Практикум по теории управления в среде MATLAB. Никульчев Е.В. - 75 стр.

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Варианты заданий
Модель системы Функционал качества управления
1.
∑
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++=
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0
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++=
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=
1.
3)0(,1)0(
)()()()1(
)(3)()(3)1(
21
1212
2111
==
+−=+
+−=+
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3.
)(12)(8)(4)(2
2
1
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1
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∑
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1.
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)(5)()(7
2
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2.
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2
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2.
2)0(,1)0(
)()(4)()(2)1(
)(3)(2)(2)()1(
21
21212
21211
==
−−+=+
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3.
)(8)(6)(4)(
2
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1.
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1
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2
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)()(4)(2
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1
k
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2.
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2
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1
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k
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3.
1)0(,1)0(
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)()()(2)1(
21
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1121
==
−+=+
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3.
)(7)(3)(2)(
2
1
1
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1
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2
2
2
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1
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1.
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+=
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2
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2
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4.
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)(2)()(2)1(
)()()()1(
21
212
211
==
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kukxkxkx
3.
)()(3)(3
2
1
1
2
1
0
1
2
2
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k
++=
∑
=
–
75 – 
                                              Варианты заданий


               Модель системы                                Функционал качества управления
1.                                                                 1 1 2                          2
x1 (k + 1) = 3 x1 (k ) − u1 (k ) + 3u2 (k )                  1. J =  ∑ 3u1 (k ) + 4u22 (k ) + x2 (k )
                                                                   2 k =0
x2 (k + 1) = x1 (k ) − x2 (k ) + u1 (k )
                                                                   1 1 2
                                                             2. J = ∑ 2u1 ( k ) + x2 ( k ) + 4 x2 (k )
x1 (0) = 1, x2 (0) = 3                                                             1            1
                                                                   2 k =0
                                                             3.

                                                                     1 1 2
                                                              J=       ∑ 2u1 (k ) + 4u22 (k ) + 8 x12 (k ) + 12 x12 (k )
                                                                     2 k =0

2.                                                                 1 1 2                          2
x1 (k + 1) = x1 ( k ) + 2 x2 ( k ) − 2u1 (k ) − 3u 2 (k )    1. J =  ∑ 7u1 (k ) + u22 (k ) + 5 x2 (k )
                                                                   2 k =0
x2 (k + 1) = 2 x1 (k ) + x2 (k ) − 4u1 (k ) − u 2 (k )
                                                                   1 1 2
                                                             2. J = ∑ 2u1 ( k ) + 9 x2 ( k ) + x2 ( k )
x1 (0) = 1, x2 (0) = 2                                                                1         1
                                                                   2 k =0
                                                                    1 1 2
                                                             3. J =   ∑u1 (k) + 4u22 (k ) + 6x12 (k) + 8x12 (k)
                                                                    2 k =0

3.                                                                 1 1 2                         2
x1 (k + 1) = 2 x2 (k ) + x1 (k ) − u1 (k )                   1. J =  ∑ 2u1 (k ) + 4u22 (k ) + x2 (k )
                                                                   2 k =0
x2 (k + 1) = x1 (k ) + 4 x2 (k ) − 2u 2 (k )
                                                                   1 1 2
                                                             2. J = ∑ 2u1 (k ) + 3 x2 ( k ) + 4 x2 ( k )
                                                                                    1            1
x1 (0) = 1, x2 (0) = 1
                                                                   2 k =0
                                                                   1 1 2
                                                             3. J = ∑u1 (k ) + 2u2 (k ) + 3x2 (k ) + 7x2 (k )
                                                                                 2          1          1
                                                                   2 k =0

4.                                                                 1 1 2               2
x1 (k + 1) = x1 (k ) − x2 (k ) − u (k )
                                                             1. J =  ∑ 4u (k ) + 8 x2 (k )
                                                                   2 k =0
x2 (k + 1) = 2 x1 (k ) + x2 (k ) − 2u (k )                         1 1 2
                                                             2. J = ∑ u ( k ) + x2 ( k ) + 4 x2 ( k )
                                                                                 1            1
x1 (0) = 1, x2 (0) = 1
                                                                   2 k =0
                                                                   1 1 2
                                                             3. J = ∑ 3u (k ) + 3x2 (k ) + x2 (k )
                                                                                  1         1
                                                                   2 k =0




                                                            – 75 –

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