In signal processing and control theory, the Bistritz criterion is a simple method to determine whether a discrete linear time invariant (LTI) system is stable proposed by Yuval Bistritz.[1] [2] Stability of a discrete LTI system requires that its characteristic polynomial
Dn(z)=d0+d1z+d2z2+ … +dn-1zn-1+dnzn
(obtained from its difference equation, its dynamic matrix, or appearing as the denominator of its transfer function) is a stable polynomial, where
Dn(z)
|zk|<1,k=1,...,n
where
Dn(z)=dn
n | |
\prod | |
k=1 |
(z-zk)
Dn(z)
|zk|<1
|zk|=1
|zk|>1
In the following, the focus is only on how to test stability of a real polynomial. However, as long as the basic recursion needed to test stability remains valid, ZL rules are also brought.
Consider
Dn(z)
Dn(1) ≠ 0
Dn(1)=0
n | |
D | |
n(z)=z |
Dn(1/z)=dn+dn-1z+dn-2
2+ … +d | |
z | |
n-1 |
zn-1+d0zn
The algorithm assigns to
Dn(z)
Tm(z)=
\sharp | |
T | |
m(z), m=n,n-1,\ldots, |
0
created by a three-term polynomial recursion. Write out the polynomials by their coefficients,
Tm(z)=
m | |
\sum | |
k=0 |
tm,kzk
symmetry means that
Tm(z)=tm,0+tm,1z+ … +tm,1zm-1+tm,0zm
so that it is enough to calculate for each polynomial only about half of the coefficients. The recursion begins with two initial polynomials driven from the sum and difference of the tested polynomial and its reciprocal, then each subsequent polynomial of reduced degree is produced from the last two known polynomials.
Initiation:
Tn(z)=
\sharp | |
D | |
n |
(z), Tn-1(z)=
| |||||||||
z-1 |
Recursion: For
m=n-1,\ldots,1
\deltam+1=
Tm+1(0) | |
Tm(0) |
Tm-1(z)=
\deltam+1(1+z)Tm(z)-Tm+1(z) | |
z |
The successful completion of the sequence with the above recursion requires
Tm(0) ≠ 0, m=n-1,...,1
Tm(0) ≠ 0, m=n,...,0
Normal conditions are necessary for stability. This means that, the tested polynomial can be declared as not stable as soon as a
Tm(0)=tm,0=tm,m=0
Theorem. If the sequence is not normal then
Dn(z)
\nu=Var\{Tn(1),Tn-1(1),\ldots,T1(1),t0,0\}
denote the count of the number of sign variations in the indicated sequence. Then
Dn(z)
\nu=0
Dn(z)
\nu
n-\nu
Violation of various necessary conditions for stability may be used advantageously as early indications that the polynomial is not stable (has at least one UC or OUC zero). The polynomial can be declared not stable as soon as a
Tm(0)=0
\deltam<0
Tm(1)
Consider the polynomial
D3(z)=2+Kz-22z2+24z3
K
Q1: For what values of
K
Construct the sequence:
T3(z)=26+(K-22)z+(K-22)z2+26z3
T2(z)=22-Kz+22z2
T1(z)=
24(22-K) | |
11 |
(1+z)
T0(z)=44+K
Use their values at z = 1 to form
\operatorname{Var}(8+2K,44-K,48(22-K)/11,44+K)
All the entries in the sequence are positive for −4 < K < 22 (and for no K are they all negative). Therefore D(z) is stable for −4 < K < 22.
Q2: Find ZL for K = 33 Var = 2 ⇒ 2 OUC, 1 IUC zeros.
Q3: Find ZL for K = −11 Var = 1 ⇒ 1 OUC, 2 IUC zeros.
(1) The test bears a remarkable similarity to the Routh test. This is best observed when the Routh test is arranged appropriately into a corresponding three-term polynomial recursion.
(2) The Bistritz test uses three-term polynomial recursion that propagates polynomials with symmetry as opposed to previously available classical tests for discrete systems that propagate polynomials with no particular structure using a two-term recursion. It stimulated the discovery of more algorithms in the area of digital signal processing (e.g. solving the linear prediction problem) and discrete systems (e.g. testing stability of higher-dimensional systems) collectively called "immittance" or "split" algorithms that adopted this technique to more efficient counterparts to also other classical so called "scattering" algorithms.[5] [6] [7] The Bistritz test forms the "immittance" counterpart of the "scattering" type classical tests of Schur–Cohn and Jury.