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:
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:
Show Hint
- 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.
- 5
- 3
- 2
- 4
The Correct Option is A
Solution and Explanation
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.
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