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

MSc in Computer Engineering, Cybersecurity and Artificial Intelligence, Fault Diagnosis and Estimation in Dynamical Systems (FDE), a.a. 2019/2020, Lecture 3 Analysis of continuous-time linear dynamical systems in the state space Prof. Mauro Franceschelli Dept. of Electrical and Electronic Engineering University of Cagliari, Italy Monday, 23rd March 2020 1 / 29

Outline Introduction The state transition matrix The Lagrange formula 2 / 29

State variable model A linear dynamical system, stationary of order n , with r inputs and p outputs, has the following standard representation (or realization) as a state variable model (SV): � ① ( t ) ˙ = ❆① ( t ) + ❇✉ ( t ) ② ( t ) = ❈① ( t ) + ❉✉ ( t ) with         x 1 ( t ) x 1 ( t ) ˙ u 1 ( t ) y 1 ( t ) x 2 ( t ) x 2 ( t ) ˙ u 2 ( t ) y 2 ( t )                 ① ( t ) =  ; ˙ ① ( t ) =  ; ✉ ( t ) =  ; ② ( t ) = . . . .         . . . . . . . .      x n ( t ) x n ( t ) ˙ u r ( t ) y p ( t ) 3 / 29

Analysis of a linear stationary SV model Our objective is to determine the evolution of the state variables ① ( t ) and the output ② ( t ) for t ≥ t 0 given: the value of the initial state ① ( t 0 ); the evolution of the inputs ✉ ( t ) per t ≥ t 0 . Also in this case we separate the natural evolution from the forced evolution: ① ( t ) = ① ℓ ( t ) + ① f ( t ) ② ( t ) = ② ℓ ( t ) + ② f ( t ) 4 / 29

Outline Introduction The state transition matrix The Lagrange formula 5 / 29

State transition matrix Matrix exponential Given a scalar z ∈ C its exponential is ∞ e z = 1 + z + z 2 2! + z 3 z k � 3! + · · · = k ! , k =0 and it can be shown that the series is always convergent. Analogously, we the same concept can be applied to square matrices. Definition Given a matrix ❆ n × n , its exponential is an n × n matrix defined as ∞ e ❆ = ■ + ❆ + ❆ 2 2! + ❆ 3 ❆ k � 3! + · · · = k ! . k =0 it can be shown that the series is always convergent. 6 / 29

State transition matrix Matrix exponential Given a scalar z ∈ C its exponential is ∞ e z = 1 + z + z 2 2! + z 3 z k � 3! + · · · = k ! , k =0 and it can be shown that the series is always convergent. Analogously, we the same concept can be applied to square matrices. Definition Given a matrix ❆ n × n , its exponential is an n × n matrix defined as ∞ e ❆ = ■ + ❆ + ❆ 2 2! + ❆ 3 ❆ k � 3! + · · · = k ! . k =0 it can be shown that the series is always convergent. 6 / 29

State transition matrix Exponential of a diagonal matrix Given a generic square matrix n × n     e λ 1 λ 1 0 0 · · · · · · . . e ❆ = . . ... ...  . .   . .  ❆ = it holds  ,  . . . . .   e λ n 0 0 · · · λ n · · · Proof. It can be shown that for each k ∈ N it holds   λ k 0 · · · 1 ❆ k = . . ...   . . . .   λ k 0 · · · n   λ k   � ∞ e λ 1 0 0 1 · · · · · · k =0 k ! ∞ ❆ k � ⇒ e ❆ =  . .  . . ... ...   = k ! = . .  = . .   . . . .    e λ n k =0 � ∞ λ k 0 · · · 0 n · · · k =0 k ! 7 / 29

State transition matrix Exponential of a diagonal matrix Given a generic square matrix n × n     e λ 1 λ 1 0 0 · · · · · · . . e ❆ = . . ... ...  . .   . .  ❆ = it holds  ,  . . . . .   e λ n 0 0 · · · λ n · · · Proof. It can be shown that for each k ∈ N it holds   λ k 0 · · · 1 ❆ k = . . ...   . . . .   λ k 0 · · · n   λ k   � ∞ e λ 1 0 0 1 · · · · · · k =0 k ! ∞ ❆ k � ⇒ e ❆ =  . .  . . ... ...   = k ! = . .  = . .   . . . .    e λ n k =0 � ∞ λ k 0 · · · 0 n · · · k =0 k ! 7 / 29

State transition matrix State transition matrix Definition Given a SV model in which the matrix ❆ has dimension n × n , the state transition matrix is the matrix n × n ∞ ❆ k t k � e ❆ t = k ! k =0 Note: the elements of the state transition matrix e ❆ t are not constant but are function of time. 8 / 29

State transition matrix Properties • Derivative of the state transition matrix: d dt e ❆ t = ❆ e ❆ t = e ❆ t ❆ . • Product of two state transition matrices: e ❆ t e ❆ τ = e ❆ ( t + τ ) . Note: This result is not trivial. For instance the relationship e ❆ t e ❇ t = e ( ❆ + ❇ ) t holds if and only if matrix ❆ and ❇ commute, i.e., ❆❇ = ❇❆ . • Inverse of a state transition matrix The inverse of e ❆ t is the matrix e − ❆ t : e ❆ t e − ❆ t = e − ❆ t e ❆ t = ■ . Note: This implies that the inverse always exists! 9 / 29

State transition matrix Properties • Derivative of the state transition matrix: d dt e ❆ t = ❆ e ❆ t = e ❆ t ❆ . • Product of two state transition matrices: e ❆ t e ❆ τ = e ❆ ( t + τ ) . Note: This result is not trivial. For instance the relationship e ❆ t e ❇ t = e ( ❆ + ❇ ) t holds if and only if matrix ❆ and ❇ commute, i.e., ❆❇ = ❇❆ . • Inverse of a state transition matrix The inverse of e ❆ t is the matrix e − ❆ t : e ❆ t e − ❆ t = e − ❆ t e ❆ t = ■ . Note: This implies that the inverse always exists! 9 / 29

State transition matrix Properties • Derivative of the state transition matrix: d dt e ❆ t = ❆ e ❆ t = e ❆ t ❆ . • Product of two state transition matrices: e ❆ t e ❆ τ = e ❆ ( t + τ ) . Note: This result is not trivial. For instance the relationship e ❆ t e ❇ t = e ( ❆ + ❇ ) t holds if and only if matrix ❆ and ❇ commute, i.e., ❆❇ = ❇❆ . • Inverse of a state transition matrix The inverse of e ❆ t is the matrix e − ❆ t : e ❆ t e − ❆ t = e − ❆ t e ❆ t = ■ . Note: This implies that the inverse always exists! 9 / 29

State transition matrix Computation of e ❆ t for diagonal ❆ matrices If ❆ is a diagonal matrix of dimension n × n :     e λ 1 t λ 1 0 0 0 0 · · · · · · e λ 2 t 0 0 0 0 λ 2  · · ·   · · ·  e ❆ t =     ❆ = vale  . . . . . . . ... ...  . . .   . . .  . . . . . .    e λ n t 0 0 λ n 0 0 · · · · · · Proof . It follows from the definition of matrix exponential. � e − t � − 1 � � 0 0 e ❆ t = Example : Given ❆ = , it holds e − 2 t 0 − 2 0 For a general matrix we can compute it via the so-called Sylvester formula, by diagonalization of the SV model or numerically. More on this in the next lecture. 10 / 29

Outline Introduction The state transition matrix The Lagrange formula 11 / 29

The Lagrange formula Theorem Theorem [Lagrange formula] The evolution of the state and output of a stationary, linear, SV, continuous-time dynamical system given the initial state ① ( t 0 ) and input ✉ ( t ) (for t ≥ t 0 ), is equal to:  ① f ( t )  ① ℓ ( t )   � �� �  � t � �� �   e ❆ ( t − t 0 ) ① ( t 0 ) e ❆ ( t − τ ) ❇✉ ( τ ) d τ   ① ( t ) = +     t 0 � t   ② ( t ) = ❈ e ❆ ( t − t 0 ) ① ( t 0 ) e ❆ ( t − τ ) ❇✉ ( τ ) d τ + ❉✉ ( t )  + ❈     � �� � t 0   ② ℓ ( t ) � �� �    ② f ( t ) In this formula we can recognize: The natural evolution of the state ① ℓ ( t ) and the output ② ℓ ( t ) The forced evolution of the state ① f ( t ) and the output ② f ( t ) 12 / 29

The Lagrange formula Theorem In the particular case t 0 = 0  � t ① ( t ) = e ❆ t ① (0) 0 e ❆ ( t − τ ) ❇✉ ( τ ) d τ +  � t ② ( t ) = ❈ e ❆ t ① (0) 0 e ❆ ( t − τ ) ❇✉ ( τ ) d τ + ❉✉ ( t )  + ❈ 13 / 29

The Lagrange formula Example Compute for t ≥ 0 the state and output evolution of the SV model: � ˙ � 0 � − 1 � x 1 ( t ) � � � �  x 1 ( t ) 1  = + u ( t )   x 2 ( t ) ˙ 0 − 2 x 2 ( t ) 1 � � � 2 x 1 ( t ) 1 �   y ( t ) =  x 2 ( t ) � 3 4 � T . given an input u ( t ) = 2 for t ≥ 0 and an initial state ① (0) = The matrix exponential of the state transition matrix is: � e − t � e − t − e − 2 t e ❆ t = e − 2 t 0 The natural state evolution is by the Lagrange formula: � � � � � � e − t − e − 2 t 4 e − t − e − 2 t e − t 3 ① ℓ ( t ) = e ❆ t ① (0) = = . · e − 2 t e − 2 t 0 4 14 / 29