Re: Book & RGA

Dear Mr. Cameron,

Thank you for your email.

>Do you think that the steps 1,2 and 3 of 10.8.5 solve
>the problems found by Huang et al?

Yes, I think so, but you must remember to look at the dynamic RGA.
To get an idea of the RGA at high frequency, add together
all the time constants and the delay for each element into Tij, then
the element at high frequency is approximately kij/Tij.
You can then compute the RGA

For example, consider Example 3 in Huang:
G is a 3x3 matrix with elements gij whith
(gain kij, time constant tauij, and delay thetaij).

So we have:
   [ (-1.986, 66.67, 0.71) (5.24,400,60) (5.984,14.29,2.24) ]
G= [ (0.0204, 5, -4.199) (-0.33, 3.904, 1.883) (2.38, 10, 1.143) ]
   [ (0.374, 22.22, 7.75) (-11.3, 35.66, 14.78) (-9.881, 11.35, 1.59) ]

and the static RGA is

>> G = [-1.986 5.24 5.984; 0.0204 -0.33 2.38; 0.374 -11.3 -9.881]
>> vrga(G)
    1.0927 -0.1044 0.0117
    0.0059 0.1047 0.8894
   -0.0986 0.9997 0.0989

At high frequency use the approximation g11 = -1.986/(66.67 + 0.71) =
-0.0294 etc. to get
>> Ghigh = [-0.0294 0.0114 0.362; 0.0022 -0.057 0.2136; 0.0125 .224
.7636]
>> vrga(Ghigh)
   0.8572 0.0036 0.1392
   0.0508 0.4906 0.4586
   0.0920 0.5058 0.4022
At high frequencies the pairing 1-1/2-3/3-2/ is the one with
RGA-elements closest to 1. Indeed, this is also the best pairing
at low frequency. This confirms the more detailed analysis
in Huang et al.

Huang uses a much more detailed analysis where also controller
design is included, so their results should be more accurate
(but a lot more time consuming). I say "should" because the problem is that
during implementation one may not use the controllers which
were used in the analysis, and then the conclusion may not hold.
Therefore, I prefer controller independent tools.

Huang Example 1. Process 1 and 2. The choice of the offdiagonal
pairing, chosen by Huang, would also result from the dynamic RGA
(you should look at the RGA at the frequency-range corresponding
to the closed-loop time constant, e.g. if it is 5 minutes, then
look around the frequency 1/5 = 0.2 rad/min.
The static RGA has postive elements for all pairings so it cannot
be used to eliminate any pairings).

Example 1. process 3. Here one offdiagonal element has a larger time delay
so it is not so clear what the best pairing is.

Example 2 (Process 4). There must be some misprint in the paper as
the given static RGA does not agree with the model.

Yours sincerely,

Sigurd Skogestad
Received on Thu Mar 6 10:14:25 1997