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I am trying to linearise the following expresssion.

$C(k) = B(k) e^{-d(k)}, B(k) \ge 0 , d(k) \ge 0 $

I am trying to do this by using SOS2 sets.

I set $X(k) = e^{-d(k)}$ and I get $C(k) = B(k) X(k)$

By setting

$ u_{1}(k) = 0.5 (B(k) + X(k)) \\ u_{2}(k) = 0.5 (B(k) - X(k)) $

I get

$C(k) = u_{1}(k)^2 - u_{2}(k)^2 \\ u_{1}(k) = 0.5 (B(k) + X(k)) \\ u_{2}(k) = 0.5 (B(k) - X(k)) \\ X(k) = e^{-d(k)} $

Setting $y_{1}(k) = u_{1}^2, y_{2}(k) = u_{2}^2, step = 0.1$ I can linearise for $N$ points as follows:

$ y_{1}(k) = \sum_{i=1}^N (step (i-1))^2 \lambda_{k}^{u1}(i) \\ u_{1}(k) = \sum_{i=1}^N step (i-1) \lambda_{k}^{u1}(i)\\ \sum_{i=1}^N \lambda_{k}^{u1}(i) = 1.0 $

$ y_{2}(k) = \sum_{i=1}^N (step (i-1))^2 \lambda_{k}^{u2}(i)\\ u_{2}(k) = \sum_{i=1}^N step (i-1) \lambda_{k}^{u2}(i) \\ \sum_{i=1}^N \lambda_{k}^{u2}(i) = 1.0 $

$X(k) = e^{-d(k)}$ is linearised as follows:

$ X(k) = \sum_{i=1}^N e^{step (i-1)} \lambda_{k}^{d}(i) \\ d(k) = \sum_{i=1}^N step (i-1) \lambda_{k}^{d}(i) \\ \sum_{i=1}^N \lambda_{k}^{d}(i) = 1.0 $

So finally, the model becomes linearised to:

$ C(k) = y_{1}(k) - y_{2}(k) \\ y_{1}(k) = \sum_{i=1}^N (step (i-1))^2 \lambda_{k}^{u1}(i) \\ u_{1}(k) = \sum_{i=1}^N step (i-1) \lambda_{k}^{u1}(i)\\ \sum_{i=1}^N \lambda_{k}^{u1}(i)) = 1.0 \\ y_{2}(k) = \sum_{i=1}^N (step (i-1))^2 \lambda_{k}^{u2}(i)\\ u_{2}(k) = \sum_{i=1}^N step (i-1) \lambda_{k}^{u2}(i) \\ \sum_{i=1}^N \lambda_{k}^{u2}(i)) = 1.0 \\ u_{1}(k) = 0.5 (B(k) + X(k)) \\ u_{2}(k) = 0.5 (B(k) - X(k)) \\ X(k) = \sum_{i=1}^N e^{step (i-1)} \lambda_{k}^{d}(i) \\ d(k) = \sum_{i=1}^N (step (i-1)) \lambda_{k}^{d}(i) \\ \sum_{i=1}^N \lambda_{k}^{d}(i) = 1.0 $

My question is,

1) Is what I have done correct?

2) Do I have to distinguish between different lamdas or do they have to be the same? ($\lambda_{k}^{u2}(i)$ the same as $\lambda_{k}^{u1}(i)$ the same as $\lambda_{k}^{d}(i)$)

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