ECE 602 Fall 2011

A Glimpse at Linear Time-Varying (LTV) Systems

Stanislaw H. Żak

”As we shall see, the concept of time has no meaning before the beginning of the universe. This was first pointed out by St. Augustine. When asked: What did God do before he created the universe? Augustine didn’t reply: He was preparing Hell for people who asked such questions. Instead, he said that time was a property of the universe that God created, and that time did not exist before the beginning of the universe.” Stephen W. Hawking [5, page 8]

1 Motivating Example

Of considerable importance in many applications are linear systems in which the elements of A and B are continuous functions of time. We present a linear time-varying model of a battery as a motivation for our discussion on linear time-varying systems. Our derivation follows that of Chiasson and Vairamohan [3, 4].

The simplest battery model is shown in Figure 1, which is called a practical voltage source. This model consists of an ideal voltage source that produces voltage V_{oc}. We refer to

Figure 1: Practical voltage source as a simple battery model.

Figure 2: Battery model accounting for the different charging, R_{c}, and discharging, R_{d}, resistances.

V_{oc} as the open-circuit voltage. Thus the open-circuit voltage is obtained by measuring the voltage across the open-circuit, that is, when I_{b} = 0. The internal resistance of the practical voltage source is denoted R_{int}. The battery terminal voltage is labeled as V_{b}. This model, however, does not take into account the fact that the internal resistance is different under discharge and charge conditions. To account for this, the circuit of Figure 1 is modified by introducing two different internal resistances; R_{c}—the charging resistance and R_{d}—the discharging resistance. These resistances are connected to the diodes to model charging and discharging modes of the battery operation. The modified circuit is shown in Figure 2. This model, on the other hand, does not take into account the diffusion of the electrolytic through the battery. To model this effect, a polarization capacitor is added to model electro-chemical reaction taking place in the battery. The resulting model is depicted in Figure 3.

In electric vehicles, the state of charge (SOC) of the battery corresponds to a fuel gauge in a conventional, internal combustion, cars. To define the state of charge, consider a completely charged battery. Let I_{b}(t) be the discharging current. Then, the charge delivered by the battery is

Q(t) = ∫_{t_{0}}^{t} I_{b}(τ)dτ.

Let Q_{0} = ∫_{t_{0}}^{∞} I_{b}(τ)dτ be the total charge that battery can hold. Then the state of charge Figure 3: Battery model accounting for the different charging, R_{c}, and discharging, R_{d}, resistances and polarization capacitance, C.

(SOC) of the battery is

SOC(t) = \frac{Q_{0} − Q(t)}{Q_{0}} × 100
= \frac{Q_{0} − ∫_{t_{0}}^{t} I_{b}(τ)dτ}{Q_{0}} × 100.  (1)

Direct application of (1) to estimate the battery SOC is not accurate. An alternative approach is to use an estimate of the open circuit voltage, V_{oc} of the battery. Then, the SOC can be evaluated from the relation,

SOC(t) = \frac{V_{oc}(t) − a_{0}}{a_{1}} ,  (2)

where a_{0} is the battery terminal voltage when SOC(t) = 0% and a_{1} is obtained using a_{0} and V_{oc} at SOC(t) = 100%. Thus estimating the SOC is equivalent to estimating the battery open-circuit voltage V_{oc}(t). Now, in order to measure V_{oc}, the battery must be disconnected from its load, which is not possible during the electric vehicle operation that uses the battery whose SOC we wish to estimate. The approach proposed by Chiasson and Vairamohan [3, 4] is to construct the open-circuit voltage estimator. This estimator construction is based on the battery model that we develop next.

We analyze the condition when the battery is in the discharging mode, that is, when I_{b} ≥ 0. The charging mode is analyzed in a similar way. The battery model in the discharging mode is depicted in Figure 4. To proceed, we apply Kirchhoff’s current law (KCL) at the node between the resistances R_{d} and R_{b} to obtain

I_{c} + I_{b} − I_{d} = 0,

Figure 4: A battery model in the discharging mode of operation.

that is,

I_{c} = I_{d} − I_{b}.

Applying Ohm’s law gives

I_{d} = \frac{V_{oc} − V_{p}}{R_{d}} and I_{b} = \frac{V_{p} − V_{b}}{R_{b}} .  (3)

We also have

I_{c} = CV̇_{p}.

Combining the above equations yields

CV̇_{p} = I_{d} − I_{b}
= \frac{V_{oc} − V_{p}}{R_{d}} − I_{b}
= −\frac{V_{p}}{R_{d}} + \frac{V_{oc}}{R_{d}} − I_{b}.  (4)

Dividing both sides by C, we obtain

V̇_{p} = −\frac{V_{p}}{R_{d}C} + \frac{V_{oc}}{R_{d}C} − \frac{1}{C} I_{b}.  (5)

The output equation is obtained from the second of the equations in (3),

V_{b} = V_{p} − R_{b}I_{b}  (6)

We define the state variables,

x_{1} = V_{p}, x_{2} = \frac{1}{R_{d}C} , x_{3} = \frac{V_{oc}}{R_{d}C} , x_{4} = \frac{1}{C} , and x_{5} = R_{b}.

Applying the state variable definitions to (5) and (6), and taking into account that x_{2}, x_{3}, x_{4}, and x_{5} are constant, we obtain a nonlinear time-varying state-space model,

\begin{matrix}ẋ_{1} & = & −x_{1}x_{2} + x_{3} − I_{b}(t)x_{4} \\ ẋ_{2} & = & 0 \\ ẋ_{3} & = & 0 \\ ẋ_{4} & = & 0 \\ ẋ_{5} & = & 0 \\ y & = & V_{b} = x_{1} − I_{b}(t)x_{5} \end{matrix} \}  (7)

The initial state vector is unknown to us. The state variable x_{3} depends on V_{oc}. Recall that the state variables, x_{2}, x_{3}, x_{4} and x_{5} are constant. Our objective is to compute the open-circuit voltage V_{oc}, which is a constant. We note that

x_{3} = \frac{V_{oc}}{R_{d}C} = V_{oc}x_{2}.

Therefore

V_{oc} = \frac{x_{3}}{x_{2}} = \frac{x_{30}}{x_{20}}

because V_{oc} is constant. Hence, we can determine V_{oc} if we can estimate the initial values of the state variables x_{20} and x_{30}. We proceed by redefining the state variables as

z_{1} = x_{1}, z_{2} = x_{3}, z_{3} = x_{4}, z_{4} = x_{5}.

With the above in mind, and using the fact that x_{2} = constant = x_{20}, we re-write (7) as

[ ż_{1} ; ż_{2} ; ż_{3} ; ż_{4} ] = [ \begin{matrix}−x_{20} & 1 & −I_{b}(t) & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ] [ \begin{matrix}z_{1} \\ z_{2} \\ z_{3} \\ z_{4} \end{matrix} ]  (8)

y = [ 1 0 0 −I_{b}(t) ] [ \begin{matrix}z_{1} \\ z_{2} \\ z_{3} \\ z_{4} \end{matrix} ] .  (9)

The above is a linear LTV systemn because the current I_{b}(t) is time-varying. In summary, we have a homogenous, or free, or unforced, LTV system model

ż = A(t, x_{20})z, z(t_{0})
y = c(t)z.

Our objective is to estimate z(t_{0}) from knowledge of the output y(t) and the current I_{b}(t) on some time interval [t_{0}, t_{f}].

We now outline a solution to the problem of estimating z(t_{0}). The first step is to compute the state-transition matrix, Φ(t, t_{0}), which relates the state at time t to the initial state,

z(t) = Φ(t, t_{0})z(t_{0}).

We then use the output equation to obtain

y(t) = cz(t)
= cΦ(t, t_{0})z(t_{0}).

Premultiply both sides of the above by Φ(t, t_{0})^{⊤}c^{⊤} and then integrate both sides to get

∫_{t_{0}}^{t_{f}} Φ(t, t_{0})^{⊤}c^{⊤}y(t)dt = ∫_{t_{0}}^{t_{f}} Φ(t, t_{0})^{⊤}c^{⊤}cΦ(t, t_{0})dt z(t_{0}).

Let

W_{o}(t_{0}, t_{f}) = ∫_{t_{0}}^{t_{f}} Φ(t, t_{0})^{⊤}c^{⊤}cΦ(t, t_{0})dt,

which, as in the LTI case, we refer to as the observability gramian. If W_{o}(t_{0}, t_{f}) is invertible then

z(t_{0}) = W_{o}(t_{0}, t_{f})^{−1} ∫_{t_{0}}^{t_{f}} Φ(t, t_{0})^{⊤}c^{⊤}y(t)dt.

Having obtained z(t_{0}) we can compute V_{oc} and then the SOC. We refer to Chiasson and Vairamohan [3, 4] for a complete solution to this problem; see also Pang et al. [10], and McIntyre et al. [8] for alternative solutions.

To be able to solve this problem, we need to acquaint ourselves with the state transition matrix for LTV systems. We start with the LTV uncontrolled system.

2 Solving the Uncontrolled LTV System

In general, it is not possible to give explicit expressions for solutions of linear time-varying systems. We obtain some general properties that for some cases will be used to obtain explicit solutions. We first consider the uncontrolled system

ẋ(t) = A(t)x(t), x(t_{0}) = x_{0}.  (10)

Theorem 1 If A(t) is continuous for t ≥ t_{0} then the uncontrolled system modeled by (10) has a unique solution for t ≥ t_{0} given by

x(t) = X(t)x_{0},

where X(t) is the unique n × n matrix satisfying

\frac{dX}{dt} = A(t)X, X(t_{0}) = I_{n}.

Proof See Rugh [11, Chp. 3] or Antsaklis and Michel [1, Section 1.8]. See also Exercise 2.

□

Note that it follows from the above theorem that for every initial state x^{(i)}(t_{0}), i = 1, 2, …, n, there exists a unique solution x^{(i)}(t), i = 1, 2, …, n. We arrange these n solutions as

X(t) = [ x^{(1)}(t) x^{(2)}(t) ⋯ x^{(n)}(t) ] .

Because every x^{(i)}(t) satisfies (10), we have

Ẋ(t) = A(t)X(t).

If X(t_{0}) is nonsingular, then X(t) is called a fundamental matrix of ẋ = A(t)x. Because the initial states are linearly independent, the fundamental matrix is nonsingular—see Exercise 1. The fundamental matrix is not unique because we can choose initial states arbitrarily.

We now analyze a scalar LTV case,

ẋ = a(t)x, x(0) = x_{0}.  (11)

Recall that when a(t) = a =constant, then the solution has the form

x(t) = e^{at}x(0) = e^{∫_{0}^{t} adτ}x(0).

The above formula works for the scalar LTV case, that is,

x(t) = e^{∫_{0}^{t} a(τ)dτ}x(0).  (12)

It is easy to verify that the above is the solution of (11) because

\frac{d}{dt}e^{∫_{0}^{t} a(τ)dτ} = e^{∫_{0}^{t} a(τ)dτ}a(t)
= a(t)e^{∫_{0}^{t} a(τ)dτ}.

That is, a(t) and ∫_{0}^{t} a(τ)dτ commute.

Example 1 We use the solution formula (12) to solve the equation,

ẋ = (sin t)x, x(0) = x_{0}.

We have

x(t) = e^{∫_{0}^{t} a(τ)dτ}x_{0}
= x_{0}e^{∫_{0}^{t} sin(τ)dτ}
= x_{0} e^{− cos t}|_{0}^{t}
= e^{− cos t+1}x_{0}
= e × e^{− cos t}x_{0}.

The solution formula (12) does not generalize to the multi-dimensional case because, in general,

\frac{d}{dt}e^{∫_{0}^{t} A(τ)dτ} ≠ A(t)e^{∫_{0}^{t} A(τ)dτ}.

If, however, A(t) and ∫_{0}^{t} A(τ)dτ commute, then the solution to (10) can be written as

x(t) = e^{∫_{0}^{t} A(τ)dτ}x(0).

We note that A(t) and ∫_{0}^{t} A(τ)dτ commute if

A(t)A(s) = A(s)A(t).

We now analyze a general case when t_{0} is not necessarily equal to 0. If A(t) = A in (10), then the solution is

x(t) = e^{A(t−t_{0})}x(t_{0}).

We re-write the above as

x(t) = e^{At}e^{−At_{0}}x(t_{0}).

Let X(t) = e^{At}, then

X(t_{0})^{−1} = e^{−At_{0}}

and we represent the solution as

x(t) = X(t)X(t_{0})^{−1}x(t_{0})
= Φ(t, t_{0})x(t_{0}),  (13)

where Φ(t, t_{0}) is the state transition matrix.

Definition 1 Let X(t) be a fundamental matrix of (10). Then

Φ(t, t_{0}) = X(t)X(t_{0})^{−1}

is the state transition matrix of (10). The state transition matrix is the unique solution of

\frac{∂}{∂t}Φ(t, t_{0}) = A(t)Φ(t, t_{0}), Φ(t_{0}, t_{0}) = I_{n}.

As in the LTI case, the state transition matrix in the LTV case has the following properties (see Exercise 5):

Φ(τ, τ) = I_{n}  (14)
Φ(t, τ)^{−1} = Φ(τ, t)  (15)
Φ(t, t_{0}) = Φ(t, t_{1})Φ(t_{1}, t_{0})  (16)
\frac{∂}{∂τ} Φ(t, τ) = −Φ(t, τ)A(τ), for all t, τ  (17)

The third of the above properties is referred to as the semigroup property.

Example 2 The state transition matrix of

ẋ = (sin t)x, x(t_{0}) = x_{0},

has the form

Φ(t, t_{0}) = X(t)X(t_{0})^{−1}
= \frac{e^{− cos t+1}}{e^{− cos t_{0}+1}} .

Example 3 We construct the state transition matrix for the LTV system,

ẋ = [ \begin{matrix}0 & −t \\ 0 & 0 \end{matrix} ] x, x(0) = x(0).

The above system consists of two first-order linear differential equations,

ẋ_{1} = −tx_{2}, and ẋ_{2} = 0.

We can easily obtain the solution to the second of the above equations,

x_{2}(t) = x_{2}(0).  (18)

We next obtain the solution to ẋ_{1} = −tx_{2} by integrating its both sides,

x_{1}(t) = x_{1}(0) − ∫_{0}^{t} τx_{2}(0)dτ
= x_{1}(0) − \frac{t^{2}x_{2}(0)}{2}.  (19)

We can now construct a fundamental matrix. Let

x^{(1)}(0) = [ \begin{matrix}1 \\ 1 \end{matrix} ] and x^{(2)}(0) = [ \begin{matrix}0 \\ 1 \end{matrix} ] .

Then using (19) and (18), we obtain the fundamental matrix corresponding to the above initial conditions is

X(t) = [ x^{(1)}(t) x^{(2)}(t) ]
= [ \begin{matrix}1 − \frac{t^{2}}{2} & −\frac{t^{2}}{2} \\ 1 & 1 \end{matrix} ] .

If we select the initial conditions

x^{(1)}(0) = [ \begin{matrix}1 \\ 0 \end{matrix} ] and x^{(2)}(0) = [ \begin{matrix}0 \\ 1 \end{matrix} ]

then we obtain a different fundamental matrix,

X(t) = [ \begin{matrix}1 & −\frac{t^{2}}{2} \\ 0 & 1 \end{matrix} ] .

The state transition matrix, however, is unique and has the form

Φ(t, t_{0}) = X(t)X(t_{0})^{−1}
= [ \begin{matrix}1 − \frac{t^{2}}{2} & −\frac{t^{2}}{2} \\ 1 & 1 \end{matrix} ] [ \begin{matrix}1 − \frac{t_{0}^{2}}{2} & −\frac{t_{0}^{2}}{2} \\ 1 & 1 \end{matrix} ]^{−1}
= [ \begin{matrix}1 − \frac{t^{2}}{2} & −\frac{t^{2}}{2} \\ 1 & 1 \end{matrix} ] [ \begin{matrix}1 & \frac{t_{0}^{2}}{2} \\ −1 & 1 − \frac{t_{0}^{2}}{2} \end{matrix} ]
= [ \begin{matrix}1 & \frac{1}{2}(t_{0}^{2} − t^{2}) \\ 0 & 1 \end{matrix} ] .

Example 4 We construct the state-transition matrix for the LTV system model found in Luenberger [7, p. 115],

ẋ = [ \begin{matrix}0 & 0 \\ t & \frac{1}{t} \end{matrix} ] x, t ≥ 1.

Then, for the given x(2) = [ 2 −1 ]^{⊤}, we compute x(1).

First, we solve for x_{1} to obtain,

x_{1}(t) = x_{1}(1), t ≥ 1

Substituting the above into the equation for ẋ_{2} gives

ẋ_{2} = tx_{1}(t) + \frac{x_{2}(t)}{t}.

Next, we make the substitution

z(t) = \frac{x_{2}(t)}{t}.

Computing the derivative of z(t) yields

ż = \frac{ẋ_{2}t − x_{2}}{t^{2}}
= \frac{ẋ_{2}}{t} − \frac{x_{2}}{t^{2}} .

Multiplying both sides of the above by t, we obtain

tż = ẋ_{2} − \frac{x_{2}}{t^{2}}
= ẋ_{2} − z.  (20)

We substitute ẋ_{2} = tx_{1}(1) + z into 20 to obtain

tż + z = tx_{1}(1) + z,

that is,

ż = x_{1}(1).

Integrating the above from 1 to t gives

z(t) = z(1) + x_{1}(1)(t − 1).

Therefore,

z = \frac{x_{2}}{t}
= z(1) + x_{1}(1)(t − 1),

and multiplying both sides by t, we obtain

x_{2}(t) = tx_{2}(1) + tx_{1}(1)(t − 1).

We are now ready to generate a fundamental matrix. Let

x^{(1)}(1) = [ \begin{matrix}1 \\ 0 \end{matrix} ] and x^{(2)}(1) = [ \begin{matrix}0 \\ 1 \end{matrix} ] ,

then we have

x^{(1)}(t) = [ \begin{matrix}1 \\ t^{2} − t \end{matrix} ] and x^{(2)}(t) = [ \begin{matrix}0 \\ t \end{matrix} ] .

The fundamental matrix corresponding to the above initial conditions is

X(t) = [ x^{(1)}(t) x^{(2)}(t) ] = [ \begin{matrix}1 & 0 \\ t^{2} − t & t \end{matrix} ] .

The state transition matrix is

Φ(t, τ) = X(t)X(τ)^{−1}
= [ \begin{matrix}1 & 0 \\ t^{2} − t & t \end{matrix} ] [ \begin{matrix}1 & 0 \\ τ^{2} − τ & τ \end{matrix} ]^{−1}

= [ \begin{matrix}1 & 0 \\ t^{2} − t & t \end{matrix} ] [ \begin{matrix}1 & 0 \\ −τ + 1 & \frac{1}{τ} \end{matrix} ]

= [ \begin{matrix}1 & 0 \\ t(t − τ) & \frac{t}{τ} \end{matrix} ] .

Note that

x(t) = Φ(t, τ)x(τ).

Thus, for example, if we are given x(2) = [ 2 −1 ]^{⊤} and we wish to find x(1), we have,

x(1) = Φ(1, 2)x(2)
= [ \begin{matrix}1 & 0 \\ −1 & \frac{1}{2} \end{matrix} ] [ \begin{matrix}2 \\ −1 \end{matrix} ]

= [ \begin{matrix}2 \\ −\frac{5}{2} \end{matrix} ] .

3 Solving the LTV Controlled System

We now discuss the problem of obtaining the solution to the LTV controlled system

ẋ = A(t)x + B(t)u, x(t_{0}) = x_{0}.  (21)

We claim that, by analogy with the LTI controlled system, the solution to (21) is

x(t) = Φ(t, t_{0})x_{0} + ∫_{t_{0}}^{t} Φ(t, τ)B(τ)u(τ)dτ.  (22)

We can prove our claim by verification utilizing Leibniz’s rule (see Exercise 7),

\frac{∂}{∂t} ∫_{t_{0}}^{t} f(t, τ)dτ = ∫_{t_{0}}^{t} (\frac{∂}{∂t}f(t, τ)) dτ + f(t, τ)|_{τ=t}  (23)

Example 5 (R. A. DeCarlo) The LTV state model,

ẋ(t) = A(t)x(t) + [ \begin{matrix}πe^{−α(t)^{2}} \\ 0 \end{matrix} ] u(t),

has associated state transition matrix

Φ(t, τ) = e^{α(τ)^{2}−α(t)^{2}} [ \begin{matrix}1 & 0 \\ cos(πt) − cos(πτ) & 1 \end{matrix} ] ,

where α(·) is a continuous function.

1. Compute A(t) using the given state transition matrix.

2. Find the zero-state response when the input u(t) is the unit step.

We first compute A(t) using the given state transition matrix. We use the state transition matrix properties to accomplish this. Recall that the state transition matrix satisfies the equation

\frac{∂}{∂t}Φ(t, τ) = A(t)Φ(t, τ), Φ(τ, τ) = I_{n}.

Note that, we can also write.

Φ(t, τ)|_{τ=t} = I_{n}

Taking this into account, we obtain

A(t) = \frac{∂}{∂t}Φ(t, τ)|_{τ=t} .

To proceed, we first evaluate

\frac{∂}{∂t}e^{α(τ)^{2}−α(t)^{2}} |_{τ=t} = e^{α(τ)^{2}−α(t)^{2}}(−2α(t)α̇(t))|_{τ=t}
= −2α(t)α̇(t).

We next evaluate the derivative

\frac{∂}{∂t} (e^{α(τ)^{2}−α(t)^{2}} (cos(πt) − cos(πτ)))|_{τ=t} = −2α(t)α̇(t)(cos(πt) − cos(πt)) − π sin(πt)
= −π sin(πt).

Hence

A(t) = \frac{∂}{∂t}Φ(t, τ)|_{τ=t}
= [ \begin{matrix}−2α(t)α̇(t) & 0 \\ −π sin(πt) & −2α(t)α̇(t) \end{matrix} ] .

We now compute the zero-state response. The formula for the zero-state response is obtained from (22) by setting x_{0} = 0,

x(t) = ∫_{t_{0}}^{t} Φ(t, τ)B(τ)u(τ)dτ.

We first compute the integrand

Φ(t, τ)B(τ) = e^{α(τ)^{2}−α(t)^{2}} [ \begin{matrix}1 & 0 \\ cos(πt) − cos(πτ) & 1 \end{matrix} ] [ \begin{matrix}πe^{−α(τ)^{2}} \\ 0 \end{matrix} ]
= πe^{−α(t)^{2}} [ \begin{matrix}1 \\ cos(πt) − cos(πτ) \end{matrix} ] .

Therefore, the zero-state response when u(t) = 1(t) is

x(t) = ∫_{t_{0}}^{t} Φ(t, τ)B(τ)u(τ)dτ
= πe^{−α(t)^{2}} ∫_{0}^{t} [ \begin{matrix}1 \\ cos(πt) − cos(πτ) \end{matrix} ] dτ
= πe^{−α(t)^{2}} [ \begin{matrix}t \\ t cos(πt) − \frac{sin(πτ)}{π} \end{matrix} ] .

4 Notes

For additional information on the fundamentals of the LTV systems, we recommend Kamen [6]. An in-depth treatment of the LTV systems can be found in Rugh [12], and in Antsaklis and Michel [1]. For results on observers for LTV systems, see O’Reilly [9]. A novel approach to the analysis and control of LTV systems is proposed by Bourlès and Marinescu [2].

5 Exercises

Exercise 1 Prove that a fundamental matrix is nonsingular for any t.

Exercise 2 Use the Contraction Map Theorem ?? to prove that (10) has a unique solution going through x_{0} at t = t_{0}, where t_{0} is an element of a finite time interval [a, b]. (Start with the Lipschitz condition.)

Exercise 3 (a) Find the state-transition matrix Φ(t, τ) associated with a homogenous LTI system with

A = T [ \begin{matrix}−2 & 0 \\ 0 & 0 \end{matrix} ] T^{−1},

where

T = [ \begin{matrix}1 & 0 \\ −3 & 1 \end{matrix} ] .

(b) Find the state-transition matrix Φ(t, τ) associated with a homogenous LTV system

with

A(t) = T [ \begin{matrix}−2 & 0 \\ 0 & cos t \end{matrix} ] T^{−1},

where

T = [ \begin{matrix}1 & 0 \\ −3 & 1 \end{matrix} ] .

Exercise 4 Compute the state transition matrix Φ(t, τ) for the linear time-varying (LTV) system model,

ẋ(t) = [ \begin{matrix}0 & 0 \\ 1 & \frac{1}{t} \end{matrix} ] x(t)

defined for t ≥ 1. Then, for the given x(2) = [ 0 4 ]^{⊤}, compute x(1).

Exercise 5 Use the definition of the state transition matrix to prove its properties (14), (15), (16), and (17). Exercise 6 (a) The value of the state-transition matrix for a homogeneous linear system

is

Φ(1, 3) = [ \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ] .

Find x(3) when

x(1) = [ \begin{matrix}1 \\ 3 \end{matrix} ] .

(b) The values of the state-transition matrix for a homogeneous linear system is

Φ(1, 3) = [ \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ] and Φ(2, 3) = [ \begin{matrix}2 & 3 \\ 4 & 5 \end{matrix} ] .

If

x(1) = [ \begin{matrix}2 \\ −3 \end{matrix} ] ,

find x(2). Exercise 7 Derive Leibniz’s rule given by (23). Exercise 8 (a) The state-transition matrix of a homogenous LTV system is

Φ(t, τ) = [ \begin{matrix}e^{cos t−cos τ} & 0 \\ 0 & e^{− sin t+sin τ} \end{matrix} ] .

Find the associated A(t)-matrix.

(b) A fundamental matrix associated with a homogenous continuous-time LTV state model

has the form,

X(t) = [ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ] [ \begin{matrix}1 & 0 \\ t^{2} − t & t \end{matrix} ] .

Compute the state-transition matrix Φ(t, τ). Exercise 9 (J. Hu) For the linear time-invariant (LTV) system model,

ẋ(t) = [ \begin{matrix}−e^{−t} & \frac{1}{t+1} \\ 0 & −e^{−t} \end{matrix} ] x(t),

• Find the state transition matrix Φ(t, τ),

• Is the system asymptotically stable?

• Let x(0) = [ 1 0 ]^{⊤}. Will the system trajectory x(t) stay bounded as t → ∞?

Exercise 10 (R. Balakrishnan) For the linear time-varying (LTV) system model,

ẋ(t) = [ \begin{matrix}0 & −2t \\ 0 & −t \end{matrix} ] x(t) + [ \begin{matrix}0 \\ t \end{matrix} ] u(t),

• Find the state transition matrix Φ(t, τ);

• Let x(1) = [ 1 1 ]^{⊤} and u(t) = 0 for t ≥ 0. Find x(0);

• Let x(0) = 0 and u(t) = 1 for t ≥ 0. Find x(t) for t ≥ 0.

References

[1] P. J. Antsaklis and A. N. Michel. A Linear Systems Primer. Birkhäuser, Boston, 2007.

[2] H. Bourlès and B. Marinescu. Linear Time-Varying Systems: Algebraic-Analytic Approach. Springer-Verlag, Berlin Heidelberg, 2011. Lecture Notes in Control and Information Sciences 410.

[3] J. Chiasson and B. Vairamohan. Estimating the state of charge of a battery. In Proceedings of the 2003 American Control Conference, Denver, Colorado, pages 2863–2868, June 4–6, 2003.

[4] J. Chiasson and B. Vairamohan. Estimating the state of charge of a battery. IEEE Transactions on Control Systems Technology, 13(3):465–470, May 2005.

[5] S. W. Hawking. A Brief History of Time: From the Big Bang to Black Holes. Bantam Books, New York, 1988.

[6] E. W. Kamen. Fundamentals of linear time-varying systems. In W. S. Levine, editor, The Control Handbook, chapter 25, pages 451–468. CRC Press in cooperation with IEEE Press, Boca Raton, Florida, 1996.

[7] D. G. Luenberger. Introduction to Dynamic Systems: Theory, Models, and Applications. John Wiley & Sons, New York, 1979.

[8] M. McIntyre, T. Burg, D. Dawson, and B. Xian. Adaptive state of charge (SOC) estimator for a battery. In Proceedings of the 2006 American Control Conference, Minneapolis, Minnesota, pages 5740–5744, June 14–16, 2006.

[9] J. O’Reilly. Observers for Linear Systems. Academic Press, London, 1983.

[10] S. Pang, J. Farrell, J. Du, and M. Barth. Battery state-of-charge estimation. In Proceedings of the 2001 American Control Conference, Arlington, VA, pages 1644–1649, June 25–27, 2001.

[11] W. J. Rugh. Linear System Theory. Prentice Hall, Englewood Cliffs, New Jersey 07632, 1993.

[12] W. J. Rugh. Linear System Theory. Prentice Hall, Upper Saddle River, New Jersey 07458, second edition, 1996.