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:[[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]] | :[[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]] | ||
| − | |||
<br> '''Solution: ''' <br> | <br> '''Solution: ''' <br> | ||
<math>x(3) = x(2) + 2u(2) = x(1) + 2u(1) + 2u(2) = x(0) + 2u(0) + 2u(1) + 2u(2) </math> | <math>x(3) = x(2) + 2u(2) = x(1) + 2u(1) + 2u(2) = x(0) + 2u(0) + 2u(1) + 2u(2) </math> | ||
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<math>\therefore \ Constraint: 2u(0) + 2u(1) + 2u(2) = 6 </math> | <math>\therefore \ Constraint: 2u(0) + 2u(1) + 2u(2) = 6 </math> | ||
| − | Let <math>f(u) = u^2(0) + u^2(1) + u^2(2), h(u) = 2u(0) + 2u(1) + 2u(2) - 6 </math> | + | Let <math>f(u) = u^2(0) + u^2(1) + u^2(2), h(u) = 2u(0) + 2u(1) + 2u(2) - 6 </math> (1/2 in the objective function can be ignored for now) |
Let u* be a local minimizer. Lagrange theorem says there exists a λ such that: | Let u* be a local minimizer. Lagrange theorem says there exists a λ such that: | ||
| − | <math>\nabla f(u*) + \lambda \nabla h(u*) = 0 \\ | + | <math>\nabla f(u^*) + \lambda \nabla h(u^*) = 0 \\ |
| − | h(u*) = 0 </math> | + | h(u^*) = 0 </math> |
| + | |||
| + | Therefore, | ||
| + | |||
| + | <math>2 u^*(0) + 2 \lambda = 0 \\ | ||
| + | 2 u^*(1) + 2 \lambda = 0 \\ | ||
| + | 2 u^*(2) + 2 \lambda = 0 \\ | ||
| + | 2 u^*(0) + 2 u^*(1) + 2 u^*(2) - 6 = 0 \\ \\ | ||
| + | |||
| + | \therefore u^*(0) = u^*(1) = u^*(2) = 1</math> | ||
| + | |||
| + | Optimal performance index is <math>\frac{3}{2}</math> | ||
| + | |||
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | [[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | ||
Revision as of 10:54, 27 January 2015
QE2013_AC-3_ECE580-4
Solution:
$ x(3) = x(2) + 2u(2) = x(1) + 2u(1) + 2u(2) = x(0) + 2u(0) + 2u(1) + 2u(2) $
$ \because x(0) = 3, x(3) = 9 $
$ \therefore \ Constraint: 2u(0) + 2u(1) + 2u(2) = 6 $
Let $ f(u) = u^2(0) + u^2(1) + u^2(2), h(u) = 2u(0) + 2u(1) + 2u(2) - 6 $ (1/2 in the objective function can be ignored for now)
Let u* be a local minimizer. Lagrange theorem says there exists a λ such that:
$ \nabla f(u^*) + \lambda \nabla h(u^*) = 0 \\ h(u^*) = 0 $
Therefore,
$ 2 u^*(0) + 2 \lambda = 0 \\ 2 u^*(1) + 2 \lambda = 0 \\ 2 u^*(2) + 2 \lambda = 0 \\ 2 u^*(0) + 2 u^*(1) + 2 u^*(2) - 6 = 0 \\ \\ \therefore u^*(0) = u^*(1) = u^*(2) = 1 $
Optimal performance index is $ \frac{3}{2} $
