[16.46° At 377 A=-0.25+ j7.9852 Stable -0.1693 -0.50
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How is lambda calculated?
![**Matrix Analysis and Stability**
In this example, we analyze the stability of a system by examining its Jacobian matrix \( J \).
### Jacobian Matrix \( J \)
\[
J = \begin{bmatrix}
0 & 377 \\
-0.1765 \cos \delta & -0.5
\end{bmatrix}
\]
### Case Analysis
1. **At \( \delta = 16.46^\circ \)**
\[
J = \begin{bmatrix}
0 & 377 \\
-0.1693 & -0.50
\end{bmatrix}
\]
The eigenvalues are calculated as:
\[
\lambda = -0.25 \pm j 7.9852
\]
**Conclusion:** This configuration is **Stable** since the real part of the eigenvalues is negative.
2. **At \( \delta = 163.54^\circ \)**
\[
J = \begin{bmatrix}
0 & 377 \\
0.1693 & -0.50
\end{bmatrix}
\]
The eigenvalues are calculated as:
\[
\lambda_1 = 7.743, \quad \lambda_2 = -8.243
\]
**Conclusion:** This configuration is **Unstable**, as \( \lambda_1 > 0 \) indicates a positive real part, leading to instability.
### Explanation
- The stability of a system is determined by examining the eigenvalues of its Jacobian matrix.
- If the real parts of all eigenvalues are negative, the system is considered stable.
- If any eigenvalue has a positive real part, the system is deemed unstable.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5312a15-860e-407c-ae9e-8729c86c8316%2F6e982fe0-588a-4ec6-a1de-13646fbe9b25%2Fyfdb7j8_processed.png&w=3840&q=75)
Transcribed Image Text:**Matrix Analysis and Stability**
In this example, we analyze the stability of a system by examining its Jacobian matrix \( J \).
### Jacobian Matrix \( J \)
\[
J = \begin{bmatrix}
0 & 377 \\
-0.1765 \cos \delta & -0.5
\end{bmatrix}
\]
### Case Analysis
1. **At \( \delta = 16.46^\circ \)**
\[
J = \begin{bmatrix}
0 & 377 \\
-0.1693 & -0.50
\end{bmatrix}
\]
The eigenvalues are calculated as:
\[
\lambda = -0.25 \pm j 7.9852
\]
**Conclusion:** This configuration is **Stable** since the real part of the eigenvalues is negative.
2. **At \( \delta = 163.54^\circ \)**
\[
J = \begin{bmatrix}
0 & 377 \\
0.1693 & -0.50
\end{bmatrix}
\]
The eigenvalues are calculated as:
\[
\lambda_1 = 7.743, \quad \lambda_2 = -8.243
\]
**Conclusion:** This configuration is **Unstable**, as \( \lambda_1 > 0 \) indicates a positive real part, leading to instability.
### Explanation
- The stability of a system is determined by examining the eigenvalues of its Jacobian matrix.
- If the real parts of all eigenvalues are negative, the system is considered stable.
- If any eigenvalue has a positive real part, the system is deemed unstable.
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