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[Studies in Computational Intelligence 481] Artur Babiarz, Robert Bieda, Karol Jędrasiak, Aleksander Nawrat (auth.), Aleksander Nawrat, Zygmunt Kuś (eds.) - Vision Based Systemsfor UAV Applications (2013, Sprin

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Estimation of Solution of Discrete L<strong>in</strong>ear Time-Vary<strong>in</strong>g System 323<br />

∞<br />

It is well known (see [17]), that an <strong>in</strong>f<strong>in</strong>ite product ∏ (x(p)+1) with positive<br />

p=0<br />

x(p) converges if and only if the series ∑<br />

∞ x(p) converges. Also similarly, if 0 <<br />

p=0<br />

x(p) < 1 holds, then ∏<br />

∞ (1 − x(p)) approaches a non-zero limit if and only if the<br />

p=0<br />

series ∑<br />

∞ x(p) converges. Therefore, from Theorem 2 we obta<strong>in</strong> the follow<strong>in</strong>g result<br />

p=0<br />

about stability.<br />

Corollary 1. If the series ∑<br />

∞ b(p), where b(p) are given by (6), converges, then (1)<br />

p=0<br />

is stable.<br />

It is clearly that <strong>in</strong> Example (4) the series ∑<br />

∞ b(p) is convergent, and therefore<br />

p=0<br />

system (1) is stable.<br />

Observe that from Theorem 1 we know that Lyapunov exponents of system (1)<br />

with coefficients given by (9) are all equal to zero. But it doesn’t imply that the<br />

system is stable.<br />

Bound of norm of a matrix power by coefficients is a particular case of our estimate.<br />

Our result is also valid for stationary systems and we will ilustrate it by an<br />

example of 2-by-2 matrix.<br />

Example 5. Consider system (1) with all A(n) equal to A,where<br />

[ ] ef<br />

A = .<br />

gh<br />

Namely, its elements are as follows<br />

a 11 (n)=ea 12 (n)= f<br />

a 21 (n)=ga 22 (n)=h .<br />

Apply<strong>in</strong>g the above-mentioned elements of A to (5) and (3) we obta<strong>in</strong> (5) <strong>in</strong> the<br />

follow<strong>in</strong>g form<br />

a (1)<br />

11 (n) =a 11(n)ã 11 (n)=(a 11 (n)) 2 − 1 = e 2 − 1<br />

a (1)<br />

12 (n) =a 11(n)ã 12 (n)=(a 11 (n) − 1) a 12 (n)=(e − 1) f<br />

a (1)<br />

21 (n) =a 12(n)ã 11 (n)=a 12 (n)((a 11 (n)) + 1)=(e + 1) f<br />

a (1)<br />

22 (n) =a 12(n)ã 12 (n)=(a 12 (n)) 2 = f 2<br />

a (2)<br />

11 (n) =a 21(n)ã 21 (n)=(a 21 (n)) 2 = g 2<br />

a (2)<br />

12 (n) =a 21(n)ã 22 (n)=a 21 (n)(a 22 (n)+1) =g(h + 1)<br />

a (2)<br />

21 (n) =a 22(n)ã 21 (n)=(a 22 (n) − 1) a 21 (n)=g(h − 1)<br />

a (2)<br />

22 (n) =a 22(n)ã 22 (n)=(a 22 (n)) 2 − 1 = h 2 − 1

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