Relative Gain Array(RGA) for Laplacian matrix ( which have zero eigenvalues )

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Namjin Park
Namjin Park 2019 年 8 月 23 日
コメント済み: Namjin Park 2019 年 8 月 28 日
Thanks to seeing my question,
I have problem to get Relative Gain Array (RGA) matrix for Laplacian matrix
Laplacian matrix contion : eigenvalue <= 0 alway has 0 eigenvalue,
I already get RGA matrix under the condition of matrix which is Positive definite, (eigenvalues >0 case - Not Laplacian)
RGA = G(0) * trans( G(0)^(-1) ).
when there is zero eigenvalue in transfer functuin G,
I can't calculate G(0), that goes to infinie,
So how can i get RGA when there is zero eigenvalue in Transfer function ???
Thanks,

採用された回答

Shashank Gupta
Shashank Gupta 2019 年 8 月 28 日
Generally, RGA is calculated for non-singular matrices but in some specific cases similar idea can be expanded to singular matrices using the concept of Moore-Penrose pseudoinverse, that can be used to approximate for some properties of inverse.
MATLAB has a function to calculate Moore Pseudoinverse “pinv” but make sure to cross check the pseudo inverse, does it satisfy all RGA properties. Many times, it doesn’t satisfy one essential property of RGA – sum of rows is unity, but still worth giving a try.

その他の回答 (1 件)

Namjin Park
Namjin Park 2019 年 8 月 28 日
Thanks for reply Gupta!
I understood your explanation,
So Laplacian matrix is alway singular, that it doesn't have inverse matrix,
So i used 'pinv' what you said,
clear all;
clc;
L = 1*[-28 11 12 0 0 0 5; 11 -14.5 1 0 0 0 2.5; 12 1 -63 13 0 12 25; 0 0 13 -24 1 0 10; 0 0 0 1 -17.5 11.5 5; 0 0 12 0 11.5 -23.5 0; 5 2.5 25 10 5 0 -47.5];
Co = diag([1 1 1 1 1 1 1]);
Q_in = [1 1 1 1 1 1 1]';
A = Co\L;
B = diag(Co\Q_in);
C = diag([1 1 1 1 1 1 1]);
D = diag([0 0 0 0 0 0 0]);
sys_mimo = ss(A,B,C,D);
G = tf(sys_mimo)
%G_0 = evalfr(G,0);
G_0 = dcgain(G);
inv_G = pinv(G_0);
tr_G = transpose(inv_G);
%
rga_G = G_0.*tr_G
when i calculate rga,
rga_G =
1.0e+13 *
0.0737 -1.6432 0.3501 0.5031 0.2864 0.1881 0.2418
-0.1810 4.0352 -0.8598 -1.2354 -0.7032 -0.4619 -0.5939
0.0062 -0.1373 0.0293 0.0420 0.0239 0.0157 0.0202
0.0384 -0.8562 0.1824 0.2621 0.1492 0.0980 0.1260
0.0280 -0.6250 0.1332 0.1913 0.1089 0.0715 0.0920
0.0169 -0.3775 0.0804 0.1156 0.0658 0.0432 0.0556
0.0178 -0.3960 0.0844 0.1212 0.0690 0.0453 0.0583
It doesn't satisfy row sum 1, as you said, and all elements are too big,
So what i mean is, is this rga matrix is valid ??
thanks to reading,
Namjin,
  2 件のコメント
Shashank Gupta
Shashank Gupta 2019 年 8 月 28 日
Conventionally we use Moore-Penrose psuedoinverse but as you mentioned sometimes it fails to preserve critical propertiy of RGA(which is row sum property in your case). In such scenerio you can either say the rga_G matrix which you got can act as a approximate RGA or if you want more precise matrix then there are some recent work, which has been done on the RGA for singular and rectangular metrices, you can refer to this link for more information.
I hope it helps
Namjin Park
Namjin Park 2019 年 8 月 28 日
I'm so appriciate you,
Finally I find it !
Thanks a lots,
god bless you !

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