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 using inputs.

Step 1: Define the Controllability Matrix

For a second-order discrete-time system \( \mathbf{x}(k+1) = \mathbf{A}\mathbf{x}(k) + \mathbf{b}u(k) \), the system is completely controllable if and only if the controllability matrix \( \mathcal{C} \) has full rank (rank equal to the dimension of the state space, which is 2).

The controllability matrix is defined as:
\[ \mathcal{C} = \begin{bmatrix} \mathbf{b} & \mathbf{A}\mathbf{b} \end{bmatrix} \]

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

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

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

First, 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 \( \mathcal{C}_1 \):
\[ \mathcal{C}_1 = \begin{bmatrix} \mathbf{b}_1 & \mathbf{A}\mathbf{b}_1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \]

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

Since the determinant is non-zero, \( \mathcal{C}_1 \) has full rank (rank = 2).
Thus, the system is **completely controllable** for \( \mathbf{b}_1 \).

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

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

First, 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 \( \mathcal{C}_2 \):
\[ \mathcal{C}_2 = \begin{bmatrix} \mathbf{b}_2 & \mathbf{A}\mathbf{b}_2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} \]

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

Since the determinant is non-zero, \( \mathcal{C}_2 \) has full rank (rank = 2).
Thus, the system is **completely controllable** for \( \mathbf{b}_2 \).

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

We want to find the shortest sequence of inputs \( u(0), u(1), \dots, u(N-1) \) such that the state at step \( N \) is zero:
\[ \mathbf{x}(N) = \mathbf{0} \]

The state equation propagated over \( N \) steps is:
\[ \mathbf{x}(N) = \mathbf{A}^N \mathbf{x}(0) + \sum_{i=0}^{N-1} \mathbf{A}^{N-1-i} \mathbf{b} u(i) \]

Since the system is controllable and of dimension 2, any state can be driven to the origin in at most \( N = 2 \) steps. Let's check if it can be done in \( N = 1 \) step first.

Case 1: For \( N = 1 \) step

\[ \mathbf{x}(1) = \mathbf{A}\mathbf{x}(0) + \mathbf{b}u(0) = \mathbf{0} \]
\[ \mathbf{b}u(0) = -\mathbf{A}\mathbf{x}(0) \]

Let's compute \( -\mathbf{A}\mathbf{x}(0) \):
\[ -\mathbf{A}\mathbf{x}(0) = -\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \end{bmatrix} = -\begin{bmatrix} 4 + 2 \\ 4 - 2 \end{bmatrix} = \begin{bmatrix} -6 \\ -2 \end{bmatrix} \]

For \( \mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \):
\[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} u(0) = \begin{bmatrix} -6 \\ -2 \end{bmatrix} \]
There is no scalar \( u(0) \) that satisfies this.

For \( \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \):
\[ \begin{bmatrix} 1 \\ -1 \end{bmatrix} u(0) = \begin{bmatrix} -6 \\ -2 \end{bmatrix} \]
There is no scalar \( u(0) \) that satisfies this.

Thus, the shortest input sequence length is \( N = 2 \).

Case 2: For \( N = 2 \) steps

The state equation at \( N = 2 \) is:
\[ \mathbf{x}(2) = \mathbf{A}^2 \mathbf{x}(0) + \mathbf{A}\mathbf{b}u(0) + \mathbf{b}u(1) = \mathbf{0} \]
\[ \begin{bmatrix} \mathbf{b} & \mathbf{A}\mathbf{b} \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = -\mathbf{A}^2 \mathbf{x}(0) \]

Let's compute \( -\mathbf{A}^2 \mathbf{x}(0) \):
\[ \mathbf{A}^2 = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \]
\[ -\mathbf{A}^2 \mathbf{x}(0) = -\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \end{bmatrix} = -\begin{bmatrix} 4 \\ 8 \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

Now we solve the linear system for each \( \mathbf{b} \):

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

\[ \mathcal{C}_1 \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

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 shortest input sequence for \( \mathbf{b}_1 \) is:
\[ u(0) = -2, \quad u(1) = -4 \]

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

\[ \mathcal{C}_2 \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} u(1) \\ u(0) \end{bmatrix} = \begin{bmatrix} -4 \\ -8 \end{bmatrix} \]

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 shortest input sequence for \( \mathbf{b}_2 \) is:
\[ u(0) = -6, \quad u(1) = 8 \]

Answer:

(i) Controllability Analysis

• For \( \mathbf{b}_1 \): The system is completely controllable because \( \det(\mathcal{C}_1) = 2

eq 0 \).

• For \( \mathbf{b}_2 \): The system is completely controllable because \( \det(\mathcal{C}_2) = 2

eq 0 \).

---

(ii) Shortest Input Sequence to Drive State to Zero

The shortest sequence length is \( N = 2 \) steps for both cases:

• For \( \mathbf{b}_1 \):

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

• For \( \mathbf{b}_2 \):

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