Jury stability criterion
method of determining the stability of a discrete-time linear system
In signal processing and control theory, the Jury stability criterion is a method of determining the stability of a discrete-time, linear system by analysis of the coefficients of its characteristic polynomial. It is the discrete time analogue of the Routh–Hurwitz stability criterion. The Jury stability criterion requires that the system poles are located inside the unit circle centered at the origin, while the Routh-Hurwitz stability criterion requires that the poles are in the left half of the complex plane. The Jury criterion is named after Eliahu Ibraham Jury.
Method
If the characteristic polynomial of the system is given by
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=
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{\displaystyle f(z)=a_{n}+a_{n-1}z^{1}+a_{n-2}z^{2}+\dots +a_{1}z^{n-1}+a_{0}z^{n}}
then the table is constructed as follows:
row
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{\displaystyle {\begin{array}{lcccccc}{\underline {\text{row}}}&\ {\underline {z^{n}}}\ &\ {\underline {z^{n-1}}}\ &\ {\underline {z^{n-2}}}\ &\ \cdots \ &\ {\underline {z^{1}}}\ &\ {\underline {z^{0}}}\ \\[8pt]1&a_{0}&a_{1}&a_{2}&\cdots &a_{n-1}&a_{n}\\[4pt]2&a_{n}&a_{n-1}&a_{n-2}&\cdots &a_{1}&a_{0}\\[10pt]3&b_{0}&b_{1}&\cdots &b_{n-2}&b_{n-1}\\[4pt]4&b_{n-1}&b_{n-2}&\cdots &b_{1}&b_{0}\\[10pt]5&c_{0}&c_{1}&\cdots &c_{n-2}&\\[4pt]6&c_{n-2}&c_{n-3}&\cdots &c_{0}&&\\[10pt]\ \!\vdots &\vdots &\vdots &\vdots &\vdots &&\\[10pt]2n-5\quad &p_{0}&p_{1}&p_{2}&p_{3}&&\\[4pt]2n-4&p_{3}&p_{2}&p_{1}&p_{0}&&\\[10pt]2n-3&q_{2}&q_{1}&q_{0}&&&\end{array}}}
That is, the first row is constructed of the polynomial coefficients in order, and the second row is the first row in reverse order and conjugated.
The third row of the table is calculated by subtracting
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{\displaystyle {\tfrac {a_{n}}{a_{0}}}}
times the second row from the first row, and the fourth row is the third row with the first n elements reversed (as the final element is zero).
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{\displaystyle {\begin{array}{ccccc}a_{0}&a_{1}&\dots &a_{n-1}&\quad a_{n}\quad \\[4pt]a_{n}&a_{n-1}&\dots &a_{1}&a_{0}\\[4pt]a_{0}-a_{n}{\frac {a_{n}}{a_{0}}}&a_{1}-a_{n-1}{\frac {a_{n}}{a_{0}}}&\dots &a_{n-1}-a_{1}{\frac {a_{n}}{a_{0}}}&0\\[4pt]a_{n-1}-a_{1}{\frac {a_{n}}{a_{0}}}&\dots &a_{1}-a_{n-1}{\frac {a_{n}}{a_{0}}}&a_{0}-a_{n}{\frac {a_{n}}{a_{0}}}&0\end{array}}}
The expansion of the table is continued in this manner until a row containing only one non-zero element is reached.
Note the
a
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a
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{\displaystyle {\tfrac {a_{n}}{a_{0}}}}
is
a
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{\displaystyle a_{n}}
for the 1st two rows. Then for 3rd and 4th row the coefficient changes (i.e.
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−
1
b
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{\displaystyle {\tfrac {b_{n-1}}{b_{0}}}}
).
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