Simple Pole Approximation

The IOP equations (i)-(iii) are hard to solve because $W[z],X[z],V[z]$ lie in an infinite-dimensional vector space. To make the problem tractable, we make a finite dimensional approximation of this infinite dimensional space. In particular, we choose the simple pole approximation (SPA).

We choose $\{p_i{\}}_{i=1}^m$ as part of our control design. We approximate:

$$W[z]=\sum _{i=1}^m\frac{w_i}{z-p_i}$$

• $\{w_i{\}}_{i=1}^m$ are variable coefficients in $\mathbb{C}$

Assumption: Our plant $G[z]$ has no repeated plots. (But this doesn’t actually matter; we can have multiple poles, it just makes it more messy.)

Then, we have

$$G[z]=\sum _{k=1}^n\frac{c_k}{z-q_k}$$

• $\{q_k{\}}_{k=1}^n$ are the plant poles and $\{c_k{\}}_{k=1}^n$ are the coefficients in $\mathbb{C}$. (These are given, since the plant is known.)

Then, IOP equation (i) becomes:

$$\begin{array}{rl}X[z]& +G[z]W[z]=1\\ X[z]& =1-G[z]W[z]\end{array}$$

Then, we can write $G[z]W[z]$

$$\begin{array}{rl}G[z]W[z]& =\left(\sum _{i=1}^m\frac{w_i}{z-p_i}\right)\left(\sum _{k=1}^n\frac{c_k}{z-q_k}\right)\\ & =\sum _{i=1}^m\sum _{k=1}^n\frac{c_k{w}_i}{(z-q_k)(z-p_i)}\\ & =\sum _{i=1}^m\sum _{k=1}^n{c}_k{w}_i\left(\frac{1}{p_i-q_k}\frac{1}{z-p_i}+\frac{1}{q_k-p_i}\frac{1}{z-q_k}\right)\\ & =\sum _{i=1}^m\underset{=:{\alpha }_i(\text{not a function of }k)}{\underbrace{\left(\sum _{k=1}^n\frac{c_k}{p_i-q_k}\right)}}{w}_i\frac{1}{z-p_i}+\sum _{k=1}^n\left(\sum _{i=1}^m\underset{:={\beta }_{k,i}}{\underbrace{\frac{c_k}{q_k-p_i}}}{w}_i\right)\frac{1}{z-q_k}\\ & =\sum _{i=1}^m{\alpha }_i{w}_i\frac{1}{z-p_i}+\sum _{k=1}^n\left(\sum _{i=1}^m{\beta }_{k,i}{w}_i\right)\frac{1}{z-q_k}\end{array}$$

• Note that we can write $\frac{1}{(z-p)(z-q)}=\frac{1}{p-q}\frac{1}{z-p}+\frac{1}{q-p}\frac{1}{z-q}$

Substituting back:

$$\begin{array}{r}X[z]=1-\sum _{i=1}^m{\alpha }_i{w}_i\frac{1}{z-p_i}-\sum _{k=1}^n\left(\sum _{i=1}^m{\beta }_{k,i}{w}_i\right)\frac{1}{z-q_k}\end{array}$$

For our $X[z]=1-G[z]W[z]$, what if $G[z]$ has an unstable pole? Will that also be a pole of $X[z]$? The coefficient of each unstable pole in $X[z]$ must be zero for $X[z]$ to be stable.

We can then re-order the poles of the plant $G[z]$:

$$\{q_k{\}}_{k=1}^n=\underset{\text{stable poles}}{\underbrace{\{q_k{\}}_{k=1}^{\hat{n}}}}\cup \underset{\text{unstable poles}}{\underbrace{\{q_k{\}}_{\hat{n}+1}^n}}$$

Then, the poles of $X$ are contained in

$$\text{poles}(X)\subset \{p_i{\}}_{i=1}^m\cup \{q_k{\}}_{k=1}^{\hat{n}}$$

because $X$ is stable.

So, we can write

$$X[z]=1+\sum _{i=1}^m\frac{x_i}{z-p_i}+\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{z-q_k}$$

where $\{x_i{\}}_{i=1}^m$ and $\{{\hat{x}}_k{\}}_{k=1}^{\hat{n}}$ are variable coefficients in $\mathbb{C}$.

Matching coefficients of the above equation with $X[z]=1-G[z]W[z]$ gives us:

$$\begin{array}{rl}{x}_i& =-{\alpha }_i{w}_i\forall i\in \{1,\dots ,m\}\\ {\hat{x}}_k& =\sum _{i=1}^m-{\beta }_{k,i}{w}_i\forall k\in \{1,\dots ,\hat{n}\}\\ 0& =\sum _{i=1}^m-{\beta }_{k,1}{w}_i\forall k\in \{\hat{n}+1,\dots ,n\}(\ast )\end{array}$$

• The final equation ensures all unstable poles have zero coefficients.
• These 3 are essentially another representation of IOP eq. (i)

IOP equation (ii):

$$V=-GX$$

Since $X$ only contains simple poles like $W$, we can u se the same procedure as above for calculating $GW$ to find $GX$ and then $V$.

However, as $D=\frac{W}{X}$ and $T_{ry}=1-X$, we do not need $V$ to satisfy our desired specs or recover our controller. We just need to ensure that $V$ is stable.

Thus, we just reuse the result from Eq. ($\ast \ast$) applied to $V$ to ensure all coefficients of the unstable poles in $V$ are zero:

$$\begin{array}{r}0=-c_j+\sum _{i=1}^m-{\gamma }_{j,i}{x}_i+\sum _{k=1}^{\hat{n}}-{\hat{\gamma }}_{j,k}{\hat{x}}_k\forall j\in \{\hat{n}+1,n\}(\ast \ast )\end{array}$$

• ${\gamma }_{j,i}=\frac{c_j}{q_j-p_i}$,
• ${\hat{\gamma }}_{j,k}=\frac{c_j}{q_j-q_k}$

The first sum corresponds to the poles $\{p_i{\}}_{i=1}^m$ in $X$, the second sum to the poles $\{q_k{\}}_{k=1}^{\hat{n}}$ in $X$, and the $-c_j$ term comes from the $1$ in $X$ that is not in $W$.