Understanding Discrete System Pole Locations

This problem asks us to find the pole locations of a discrete system represented by a state-space difference equation. The system is defined as:

$$ \begin{bmatrix} X_1(k + 1) \\ X_2(k + 1) \end{bmatrix} = \begin{bmatrix} a & a - 1 \\ a + 1 & a \end{bmatrix} \begin{bmatrix} X_1(k) \\ X_2(k) \end{bmatrix} $$

We are given initial conditions $X_1(0) = 1$ and $X_2(0) = 0$, and we need to find the pole locations specifically when the parameter a = 1.

Identifying the State-Transition Matrix

The general form of a discrete-time linear time-invariant (LTI) system in state-space is given by $X(k+1) = A X(k)$, where $X(k)$ is the state vector and $A$ is the state-transition matrix. From the provided difference equation, we can identify the state-transition matrix $A$ as:

$$ A = \begin{bmatrix} a & a - 1 \\ a + 1 & a \end{bmatrix} $$

Calculating Pole Locations for a = 1

To find the pole locations, we need to evaluate the state-transition matrix $A$ at the given parameter value, a = 1.

Substituting a = 1 into the matrix $A$:

$$ A \Big|_{a=1} = \begin{bmatrix} 1 & 1 - 1 \\ 1 + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} $$

The pole locations of a system are the eigenvalues of its state-transition matrix. To find the eigenvalues ($\lambda$), we solve the characteristic equation:

det(A - λI) = 0

where $I$ is the identity matrix.

First, we form the matrix $(A - \lambda I)$:

$$ A - \lambda I = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 0 \\ 2 & 1 - \lambda \end{bmatrix} $$

Now, we calculate the determinant:

$$ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (0)(2) $$ $$ \det(A - \lambda I) = (1 - \lambda)^2 $$

Setting the determinant to zero to find the eigenvalues:

$$ (1 - \lambda)^2 = 0 $$

Solving for $\lambda$ gives:

$$ 1 - \lambda = 0 \implies \lambda = 1 $$

This equation indicates a repeated eigenvalue at $\lambda = 1$. Therefore, the pole locations are $1$ and $1$. These can be expressed in the form $p \pm jq$ as $1 \pm j0$. The initial conditions provided ($X_1(0)=1$, $X_2(0)=0$) are not needed to determine the pole locations, which are intrinsic properties of the system matrix $A$. The initial conditions affect the specific response trajectory of the system, but not the locations of its poles.

Matching with Options

Comparing our calculated pole locations ($1, 1$ or $1 \pm j0$) with the given options:

• Option 1: $1 \pm j0$ (matches our result)
• Option 2: $-1 \pm j0$ (corresponds to eigenvalues -1, -1)
• Option 3: $\pm1 + j0$ (corresponds to eigenvalues 1, -1)
• Option 4: $0 \pm j1$ (corresponds to eigenvalues $j$, $-j$)

The calculated pole locations are $1 \pm j0$, which corresponds to Option 1.