Specs for Control Design

How do we design controllers to meet performance specifications using IOP/SPA?

Closed-loop stability

Closed loop stability $⟺$ $W,X,V$ stable $⟹$ already guaranteed by choosing $\{p_i{\}}_{i=1}^m\subset \mathbb{D}$ and satisfying Equations ($\ast$) and ($\ast \ast$) from Simple Pole Approximation.

Steady-state error

($e_{ss}$) is given by

$$\begin{array}{rl}{e}_{ss}& =T_{re}[1]=X[1][\text{IOP theorem part b}]\\ & =1+\sum _{i=1}^m\frac{x_i}{1-p_i}+\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{1-q_k}\end{array}$$

The $p_i$ terms are fixed because we choose them in advance and $q_k$ are fixed because they are from the plant.

Case 1: We want the steady state error to be zero such that $e_{ss}=0$

$$e_{ss}=1+\sum _{i=1}^m\frac{x_i}{1-p_i}+\frac{{\sum }_{k=1}^{\hat{n}}{\hat{x}}_k}{1-q_k}=0$$

Case 2: We want the steady-state error to be bounded such that $|e_{ss}|\le C$

$$\begin{array}{rl}{e}_{ss}& \le C\\ e_{ss}& \ge -C\end{array}$$

Vector Form

We have:

$$\begin{array}{rl}{e}_{ss}& =1+\sum _{i=1}^m\frac{x_i}{1-p_i}+\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{1-q_k}=0\\ & =1+\underset{\text{Steady state}}{\underbrace{\left[\begin{array}{cccc}\frac{1}{1-p_1}\dots \frac{1}{1-p_m}& \frac{1}{1-q_1}& \dots & \frac{1}{1-q_{\hat{n}}}\end{array}\right]}}\left[\begin{array}{c}{x}_1\\ ⋮\\ x_m\\ {\hat{x}}_1\\ ⋮\\ {\hat{x}}_{\hat{n}}\end{array}\right]\\ & =\boxed{1+\text{Steady state}\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]}\end{array}$$

which we can set to $=0$, $\le C$, $\ge -C$, etc.

Limit on control effort

$$\begin{array}{rlr}u[k]\le C\forall k>0& ⟹\text{step}(T_{ru})[k]\le C& \forall k\ge 0\\ & ⟹\text{step}(W)[k]\le C& \forall k\ge 0\\ & ⟺\text{step}\left(\sum _{i=1}^m\frac{w_i}{z-p_i}\right)[k]\le C& \forall k\ge 0\\ & ⟺\sum _{i=1}^m{w}_i\text{step}\left(\frac{1}{z-p_i}\right)[k]\le C& \\ & ⟺\sum _{i=1}^m\frac{1-p_i^k}{1-p_i}{w}_i\le C& \forall k\ge 0\end{array}$$

$T_{ry}$

Note we have

$$\begin{array}{rl}e=r-y& ⟹y=r-e\\ & ⟹Y=R-E=R-T_{re}R=(1-T_{re})R\end{array}$$

Then:

$$T_{ry}=\frac{Y}{R}=1-T_{re}=1-X$$

Vector form

We are consider the time horizon where $k\ge 0$. There are infinitely many such time steps. This makes the computation hard to do on a computer. At the same time, for a practical stable system, the step response will settle down to a steady-state and we don’t really need to keep enforcing this limit. Thus, we set some practical limit $K>0$ to give us a finite number of timesteps.

$$\underset{0<k\le K}{\max }\left(\sum _{i=1}^K\frac{1-p_i^k}{1-p_i}{w}_i\right)\le C$$

Supposing some point in time $k$, we have

$$u[k]=\text{step}(T_{ru})[k]=\sum _{i=1}^m\frac{1-p_i^k}{1-p_i}{w}_i=\left[\begin{array}{ccc}\frac{1-p_1^k}{1-p_1}& \dots & \frac{1-p_m^k}{1-p_m}\end{array}\right]\left[\begin{array}{c}{w}_1\\ ⋮\\ w_m\end{array}\right]$$

Extending this to the full time horizon $0<k\le K$:

$$\begin{array}{r}\left[\begin{array}{c}\text{step}(T_{ru})[1]\\ ⋮\\ \text{step}(T_{ru})[K]\end{array}\right]=\underset{\text{step\_ru}}{\underbrace{\left[\begin{array}{ccc}\frac{1-p_1^1}{1-p_1}& \dots & \frac{1-p_m^1}{1-p_m}\\ & \ddots & \\ \frac{1-p_1^k}{1-p_1}& \dots & \frac{1-p_m^k}{1-p_m}\end{array}\right]}}\left[\begin{array}{c}{w}_1\\ ⋮\\ w_m\end{array}\right]\end{array}$$

We can then write this constraint as:

$$\boxed{\text{max}(\text{step\_ru}\ast w)\le C}$$

Overshoot

We want some $\%\text{OS}\le C$.

Then, we have

$$\begin{array}{rl}\text{step}(T_{ry})[j]& \le (1+C)y_{ss}\\ & =(1+C)(1-e_{ss})\\ & =(1+C)(1-X[1])\end{array}$$

Note that $T_{ry}=1-T_{re}=1-X$. Then, we can write the above as

$$\underset{j\ge 0}{\max }\left(-\sum _{i=1}^m\frac{1-p_i^j}{1-p_i}{x}_i-\sum _{k=1}^{\hat{n}}\frac{1-q_k^j}{1-q_k}{\hat{x}}_k\right)\le (1+C)\left(-\sum _{i=1}^m\frac{x_i}{1-p_i}-\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{1-q_k}\right)$$

Vector Form

Similar to control effort, we can write

$$\begin{array}{r}\left[\begin{array}{c}\text{step}(T_{ry})[1]\\ ⋮\\ \text{step}(T_{ru})[K]\end{array}\right]=\underset{\text{step\_ry}}{\underbrace{\left[\begin{array}{cccccc}-\frac{1-p_1^1}{1-p_1}& \dots & -\frac{1-p_m^1}{1-p_m}& -\frac{1-q_1^1}{1-q^1}& \dots & -\frac{1-q_{\hat{n}}^1}{1-q_{\hat{n}}}\\ & \ddots & & & & \\ \frac{1-p_1^k}{1-p_1}& \dots & \frac{1-p_m^k}{1-p_m}& -\frac{1-q_k^1}{1-q^1}& \dots & -\frac{1-q_{\hat{n}}^k}{1-q_{\hat{n}}}\end{array}\right]}}\left[\begin{array}{c}{x}_1\\ ⋮\\ x_m\\ {\hat{x}}_1\\ ⋮\\ {\hat{x}}_{\hat{n}}\end{array}\right]\end{array}$$

We can then write this as:

$$\boxed{\text{max}\left(\text{step\_ry}\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)\le (1+C)\left(-\text{steady\_state}\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)}$$

Settling Time (within 2%)

Let $T$ be the sample time and let $\hat{j}=\text{min}\{j:jT\ge C\}$.

Then:

$$\begin{array}{r}\text{step}(T_{ry})[j]\le 1.02y_{ss}\\ \text{step}(T_{ry})[j]\ge 1.02y_{ss}\end{array}$$

• where $T_{ry}=1-X$

Thus, we have

$$\begin{array}{rl}\underset{j\ge \hat{j}}{\max }(\text{step}(T_{ry}))& \le 1.02y_{ss}\\ \underset{j\ge \hat{j}}{\max }\left(-\sum _{i=1}^m\frac{1-p_i^j}{1-p_i}{x}_i-\sum _{k=1}^{\hat{n}}\frac{1-q_k^j}{1-q_k}{\hat{x}}_k\right)& \le 1.02\left(-\sum _{i=1}^m\frac{x_i}{1-p_i}-\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{1-q_k}\right)\end{array}$$

and also

$$\begin{array}{rl}\underset{j\ge \hat{j}}{\min }(\text{step}(T_{ry}))& \ge 0.98y_{ss}\\ \underset{j\ge \hat{j}}{\min }\left(-\sum _{i=1}^m\frac{1-p_i^j}{1-p_i}{x}_i-\sum _{k=1}^{\hat{n}}\frac{1-q_k^j}{1-q_k}{\hat{x}}_k\right)& \ge 0.98\left(-\sum _{i=1}^m\frac{x_i}{1-p_i}-\sum _{k=1}^{\hat{n}}\frac{{\hat{x}}_k}{1-q_k}\right)\end{array}$$

Vector Form

We can borrow our result from the overshoot spec:

$$\boxed{\begin{array}{r}\text{max}\left(\text{step\_ry}(\hat{j}:\text{end, :})\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)\le 1.02\left(-\text{steady state}\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)\\ \text{min}\left(\text{step\_ry}(\hat{j}:\text{end, :})\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)\ge 0.98\left(-\text{steady state}\left[\begin{array}{c}x\\ \hat{x}\end{array}\right]\right)\end{array}}$$