Question

If the eigenvalues of the system matrix A are -2 and -5, the system is:

• A. Stable
• B. Unstable
• C. Oscillatory
• D. Critically stable

Hint:

Eigenvalues of matrix \( A \) are the same as closed-loop poles.
Negative real parts \( \implies \) Stable system.
If any eigenvalue is positive, the system is unstable.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to determine the stability of a system given the eigenvalues of its system matrix \( A \).
The eigenvalues of the state matrix \( A \) determine the natural response modes of a state-space system.

Step 2: Key Formula or Approach:
The eigenvalues (\( \lambda \)) of the system matrix \( A \) are obtained by solving the characteristic equation:
\[ \det(\lambda I - A) = 0 \] These eigenvalues are mathematically identical to the poles of the system's closed-loop transfer function.
The absolute stability of a continuous-time system depends on the location of these eigenvalues in the complex s-plane:
- Stable: All eigenvalues have strictly negative real parts (\( \text{Re}(\lambda) < 0 \)).
- Unstable: At least one eigenvalue has a positive real part (\( \text{Re}(\lambda) > 0 \)).
- Marginally/Critically Stable: Zero real part eigenvalues with no repeated roots on the imaginary axis.

Step 3: Detailed Explanation:
Let us analyze the given eigenvalues:

• Given Eigenvalues:
- The eigenvalues are \( \lambda_1 = -2 \) and \( \lambda_2 = -5 \).
- Both of these values are real numbers with strictly negative signs.

• Transient Response Analysis:
- The zero-input response of the system is composed of exponential terms corresponding to these eigenvalues:
\[ x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} = c_1 e^{-2t} + c_2 e^{-5t} \] - As time \( t \) approaches infinity, both exponential terms decay to zero:
\[ \lim_{t \to \infty} x(t) = \lim_{t \to \infty} \left( c_1 e^{-2t} + c_2 e^{-5t} \right) = 0 \] - Because the system's natural response decays to zero over time, the system is asymptotically stable.
- Since there are no imaginary components, the response is non-oscillatory (overdamped), making Option C incorrect.

Step 4: Final Answer:
Since both eigenvalues are real and negative, the system is stable.