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<math>\color{blue}\text{Solution 1:}</math> | <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> | ||
| + | <math>\text{The problem is equivalent to minimize }</math> | ||
| − | + | <br> | |
---- | ---- | ||
| − | <math>\color{blue}\text{Solution 2:}</math> | + | <math>\color{blue}\text{Solution 2:}</math> |
<math>x\left ( 1 \right )=\mu\left ( 0 \right )</math><br> | <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 ( 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>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{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{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>\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} | <math>\left\{\begin{matrix} | ||
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\mu\left(2 \right)=\frac{1}{3}\\ | \mu\left(2 \right)=\frac{1}{3}\\ | ||
\lambda=-\frac{1}{3} | \lambda=-\frac{1}{3} | ||
| − | \end{matrix}\right.</math> | + | \end{matrix}\right.</math> |
<math>\text{Check SOSC: } L\left( \mu,\lambda \right)=D^{2}l\left( \mu,\lambda \right)=\begin{bmatrix} | <math>\text{Check SOSC: } L\left( \mu,\lambda \right)=D^{2}l\left( \mu,\lambda \right)=\begin{bmatrix} | ||
| Line 55: | Line 67: | ||
0 & 1 & 0\\ | 0 & 1 & 0\\ | ||
0 & 0 & 1 | 0 & 0 & 1 | ||
| − | \end{bmatrix}>0</math> | + | \end{bmatrix}>0</math> |
| − | <math>\therefore \text{For all y, } y^{T}Ly\geq 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> | + | <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> |
<br> | <br> | ||
Revision as of 17:17, 27 June 2012
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 } $
$ \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.} $
