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December 10th, 2017, 01:12 AM   #1
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Joint probability of dependent variables

I have multiple-input-muliple-output (MIMO) problem formulated as

$\displaystyle T(t) = N*(W(t) - F(t) ) $
$\displaystyle F(t) = K*∑_i{ W_i(t) } $

that I need help in finding which math tools to apply. NOTE! I have very little experience with the type of statistical tools that I'm looking for, so be patient in how the problem is presented...

The problem stems from power system analysis where

$\displaystyle nT = $number of rows for $\displaystyle T$
$\displaystyle nW = $number of rows for $\displaystyle W$
$\displaystyle t_{tot} = $total time = number of columns

$\displaystyle T(t) = [nT,t_{tot}], $ tie-line power flow between two areas (i,j)
$\displaystyle N = [nT,nW], $ DC-power flow linearization matrix between areas (i,j)
$\displaystyle W(t) = [nW,t_{tot}] $ disturbance in area (i)
$\displaystyle F(t) = [nW,t_{tot}] $ primary frequency control activation in area (i)
$\displaystyle K = [nW,1] $ primary frequency control distribution among the different areas

----

$\displaystyle W$ - The known variable beforehand that have been logged is

$\displaystyle N$ - Known variable, calculated...

$\displaystyle K$ - Variable that can be chosen at free will, 0 < k_i < 1, $\displaystyle ∑ k_i = 1$

----

What I'm searching for is a way, that after the probability distribution for $\displaystyle T$ has been calculated, I want to how much $\displaystyle W$ and $\displaystyle F$ have contributed to this probability distribution.

I guess this can be done by calculating the correlation between $\displaystyle W, F$ and $\displaystyle T$.

But more specifically:
I want to be able to remove the impact of $\displaystyle W$ from the probability density of $\displaystyle T$. How do I do this?

The reason is that we want to find the impact of activation of the primary frequency control activation on tie-line power flows and not the individual netting of two disturbances on either side of a tie-line.

Last edited by skipjack; December 10th, 2017 at 03:41 AM.
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