# Linearisation using SOS2

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 $$\lambda_{k}^{d}(i)$$)