Introduction

Eigenvalues play a crucial role in understanding the stability analysis of systems. In this blog post, we will explore the concept of eigenvalues and their significance in stability analysis. We will then dive into a practical example to illustrate how these concepts can be applied to real-world problems.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra, which are used to characterize linear transformations.

Given a square matrix A of dimensions n x n, a scalar λ and a non-zero vector v, the following relationship holds:

Av = λv

In this equation, λ is called the eigenvalue, and v is the corresponding eigenvector of the matrix A. There can be multiple eigenvalues, and each has its corresponding eigenvectors.

Stability Analysis

Stability analysis is the study of how a system behaves over time, particularly when subjected to perturbations or disturbances. It helps us determine whether a system is stable, meaning it returns to its equilibrium state, or unstable, meaning it diverges from its equilibrium state.

Eigenvalues play a key role in stability analysis because they reveal how different modes of the system evolve over time. When analyzing the stability of a linear system, the eigenvalues of its characteristic matrix determine the behavior of the system.

Stability Criteria

For continuous-time linear systems, the general rule for stability is:

• If all eigenvalues have negative real parts, the system is stable.
• If at least one eigenvalue has a positive real part, the system is unstable.

• If all eigenvalues have non-positive real parts, and any eigenvalue with a real part equal to zero has an associated eigenvector with no purely imaginary components, the system is marginally stable.

For discrete-time linear systems, the stability criteria are based on the magnitudes of eigenvalues:

• If all eigenvalues have magnitudes less than 1, the system is stable.
• If at least one eigenvalue has a magnitude greater than or equal to 1, the system is unstable.

Example: Stability Analysis of a Continuous-Time System

Let’s consider a continuous-time linear system represented by the following state-space equation:

ẋ(t) = Ax(t)

where ẋ(t) is the derivative of the state vector x(t), and A is a constant matrix. Suppose the matrix A is given as:

A = | -3 0 | | 0 -1 |

To analyze the stability of this system, we need to find the eigenvalues of matrix A.

Step 1: Compute the characteristic equation

det(A – λI) = 0

where λ is an eigenvalue and I is the identity matrix. For our given matrix A:

| -3-λ 0 |

| 0 -1-λ |

Expanding the determinant, we get:

(-3-λ)((-1)-λ) = λ^2 + 4λ + 3 = 0

Step 2: Solve for eigenvalues

The characteristic equation is a quadratic equation, which can be solved using the quadratic formula:

λ = (-B ± √(B^2 – 4AC)) / 2A

In this case, A = 1, B = 4, and C = 3. Plugging in these values, we obtain:

λ = (-4 ± √(4^2 – 4(1)(3))) / 2(1) λ = (-4 ± √(16 – 12)) / 2 λ = (-4 ± √4) / 2

Solving for the two eigenvalues, we find:

λ1 = (-4 + 2) / 2 = -1 λ2 = (-4 – 2) / 2 = -3

Step 3: Determine stability

Since both eigenvalues have negative real parts, the continuous-time linear system is stable.

Stability Analysis of a Mass-Spring System

Consider a mass-spring system with a mass m and spring constant k. The mass is attached to a spring and is free to move horizontally without friction. The governing equation for this system is given by Hooke’s law:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the mass from its equilibrium position. The force F is also equal to the product of mass and acceleration (Newton’s second law):

F = ma

where m is the mass and a is the acceleration. By combining these two equations, we obtain the second-order differential equation for the mass-spring system:

mx”(t) = -kx(t)

To analyze the stability of this system, we first convert the second-order differential equation into a system of first-order differential equations using state-space representation. Let x₁(t) = x(t) and x₂(t) = x'(t). Then, we have:

x₁'(t) = x₂(t) x₂'(t) = -k/m * x₁(t)

This results in the following state-space representation:

ẋ(t) = Ax(t)

where ẋ(t) is the derivative of the state vector x(t) = [x₁(t), x₂(t)]ᵀ, and A is a constant matrix:

A =[ [0 1 ] [ -k/m 0 ] ]

Now, we analyze the stability of the system by finding the eigenvalues of matrix A.

Step 1: Compute the characteristic equation

det(A – λI) = 0

where λ is an eigenvalue and I is the identity matrix. For our given matrix A:

| -λ 1 | | -k/m -λ |

Expanding the determinant, we get:

(-λ)(-λ) – (-k/m)(1) = λ^2 + k/m = 0

Step 2: Solve for eigenvalues

The characteristic equation is a quadratic equation:

λ^2 + k/m = 0

This equation can be factored as:

λ^2 = -k/m

Taking the square root of both sides, we find two eigenvalues:

λ1 = i√(k/m) λ2 = -i√(k/m)

Step 3: Determine stability

Since both eigenvalues have zero real parts and are purely imaginary, the mass-spring system is marginally stable. This means that the system oscillates without damping, and the oscillations neither grow nor decay over time.

Conclusion

Eigenvalues play a crucial role in understanding the stability of linear systems. By analyzing the eigenvalues of the system’s characteristic matrix, we can determine whether a system is stable, unstable, or marginally stable. This knowledge is invaluable in various fields such as control systems, engineering, and physics, where stability analysis is critical for predicting and controlling the behavior of complex systems.