW=1-X$ from equation (i) re-arranged
• $W=DX=D{T}_{re}=D\frac{1}{1+GD}=T_{ru}$
• $V=-GX=-G{T}_{re}=-G\frac{1}{1+GD}=T_{de}$
  • $V=-GX$ from equation (iii) re-arranged