Finding System Matrix with Pole at Origin

The poles of a linear system described by the state-space equation $ \dot{x} = Ax + Bu $ are the eigenvalues of the system matrix $A$. A pole located at the origin signifies that one of the eigenvalues of the matrix $A$ must be equal to zero.

Eigenvalue Calculation Method

To determine which matrix $A$ has a pole at the origin, we calculate its eigenvalues. This is done by finding the roots of the characteristic equation: $ det(A - \lambda I) = 0 $ where $ \lambda $ represents the eigenvalues and $I$ is the identity matrix. A zero eigenvalue ($ \lambda = 0 $) indicates a pole at the origin.

Analysis of Matrix Options

We will now analyze the eigenvalues for each provided matrix $A$.

Option 1: $ \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} $

The characteristic equation is: $ det \begin{bmatrix} 0-\lambda & 1 \\ -2 & -3-\lambda \end{bmatrix} = (-\lambda)(-3-\lambda) - (1)(-2) $ $ = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) $ The eigenvalues are $ \lambda = -1 $ and $ \lambda = -2 $. This matrix does not have a pole at the origin.

Option 2: $ \begin{bmatrix} 1 & -1.5 \\ -2 & 3 \end{bmatrix} $

The characteristic equation is: $ det \begin{bmatrix} 1-\lambda & -1.5 \\ -2 & 3-\lambda \end{bmatrix} = (1-\lambda)(3-\lambda) - (-1.5)(-2) $ $ = (3 - 4\lambda + \lambda^2) - 3 $ $ = \lambda^2 - 4\lambda $ $ = \lambda(\lambda - 4) $ The eigenvalues are $ \lambda = 0 $ and $ \lambda = 4 $. Since $ \lambda = 0 $ is an eigenvalue, this matrix has a pole at the origin.

Option 3: $ \begin{bmatrix} 1 & 1.5 \\ 2 & -3 \end{bmatrix} $

The characteristic equation is: $ det \begin{bmatrix} 1-\lambda & 1.5 \\ 2 & -3-\lambda \end{bmatrix} = (1-\lambda)(-3-\lambda) - (1.5)(2) $ $ = (-3 + 2\lambda + \lambda^2) - 3 $ $ = \lambda^2 + 2\lambda - 6 $ Using the quadratic formula, $ \lambda = \frac{-2 \pm \sqrt{2^2 - 4(1)(-6)}}{2(1)} = \frac{-2 \pm \sqrt{28}}{2} $. The eigenvalues are not zero.

Option 4: $ \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} $

The characteristic equation is: $ det \begin{bmatrix} 0-\lambda & 1 \\ -2 & 3-\lambda \end{bmatrix} = (-\lambda)(3-\lambda) - (1)(-2) $ $ = \lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2) $ The eigenvalues are $ \lambda = 1 $ and $ \lambda = 2 $. This matrix does not have a pole at the origin.

Result

The matrix $ A = \begin{bmatrix} 1 & -1.5 \\ -2 & 3 \end{bmatrix} $ has eigenvalues $ \lambda = 0 $ and $ \lambda = 4 $, confirming it possesses a pole at the origin.