MSc in Computer Engineering, Cybersecurity and Artificial - PowerPoint PPT Presentation

MSc in Computer Engineering, Cybersecurity and Artificial Intelligence Course FDE , a.a. 2019/2020, Lecture 4 Review of matrix algebra and state-space transformations Prof. Mauro Franceschelli Dept. of Electrical and Electronic Engineering University of Cagliari, Italy Wednesday, 25th March 2020 1 / 35

Outline Diagonalization Jordan form State transition matrix and modes Transient and steady state behavior 2 / 35

Diagonalization Brief review of matrix algebra Definition. Given a square matrix ❆ of order n , let λ ∈ R be a scalar and let ✈ � = 0 be a column vector with dimensions n × 1. If it holds ❆ ✈ = λ ✈ then λ is said to be an eigenvalue (”autovalore” in italian) of matrix ❆ associated to the eigenvector (”autovettore”) ✈ . In practice: The eigenvalues are the roots of the characteristic polynomial: P ( λ ) = det ( λ ■ − ❆ ) = 0 . If λ is an eigenvalue of ❆ , then the corresponding eigenvector ✈ is a non-zero solution of the linear system of equations: ( λ ■ − ❆ ) ✈ = 0 where 0 is a column vector n × 1 the elements of which are all zeros. 3 / 35

Diagonalization Diagonalization procedure The diagonalization procedure computes a similarity transformation which allows to transform a general matrix ❆ to a diagonal matrix ❆ ′ = P − 1 ❆P . If matrix ❆ has eigenvalues λ 1 , λ 2 , . . . , λ n , then matrix ❆ ′ is:   λ 1 0 · · · 0 0 0 λ 2 · · ·   ❆ ′ =    . . . . ...  . . .  . . .  0 0 · · · λ n A sufficient condition for a matrix ❆ of order n to be diagonalizable is that it has distinct eigenvalues: λ i � = λ j if i � = j . 4 / 35

Diagonalization Advantages of a diagonal representation 1 - The state transition matrix can be immediately computed:     e λ 1 t · · · 0 · · · 0 λ 1 . . ′ t = . . ❆ ′ = ... e ❆ ...  . .   . .  = ⇒ . . . .     e λ n t 0 0 · · · λ n · · · 2 - The analysis is easier because all state variables are decoupled. For instance, for a system with single input:  x 1 ( t ) ˙ = λ 1 x 1 ( t ) + b 1 u ( t )   . . . . . .   x n ( t ) ˙ = λ n x n ( t ) + b n u ( t ) 5 / 35

Diagonalization Advantages of a diagonal representation 1 - The state transition matrix can be immediately computed:     e λ 1 t · · · 0 · · · 0 λ 1 . . ′ t = . . ❆ ′ = ... e ❆ ...  . .   . .  = ⇒ . . . .     e λ n t 0 0 · · · λ n · · · 2 - The analysis is easier because all state variables are decoupled. For instance, for a system with single input:  x 1 ( t ) ˙ = λ 1 x 1 ( t ) + b 1 u ( t )   . . . . . .   x n ( t ) ˙ = λ n x n ( t ) + b n u ( t ) 5 / 35

Diagonalization The modal matrix Definition. Given a matrix ❆ of dimension n × n , let ✈ 1 , ✈ 2 , . . . , ✈ n , be a set of linearly independent eigenvectors corresponding to the eigenvalues λ 1 , λ 2 , . . . , λ n . We denote modal matrix ❆ the matrix n × n � � ❱ = ✈ 1 ✈ 2 · · · ✈ n . Theorem. If a matrix ❆ has distinct eigenvalues λ 1 , . . . , λ n , then the corresponding eigenvectors ✈ 1 , . . . , ✈ n are linearly indepen- dent. If ❆ does not have n distinct eigenvalues, then the modal matrix exists if and only if to each eigenvalue with algebraic multiplicity ν correspond ν linearly independent eigenvectors. 6 / 35

Diagonalization The modal matrix Definition. Given a matrix ❆ of dimension n × n , let ✈ 1 , ✈ 2 , . . . , ✈ n , be a set of linearly independent eigenvectors corresponding to the eigenvalues λ 1 , λ 2 , . . . , λ n . We denote modal matrix ❆ the matrix n × n � � ❱ = ✈ 1 ✈ 2 · · · ✈ n . Theorem. If a matrix ❆ has distinct eigenvalues λ 1 , . . . , λ n , then the corresponding eigenvectors ✈ 1 , . . . , ✈ n are linearly indepen- dent. If ❆ does not have n distinct eigenvalues, then the modal matrix exists if and only if to each eigenvalue with algebraic multiplicity ν correspond ν linearly independent eigenvectors. 6 / 35

Diagonalization The modal matrix Given a matrix ❆ of dimension n × n let ❱ be its modal matrix. The matrix ❆ ′ is obtained by the similarity transformation ❆ ′ = ❱ − 1 ❆❱ is a diagonal matrix. Proof. By the definition of eigenvalue and eigenvector it holds λ i ✈ i = ❆✈ i ( i = 1 , . . . , n ) � � � � = ⇒ · · · = · · · λ 1 ✈ 1 λ 2 ✈ 2 λ n ✈ n ❆✈ 1 ❆✈ 2 ❆✈ n thus   0 · · · 0 λ 1 0 λ 2 · · · 0   � � � �   · · ·  = ❆ · · · ✈ 1 ✈ 2 ✈ n . . . ✈ 1 ✈ 2 ✈ n ...  . . .  . . .  0 0 · · · λ n therefore ❱ ❆ ′ = ❆❱ ❆ ′ = ❱ − 1 ❆❱ = ⇒ 7 / 35

Diagonalization The modal matrix Given a matrix ❆ of dimension n × n let ❱ be its modal matrix. The matrix ❆ ′ is obtained by the similarity transformation ❆ ′ = ❱ − 1 ❆❱ is a diagonal matrix. Proof. By the definition of eigenvalue and eigenvector it holds λ i ✈ i = ❆✈ i ( i = 1 , . . . , n ) � � � � = ⇒ · · · = · · · λ 1 ✈ 1 λ 2 ✈ 2 λ n ✈ n ❆✈ 1 ❆✈ 2 ❆✈ n thus   0 · · · 0 λ 1 0 λ 2 · · · 0   � � � �   · · ·  = ❆ · · · ✈ 1 ✈ 2 ✈ n . . . ✈ 1 ✈ 2 ✈ n ...  . . .  . . .  0 0 · · · λ n therefore ❱ ❆ ′ = ❆❱ ❆ ′ = ❱ − 1 ❆❱ = ⇒ 7 / 35

Diagonalization Example (homework) Given the representation:  � � � � � � � � x 1 ( t ) ˙ − 1 1 x 1 ( t ) 0  = + u ( t )   x 2 ( t ) ˙ 0 − 2 x 2 ( t ) 1   � � � � � � � �  y 1 ( t ) 2 1 x 1 ( t ) 1 . 5   = + u ( t )   y 2 ( t ) 0 2 x 2 ( t ) 0 show that the a change of variables x ( t ) = ❱ z ( t ) based on the modal matrix � 1 � 1 ❱ = transforms the system into a diagonal form 0 − 1 � ˙ � − 1 � z 1 ( t )  � � � � � z 1 ( t ) 0 1  = + u ( t )   z 2 ( t ) ˙ 0 − 2 z 2 ( t ) − 1   � � � � � � � �  y 1 ( t ) 2 1 z 1 ( t ) 1 . 5   = + u ( t )   y 2 ( t ) 0 − 2 z 2 ( t ) 0 8 / 35

Diagonalization System analysis via diagonalization Given a representation in which matrix ❆ has modal matrix ❱ , consider the diagonal system obtained by the diagonalization procedure: � � ❆ ′ ③ ( t ) + ❇ ′ ✉ ( t ) ① ( t ) ˙ = ❆① ( t ) + ❇✉ ( t ) ③ ( t ) ˙ = = ⇒ ❈ ′ ① ( t ) + ❉✉ ( t ) ② ( t ) = ❈① ( t ) + ❉✉ ( t ) ② ( t ) = By the properties of the state transition matrix and the Lagrange formula, it holds: e ❆ t = ❱ e ❆ ′ t ❱ − 1 The natural state evolution ① n ( t ) starting from the initial state ① (0) is ′ t ③ (0) = ❱ e ❆ ′ t ❱ − 1 ① (0) ① n ( t ) = ❱ ③ n ( t ) = ❱ e ❆ The forced evolution ① f ( t ) for a given input signal ✉ ( t ) is � t e ❆ ′ ( t − τ ) ❇ ′ ✉ ( τ ) d τ ① f ( t ) = ❱ ③ f ( t ) = ❱ 0 9 / 35

Diagonalization System analysis via diagonalization Given a representation in which matrix ❆ has modal matrix ❱ , consider the diagonal system obtained by the diagonalization procedure: � � ❆ ′ ③ ( t ) + ❇ ′ ✉ ( t ) ① ( t ) ˙ = ❆① ( t ) + ❇✉ ( t ) ③ ( t ) ˙ = = ⇒ ❈ ′ ① ( t ) + ❉✉ ( t ) ② ( t ) = ❈① ( t ) + ❉✉ ( t ) ② ( t ) = By the properties of the state transition matrix and the Lagrange formula, it holds: e ❆ t = ❱ e ❆ ′ t ❱ − 1 The natural state evolution ① n ( t ) starting from the initial state ① (0) is ′ t ③ (0) = ❱ e ❆ ′ t ❱ − 1 ① (0) ① n ( t ) = ❱ ③ n ( t ) = ❱ e ❆ The forced evolution ① f ( t ) for a given input signal ✉ ( t ) is � t e ❆ ′ ( t − τ ) ❇ ′ ✉ ( τ ) d τ ① f ( t ) = ❱ ③ f ( t ) = ❱ 0 9 / 35

Diagonalization System analysis via diagonalization Given a representation in which matrix ❆ has modal matrix ❱ , consider the diagonal system obtained by the diagonalization procedure: � � ❆ ′ ③ ( t ) + ❇ ′ ✉ ( t ) ① ( t ) ˙ = ❆① ( t ) + ❇✉ ( t ) ③ ( t ) ˙ = = ⇒ ❈ ′ ① ( t ) + ❉✉ ( t ) ② ( t ) = ❈① ( t ) + ❉✉ ( t ) ② ( t ) = By the properties of the state transition matrix and the Lagrange formula, it holds: e ❆ t = ❱ e ❆ ′ t ❱ − 1 The natural state evolution ① n ( t ) starting from the initial state ① (0) is ′ t ③ (0) = ❱ e ❆ ′ t ❱ − 1 ① (0) ① n ( t ) = ❱ ③ n ( t ) = ❱ e ❆ The forced evolution ① f ( t ) for a given input signal ✉ ( t ) is � t e ❆ ′ ( t − τ ) ❇ ′ ✉ ( τ ) d τ ① f ( t ) = ❱ ③ f ( t ) = ❱ 0 9 / 35