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Lecture 05 Analysis (I)

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Title: Lesson 6 Author: marilyn Last modified by: lcs Created Date: 10/23/2003 5:19:00 AM Document presentation format: Company: – PowerPoint PPT presentation

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Title: Lecture 05 Analysis (I)


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Lecture 05 Analysis (I) Time Response and State
Transition Matrix
5.1 State Transition Matrix 5.2 Modal
decomposition --Diagonalization 5.3
Cayley-Hamilton Theorem
2
The behavior of x(t) et y(t)
  1. Homogeneous solution of x(t)
  2. Non-homogeneous solution of x(t)

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Homogeneous solution
State transition matrix
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Properties
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Non-homogeneous solution
Convolution
Homogeneous
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Zero-input response
Zero-state response
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Example 1
Ans
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Using Maisons gain formula
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Methode 1
Methode 2
Methode 3 Cayley-Hamilton Theorem
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Methode 1
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Method 2 Diagonalization
Example 4.5
diagonal matrix
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Diagonalization via Coordinate Transformation
Plant
Eigenvalue of A
Assume that all the eigenvalues of A are
distinct, i.e.
Then eigenvectors,
are independent.
Coordinate transformation matrix
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where
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New coordinate
(4.1)
Solution of (4.1)
The above expansion of x(t) is called modal
decomposition.
Hence, system asy. stable ? all the eigenvales of
A lie in LHP
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Example
Find eigenvector
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(4.2)
Solution of (4.1)
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In the case of A matrix is phase-variable form
and
Vandermonde matrix for phase-variable form
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Example
depend
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Case 3
Jordan form
Generalized eigenvectors
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Example
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Cayley-Hamilton Theorem
Method 3
Theorem Every square matrix satisfies its char.
equation.
Given a square matrix A, . Let
f(?) be the char. polynomial of A.
Char. Equation
By Caley-Hamilton Theorem
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any
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Example
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Example
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Example

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Example
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