Driving the state of a discrete system to zero in one step

Question

I have the following system of difference equations:

$\textbf{x}(k+1) = A \textbf{x}(k) + \textbf{b} u(k)$

where:

$A = \begin{bmatrix} 1 & 2 \\ 3 & \alpha \end{bmatrix} $ and, $\mathbf{b} = \begin{bmatrix} 1\\ 1 \end{bmatrix}$

.. and am asked to find the values of $\alpha$ for which the state of the system can be driven to zero in just one step if $x(0) = \begin{bmatrix} 1\\ -2 \end{bmatrix}$

I have tried substituting $x(0)$ into $x(k)$ and $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ into $x(k+1)$ to give me two equations. Whilst the first equation results in a contradiction, i.e. the $LHS \neq RHS$, the second equation gives me $\alpha = 0.5$. Is my method for determining $\alpha$ correct?

Answer 1

You have a discrete system:

$x(k+1) = A x(k) + b u(k)$

where

$A = \begin{bmatrix} 1 & 2 \\ 3 & \alpha \end{bmatrix} $ and $b = \begin{bmatrix} 1\\ 1 \end{bmatrix}$

and you want to find $u(k)$ such that $x(k + 1) = 0$ from the initial condition

$$ x(0) = \begin{bmatrix}1 \\ -2 \end{bmatrix} $$

So just insert everything:

$$ \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & \alpha \end{bmatrix} \begin{bmatrix}1 \\ -2 \end{bmatrix} + \begin{bmatrix}1 \\ 1\end{bmatrix} u(k) $$

Multiply these out you get:

$$ \begin{align} u(k) - 3 &= 0 \\ u(k) - 2 \alpha + 3 &= 0 \end{align} $$

So from the first equation we see that $u(k) = 3$. Insert that into the second equation you get:

$$ 6 - 2\alpha = 0 $$

So you can see that your problem can be solved if and only if $\alpha = 3$.