Lecture 27 - Characteristic Equation, Eigenvalues and Eigenvectors - Electrical Engineering (EE) PDF Download

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Lecture 27 - Characteristic Equation, eigenvalues and eigen vectors, Control Systems

1 Characteristic Equation, eigenvalues and eigen vectors

For a discrete state space model, the characteristic equation is defined as
|zI − A| = 0
The roots of the characteristic equation are the eigenvalues of matrix A
1. If det(A) = 0, i.e., A is nonsingular and λ₁, λ₂, · · · , λn are the eigenvalues of A, then, will be the eigenvalues of A⁻¹.

2. Eigenvalues of A and A^T are same when A is a real matrix.

3. If A is a real symmetric matrix then all its eigenvalues are real.

The n × 1 vector vi which satisfies the matrix equation

Av_i = λ_iv_i                                     (1)

where λ_i, i = 1, 2, · · · , n denotes the ith eigenvalue, is called the eigen vector of A associated with the eigenvalue λ_i. If eigenvalues are distinct, they can be solved directly from equation (1).

Properties of eigen vectors 

1. An eigen vector cannot be a null vector.

2. If vi is an eigen vector of A then mvi is also an eigen vector of A where m is a scalar.

3. If A has n distinct eigenvalues, then the n eigen vectors are linearly independent.

Eigen vectors of multiple order eigenvalues When the matrix A an eigenvalue λ of multiplicity m, a full set of linearly independent may not exist. The number of linearly independent eigen vectors is equal to the degeneracy d of λI − A.

The degeneracy is defined as

d = n − r

where n is the dimension of A and r is the rank of λI − A. Furthermore,

1 ≤ d ≤ m

2 Similarity Transformation and Diagonalization 

Square matrices A and  are similar if

The non-singular matrix P is called similarity transformation matrix. It should be noted that eigenvalues of a square matrix A are not altered by similarity transformation.
Diagonalization:

If the system matrix A of a state variable model is diagonal then the state dynamics are decoupled from each other and solving the state equations become much more simpler.

In general, if A has distinct eigenvalues, it can be diagonalized using similarity transformation. Consider a square matrix A which has distinct eigenvalues λ₁, λ₂, . . . λ_n. It is required to find a transformation matrix P which will convert A into a diagonal form

through similarity transformation AP = P Λ. If v₁, v₂, . . . , v_n are the eigenvectors of matrix A corresponding to eigenvalues λ₁, λ₂, . . . λ_n, then we know Av_i = λ_iv_i. This gives

A [v₁ v₂ . . . v_n ] = [v₁ v₂ . . . v_n ]

Thus P = [v₁ v₂ . . . v_n].

Consider the following state model.

x(k + 1) = Ax(k) + Bu(k)

If P transforms the state vector x(k) to z(k) through the relation

x(k) = P z(k), or, z(k) = P ⁻¹x(k)

then the modified state space model becomes

z(k + 1) = P ⁻¹AP z(k) + P ⁻¹Bu(k)

where P ⁻¹AP = Λ.

3 Computation of Φ(t) 

We have seen that to derive the state space model of a sampled data system, we need to know the continuous time state transition matrix Φ(t) = e^At.

3.1 Using Inverse Laplace Transform For the system

 = Ax(t) + B u(t), the state transition matrix e^At can be computed as,

3.2 Using Similarity Transformation

If Λ is the diagonal representation of the matrix A, then Λ = P ⁻¹AP . When a matrix is in diagonal form, computation of state transition matrix is straight forward:

Given eΛt, we can show that

e^At = P e^ΛtP ⁻¹

3.3 Using Caley Hamilton Theorem 

Every square matrix A satisfies its own characteristic equation. If the characteristic equation is

then,

Application: Evaluation of any function f (λ) and f (A)  

If A has distinct eigenvalues λ₁, · · · , λ_n , then,

f (λ_i ) = g(λ_i ), i = 1, · · · , n

The solution will give rise to β₀ , β₁ , · · · , β_n−1 , then

f (A) = β₀I + β₁A + · · · + β_n−1Aⁿ⁻¹ 

If there are multiple roots (multiplicity = 2), then

f (λ_i) = g(λ_i)                                                 (2)

                          (3)

Example 1:

then compute the state transition matrix using Caley Hamilton Theorem.

= (λ−1)²(λ−2) = 0 ⇒ λ₁ = 1 (with multiplicity 2), λ₂ = 2

Let f (λ) = e^λt and g(λ) = β₀ + β₁λ + β₂λ₂ Then using (2) and (3), we can write

This implies

Solving the above equations  

Then

Example 2 For the system  = Ax(t) + B u(t), where A = compute e^At using 3 different techniques.

Solution: Eigenvalues of matrix A are 1 ± j 1.

Method 1

Method 2

e^At = P e^ΛtP ⁻¹ where e^Λt = Eigen values are 1 ± j . The corresponding eigenvectors are found by using equation Av_i = λ_iv_i as follows

Taking v₁ = 1, we get v₂ = j . So, the eigenvector corresponding to 1 + j is  and the one corresponding to 1 − j is   The transformation matrix is given by

Now,

Method 3: Caley Hamilton Theorem The eigenvalues are λ₁,2 = 1 ± j .

Solving,

Hence,

We will now show through an example how to derive discrete state equation from a continuous one.

Example: Consider the following state model of a continuous time system.

If the system is under a sampling process with period T , derive the discrete state model of the system.

To derive the discrete state space model, let us first compute the state transition matrix of the continuous time system using Caley Hamilton Theorem.

This implies

Solving the above equations

Then

Thus the discrete state matrix A is given as

The discrete input matrix B can be computed as

The discrete state equation is thus described by

When T = 1, the state equations become