The problem is that i am getting negative values of kp and ki, how i can get value in specific range?

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zain yousaf 2017-7-9

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评论： zain yousaf 2017-7-12

PSO.m

    function F = tracklsq(pid)
         syms x;
         limit_1 = 0;
         limit_2 = 10;
         Area = int(x,limit_1, limit_2);
         Kp = pid(1);
         Ki =  pid(2);
         z=Kp*0.1+Ki*0.1*Area; %objective function
         sys_overshoot=z-1; % compute the overshoot
         alpha=10;beta=10;
         F=0.1*2*beta+sys_overshoot*alpha;
    end

PSO code is attached

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Answer 1

Walter Roberson 2017-7-10

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Your pso code has

current_position = 10*(rand(dim, n)-.5);

rand() is going to be in the range 0 to 1, so rand() - 0.5 is going to be in the range -0.5 to +0.5; 10 times that would be -5 to +5 .

If you do not want negative inputs, then do not subtract .5 in this statement.

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Walter Roberson 2017-7-11

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In your update loop you have

velocity = w *velocity + c1*(R1.*(local_best_position-current_position)) + c2*(R2.*(globl_best_position-current_position));

Suppose that current_position is a fair bit more positive than local_best_position or globl_best_position, then (local_best_position-current_position) or (globl_best_position-current_position) could get fairly negative. The R2 values are rand 0 to 1, but c2 is 1.2, so the last part of the calculation could be negative up to 1.2 times distance the current position is relative to the global best, and the c1*(R1.*(local_best_position-current_position)) could be contributing even more negative. The previous velocity could have been negative as well. Put these three negative components together and you can easily get a velocity that is more negative than the current position is positive, leading to a negative after the velocity is added to the current position.

Within tracklsq you calculate Area as a symbolic integral between fixed limits. The result would be a constant (in particular, 50), but in symbolic form. That forces other calculations into symbolic form, which is leading to a lot of performance loss. It is not obvious why you use a constant calculation for Area there.

At the end of your program you have

            xx=fitness(:,n);
            [Y,I] = min(xx);
            current_position(:,I)

However, you never updated fitness and so the array is all 0. You have already calculated the best value: it is in global_best_fitness, and the corresponding point is any entry in globl_best_position (all of the entries are the same for that array, by design.)

Walter Roberson 2017-7-12

? How do you get from those minimizations in the paper to your u expression ??

zain yousaf 2017-7-12

i am choosing some random values of error, in the case of I-Iref, i set that equal to 0.1,and interms of integration, i am setting some limits and using integrating function to integrate t, in my code area=t. i hope u get my point. in this paper there is no transfer function, so i am not sure that should i use transfer function or not.

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