Question

A linear time invariant (LTI) system is described by:
\[ \begin{bmatrix}\dot{x}_1\\ \dot{x}_2\end{bmatrix}=\begin{bmatrix}0 & 1\\ -10 & -7\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}+\begin{bmatrix}1\\ 1\end{bmatrix}u \]
\[ y=\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix} \]
The poles of the system are ______.

• A. \(-2,\ -5\)
• B. \(2,\ 5\)
• C. \(-2,\ 5\)
• D. \(2,\ -5\)

Hint:

The poles are the eigenvalues of the state matrix \(A\): solve \(\det(sI-A)=0\).

Answer 1

The Correct Option is A

Step 1: Recall how poles relate to the state matrix.
For a state-space model \(\dot{x}=Ax+Bu\), \(y=Cx\), the poles of the transfer function are the roots of the characteristic equation of the state matrix \(A\), that is, the eigenvalues of \(A\), found from
\[ \det(sI-A)=0 \]
This holds as long as there is no pole-zero cancellation between the state equation and the output equation, which we will check at the end.

Step 2: Write \(sI-A\).
Here \(A=\begin{bmatrix}0 & 1\\ -10 & -7\end{bmatrix}\), so
\[ sI-A=\begin{bmatrix}s & -1\\ 10 & s+7\end{bmatrix} \]

Step 3: Take the determinant.
\[ \det(sI-A)=s(s+7)-(-1)(10)=s^2+7s+10 \]

Step 4: Solve the characteristic equation.
\[ s^2+7s+10=0 \]
This factors as
\[ (s+2)(s+5)=0 \]
giving
\[ s=-2,\ s=-5 \]

Step 5: Confirm there is no cancellation with \(C\).
With \(B=\begin{bmatrix}1\\1\end{bmatrix}\) and \(C=\begin{bmatrix}1 & 1\end{bmatrix}\), the transfer function works out to
\[ G(s)=C(sI-A)^{-1}B=\frac{2(s-1)}{(s+2)(s+5)} \]
The zero is at \(s=1\), which does not coincide with either pole, so both \(s=-2\) and \(s=-5\) remain genuine poles of the system.

Step 6: Analyze the options.
Option (B) drops the minus signs entirely, which would come from mixing up the signs while solving the quadratic. Options (C) and (D) keep only one sign flipped, which does not match a stable, correctly factored quadratic with two negative roots.

Final Answer:
The poles of the system are \(-2\) and \(-5\).
\[ \boxed{-2,\ -5} \]