Chapter 1

Linear Systems Theory 1.1 1.1.1

Classification of systems Continous time versus discrete time

• Continuous time systems evolve with time indices t ∈ ℜ. • Discrete time systems evolve with time indices t ∈ Z = {. . . , −3, −2, −1, 0, 1, 2, . . .}. Usually, the symbol k instead of t is used to denote discrete time indices.

1.1.2

Static versus dynamic systems

• A system with input u and output y is called a static system if ∃ (there existis) a function f (u, t) such that for all t ∈ T , y(t) = f (u(t), t). (1.1) • A static time invariant system is one with y(t) = f (u(t)) for all t. • To determine the output of a static system, only the present input is needed. In contrast, a (causal) dynamic system requires past input to determine the system output. i.e. to determine y(t) one needs to know u(τ ), τ ∈ (−∞, t].

1.2

A state determined dynamic system

The state of a dynamic system at time t0 , x(t0 ), is the extra piece of information needed, so that given the input trajectory u(τ ), τ ≥ t0 , one is able to determine the behavior of the system for all times t ≥ t0 . The behaviors are usually captured by defining appropriate outputs y(t). • State is not unique. Two different pieces of information can both be valid states of the system. • What constitutes a state depends on what behaviors are of interest. • Some authors require a state to be a minimal piece of information. In these notes, we do not require this to be so. An Example: Consider a car with input u(t) being its acceleration. Let y(t) be the position of the car. 5

1. If the behavior of interest is just the speed of the car, then x(t) = y(t) ˙ can be used as the state. It is qualified to be a state because given u(τ ), τ ∈ [t0 , t], the speed at t is obtained by: Z t v(t) = y(t) ˙ = x(t0 ) + u(τ )dτ. t0

  y(t) 2. If the behavior of interest is the position of the car, then xa (t) = ∈ ℜ2 can be used as y(t) ˙ the state.   y(t) + 2y(t) ˙ 3. An alternate state might be xb (t) = . Obviously, since we can determine the y(t) ˙ old state vector xa (t) from this alternate one xb (t), and vice versa, both are valid state vectors. This illustrates the point that state vectors are not unique. Remark 1.2.1 1. If y(t) is defined to be the behavior of interest, then by taking t = t0 , the definition of a state determined system implies that one can determine y(t) from the state x(t) and input u(t), at time t. i.e. there is a static output function h(·, ·, ·) so that the output y(t) is given by: y(t) = h(x(t), u(t), t) h : (x, u, t) 7→ y(t) is called the output readout map. 2. The usual representation of continuous time dynamical system is given by the form: x˙ = f (x, u, t) y = h(x, u, t) and for discrete time system, x(k + 1) = f (x(k), u(k), k) y(k) = h(x(k), u(k), k) 3. Notice that a state determined dynamic system defines, for every pair of initial and final times, t0 and t1 , a mapping (or transformation) of the initial state x(t0 ) = x0 and input trajectory u(τ ), τ ∈ [t0 , t1 ] to the state at a time t1 , x(t1 ). In these notes, we shall use the notation: s(t1 , t0 , x0 , u(·)) to denote this state transition mapping. i.e. x(t1 ) = s(t1 , t0 , x0 , u(·)) if the initial state at time t0 is x0 , and the input trajectory is given by u(·). A state transition map must satisfy two important properties: • State transition property For any t0 ≤ t1 , if two input signals u1 (t) and u2 (t) are such that u1 (t) = u2 (t) ∀t ∈ [t0 , t1 ], then s(t1 , t0 , x0 , u1 (·)) = s(t1 , t0 , x0 , u2 (·)) i.e. if x(t0 ) = x0 , then the final state x(t1 ) depends only on past inputs (from t1 ) that occur after t0 , when the initial state is specified. Systems like this is called causal because the state does not depend on future inputs.

• Semi-group property For all t2 ≥ t1 ≥ t0 ∈ T , for all x0 , and for all u(·), s(t2 , t1 , x(t1 ), u) = s(t2 , t1 , s(t1 , t0 , x0 , u), u) = s(t2 , t0 , x0 , u) Thus, when calculating the state at time t2 , we can first calculate the state at some intermediate time t1 , and then utilize this result to calculate the state at time t2 in terms of x(t1 ) and u(t) for t ∈ [t1 , t2 ].

1.3

Linear Differential Systems

The most common systems we will study are of the form: x˙ = A(t)x + B(t)u

( = f (x, u, t) )

(1.2)

y = C(t)x + D(t)u

( = h(x, u, t) )

(1.3)

where x(t) ∈ ℜn , u : [0, ∞) → ℜm and y : [0, ∞) → ℜp . A(t), B(t), C(t), D(t) are matrices with compatible dimensions of real-valued functions. This system may have been obtained from the Jacobian linearization of a nonlinear system x˙ = f (x, t),

y = h(t, x, u)

about a pair of nominal input and state trajectories (¯ u(t), x ¯(t)). Assumption 1.3.1 We assume that A(t), B(t) and the input u(t) are piecewise continuous functions of t. In other words, a function f (t) is piecewise continuous in t if it is continuous in t except possibly at points of a set which contains at most a finite number of points per unit interval. Also, at points of discontinuity, both the left and right limits exist. Remark 1.3.1 • Assumption 1.3.1 allows us to claim that given an initial state x(t0 ), and a input function u(τ ), τ ∈ [t0 , t1 ], the solution, Z t x(t) = s(t, t0 , x(t0 ), u(·)) = x( t0 ) + [A(τ )x(τ ) + B(τ )u(τ )]dτ t0

exists and is unique via the Fundamental Theorem of Ordinary Differential Equations (see below) often proved in a course on nonlinear systems. • Uniqueness and existence of solutions of nonlinear differential equations are not generally guaranteed. Consider the following examples. An example in which existence is a problem: x˙ = 1 + x2 with x(t = 0) = 0. This system does not have a solution for t ≥ π/2 (x(t) = tan t otherwise. This phenomenon is called finite esacpe. The other issue is whether if a differential equation can have more than one solution for the same initial condition (non-uniqueness). e.g. for the system: 2

x˙ = 3x 3 ,

x(0) = 0.

both x(t) = 0, ∀t ≥ 0 and x(t) = t3 are both valid solutions.

Theorem 1.3.1 (Fundamental Theorem of ODE’s) Consider the following ODE x˙ = f (x, t)

(1.4)

where x(t) ∈ ℜn , t ≥ 0, and f : ℜn × ℜ+ → ℜn . On a time interval [t0 , t1 ], if the function f (x, t) satisfies 1. For any x ∈ ℜn , the function t 7→ f (x, t) is piecewise continuous, 2. There exists a piecewise continuous function, k(t), such that for all x1 , x2 ∈ ℜn , kf (x1 , t) − f (x2 , t)k ≤ k(t)kx1 − x2 k . Then, 1. there exists a solution to the differential equation x˙ = f (x, t) for all t, meaning that: a) For each initial state xo ∈ Rn , there is a continuous function φ : R+ → Rn such that φ(to ) = x0 and ˙ = f (φ, t) φ(t)

∀t∈ℜ

2. Moreover, the function φ is unique. [If φ1 (·) and φ2 (·) have the same properties above, then they must be the same.] Remark 1.3.2 This version of the fundamental theorem of ODE is taken from [Callier and Desoer, 91]. A weaker condition (and result) is that on the interval [t0 , t1 ], the second condition says that there exists a constant k[t0 ,t1 ] such that for any t ∈ [t0 , t1 ], the function x 7→ f (x, t), for all x1 , x2 ∈ Rn , satisfies kf (x1 , t) − f (x2 , t)k ≤ k[t0 ,t1 ] kx1 − x2 k . In this case, the solution exists and is unique on [t0 , t1 ]. The proof of this result can be found in many books on ODE’s or dynamic systems usually proved in details in a Nonlinear Systems Analysis class.

1.3.1

1

and is

Linearity Property

The reason why the system (1.2)-(1.3) is called a linear differential system is because of the following linearity property. Theorem 1.3.2 For any pairs of initial and final times t0 , t1 ∈ ℜ, the state transition map s : (t0 , t1 , x(t0 ), u(·)) 7→ x(t1 ) of the linear differential system (1.2)-(1.3) is a linear map of the pair of the initial state x(t0 ) and the input u(τ ), τ ∈ [t0 , t1 ]. 1

e.g. Vidyasagar, Nonlinear Systems Analysis, 2nd Prentice Hall, 93, or Khalil Nonlinear Systems, McMillan, 92

′ n In order words, for any u(·), u′ (·) ∈ ℜm [t0 ,t1 ] , x, x ∈ ℜ and α, β ∈ ℜ,

s(t, t0 , (αx + βx′ ), (αu(·) + βu′ (·))) = α · s(t, t0 , x, u(·)) + β · s(t, t0 , x′ , u′ (·)).

(1.5)

Similarly, for each pair of t0 , t1 , the mapping ρ : (t0 , t1 , x(t0 ), u(·)) 7→ y(t1 ) from the initial state x(t0 ) and the input u(τ ), τ ∈ [t0 , t1 ], to the output y(t1 ) is also a linear map, i.e. ρ(t, t0 , (αx + βx′ ), (αu(·) + βu′ (·))) = α · ρ(t, t0 , x, u(·)) + β · ρ(t, t0 , x′ , u′ (·)). (1.6) Before we prove this theorem, let us point out a very useful principle for proving that two time functions x(t), and x′ (t), t ∈ [t0 , t1 ], are the same. Lemma 1.3.3 Given two time signals, x(t) and x′ (t). Suppose that • x(t) and x′ (t) satisfies the same differential equation, p˙ = f (t, p)

(1.7)

• they have the same initial conditions, i.e. x(t0 ) = x′ (t0 ) • The differential equation (1.7) has unique solution on the time interval [t0 , t1 ], then x(t) = x′ (t) for all t ∈ [t0 , t1 ]. Proof: (of Theorem 1.3.2) We shall apply the above lemma (principle) to (1.5). Let t0 be an initial time, (x0 , u(·)) and (x′ , u′ (·)) be two pairs of initial state and input, producing state and output trajectories x(t), y(t) and x′ (t), y ′ (t) respectively. We need to show that if the initial state is x′′ (t0 ) = αx0 + βx0 ′, and input u′′ (t) = αu(t) + βu′ (t) for all t ∈ [t0 , t1 ], then at any time t, the response y ′′ (t) is given by the function y ′′ (t) := αy(t) + βy ′ (t).

(1.8)

We will first show that for all t ≥ t0 , the state trajectory x′′ (t) is given by: x′′ (t) = αx(t) + βx′ (t).

(1.9)

Denote the RHS of (1.9) by g(t). To prove (1.9), we use the fact that (1.2) has unique solutions. Clearly, (1.9) is true at t = t0 , x′′ (t0 ) = αx0 + βx′0 = αx(t0 ) + βx′ (t0 ) = g(t0 ). By definition, if x′′ is a solution to (1.2), x˙ ′′ (t) = A(t)x′′ (t) + (αu(t) + βu′ (t)). Moreover, g(t) ˙ = αx(t) ˙ + β x˙ ′ (t)   = α [A(t)x(t) + B(t)u(t)] + β A(t)x′ (t) + B(t)u′ (t)

= A(t)[αx(t) + βx′ (t)] + [αu(t) + βu′ (t)] = A(t)g(t) + [αu(t) + βu′ (t)].

Hence, g(t) and x′′ (t) satisfy the same differential equation (1.2). Thus, by the existence and uniqueness property of the linear differential system, the solution is unique for each initial time t0 and initial state. Hence x′′ (t) = g(t). ⋄

1.4

Decomposition of the transition map

Because of the linearity property, the transition map of the linear differential system (1.2) can be decomposed into two parts: s(t, t0 , x0 , u) = s(t, t0 , x0 , 0u ) + s(t, t0 , 0, u) where 0u denotes the identically zero input function (u(τ ) = 0 for all τ ). It is so because we can decompose (x0 , u) ∈ ℜn × U into (x0 , u) = (x0 , 0u ) + (0, u) and then apply the defining property of a linear dynamical system to this decomposition. Because of this property, the zero-state response (i.e. the response of the system when the inital state is x(t0 ) = 0) satisfies the familiar superposition property: ρ(t, t0 , x = 0, αu + βu′ ) = αρ(t, t0 , x = 0, u) + βρ(t, t0 , x = 0, u′ ). Similarly, the zero-input response satisfies a superposition property: ρ(t, t0 , αx + βx′ , 0u ) = αρ(t, t0 , x, 0u ) + βρ(t, t0 , x′ , 0u ).

1.5

Zero-input transition and the Transition Matrix

From the proof of linearity of the Linear Differential Equation, we actually showed that the state transition function s(t, t0 , x0 , u) is linear with respect to (x0 , u). In particular, for zero input (u = 0u ), it is linear with respect to the initial state x0 . We call the transition when the input is identically 0, the zero-input transition. It is easy to show, by choosing x0 to be columns in an identity matrix successively (i.e. invoking the so called 1st representation theorem - Ref: Desoer and Callier, or Chen), that there exists a matrix, Φ(t, t0 ) ∈ ℜn×n so that s(t, t0 , x0 , 0u ) = Φ(t, t0 )x0 . This matrix function is called the transition matrix. Claim: Φ(t, t0 ) satisfies (1.2). i.e. ∂ Φ(t, t0 ) = A(t)Φ(t, t0 ) ∂t

(1.10)

and Φ(t0 , t0 ) = I. Proof: Consider an arbitrary initial state x0 ∈ ℜn and zero input. By definition, x(t) = Φ(t, t0 )x0 = s(t, t0 , x0 , 0u ). Differentiating the above,

∂ Φ(t, t0 )x0 = A(t)Φ(t, t0 )x0 . ∂t Now pick sucessively n different initial conditions x0 = e1 , x0 = e2 , · · · , x0 = en so that {e1 , · · · , en } form a basis of ℜn . (We can take for example, ei to be the i − th column of the identity matrix). Thus,

∂ Φ(t, t0 ) e1 e2 · · · en = A(t)Φ(t, t0 ) e1 e2 · · · en ∂t
Since e1 e2 · · · en ∈ ℜn×n is invertible, we multiply both sides by its inverse to obtain the required answer. ⋄ x(t) ˙ =

Definition 1.5.1 A n × n matrix X(t) that satisfies the system equation, ˙ X(t) = A(t)X(t) and X(τ ) is non-singular for some τ , is called a fundamental matrix. Remark 1.5.1 The transition matrix is a fundamental matrix. It is invertible at at least t = t0 . Proposition 1.5.1 Let X(t) ∈ ℜn×n be a fundamental matrix. Then, X(t) is non-singular at all t ∈ ℜ. Proof: Since X(t) is a fundamental matrix, X(t1 ) is nonsingular for some t1 . Suppose that X(τ ) is singular, then ∃ a non-zero vector k ∈ ℜn s.t. X(τ )k = 0, the zero vector in ℜn . Consider now the differential equation: x(τ ) = X(τ )k = 0 ∈ ℜn .

x˙ = A(t)x,

Then, x(t) = 0 for all t is the unique solution to this system with x(τ ) = 0. However, the function x′ (t) = X(t)k satisfies x˙ ′ (t) = A(t)X(t)k = A(t)x′ (t), and x′ (τ ) = x(τ ) = 0. Thus, by the uniqueness of differential equation, x(t) = x′ (t) for all t. Hence x′ (t1 ) = 0 extracting a contradiction because X(t1 ) is nonsingular so x′ (t1 ) = X(t1 )k is non-zero. ⋄

1.5.1

Properties of Transition matrices

1. (Existence and Uniqueness) Φ(t, t0 ) exists and is unique for each t, t0 ∈ ℜ (t ≥ t0 is not necessary). 2. (Group property) For all t, t1 , t0 (not necessarily t0 ≤ t ≤ t1 ) Φ(t, t0 ) = Φ(t, t1 )Φ(t1 , t0 ). 3. (Non-singularity) Φ(t, t0 ) = [Φ(t0 , t)]−1 . 4. (Spliting property) If X(t) is any fundamental matrix, then Φ(t, t0 ) = X(t)X(t0 )−1 5.

∂ ∂t Φ(t, t0 )

6.

∂ ∂t0 Φ(t, t0 )

= A(t)Φ(t, t0 ), and Φ(t0 , t0 ) = I for each t0 ∈ ℜ. = −Φ(t, t0 )A(t0 ).

Proof: 1. This comes from the fact that the solution s(t, t0 , x0 , 0u ) exists for all t, t0 ∈ ℜ and for all x0 ∈ ℜn (fundamental theorem of differential equation) and that s(t, t0 , x0 , 0u ) is linear in x0 .

2. Differentiate both sides with respect to t to find that RHS and LHS satisfy the same differential equation and have the same value at t = t1 . So apply existence and uniqueness theorem. 3. From 2), take t = t0 !. 4. Use the existence and uniquenss theorem again : For each t0 , consider both sides as functions d of t. So, LHS(t = t0 ) = RHS(t = t0 ). Now, the LHS satisfies, dt Φ(t, t0 ) = A(t)Φ(t, t0 ). The RHS satisfies:   d d [X(t)X(t0 )] = X(t) X(t0 ) = A(t)X(t)X(t0 )−1 dt dt d RHS(t) = A(t)RHS(t). So LHS and RHS satisfy same differential equation and Hence dt agree at t = t0 . Hence they must be the same at all t.

5. Already shown. 6. From 3) we have Φ(t, t0 )Φ(t0 , t) = I, the identity matrix. Differentiate both sides with respect to t0 ,     ∂ ∂ Φ(t, t0 ) Φ(t0 , t) + Φ(t, t0 ) Φ(t0 , t) = 0 ∂t0 ∂t0 since dtd0 [X(t0 )Y (t0 )] = dtd0 X(t0 )Y (t0 ) + X(t0 ) dtd0 Y (t0 ) for X(·), Y (t0 ) a matrix of functions of t0 (verify this!). Hence,     ∂ ∂ Φ(t, t0 ) Φ(t0 , t) = −Φ(t, t0 ) Φ(t0 , t) ∂t0 ∂t0   ∂ ⇒ Φ(t, t0 ) Φ(t0 , t) = −Φ(t, t0 )A(t0 )Φ(t0 , t) ∂t0 ∂ Φ(t, t0 ) = −Φ(t, t0 )A(t0 )Φ(t0 , t)Φ(t0 , t)−1 = −Φ(t, t0 )A(t0 ) ⇒ ∂t0 ⋄

1.5.2

Explicit formula for Φ(t, t0 )

The Peano-Baker formula is given by: Φ(t, t0 ) = I +

Z

t

t0

A(σ1 )dσ1 +

Z

t



σ1

A(σ2 )dσ2 dσ1 Z Z σ 1 Z t + A(σ2 ) A(σ1 ) A(σ1 )

t0

Z

t0

t0

t0

σ2





A(σ3 )dσ3 dσ2 dσ1 + ..... (1.11)

t0

This formula can be verified formally (do it) by checking that the RHS(t = t0 ) is indeed I and that the RHS satisfies the differential equation ∂ RHS(t, t0 ) = A(t)RHS(t, t0 ). ∂t Let us define the exponential of a matrix, A ∈ ℜn+n using the power series, exp(A) = I +

A A2 Ak + + ··· + ··· 1 2! k!

by mimicking the power series expansion of the exponential function of a real number. If A(t) = A is a constant, then the Peano-Bakar formula reduces to (prove it!) Φ(t, t0 ) = exp [(t − t0 ) · A] For example, examining the 4th term, Z Z t A(σ1 ) t0 3

=A

=A3

Z tZ

t0 σ1 Z σ2

t0 t0 Z t Z σ1 t0 Z t

σ1

A(σ2 )

Z

σ2





A(σ3 )dσ3 dσ2 dσ1 t0

dσ3 dσ2 dσ1

t0

(σ2 − t0 )dσ2 dσ1

t0

(σ1 − t0 )2 (t − t0 )3 dσ1 = A3 2 3! t0 Rt A more general case is that if A(t) and t0 A(τ )dτ , commute for all t,  Z t A(τ )dτ Φ(t, t0 ) = exp =A3

(1.12)

t0

An intermediate step is to show that: k k Z t Z t k+1 Z t 1 d A(τ )dτ A(t). A(τ )dτ = = A(t) A(τ )dτ k + 1 dt 0 t0 t0 Notice that (1.12) does not in general apply unless the matrices commute. i.e. k k Z t Z t A(τ )dτ A(t). A(τ )dτ = A(t) t0

t0

Each of the following situations will guarantee that the condition A(t) and for all t is satisfied:

Rt

t0

A(τ )dτ commute

1. A(t) is constant. 2. A(t) = α(t)M where α(t) is a scalar function, and M ∈ ℜn×n is a constant matrix; P 3. A(t) = i αi (t)Mi where {Mi } are constant matrices that commute: Mi Mj = Mj Mi , and αi (t) are scalar functions; 4. A(t) has a time-invariant basis of eigenvectors spanning ℜn .

1.5.3

Computation of Φ(t, 0) = exp(tA) for constant A

The definining computational method for Φ(t, t0 ) not necessarily for time varying A(t) is: ∂ Φ(t, t0 ) = A(t)Φ(t, t0 ); Φ(t0 , t0 ) = I ∂t When A(t) = constant, there are algebraic approaches available: Φ(t1 , t0 ) = exp(A(t1 − t0 )) := I + Matlab provides expm(A ∗ (t1 − t0 )).

A(t1 − t0 ) A2 (t1 − t0 )2 + + ... 1! 2!

Laplace transform approach L(Φ(t, 0)) = [sI − A]−1 Proof: Since Φ(0, 0) = I, and ∂ Φ(t, 0) = A(t)Φ(t, 0) ∂t take Laplace transform of the (Matrix) ODE: ˆ 0) − Φ(0, 0) = AΦ(s, ˆ 0) sΦ(s, ˆ 0) is the Laplace transform of Φ(t, 0) when treated as function of t. This gives: where Φ(s, ˆ 0) = I (sI − A)Φ(s, Example

  −1 0 A= 1 −2 −1

(SI − A)

 −1   1 s+1 0 s+2 0 = = −1 s + 2 1 s+1 (s + 1)(s + 2)   1/(s + 1) 0 = 1/(s + 1)(s + 2) 1/(s + 2)

Taking the inverse Laplace transform (using e.g. a table) term by term,   e−t 0 Φ(t, 0) = −t e − e−2t e−2t Similarity Transform (decoupled system) approach Let A ∈ ℜn×n , if v ∈ C n and λ ∈ C satisfy A · v = λv then, v is an eigenvector and λ its associated eigenvalue. If A ∈ ℜn×n has n distinct eigenvalues λ1 , λ2 , . . . λn , then A is called simple. If A ∈ ℜn×n has n independent eigenvectors then A is called semi-simple (Notice that A is necessarily semi-simple if A is simple). Suppose A ∈ ℜn×n is simple, then let
T = v1 v2 . . . vn ; Λ = diag(λ1 , λ2 , . . . , λn )

be the collection of eigenvectors, and the associated eigenvalues. Then, the so-called similarity transform is: A = T ΛT −1 (1.13) Remark • A sufficient condition for A being semi-simple is if A has n distinct eigenvalues (i.e. it is simple);

• When A is not semi-simple, a similar decomposition as (1.13) is available except that Λ will be in Jordan form (has 1’s in the super diagonals) and T consists of eigenvectors and generalized eigenvectors). This topic is covered in most Linear Systems or linear algebra textbook Now exp(At) := I +

At A2 t2 + + ... 1! 2!

Notice that Ak = T Λk T −1 , thus,  Λt Λ2 t2 + + . . . T −1 = T exp(Λt)T −1 exp(tA) = T I + 1! 2! 

The above formula is valid even if A is not semi-simple. In the semi-simple case, h i Λk = diag λk1 , λk2 , . . . , λkn

so that

If we write T −1

exp(Λt) = diag [exp(λ1 t), exp(λ2 t), . . . , exp(λn t)]  T w1 w T   2 =  .  where wiT is the i-th row of T −1 , then,  ..  wnT exp(tA) = T exp(Λt)T

−1

=

n X

exp(λi t)vi wiT .

i=1

This is the dyadic expansion of exp(tA). Can you show that wiT A = λi wiT ? This means that wi are the left eigenvectors of A. The expansion shows that the system has been decomposed into a set of simple, decoupled 1st order systems. Example   −1 0 A= 1 −2 The eigenvalues and eigenvectors are:

  1 λ1 = −1, v1 = ; 1

  0 λ2 = −2, v2 = . 1

Thus,       1 0 e−t 1 0 0 e−t 0 exp(At) = , = −t 1 1 −1 1 0 e−2t e − e−2t e−2t same as by the Laplace transform method. Digression: Modal decomposition The (generalized 2 ) eigenvectors are good basis for a coordinate system. 2

for the case the A is not semi-simple, additional “nice” vectors are needed to form a basis, they are called generalized eigenvectors

Suppose that A = T ΛT −1 is the similarity transform. Let x ∈ ℜn , z = T −1 x, and x = T z, Thus, x = z1 v1 + z2 v2 + . . . zn vn . where T = [v1 , v2 , . . . , vn ] ∈ ℜn×n . Hence, x is decomposed into the components in the direction of the eigenvectors with zi being the scaling for the i−th eigenvector. The original linear differential equation is written as: x˙ = Ax + Bu T z˙ = T Λz + Bu z˙ = Λz + T −1 Bu ¯ = T −1 B, then, since Λ is diagonal, If we denote B ¯i u z˙i = λi zi + B ¯i is the i − th row of B. ¯ where zi is the i−th component of z and B This is a set of n decoupled first order differential equations that can be analyzed independently. Dynamics in each eigenvector direction is called a mode. If desired, z can be used to re-constitute x via x = T z.

1.6

Zero-State transition and response

Recall that for a linear differential system, x(t) ˙ = A(t)x(t) + B(t)u(t)

x(t) ∈ ℜn , u(t) ∈ ℜm

the state transition map can be decomposed into the zero-input response and the zero-state response: s(t, t0 , x0 , u) = Φ(t, t0 )x0 + s(t, t0 , 0x , u) Having figured out the zero-input state component, we now derive the zero-state response.

1.6.1

Heuristic guess

We first decompose the inputs into piecewise continuous parts {ui : ℜ → ℜm } for i = · · · , −2, −1, 1, 0, 1, · · · , ( u(t0 + h · i) t0 + h · i ≤ t < t0 + h · (i + 1) ui (t) = 0 otherwise where h > 0 is a small positive number. Intuitively we can see that as h → 0, u(t) =

∞ X

ui (t) as h → 0.

i=−∞

Let u ¯(t) =

P∞

i=−∞ ui (t).

By linearity of the transition map, X s(t, t0 , 0, ui ). s(t, t0 , 0, u ¯) = i

Now we figure out s(t, t0 , 0, ui ).

• Step 1: t0 ≤ t < t0 + h · i. Since u(τ ) = 0, τ ∈ [t0 , t0 + h · i) and x(t0 ) = 0, x(t) = 0

for t0 ≤ t < t0 + h · i

• Step 2: t ∈ [t0 + h · i, t0 + h(i + 1)). Input is active: x(t) ≈x(t0 + h · i) + [A(t0 + h · i) + B(t0 + h · i)u(t0 + h · i)] · ∆T = [B(t0 + h · i)u(t0 + h · i)] · ∆T where ∆T = t − t0 + h · i. • Step 3: t ≥ t0 + h · (i + 1). Input is no longer active, ui (t) = 0. So the state is again given by the zero-input transition map: Φ (t, t0 + h · (i + 1)) B(t0 + i · h)u(t0 + h · i) | {z } ≈x(t0 +(i+1)·h)

Since Φ(t, t0 ) is continuous, if we make the approximation

Φ(t, t0 + (h + 1)i) ≈ Φ(t, t0 + h · i) we only introduce second order error in h. Hence, s(t, t0 , x0 , ui ) ≈ Φ(t, t0 + h · i)B(t0 + h · i)u(t0 + h · i). The total zero-state state transition due to the input u(·) is therefore given by: (t−t0 )/h

s(t, t0 , 0, u) ≈

X

Φ (t, t0 + h · i) B(t0 + i · h)u(t0 + h · i)

i=0

As h → 0, the sum becomes an integral so that: Z t s(t, t0 , 0, u) = Φ(t, τ )B(τ )u(τ )dτ.

(1.14)

t0

In this heuristic derivation, we can see that Φ(t, τ )B(τ )u(τ ) is the contribution to the state x(t) due to the input u(τ ) for τ < t.

1.6.2

Formal Proof of zero-state transition map

We will show that for all t, t0 ∈ ℜ+ , (x(t) = s(t, t0 , 0x , u)) =

Z

t

Φ(t, τ )B(τ )u(τ )dτ.

(1.15)

t0

Clearly (1.15) is correct for t = t0 . We will now show that the LHS of (1.15), i.e. x(t) and the RHS of (1.15), which we will denote by z(t) satisfy the same differential equation.

11111111111 00000000000 00000000000 sigma 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 sigma = tau 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111

tau Figure 1.1: Changing the order of integral in the proof of zero-state transition function We know that x(t) satisfies, x(t) ˙ = A(t)x(t) + B(t)u(t). Observe first that since

d dt Φ(t, t0 )

= A(t)Φ(t, t0 ), Z t A(σ)Φ(σ, τ )dτ. Φ(t, τ ) = I + τ

Now for the RHS of (1.15) which we will call z(t), Z t z(t) := Φ(t, τ )B(τ )u(τ )dτ t0  Z t Z t Z t A(σ)Φ(σ, τ )dσ B(τ )u(τ )dτ = B(τ )u(τ )dτ + t0

τ

t0

let f (σ, τ ) := A(σ)Φ(σ, τ )B(τ )u(τ ) and then changing the order of the integral (see fig 1.1) = = =

Z

t

B(τ )u(τ )dτ +

t0 Z t

t0 Z t

B(τ )u(τ )dτ + B(τ )u(τ )dτ +

t0

Z tZ

t0 Z t

t0 Z t

σ

f (σ, τ )dτ dσ t0

A(σ)

Z

σ

Φ(σ, τ )B(τ )u(τ )dτ dσ t0

A(σ)z(σ)dσ

t0

which on differentiation w.r.t. t gives z(t) ˙ = B(t)u(t) + A(t)z(t). Hence, both z(t) and x(t) satisfy the same differential equation and have the same values at t = t0 . Therefore, x(t) = z(t) for all t.

1.6.3

Output Response function

The combined effect of initial state x0 and input function u(·) on the state is given by: Z t0 Φ(t, τ )B(τ )u(τ )dτ. s(t, t0 , x0 , u) = Φ(t, t0 )x0 + t

Deriving the output response is simple since: y(t) = C(t)x(t) + D(t)u(t). where C(t) ∈ ℜp×n , D(t) ∈ ℜp×m . Hence, the output response map y(t) = ρ(t, t0 , x0 , u) is simply Z t y(t) = ρ(t, t0 , x0 , u) = C(t)Φ(t, t0 )x0 + C(t) Φ(t, τ )B(τ )u(τ )dτ + D(t)u(t). (1.16) t0

1.6.4

Impulse Response Matrix

When the initial state is 0, let the input be an Dirac impulse occurring at time τ in the j−th input,



uj (t) = ǫj δ(t − τ )   0

  ..   .   0    where ǫj is the j−th unit vector  ǫj =  1 j-th row   0   .   .. 0

    . The output response is:    

yj (t) = [C(t)Φ(t, τ )B(τ ) + D(t)δ(t − τ )]ǫj

The matrix

( [C(t)Φ(t, τ )B(τ ) + D(t)δ(t − τ )] H(t, τ ) = 0

∀t ≥ τ t k ≥ k1 . 2. Existence of inverse: If k1 ≥ k0 , Φ(k1 , k0 )−1 exists and is given by Φ(k1 , k0 )−1 = A(k0 )−1 A(k0 + 1)−1 · · · A(k1 − 1)−1 if and only if A(k) is invertible for all k1 > k ≥ k0 . 3. Semi-group property: Φ(k2 , k0 ) = Φ(k2 , k1 )Φ(k1 , k0 ) for k2 ≥ k1 ≥ k0 only, unless A(k) is invertible. 4. Matrix difference equations: Φ(k + 1, k0 ) = A(k)Φ(k, k0 ) Φ(k1 , k − 1) = Φ(k1 , k)A(k − 1) Can you formulate a property for discrete time transition matrices similar to the splitting property for continuous time case?

1.7.2

Zero-initial state response

The zero-initial initial state response can be obtained easily x(k) = A(k − 1)x(k − 1) + B(k − 1)u(k − 1) = A(k − 1)A(k − 2)x(k − 2) + A(k − 1)B(k − 2)u(k − 2) + B(k − 1)u(k − 1) .. . = A(k − 1)A(k − 2) . . . A(k0 )x(k0 ) +

k−1 X

k−1 Πj=i+1 A(j)B(i)u(i)

i=k0

Thus, since x(k0 ) = 0 for the the zero-initial state response: s(k, k0 , 0x , u) =

k−1 X

i=k0 k−1 X

i=k0

Φ(k, i + 1)B(i)u(i)

k−1 Πj=i+1 A(j)B(i)u(i)