Question

1. for the discrete-time system

\\ \mathbf{x}(k+1) = \mathbf{a}\mathbf{x}(k) + \mathbf{b}u(k) \\
where
\\ \mathbf{a} = \

$$\begin{bmatrix} 1 & -1 \\\\ 1 & 1 \\end{bmatrix}$$

, \\
let
\\ \mathbf{b} = \mathbf{b}_1 = \

$$\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}$$

\quad \text{and} \quad \mathbf{b} = \mathbf{b}_2 = \

$$\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix}$$

. \\
(i) for each \\(\mathbf{b}\\), determine if the system is completely controllable.
(ii) for each \\(\mathbf{b}\\) that results in a completely controllable system, find the shortest input sequence that drives the state to zero if
\\ \mathbf{x}(0) = \

$$\begin{bmatrix} 4 \\\\ -2 \\end{bmatrix}$$

. \\

Explanation:

🆕 New Concept Discovered: Controllability of Discrete-Time Systems
*Checking if we can steer any state to any other state*

Step 1: Define the Controllability Matrix

For a second-order discrete-time system (\(n = 2\)) given by:
\[ \mathbf{x}(k+1) = \mathbf{A}\mathbf{x}(k) + \mathbf{b}u(k) \]

The system is completely controllable if and only if the controllability matrix \(\mathbf{C}\) has full rank (rank equal to \(2\)):
\[ \mathbf{C} = \begin{bmatrix} \mathbf{b} & \mathbf{A}\mathbf{b} \end{bmatrix} \]

Given:
\[ \mathbf{A} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \]

Step 2: Test Controllability for \(\mathbf{b}_1\)

Let \(\mathbf{b} = \mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Compute \(\mathbf{A}\mathbf{b}_1\):
\[ \mathbf{A}\mathbf{b}_1 = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1(1) - 1(1) \\ 1(1) + 1(1) \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix} \]

Construct the controllability matrix \(\mathbf{C}_1\):
\[ \mathbf{C}_1 = \begin{bmatrix} \mathbf{b}_1 & \mathbf{A}\mathbf{b}_1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \]

Calculate the determinant of \(\mathbf{C}_1\):
\[ \det(\mathbf{C}_1) = (1)(2) - (0)(1) = 2 eq 0 \]

Since \(\det(\mathbf{C}_1) eq 0\), the matrix \(\mathbf{C}_1\) has full rank (\(\text{rank} = 2\)).
Thus, the system is **completely controllable** for \(\mathbf{b} = \mathbf{b}_1\).

Step 3: Test Controllability for \(\mathbf{b}_2\)

Let \(\mathbf{b} = \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

Compute \(\mathbf{A}\mathbf{b}_2\):
\[ \mathbf{A}\mathbf{b}_2 = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1(1) - 1(-1) \\ 1(1) + 1(-1) \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} \]

Construct the controllability matrix \(\mathbf{C}_2\):
\[ \mathbf{C}_2 = \begin{bmatrix} \mathbf{b}_2 & \mathbf{A}\mathbf{b}_2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} \]

Calculate the determinant of \(\mathbf{C}_2\):
\[ \det(\mathbf{C}_2) = (1)(0) - (2)(-1) = 2 eq 0 \]

Since \(\det(\mathbf{C}_2) eq 0\), the matrix \(\mathbf{C}_2\) has full rank (\(\text{rank} = 2\)).
Thus, the system is also **completely controllable** for \(\mathbf{b} = \mathbf{b}_2\).

Step 4: Find the Shortest Input Sequence to Drive State to Zero

We want to find the shortest sequence of inputs \(u(0), u(1), \dots, u(N-1)\) that drives the initial state \(\mathbf{x}(0) = \begin{bmatrix} 4 \\ -2 \end{bmatrix}\) to the origin \(\mathbf{x}(N) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\).

The state equation propagates as:
\[ \mathbf{x}(1) = \mathbf{A}\mathbf{x}(0) + \mathbf{b}u(0) \]
\[ \mathbf{x}(2) = \mathbf{A}\mathbf{x}(1) + \mathbf{b}u(1) = \mathbf{A}^2\mathbf{x}(0) + \mathbf{A}\mathbf{b}u(0) + \mathbf{b}u(1) \]

Setting \(\mathbf{x}(2) = \mathbf{0}\):
\[ \mathbf{0} = \mathbf{A}^2\mathbf{x}(0) + \begin{bmatrix} \mathbf{A}\mathbf{b} & \mathbf{b} \end{bmatrix} \begin{bmatrix} u(0) \\ u(1) \end{bmatrix} \]
\[ \begin{bmatrix} \mathbf{b} & \mathbf{A}\mathbf{b} \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = -\mathbf{A}^2\mathbf{x}(0) \]
\[ \mathbf{C} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = -\mathbf{A}^2\mathbf{x}(0) \]

Since the system is controllable in 2 steps, the shortest sequence length is at most \(N = 2\). Let's check if it can be done in \(N = 1\) step:
For \(N = 1\), \(\mathbf{x}(1) = \mathbf{A}\mathbf{x}(0) + \mathbf{b}u(0) = \mathbf{0} \implies \mathbf{b}u(0) = -\mathbf{A}\mathbf{x}(0)\). This requires \(-\mathbf{A}\mathbf{x}(0)\) to be linearly dependent on \(\mathbf{b}\). Let's compute this vector:
\[ -\mathbf{A}\mathbf{x}(0) = -\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \end{bmatrix} = -\begin{bmatrix} 6 \\ 2 \end{bmatrix} = \begin{bmatrix} -6 \\ -2 \end{bmatrix} \]

Since \(\begin{bmatrix} -6 \\ -2 \end{bmatrix}\) is not a scalar multiple of either \(\mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) or \(\mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\), it cannot be solved in \(1\) step. Thus, the shortest sequence length is \(N = 2\).

First, compute the target vector \(-\mathbf{A}^2\mathbf{x}(0)\):
\[ \mathbf{A}^2 = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \]
\[ -\mathbf{A}^2\mathbf{x}(0) = -\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \end{bmatrix} = -\begin{bmatrix} 4 \\ 8 \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

Case A: For \(\mathbf{b} = \mathbf{b}_1\)

The system of equations is:
\[ \mathbf{C}_1 \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

From the first row:
\[ u(1) = -4 \]

From the second row:
\[ u(1) + 2u(0) = -8 \implies -4 + 2u(0) = -8 \implies 2u(0) = -4 \implies u(0) = -2 \]

So, the input sequence is:
\[ u(0) = -2, \quad u(1) = -4 \]

---

Case B: For \(\mathbf{b} = \mathbf{b}_2\)

The system of equations is:
\[ \mathbf{C}_2 \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

From the second row:
\[ -u(1) = -8 \implies u(1) = 8 \]

From the first row:
\[ u(1) + 2u(0) = -4 \implies 8 + 2u(0) = -4 \implies 2u(0) = -12 \implies u(0) = -6 \]

So, the input sequence is:
\[ u(0) = -6, \quad u(1) = 8 \]

Answer:

(i) Controllability

• For \(\mathbf{b} = \mathbf{b}_1 =

$$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

\): Completely controllable (since \(\det(\mathbf{C}_1) = 2
eq 0\)).

• For \(\mathbf{b} = \mathbf{b}_2 =

$$\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

\): Completely controllable (since \(\det(\mathbf{C}_2) = 2
eq 0\)).

(ii) Shortest Input Sequence to Drive State to Zero

The shortest sequence length is \(N = 2\) steps.

• For \(\mathbf{b}_1\):

$$ u(0) = -2, \quad u(1) = -4 $$

• For \(\mathbf{b}_2\):

$$ u(0) = -6, \quad u(1) = 8 $$