Let's analyze the **controllability** and **observability** of the given LTI system. Given: \[ \dot{x} = Ax + Bu \] where: \[ A = \begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \] \[ B = \begin{bmatrix} 2 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] \[ C = \begin{bmatrix} 2 & 2 & 1 & 3 & -1 & 1 & 1 \\ 1 & 1 & 1 & 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 0 \end{bmatrix} \] --- ## 1. **Controllability** The **controllability matrix** is: \[ \mathcal{C} = [B, AB, A^2B, \ldots, A^{n-1}B] \] where \( n \) is the order of the system (here, \( n = 7 \)). ### Step 1: Compute \( AB \), \( A^2B \), ..., \( A^{6}B \) But before calculation, let's note the structure of \( A \): - The first 3 rows/columns are a block upper triangular matrix (the rest are zeros). - Rows 4 and 5 are zeros (except possibly for B). - Rows 6 and 7 are lower diagonal 2x2 identity. ### Step 2: Structure Analysis - The first three states form an independent subsystem (see the block at top left). - The last two states (x6, x7) are a separate 2x2 system with eigenvalue 1. - x4 and x5 are pure integrators (since their rows are all zeros). Let's see if the system is controllable by checking the **rank of the controllability matrix**. #### **Controllability for x4 and x5:** Since the 4th and 5th rows of \( A \) are all zeros, the only way to affect \( x_4 \) and \( x_5 \) is through \( B \): Check the 4th and 5th rows of \( B \): - Row 4 of B: [-1, 0, 1] - Row 5 of B: [1, 0, 1] So both have non-zero entries. Thus, you can directly control x4 and x5. #### **Controllability for x6 and x7:** - The subsystem: \[ \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \] is controllable **if and only if** the corresponding elements in B (i.e., rows 6 and 7) span the space. Row 6 of B: [1, 0, 0] Row 7 of B: [0, 0, 0] So only the first input affects x6; x7 is not directly affected by any input. But because x6 feeds into x7 (due to the 1 in (6,7) of A), and with repeated application, you can reach x7. (This is a canonical controllable form.) #### **Controllability for x1, x2, x3:** - The subsystem: \[ \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] is upper-triangular with eigenvalue 2. Check if B (rows 1 to 3) is full rank: \[ \begin{bmatrix} 2 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \] Determinant: \[ \begin{vmatrix} 2 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{vmatrix} = 2(1*1 - 1*2) - 1(2*1 - 1*3) + 0(2*2 - 1*3) = 2(1-2) - 1(2-3) + 0 = 2(-1) - 1(-1) = -2 + 1 = -1 \neq 0 \] So, the subsystem is **controllable.** ### Step 3: **Rank of Controllability Matrix** - Each subsystem is controllable. - The entire system's controllability matrix will be full rank (rank 7). **Conclusion:** **The system is controllable.** --- ## 2. **Observability** The **observability matrix** is: \[ \mathcal{O} = \begin{bmatrix} C \\ CA \\ CA^2 \\ \vdots \\ CA^{n-1} \end{bmatrix} \] ### Step 1: Analyze C - C is 3x7, so three outputs. - Each state appears with a nonzero entry in C, except possibly x7 (check: x7 has 1, 0, 0 in columns 6,7). ### Step 2: Block Structure - The first three states are observed by the first two rows of C. - x4, x5 are observed by the first row (columns 4,5). - x6 is observed by the first and third rows. - x7 is observed only by the first row. ### Step 3: Observability of Subsystems - For the last two states, x6 and x7, check if their observability matrix is full rank: Take the 2x2 subsystem for x6, x7, and the corresponding columns of C: \[ A_{sub} = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \] \[ C_{sub} = \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} \] The observability matrix for this subsystem is: \[ \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ 1 & 0 \\ \hline C_{sub}A_{sub} = [1 1; 0 0; 1 0] \cdot \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \end{bmatrix} \] But notice that the only nonzero entries are in the first and third rows, but both have the same pattern (1,1) and (1,0). That is sufficient to distinguish both states. ### Step 4: Full Rank - The block structure and the presence of nonzero entries in each column of C means that each state is observed at least once. - Therefore, **the observability matrix is full rank**. --- ## **Final Answers** ### **Controllability:** The system is **controllable** (the controllability matrix is full rank). ### **Observability:** The system is **observable** (the observability matrix is full rank). --- **Summary Table:** | Property | Result | |------------------------|---------------| | **Controllability** | Controllable | | **Observability** | Observable | --- **If you need the explicit controllability or observability matrices or further details, let me know!**