Do "lookfor parameter". Looks like something which should be done in matlab. If you can linearize that equation and change into matrix form, that could be one way.

GuitarDaku · Sep 8, 2009 12:49 PM

I think you need the Curve Fitting Toolbox, or may be try statistical model fitting in JMP or Minitab

f22 · Sep 8, 2009 1:04 PM

Aiya!!! This is a linear combination of exponential functions! I wonder if the fminsearch function will come in handy. I've never used it though. You'll have to set up your function as follows: Y = X1 (a1 * V^{n1} + a2 * V^{n2} + a3 * V^{n3}) Let P=Y./X1; So now you'll have: P=a1*V^{n1} + a2*V^{n2}+a3*V^{n3}; Beyond that, I can't say much. If you could provide the scenario in which you're trying to estimate this, perhaps it would help?

Samayayatri · Sep 8, 2009 1:35 PM

It would still be degenerate, I think. More detail might help.

samaebajii · Sep 8, 2009 2:09 PM

Yeap, I thought the same, f22. Either CFT or system identification approach might work. Is any one using Siemens PTI PSS/E tool for power systems dynamic simulation?? -- thx guys for your inputs.

Neural · Sep 8, 2009 6:44 PM

Try least square. Here is a nonlinear solverhelp lsqnonlin LSQNONLIN solves non-linear least squares problems.    LSQNONLIN attempts to solve problems of the form:    min  sum {FUN(X).^2}    where X and the values returned by FUN can be                 x             vectors or matrices.choose FUN = Y - X1 (a1 * V^{n1} + a2 * V^{n2} + a3 * V^{n3}) and put known constants Y, X1 and V. The x vectors would be [a1, a2, a3, n1, n2, n3]. It will find the least square i.e. FUN^2 for different possible of x. Minimum FUN^2 will be 0 so you will get Y = X1 (a1 * V^{n1} + a2 * V^ + {n2}a3 * V^{n3}) by changing a's and n's.

nepaliraja · Sep 9, 2009 4:57 AM

Samayayatri · Sep 9, 2009 9:12 AM

. Thx nepaliraja for the valuable info.  (sorry for the late response). I used "polyfit" to calculate the coefficients and keeping it 3rd degree polynomial. Meanwhile, this is working. Cheers ! Happy Dashain to you all Sajha friends.

Neural · Sep 18, 2009 8:07 AM