Jury Test

Definition: Schur

A polynomial $\Delta [z]$ is Schur if $\text{roots}(\Delta )\subset \mathbb{D}$.

Lemma 1

Let $\Delta [z]={\sum }_{i=0}^n{c}_i{z}^i$. If $\Delta [z]$ is Schur, then $|c_n|>|c_0|$.

This is a necessary but not sufficient condition. So $|c_n|>|c_0|$ does not necessarily mean a polynomial is Schur.

Boundary Crossing Lemma

Let $\lambda \in [0,1]$ and $\Delta \lambda [z]={\Delta }_1[z]-\lambda {\Delta }_2[z]$. If ${\Delta }_1[z]$ is Schur and ${\Delta }_1[z]-{\Delta }_2[z]$ is not Schur (or vice versa), then $\exists {\lambda }^{\ast }\in [0,1]$ such that $\Delta {\lambda }^{\ast }[z]$ has a root on the unit circle.

Let $\Delta [z]={\sum }_{i=0}^n{c}_i{z}^i$. Let $Q[z]={\sum }_{i=0}^n{c}_{n-i}{z}^i$.

• Example: $\Delta [z]=5z^3+4z^2+1$. Then

$$Q[z]=z^3+0z^2+4z+5$$

Let

$$\begin{array}{rl}R[z]& =\frac{1}{z}\left(\Delta [z]-\frac{c_0}{c_n}Q[z]\right)\\ & =\frac{1}{z}\left(\sum _{i=0}^n{c}_i{z}^i-\frac{c_0}{c_n}\sum _{i=0}^n{c}_{n-i}{z}^i\right)\\ & =\frac{1}{z}\left(\sum _{i=0}^n\left(c_i-\frac{c_0}{c_n}{c}_{n-i}\right)z^i\right)\\ & =\frac{1}{z}\left(\sum _{i=1}^n\left(c_i-\frac{c_0}{c_n}{c}_{n-i}\right)z^i\right)\\ & =\frac{1}{z}\left(z\sum _{i=0}^{n-1}\left(c_{i+1}-\frac{c_0}{c_n}{c}_{n-i-1}\right)z^i\right)\\ & =\sum _{i=0}^{n-1}\left(c_{i+1}-\frac{c_0}{c_n}{c}_{n-i-1}\right)z^i\end{array}$$

• Line 3-4: For $i=0$ we get $c_0-\frac{c_0}{\cancel{c_n}}\cancel{c_n}=0$.

Lemma 2

Suppose $|c_n|>|c_0|$. Then $\Delta [z]$ is Schur $\iff$ $R[z]$ is Schur.

Define $R^{(0)}[z]:=\Delta [z]$:

$$\begin{array}{rl}& R^{(i+1)}[z]=\frac{1}{z}\left(R^{(i)}[z]-\frac{c_0}{c_n}{Q}^{(i)}[z]\right)\\ & R^{(0)}[z]⟶R^{(1)}[z]⟶R^{(2)}[z]⟶\dots ⟶R^{n-1}[z]\end{array}$$

• degree $n$, $n-1$, $n-2$, …, 1

For degree 1, we have $c_{1}z+c_0$, which has a root at $-\frac{c_0}{c_1}$. Then:

$$\frac{|c_0|}{|c_n|}<1(\text{aka Schur})⟺|c_n|>|c_0|$$

Jury Test Algorithm

Given $R^{(i)}[z]$ for some $i\in \{0,1,\dots ,n-1\}$.

Check if $|c_n|>|c_0|$.

• If not, then $R^{(i)}[z]$ is not Schur (lemma 1).
  • Then $R^{(i-1)}[z]$ is not Schur (lemma 2).
  • Then, $R^{(0)}[z]=\Delta [z]$ is not Schur (lemma 2).
  • (presumably only made it to this step if $|c_n|>|c_0|$ for $R^{(i-1)}[z]$)
• If yes, set $R^{(i+1)}[z]=\frac{1}{z}\left(R^{(i)}[z]-\frac{c_0}{c_n}{Q}^{(i)}[z]\right)$
  • (repeat)
  • Stop with $R^{(n-1)}[z]$
    • If Schur, then $\Delta [z]$ is Schur
    • If not Schur, then $\Delta [z]$ is not Schur

Example 10.1b

For a discrete time system, determine the range of $\tau$ that stabilizes the CL system.

$$D_2[z]=\frac{2z-\frac{1}{4}}{z+0.5},G_2[z]=\frac{z+0.5}{(z-e^{-2\tau })\left(z+\frac{1}{4}\right)}$$

Then, the closed-loop polynomial is:

$$\begin{array}{rl}{\Delta }_2[z]& =\left(2z-\frac{1}{4}\right)(z+0.5)+(z+0.5)(z-e^{-2\tau })\left(z+\frac{1}{4}\right)\\ & =(z+0.5)\left(z^2+\left(\frac{9}{4}-e^{-2\tau }\right)z-\frac{1}{4}(1+e^{-2\tau })\right)\end{array}$$

• The first term is stable (root at -0.5), but we need to check if the second term is Schur

The Jury Test:

$$\Delta [z]=z^2+\left(\frac{9}{4}-e^{-2\tau }\right)z-\frac{1}{4}(1+e^{-2\tau })$$

Then:

$$\begin{array}{rl}|c_2|& =1\\ |c_0|& =\frac{1}{4}(1+e^{-2\tau })<\frac{1}{2}\end{array}$$

Which gives $|c_2|>|c_0|$, since we must have $\tau >0$.

Then:

	$c_2$	$c_1$	$c_0$
$R^{(0)}[z]$	$1$	$\frac{9}{4}-e^{-2\tau }$	$-\frac{1}{4}(1+e^{-2\tau })$
$Q^{(0)}[z]$	$-\frac{1}{4}(1+e^{-2\tau })$	$\frac{9}{4}-e^{-2\tau }$	$1$

which gives

$$\frac{c_0}{c_n}=-\frac{1}{4}(1+e^{-2\tau })$$

Then:

$$\begin{array}{r}{R}^{(1)}[z]=1-{\left(\frac{1}{4}(1+e^{-2\tau })\right)}^2\left(\frac{9}{4}-e^{-2\tau }\right)\left(1+\frac{1}{4}(1+e^{-2\tau })\right)\end{array}$$

so:

$$\begin{array}{rl}|c_1|& =\left|\left(1+\frac{1}{4}(1+e^{-2\tau })\right)\left(1-\frac{1}{4}(1+e^{-2\tau })\right)\right|\\ |c_0|& =\left|\left(1+\frac{1}{4}(1+e^{-2\tau })\right)\left(\frac{9}{4}-e^{-2\tau }\right)\right|\end{array}$$

to get $|[c_1]|>|c_0|$, we need

$$\begin{array}{rl}1-\frac{1}{4}(1+e^{-2\tau })& >\frac{9}{4}-e^{-2\tau }\\ 3-e^{-2\tau }& >9-4e^{-2\tau }\\ 3e^{-2\tau }& >6\\ e^{-2\tau }& >2\end{array}$$

This is not possible; no value of $T>0$ would allow the system to be CL stable.

Jury Table:

Theorem: Jury Test

Assume $c_n>0$ (if not, use $-\Delta [z]$).

Then, $\Delta [z]$ is Schur if and only if

$$\begin{array}{rl}{c}_n& >0\\ b_n& >0\\ d_n& >0\\ & ⋮\\ e_n& >0\end{array}$$

We either check $|e_n|>|e_{n-1}|$ or go to $R^n[z]$.