10. LINEAR COMBINATIONS OF SOLUTIONS

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1 10. LINEAR COMBINATIONS OF SOLUTIONS Description of Method Due to the fact that m = Bu is a linear equation, two or more different solutions to this equation can be linearly combined to yield new solutions. For example, the control vectors u 1 and u 2 can be added together. Bu 1 = m (10-1) Bu 2 = m (10-2) B(c 1 u 1 + c 2 u 2 ) = c 1 Bu 1 + c 2 Bu 2 = (c 1 + c 2 )m = m for c 1 + c 2 = 1 (10-3) Or, more generally, B c i u i = m for c i = 1 (10-4) In this way, various aspects of different solutions can be combined. The combined controls are guaranteed to be admissible if the following conditions are met. Each of the control solutions to be combined is admissible: u i i (10-5) The sum of the weighting terms is equal to one, c i = 1 (10-6) Each of the weighting terms is between zero and one, 0 c i 1 (10-7) 1

2 If these conditions are met, then the combined controls will be admissible, u = c i u i (10-8) When the controls are combined in this manner, each control, u i, will range between the smallest value in any of the u i (when the corresponding c i =1 and the rest of the c i =0) and the largest value in any of the u i. For example, if two solutions are combined and the smallest value of u i is u i 2 = -5 and the largest value of u i is u i 1 = 5, the relationship between u i, c 1 and c 2 can be seen in Figure If one or more of the solutions contains admissible controls, scalar weighting terms can be found to produce an admissible final solution. If the weighting on the admissible solution is one and the other weightings are zero, then the solution is admissible. Consider combining the two solutions in Equation 10-3 when only u 2 is admissible: Bu 1 = m (10-9) Bu 2 = m (10-10) u 1, u 2 (10-11) To find the scalars c 1 and c 2 which scale c 1 u 1 + c 2 u 2 so that it is on the boundary of, examine each control. If u i is within its limits for each u i, then as long as 0 c 1 1 and c 1 + c 2 = 1, u i will be within its limits. 2

3 c2 ui c1 Figure 10-1: u i Varies Between the Minimum and Maximum Values 3

4 If u i exceeds a limit, it is possible to compute scale factors, a i and b i, which will make the sum of the control solutions equal to the limit: a i u i 1 + b i u i 2 = u * i (10-12) Where u * i is the limit which u i1 violates and, b i = 1 - a i (10-13) Substituting into Equation 10-12, a i u i 1 + (1-a i )u i 2 = u * i (10-14) Then solving for a i, a i (u i 1 -u i 2 ) = u * i - u i 2 (10-15) a i = u* i - u i 2 u i 1 - u i 2 (10-16) Because u i 1 is inadmissible and u i 2 is admissible: u i 1 > u * i u i 2, and 0 a i < 1 (10-17) After computing a i for each of saturated controls, selecting the a i which is smallest, c 1 = Min{a i }, and c 2 = 1 - c 1 will insure that all of the final controls are at or within their limits. Note that if u 2 is on the boundary of, c 1 = 0 and c 2 = 1. Example 10-1: Combining Direct Allocation and Daisy Chaining This example combines the methods of Direct Allocation and Daisy Chaining to provide a method which meets the Daisy Chaining constraint that a subset of the controls are used only when the primary controls fail to generate the desired moment, and also 4

5 allocates admissible controls for all the attainable moments. Using the F-18 HARV data, consider the ailerons, horizontal tails and rudders as the primary controls, as done in Example 6-2. B 1 = e e e e e e e e e e e e e e e-2 (10-18) The pseudo-inverse of B 1 will be used to allocate the control vector u 1. P 1 = B 1 T [B 1 B 1 T ] -1 = e e e e e e e e e e e e e e e+1 (10-19) This control solution will not use the secondary controls. To preserve this quality, P 1 is augmented with a matrix of zeros to form P. P = P 1 [0] = e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e+0 (10-20) u 1 = Pm (10-21) The second control vector, u 2, is computed using Direct Allocation. u 2 m (10-22) 5

6 When the desired moment is attainable using u 1, c 1 = 1 and c 2 = 0. Using these weightings insures that only the primary controls are used to generate these moments. For example, consider the desired moment: m d = (10-23) The controls in u 1 are allocated as follows: Pm d = u 1 = { , , , , , 0, 0, 0, 0, 0} T (10-24) Since all of the controls in u 1 are admissible, only u 1 will be used. u 1 (10-25) c 1 = 1, c 2 = 0 u = u 1 (10-26) The controls in u 1 will not be admissible for all the attainable moments, as shown in Section 6. If the desired moment is unattainable using u 1, c 1 and c 2 are found using Equations through When the desired moment is m d = (10-27) the controls in u 1 are, u 1 = { , , , , , 0, 0, 0, 0, 0} T (10-28) 6

7 The left horizontal tail, u 2, and the rudders, u 5, are saturated. The controls allocated using Direct Allocation are: u 2 = { , , , , , , , , , } T (10-29) The controls of interest are u 2 and u 5, because they are commanded to exceed their limits in u 1. Table 10-1 shows the scale factors for this moment computed using Equation The smallest a i is This value should be used as c 1, because if is used, u 5 will remain saturated. u 5 = ( ) ( ) = < (10-30) Using c 1 = , the final controls are: u = u u 2 = { , , , , , , , , , } T (10-31) Note that all the controls in Equation are admissible. Figure 10-2 shows a set of desired moments which increase in magnitude until the maximum moment in that direction is achieved. Figure 10-3 shows the controls allocated using this method for these desired moments. Note that the controls in the second group, Figure 10-3b, are not used until the rudders in u 1 saturate. 7

8 Table 10-1: Scaling factors i u i1 u i2 u * i a i

9 % of Max Moment Cl Cm Cn Figure 10-2: Desired Moments 9

10 Radians % of Max Moment Radians u1 u6 u2 u3 a) The Primary Controls % of Max Moment u7 u8 u4 b) The Secondary Controls Figure 10-3: Commanded Controls 80 u9&u10 90 u

APPENDIX A PROOFS. Proof 1: No Generalized Inverse Will Allocate u. Denote by the subset of ( ) which maps to ( ), and by u the vectors in.

APPENDIX A PROOFS. Proof 1: No Generalized Inverse Will Allocate u. Denote by the subset of ( ) which maps to ( ), and by u the vectors in. APPENDIX A PROOFS Proof 1: No Generalized Inverse Will Allocate u m Denote by the subset of ( ) which maps to ( ), and by u the vectors in. The span( ) = R m unless there are columns of B which are all