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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 315<br />

Because,<br />

a (1)<br />

11 (n) =a 11(n)ã 11 (n)=(a 11 (n)) 2 − 1 = 2s<strong>in</strong> 2 n − 1<br />

√<br />

a (1)<br />

2s<strong>in</strong>n − 1<br />

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

n 3 + 10<br />

√<br />

a (1)<br />

2s<strong>in</strong>n + 1<br />

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

n 3 + 10<br />

a (1)<br />

22 (n) =a 12(n)ã 12 (n)=(a 12 (n)) 2 1<br />

=<br />

(n 3 + 10) 2<br />

a (2)<br />

11 (n) =a 21(n)ã 21 (n)=(a 21 (n)) 2 1<br />

=<br />

(n 2 + 20) 2<br />

√<br />

a (2)<br />

2cosn + 1<br />

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

n 2 + 20<br />

√<br />

a (2)<br />

2cosn − 1<br />

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

n 2 + 20<br />

a (2)<br />

22 (n) =a 22(n)ã 22 (n)=(a 22 (n)) 2 − 1 = 2cos 2 n − 1<br />

b(n)= 1 2<br />

s<br />

∑<br />

i, j,k=1<br />

∣<br />

∣a (i)<br />

∣ ∣∣<br />

jk (n)+a(i) kj (n) =<br />

1<br />

[∣ ∣ ∣ ∣ ∣ ∣ ∣∣a (1)<br />

∣∣ ∣∣a<br />

11<br />

2<br />

(n)+a(1) 11 (n) (1)<br />

∣∣ ∣∣a +<br />

12 (n)+a(1) 21 (n) (1)<br />

∣∣+<br />

+<br />

21 (n)+a(1) 12 (n) ∣ ∣ ∣ ∣ ∣<br />

∣<br />

∣a (1)<br />

∣∣ ∣∣a<br />

22 (n)+a(1) 22 (n) (2)<br />

∣∣ ∣∣a +<br />

11 (n)+a(2) 11 (n) (2)<br />

∣∣+<br />

+<br />

12 (n)+a(2) 21 (n) ∣ ∣ ∣]<br />

∣<br />

∣a (2)<br />

∣∣ ∣∣a<br />

21 (n)+a(2) 12 (n) (2)<br />

∣∣ +<br />

22 (n)+a(2) 22 (n)<br />

[<br />

1 ∣∣2s<strong>in</strong> 2 n − 1 + 2s<strong>in</strong> 2 n − 1 ∣ ∣ ∣∣∣∣ √ √ 2s<strong>in</strong>n − 1 2s<strong>in</strong>n + 1<br />

+<br />

2<br />

n 3 +<br />

+ 10 n 3 + 10 ∣ +<br />

√ √ ∣ ∣ 2s<strong>in</strong>n + 1 2s<strong>in</strong>n − 1<br />

∣∣∣∣ ∣ n 3 +<br />

+ 10 n 3 + 10 ∣ + 1<br />

(n 3 + 10) 2 + 1 ∣∣∣∣<br />

(n 3 + 10) 2 +<br />

∣ 1<br />

∣(n 2 + 20) 2 + 1 ∣∣∣∣ √ √ 2cosn + 1 2cosn − 1<br />

(n 2 + 20) 2 +<br />

∣ n 2 +<br />

+ 20 n 2 + 20 ∣ +<br />

√ √ 2cosn − 1 2cosn + 1<br />

∣ n 2 +<br />

+ 20 n 2 + 20 ∣ + ∣ 2cos 2 n − 1 + 2cos 2 n − 1 ∣ ]<br />

=<br />

[<br />

1<br />

4<br />

2 ∣ s<strong>in</strong>2 n − 1 2∣ + 4√ 2<br />

n 3 + 10 |s<strong>in</strong>n| + 2<br />

(n 3 + 10) 2

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