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ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011

 $ \color{blue}\text{4. } \left( \text{20 pts} \right) \text{ Consider the following model of a discrete-time system, } $

$ x\left ( k+1 \right )=2x\left ( k \right )+u\left ( k \right ), x\left ( 0 \right )=0, 0\leq k\leq 2 $

$ \color{blue}\text{Use the Lagrange multiplier approach to calculate the optimal control sequence} $

$ \left \{ u\left ( 0 \right ),u\left ( 1 \right ), u\left ( 2 \right ) \right \} $

$ \color{blue}\text{that transfers the initial state } x\left( 0 \right) \text{ to } x\left( 3 \right)=7 \text{ while minimizing the performance index} $
$ J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2} $

$ \color{blue}\text{Solution 1:} $

$ \left.\begin{matrix} x\left ( 1 \right )=2x\left ( 0 \right )+\mu\left ( 0\right )\\ x\left ( 2 \right )=2x\left ( 1 \right )+\mu\left ( 1\right )\\ x\left ( 3 \right )=2x\left ( 2 \right )+\mu\left ( 2\right )\\ x\left ( 0 \right )=0 \end{matrix}\right\} \Rightarrow \left\{\begin{matrix} x\left ( 1 \right )=\mu\left ( 0 \right )\\ x\left ( 2 \right )=2\mu\left ( 0 \right )+\mu\left ( 1\right )\\ x\left ( 3 \right )=4\mu\left ( 0 \right )+2\mu\left ( 1\right )+\mu\left ( 2 \right )=7 \end{matrix}\right. $

$ \text{The problem is equivalent to minimize } J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2} $

$ \text{subject to } 4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)=7 $

$ \text{Let } h(\mu )=4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)-7 $

$ \text{FONC: } \left\{\begin{matrix} l\left(\mu,\lambda \right)=\nabla J\left ( \mu \right )+\lambda\nabla h\left( \mu \right)=\begin{pmatrix} \mu\left ( 0 \right )\\ \mu\left ( 1 \right )\\ \mu\left ( 2 \right ) \end{pmatrix}+\lambda\begin{pmatrix} 4\\ 2\\ 1 \end{pmatrix} =0\\ h(\mu )=4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)-7=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} \mu\left(0 \right)=\frac{4}{3}\\ \mu\left(1 \right)=\frac{2}{3}\\ \mu\left(2 \right)=\frac{1}{3}\\ \lambda=-\frac{1}{3} \end{matrix}\right. $

$ \text{SOSC: } L\left( \mu,\lambda \right)=\nabla l\left( \mu,\lambda \right)=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}>0 $

$ \Rightarrow \text{The sequence } \mu\left( 0 \right)=\frac{4}{3},\mu\left( 1 \right)=\frac{2}{3},\mu\left( 2 \right)=\frac{1}{3} \text{ satisfies SOSC. It's the optimal sequence.} $

$ \color{blue}\text{Solution 2:} $

$ x\left ( 1 \right )=\mu\left ( 0 \right ) $

$ x\left ( 2 \right )=2\mu\left ( 0 \right )+\mu\left ( 1 \right ) $

$ x\left ( 3 \right )=4\mu\left ( 0 \right )+2\mu\left ( 1\right )+\mu\left ( 2 \right )=7 $

$ \text{The problem transfer to min } J\left ( \mu \right )=\frac{1}{2} \mu \left ( 0 \right )^{2}+\frac{1}{2} \mu \left ( 1 \right )^{2}+\frac{1}{2} \mu \left ( 2 \right )^{2} $

$ \text{subject to } h(\mu )=4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)-7=0 $

$ \text{Apply KKT condition: } Dl\left( \mu ,\lambda \right)=DJ\left(\mu \right)+\lambda Dh\left(\mu \right)=\left[ \mu\left(0 \right)+4\lambda,\mu\left(1 \right)+2\lambda,\mu\left(2 \right)+\lambda \right]=0 $

$ \left\{\begin{matrix} \mu\left(0 \right)+4\lambda=0\\ \mu\left(1 \right)+2\lambda=0\\ \mu\left(2 \right)+\lambda=0\\ 4\mu\left(0 \right)+2\mu\left(1 \right)+\mu\left(2 \right)-7=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} \mu\left(0 \right)=\frac{4}{3}\\ \mu\left(1 \right)=\frac{2}{3}\\ \mu\left(2 \right)=\frac{1}{3}\\ \lambda=-\frac{1}{3} \end{matrix}\right. $

$ \text{Check SOSC: } L\left( \mu,\lambda \right)=D^{2}l\left( \mu,\lambda \right)=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}>0 $

$ \therefore \text{For all y, } y^{T}Ly\geq 0 $

$ \therefore \text{sequence } \left\{ \frac{4}{3},\frac{2}{3},\frac{1}{3} \right\} \text{ satisfy SOSC is a strict minimizer of the problem.} $

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