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I have widely seen this format for cost

$$ J(\boldsymbol{x},\boldsymbol{u}) = \boldsymbol{x}_f^T\boldsymbol{H}\boldsymbol{x}_f + \int_{t_0}^{t_f}{[\boldsymbol{x}^T\boldsymbol{Q}\boldsymbol{x} + \boldsymbol{u}^T\boldsymbol{R}\boldsymbol{u}]}dt $$

But I also saw this on Wikipedia

$$ J(\boldsymbol{x},\boldsymbol{u}) = \boldsymbol{x}_f^T\boldsymbol{H}\boldsymbol{x}_f + \int_{t_0}^{t_f}{[\boldsymbol{x}^T\boldsymbol{Q}\boldsymbol{x} + \boldsymbol{u}^T\boldsymbol{R}\boldsymbol{u} + 2\boldsymbol{x}^T\boldsymbol{N}\boldsymbol{u}]}dt $$

I really think that this $2\boldsymbol{x}^T\boldsymbol{N}\boldsymbol{u}$ isn't right...

Is it?

griloHBG
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  • It would be helpful if you define everything, especially the new and unfamiliar term you're interested in. In my (limited) experience, cost is a very malleable thing. You can define it so that it suites cost as you see it and interpret it, because that's the whole point. – The Count May 05 '17 at 01:10
  • Yes. Indeed the cost is really malleable. But I'm interested in the theoretical part. Not in the application of LQR itself. For the solution, the cost should become a Riccati equation to be solved and obtain the control law. The crossed term $2x^TNu$ allows it to happen? – griloHBG May 08 '17 at 15:32
  • @HenriqueGarcia According to https://www.mathworks.com/help/control/ref/lqr.html and https://en.wikipedia.org/wiki/Linear%E2%80%93quadratic_regulator the matrix $N$ goes into Algebraic Riccati's Equation like this: $A^{T}P+PA-\left(PB+N\right)R^{-1}\left(B^{T}P+N^{T}\right)+Q=0$ However none of them describes the meaning of that matrix. Maybe Kirk's Optimal Control has something bout it! – bertozzijr May 29 '17 at 13:55
  • Thank you, @bertozzijr. I understood that the N is a weight over the crossed terms of states and actuations. Seems like both forms are useful. The difference is that in the first one, the weight for crossed terms is null. – griloHBG May 30 '17 at 16:04

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