Question

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 \({\mathbb{R}^2}\).

1. 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 \(4 \times 8\).)
2. 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 \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Short answer

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 \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \).

Step-by-step solution

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 becontrollableif rank\(\left[ {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right] = n\).

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 \({\mathbb{R}^2}\).

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 \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \).