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Q 18SE

Expert-verified
Found in: Page 191

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question 18: Suppose A is a $$4 \times 4$$ matrix and B is a $$4 \times 2$$ matrix, and let $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ represent a sequence of input vectors in $${\mathbb{R}^2}$$.Set $${{\mathop{\rm x}\nolimits} _0} = 0$$, compute $${{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}$$ from equation (1), and write a formula for $${{\mathop{\rm x}\nolimits} _4}$$ involving the controllability matrix $$M$$ appearing in equation (2). (Note: The matrix $$M$$ is constructed as a partitioned matrix. Its overall size here is $$4 \times 8$$.)Suppose $$\left( {A,B} \right)$$ is controllable and v is any vector in $${\mathbb{R}^4}$$. Explain why there exists a control sequence $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ in $${\mathbb{R}^2}$$ such that $${{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits}$$.

a. The formula for $${{\mathop{\rm x}\nolimits} _4}$$ is $${{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits}$$.

b. The control sequence $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ makes $${{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits}$$.

See the step by step solution

## Step 1: State the difference equation in exercise 17

A state-space model of a control system includes a difference equation of the form

$${{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}$$ for $$k = 0,1,...$$ …(1)

The pair $$\left( {A,B} \right)$$ is said to be controllable if rank$$\left[ {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right] = n$$.

## Step 2: Write the formula for $${{\mathop{\rm x}\nolimits} _4}$$ involving the controllability matrix

a)

Consider $$A$$ as a $$m \times n$$ matrix with rank$$r$$ and $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ as a sequence of input vectors in $${\mathbb{R}^2}$$.

Use the equation $${{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}$$ for $$k = 0,1,...$$ with $${{\mathop{\rm x}\nolimits} _0} = 0$$ as shown below:

$$\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = A{{\mathop{\rm x}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _0}\\ = B{{\mathop{\rm u}\nolimits} _0}\\{{\mathop{\rm x}\nolimits} _2} = A{{\mathop{\rm x}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _1}\\ = AB{{\mathop{\rm u}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _1}\end{array}$$

$$\begin{array}{c}{{\mathop{\rm x}\nolimits} _3} = A{{\mathop{\rm x}\nolimits} _2} + B{{\mathop{\rm u}\nolimits} _2}\\ = A\left( {AB{{\mathop{\rm u}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _1}} \right) + B{{\mathop{\rm u}\nolimits} _2}\\ = {A^2}B{{\mathop{\rm u}\nolimits} _0} + AB{{\mathop{\rm u}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _2}\end{array}$$

$$\begin{array}{c}{{\mathop{\rm x}\nolimits} _4} = A{{\mathop{\rm x}\nolimits} _3} + B{{\mathop{\rm u}\nolimits} _3}\\ = A\left( {{A^2}B{{\mathop{\rm u}\nolimits} _0} + AB{{\mathop{\rm u}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _2}} \right) + B{{\mathop{\rm u}\nolimits} _3}\\ = {A^3}B{{\mathop{\rm u}\nolimits} _0} + {A^2}B{{\mathop{\rm u}\nolimits} _1} + AB{{\mathop{\rm u}\nolimits} _2} + B{{\mathop{\rm u}\nolimits} _3}\\ = \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}&{{A^3}B}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _3}}\\{{{\mathop{\rm u}\nolimits} _2}}\\{{{\mathop{\rm u}\nolimits} _1}}\\{{{\mathop{\rm u}\nolimits} _0}}\end{array}} \right)\\ = M{\mathop{\rm u}\nolimits} \end{array}$$

Matrix $$M$$ has four rows since $$B$$ does, and $$M$$ has eight columns since $$B$$ and each of the matrices $${A^k}B$$ has two columns. Vector $${\mathop{\rm u}\nolimits}$$ in the final equation is in $${\mathbb{R}^8}$$ since each $${{\mathop{\rm u}\nolimits} _k}$$ is in $${\mathbb{R}^2}$$.

## Step 3: Explain why there exists a control sequence $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ in $${\mathbb{R}^2}$$

b)

If the matrix pair $$\left( {A,B} \right)$$ is controllable, then the controllability matrix has rank 4, with a pivot in every row, and the columns of $$M$$ span $${\mathbb{R}^4}$$. Therefore, for any vector $${\mathop{\rm v}\nolimits}$$ in $${\mathbb{R}^4}$$, there exists a vector u in $${\mathbb{R}^8}$$ such that $${\mathop{\rm v}\nolimits} = M{\mathop{\rm u}\nolimits}$$.

According to part (a), $${{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits}$$ when $${\mathop{\rm u}\nolimits}$$ is partitioned into a control sequence $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$.

Thus, the control sequence $${{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}$$ makes $${{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits}$$.