View source for ECE-QE AC3-2011 solusion-4 - Rhea

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= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]]: Automatic Control (AC)- Question 3, August 2011  =

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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{4. } \left( \text{20 pts} \right) \text{ Consider the following model of a discrete-time system,  }</math></span></font> 

<math>x\left ( k+1 \right )=2x\left ( k \right )+u\left ( k \right ), x\left ( 0 \right )=0, 0\leq k\leq 2</math><br> 

<math>\color{blue}\text{Use the Lagrange multiplier approach to calculate the optimal control sequence}</math><br> 

<math>\left \{ u\left ( 0 \right ),u\left ( 1 \right ), u\left ( 2 \right ) \right \}</math> 

<math>\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}</math><br> <math>J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2}</math><br> 

<math>\color{blue}\text{Solution 1:}</math> 

<math>\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.</math><br> 

<font face="serif"><math>\text{The problem is equivalent to minimize } J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2}</math><br></font>

<font face="serif"></font><math>\text{subject to } 4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)=7</math>

<math>\text{Let } h(\mu )=4\mu \left(0 \right)+2\mu \left(1 \right)+\mu\left(2 \right)-7</math><br> 

<math>\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.</math>



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<math>\color{blue}\text{Solution 2:}</math> 

<math>x\left ( 1 \right )=\mu\left ( 0 \right )</math><br> 

<math>x\left ( 2 \right )=2\mu\left ( 0 \right )+\mu\left ( 1 \right )</math> 

<math>x\left ( 3 \right )=4\mu\left ( 0 \right )+2\mu\left ( 1\right )+\mu\left ( 2 \right )=7</math> 

<math>\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}</math> 

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

<math>\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</math> 

<math>\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.</math> 

<math>\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</math> 

<math>\therefore \text{For all y, } y^{T}Ly\geq 0</math> 

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

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