Question

Consider the state-space model:

\[ \dot{x}(t) = A x(t) + B u(t) \] \[ y(t) = C x(t) \] \[ A = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix} \]

The sum of the magnitudes of the poles is:

• A. 5
• B. 3
• C. 2
• D. 4

Hint:

To find the poles of a system represented by a state-space model:
- Compute the eigenvalues of the system matrix \( A \) by solving the characteristic equation.
- The sum of the magnitudes of the poles is the sum of the absolute values of the eigenvalues.

Answer 1

The Correct Option is A

Step 1: The poles of the system are the eigenvalues of the matrix \( A \). To find the eigenvalues, we solve the characteristic equation:

\[ \det(A - \lambda I) = 0 \]

where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix. The matrix \( A - \lambda I \) is:

\[ A - \lambda I = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -\lambda & 1 \\ -2 & -3 - \lambda \end{bmatrix} \]

Now, we compute the determinant:

\[ \det(A - \lambda I) = (-\lambda)(-3 - \lambda) - (1)(-2) = \lambda^2 + 3\lambda + 2 \]

We set this equal to zero to find the eigenvalues:

\[ \lambda^2 + 3\lambda + 2 = 0 \]

Solving this quadratic equation:

\[ \lambda = \frac{-3 \pm \sqrt{3^2 - 4(1)(2)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 8}}{2} = \frac{-3 \pm 1}{2} \]

Thus, the eigenvalues (poles) are:

\[ \lambda_1 = \frac{-3 + 1}{2} = -1, \quad \lambda_2 = \frac{-3 - 1}{2} = -2 \]

Step 2: The sum of the magnitudes of the poles is:

\[ |\lambda_1| + |\lambda_2| = |-1| + |-2| = 1 + 2 = 3 \]

Thus, the sum of the magnitudes of the poles is 3.