# Modeling linear dynamic systems

Well, that was easy to describe a dynamic model for an airplane. I suppose that you are familiar with Newton's motion equations from the high school.

In this chapter, I will generalize the model for any linear dynamic system. The following description includes integrals and differential equations. If your goal is basic understanding of the Kalman Filter and you feel uncomfortable with this math – feel free to skip this chapter. However, if you want to understand more advanced topics, such as Extended Kalman Filter you need to master this chapter.

On my side, I will try to explain as easy and clear as possible, and, of course, I will provide a real-life examples.

## Derivation of the state extrapolation equation

Our goal is to derive the state extrapolation equation in the form of:

$\boldsymbol{\hat{x}_{n+1,n} =F\hat{x}_{n,n} + G\hat{u}_{n,n} + w_{n}}$

In order to do that, we need to model the dynamic system. In other words, to figure out the state space representation of dynamic system. The following two equations are the state-space representation of the LTI system:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$ $\boldsymbol{y(t) = Cx(t) + Du(t)}$
Where:
 $$\boldsymbol{x}$$ is the state vector $$\boldsymbol{y}$$ is the output vector $$\boldsymbol{A}$$ is system dynamics matrix $$\boldsymbol{B}$$ is the input matrix $$\boldsymbol{C}$$ is the output matrix $$\boldsymbol{D}$$ is the feedthrough matrix

In order to find the state transition matrix $$\boldsymbol{F}$$ and input transition matrix $$\boldsymbol{G}$$ we need to solve the state space differential equation.

The following diagram summarizes the process of the state extrapolation equation derivation. ## The state space representation

You might be wandering, why the state space representation must be in the following form:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$ $\boldsymbol{y(t) = Cx(t) + Du(t)}$

As we will see, it is required for the differential equation solution. As well, many computer software packages that solve differential equation require this kind of representation.

The best way to describe the state space representation is by examples.

### Example: constant velocity moving body

Since there is no external force applied on the body, the system has no inputs:

$\boldsymbol{u(t)} = 0$

The state space variable $$\boldsymbol{x(t)}$$ is the body’s displacement $$p(t)$$ and the speed $$v(t)$$.

$\boldsymbol{x(t)} = \left[ \begin{matrix} p\\ v\\ \end{matrix} \right]$

Note: In order to avoid confusion of the state vector $$\boldsymbol{x(t)}$$ (bold-face font) and the body position at x axis denoted by $$x(t)$$ (normal-face font). I’ve denoted the body position by $$p(t)$$.

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \boldsymbol{A} \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \boldsymbol{B} \times 0$

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right]$

We’ve got the first equation in a differential form.

The dynamic system output $$\boldsymbol{y(t)}$$ is the body displacement $$\boldsymbol{p(t)}$$ that is also the state variable.

$\boldsymbol{y(t) = Cx(t) + Du(t)}$

$p = \boldsymbol{C} \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \boldsymbol{D} \times 0$

$p = \left[ \begin{matrix} 1 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right]$

We are done.

### Modeling high-order dynamic systems

Many dynamic systems models are described by high-order differential equations. The order of the differential equation is the number of the highest derivative in a differential equation. A differential equation of order 1 is called first order, order 2 second order, etc.

In order to tackle high order equation, we need to reduce it to the first order differential equation by defining new variables and substituting them into highest order term.

#### The algorithm of reducing the differential equation order

A general $$n$$-th order linear differential equation can be expressed as a system of $$n$$ first-order differential equations.

The high-order governing equation of the dynamic system looks like:

$a_{n}\frac{d^{n}y}{dt^{n}}+ a_{n-1}\frac{d^{n-1}y}{dt^{n-1}}+ \ldots + a_{2}\frac{d^{2}y}{dt^{2}}+a_{1}\frac{d^{1}y}{dt^{1}}+ a_{0}y=u$

The governing equation completely characterizes the dynamic state of the system.

Reducing the equation order:

1. Isolate the highest-order derivative:
2. $\frac{d^{n}y}{dt^{n}}= -\frac{a_{0}}{a_{n}}y- \frac{a_{1}}{a_{n}}\frac{d^{1}y}{dt^{1}}- \frac{a_{2}}{a_{n}}\frac{d^{2}y}{dt^{2}}- \ldots - \frac{a_{n-1}}{a_{n}}\frac{d^{n-1}y}{dt^{n-1}}+\frac{1}{a_{n}}u$

$$y$$ and its first $$n - 1$$ derivatives are the states of this system

3. Defining new variable $$x_{1}$$:
4. Setting $$x_{1} = y$$, we write:

$x_{1} \left( t \right) =y$ $x_{2} \left( t \right) = \frac{dy}{dt}$ $x_{3} \left( t \right) = \frac{d^{2}y}{dt^{2}}$ $\vdots$ $x_{n} \left( t \right) = \frac{d^{n-1}y}{dt^{n-1}}$

Now, functions $$x_{i}(t)$$ are the state variables

5. Take the derivatives of the state variables:
6. $\frac{dx_{1}}{dt}=x_{2} \left( t \right)$ $\frac{dx_{2}}{dt}=x_{3} \left( t \right)$ $\vdots$ $\frac{dx_{n-1}}{dt}=x_{n} \left( t \right)$ $\frac{dx_{n}}{dt}=\frac{d^{n}y}{dt^{n}}$

7. Plug in the isolated $$\frac{d^{n}y}{dt^{n}}$$ term (see step 1) in the last equation:
8. $\frac{dx_{1}}{dt}=x_{2} \left( t \right)$ $\frac{dx_{2}}{dt}=x_{3} \left( t \right)$ $\vdots$ $\frac{dx_{n-1}}{dt}=x_{n} \left( t \right)$ $\frac{dx_{n}}{dt}=-\frac{a_{0}}{a_{n}}x_{1}- \frac{a_{1}}{a_{n}}x_{2}- \frac{a_{2}}{a_{n}}x_{3}- \ldots - \frac{a_{n-1}}{a_{n}}x_{n}+\frac{1}{a_{n}}u$

9. Express the resulted system of equations using vector-matrix notation:
10. $\left[ \begin{matrix} \frac{dx_{1}}{dt}\\ \frac{dx_{2}}{dt}\\ \vdots \\ \frac{dx_{n-1}}{dt}\\ \frac{dx_{n}}{dt}\\ \end{matrix} \right] = \left[ \begin{matrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \vdots \\ \dot{x}_{n-1}\\ \dot{x}_{n}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1 & 0 & \cdots & 0 & 0\\ 0 & 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1\\ -\frac{a_{0}}{a_{n}} & -\frac{a_{1}}{a_{n}} & -\frac{a_{2}}{a_{n}} & \cdots & -\frac{a_{n-2}}{a_{n}} & -\frac{a_{n-1}}{a_{n}}\\ \end{matrix} \right] \left[ \begin{matrix} x_{1}\\ x_{2}\\ \vdots \\ x_{n-1}\\ x_{n}\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ 0\\ \vdots \\ 0\\ \frac{1}{a_{n}}\\ \end{matrix} \right] u$

That's it, we've got the state space equation in the form of:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$

Where:

$\boldsymbol{A} = \left[ \begin{matrix} 0 & 1 & 0 & \cdots & 0 & 0\\ 0 & 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1\\ -\frac{a_{0}}{a_{n}} & -\frac{a_{1}}{a_{n}} & -\frac{a_{2}}{a_{n}} & \cdots & -\frac{a_{n-2}}{a_{n}} & -\frac{a_{n-1}}{a_{n}}\\ \end{matrix} \right]$

$\boldsymbol{B} = \left[ \begin{matrix} 0\\ 0\\ \vdots \\ 0\\ \frac{1}{a_{n}}\\ \end{matrix} \right]$

Let us see two examples.

### Example: constant acceleration moving body

In this example, there is external force applied on the body.

The governing equation for constant acceleration moving body is the Newtons second law:

$m\ddot{p} = F$

where:

$$p$$ is the body position displacement

$$m$$ is the body mass

$$F$$ is the external force applied on the body We are going to apply order reduction algorithm on the governing equation.

1. Isolate the highest-order derivative:
2. $\ddot{p} = \frac{1}{m}F$

3. Define new variable $$x_{1}$$:
4. $x_{1} = p$ $x_{2} = \dot{p}$

$$x_{1}$$ and $$x_{2}$$ are the state variables

5. Take the derivatives of the state variables:
6. $\dot{x}_{1} = x_{2}$ $\dot{x}_{2} = \ddot{p}$

7. Plug in the isolated $$\frac{d^{n}y}{dt^{n}}$$ term (see step 1) in the last equation:
8. $\dot{x}_{1} = x_{2}$ $\dot{x}_{2} = \frac{F}{m}$

9. Express the resulted system of equations using vector-matrix notation:
10. $\left[ \begin{matrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} x_{1}\\ x_{2}\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right] F$

Remember that: $$x_{2} = \dot{p}$$, it will be more meaningful to denote $$x_{2}$$ by $$v$$, since $$x_{2}$$ is the body velocity.

We can rewrite the above equation as:

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right] F$

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ 1\\ \end{matrix} \right] a$

where $$a$$ is the body acceleration resulted by the applied force $$F$$.

We've got an equation in the form of:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$

The state-space variable $$\boldsymbol{x(t)}$$ is the body’s displacement $$p(t)$$ and the speed $$v(t)$$.

The dynamic system output $$\boldsymbol{y(t)}$$ is the body displacement $$p(t)$$ which is also an element of the state variable.

$\boldsymbol{y(t) = Cx(t) + Du(t)}$

$p = \boldsymbol{C} \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \boldsymbol{D} a$

$p = \left[ \begin{matrix} 1 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \end{matrix} \right] a$

### Example: mass-spring-damper system

The mass-spring-damper system includes three basic elements:

• mass - inertia element
• spring - elastic element
• dumper – frictional element Each of the elements has one of two possible energy behaviors:

• stores all the energy supplied to it
• dissipates all energy into heat by some kind of “frictional” effect

The mass stores energy as kinetic energy.

The spring stores energy as potential energy when it compressed from its original length.

The damper dissipates energy as a heat out of the system.

These three components, mass, spring and the damper can model any dynamic response situation in a general sense.

The forces diagram for this system is shown below. The spring force is proportional to the position displacement of the mass $$p$$.

The viscous damping force is proportional to the velocity of the mass $$\dot{p}$$.

The Newton's second law states:

$\sum _{}^{}F = ma = m\frac{d^{2}p}{dt^{2}} = m\ddot{p}$

We proceed by summing the forces and applying Newton’s second law:

$\sum _{}^{}F = F(t) - c\dot{p} - kp = m\ddot{p}$

where:

$$p$$ is the body position displacement

$$m$$ is the body mass

$$F$$ is the external force applied on the body

$$k$$ is the spring constant

$$c$$ is the damping coefficient

This equation is a governing equation that completely characterizes the dynamic state of the system.

We are going to apply order reduction algorithm on the governing equation.

1. Isolate the highest-order derivative:
2. $\ddot{p} = -\frac{k}{m}p - \frac{c}{m}\dot{p} + \frac{1}{m}F$

3. Define new variable $$x_{1}$$:
4. $x_{1} = p$ $x_{2} = \dot{p}$

$$x_{1}$$ and $$x_{2}$$ are the state variables

5. Take the derivatives of the state variables:
6. $\dot{x}_{1} = x_{2}$ $\dot{x}_{2} = \ddot{p}$

7. Plug in the isolated $$\frac{d^{n}y}{dt^{n}}$$ term (see step 1) in the last equation:
8. $\dot{x}_{1} = x_{2}$ $\dot{x}_{2} = -\frac{k}{m}p - \frac{c}{m}\dot{p} + \frac{1}{m}F$

9. Express the resulted system of equations using vector-matrix notation:
10. $\left[ \begin{matrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ -\frac{k}{m} & -\frac{c}{m}\\ \end{matrix} \right] \left[ \begin{matrix} x_{1}\\ x_{2}\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right] F$

Remember that: $$x_{2} = \dot{p}$$, it will be more meaningful to denote $$x_{2}$$ by $$v$$, since $$x_{2}$$ is the body velocity.

We can rewrite the above equation as:

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ -\frac{k}{m} & -\frac{c}{m}\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right] F$

We've got an equation in the form of:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$

The state-space variable $$\boldsymbol{x(t)}$$ is the body’s displacement $$p(t)$$ and the speed $$v(t)$$.

The dynamic system output $$\boldsymbol{y(t)}$$ is the body displacement $$p(t)$$ which is also an element of the state variable.

$\boldsymbol{y(t) = Cx(t) + Du(t)}$

$p = \boldsymbol{C} \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \boldsymbol{D} \frac{F}{m}$

$p = \left[ \begin{matrix} 1 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \end{matrix} \right] \frac{F}{m}$

### More examples

You can find many beautiful examples of the state-space modeling on the ShareTechnote site.

## Solving the differential equation

Remember, for our Kalman Filter model, we need to determine the state extrapolation equation in the form of:

$\boldsymbol{\hat{x}_{n+1,n} =F\hat{x}_{n,n} + G\hat{u}_{n,n}}$

In order to get there, we shall solve the differential equation that describes the state space representation.

We can use a computer software, that will solve the differential equation for us, or we can do it ourselves.

Let us see, how to solve the differential equation.

### Dynamic systems without input variable

LTI dynamic system without external input can be described by the first order differential equation:

$\boldsymbol{\dot{x} = Ax}$

where $$\boldsymbol{A}$$ is a system dynamics matrix.

Our goal is finding the state transition matrix $$\boldsymbol{F}$$.

We need to solve the differential equation, in order to find $$\boldsymbol{F}$$.

In a single dimension, the differential equation looks like:

$\frac{dx}{dt} = kx$ $\frac{dx}{x} = kdt$

Integrating both sides results:

$\int _{x_{0}}^{x_{1}}\frac{1}{x}dx = \int _{0}^{ \Delta t}kdt$

Solving the integrals:

$ln \left( x_{1} \right) - ln \left( x_{0} \right) =k \Delta t$ $ln \left( x_{1} \right) = ln \left( x_{0} \right) +k \Delta t$ $x_{1}= e^{ln \left( x_{0} \right) +k \Delta t}$ $x_{1}= e^{ln \left( x_{0} \right) }e^{k \Delta t}$ $x_{1}= x_{0}e^{k \Delta t}$

Similarly, in multi-dimension, for:

$\boldsymbol{\dot{x} = Ax}$

the solution is:

$x_{n+1}= x_{n}e^{\boldsymbol{A} \Delta t}$

We've found the state transition matrix $$\boldsymbol{F}$$:

$\boldsymbol{F} = e^{\boldsymbol{A} \Delta t}$

$$e^{\boldsymbol{A}t}$$ is a matrix exponential.

The matrix exponential can be computed by Taylor series expansion:

$e^{\boldsymbol{X}}= \sum _{k=0}^{\infty}\frac{1}{k!}\boldsymbol{X}^{k}$

Therefore:

$\boldsymbol{F} = e^{\boldsymbol{A} \Delta t} = \boldsymbol{I} + \boldsymbol{A} \Delta t+ \frac{ \left( \boldsymbol{A} \Delta t \right) ^{2}}{2!}+ \frac{ \left( \boldsymbol{A} \Delta t \right) ^{3}}{3!}+ \frac{ \left( \boldsymbol{A} \Delta t \right) ^{4}}{4!}+ \ldots$

#### Example continued: constant velocity moving body

Now we can find the state transition matrix $$\boldsymbol{F}$$ for the constant velocity motion equations.

The constant velocity dynamic model can be described by the following set of differential equations.

$\begin{cases} \frac{dp}{dt}=v\\ \frac{dv}{dt}=0\\ \end{cases}$

In a matrix form:

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right]$ $\boldsymbol{A} = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right]$

Let's calculate $$\boldsymbol{A}^{2}$$

$\boldsymbol{A}_{2} = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix} \right]$

Since $$\boldsymbol{A}^{2} = 0$$, higher powers of $$\boldsymbol{A}$$ are also equal to 0.

Now, we can find the state transition matrix $$\boldsymbol{F}$$ for the constant velocity model:

$F= e^{\boldsymbol{A} \Delta t} = \boldsymbol{I} + \boldsymbol{A} \Delta t = \left[ \begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix} \right] + \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \Delta t = \left[ \begin{matrix} 1 & \Delta t\\ 0 & 1\\ \end{matrix} \right]$ $x_{n+1} = \boldsymbol{F}x_{n} = \left[ \begin{matrix} 1 & \Delta t\\ 0 & 1\\ \end{matrix} \right] x_{n}$ $\left[ \begin{matrix} x_{n+1}\\ \dot{x}_{n+1}\\ \end{matrix} \right] = \left[ \begin{matrix} 1 & \Delta t\\ 0 & 1\\ \end{matrix} \right] \left[ \begin{matrix} x_{n}\\ \dot{x}_{n}\\ \end{matrix} \right]$

### Dynamic systems with input variable

For zero order hold sampling, assuming the input is piecewise constant, the general solution of state space equation in the form of:

$\boldsymbol{\dot{x}(t) = Ax(t) + Bu(t)}$

is given by:

$\boldsymbol{x ( t+ \Delta t )} = \color{red}{\underbrace{e^{\boldsymbol{A} \Delta t}}_\textrm{F}} \boldsymbol{x(t)} + \color{blue}{\underbrace{\int _{0}^{\Delta t} e^{\boldsymbol{A}t}dt\boldsymbol{B}}_\textrm{G}} \boldsymbol{u(t)}$

If we remove the input variable $$(\boldsymbol{u(t)} = 0)$$, will get the solution that we just derived in the previous chapter.

I am not going to prove this, you can find the proof in Tom M. Apostol's "Calculus" textbook (second edition), theorem 8.3 or any other calculus textbook.

Now, let's solve the state space equations for the constant acceleration moving body and the mass-spring-damper system examples.

#### Example continued: constant acceleration moving body

Recall, that the state-space representation of the constant acceleration moving body:

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ 1\\ \end{matrix} \right] a$

when:

$\boldsymbol{A} = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right]$

$\boldsymbol{B} = \left[ \begin{matrix} 0\\ 1\\ \end{matrix} \right]$

Let's solve the equation:

$\boldsymbol{x ( t+ \Delta t )} = \color{red}{\underbrace{e^{\boldsymbol{A} \Delta t}}_\textrm{F}} \boldsymbol{x(t)} + \color{blue}{\underbrace{\int _{0}^{\Delta t} e^{\boldsymbol{A}t}dt\boldsymbol{B}}_\textrm{G}} \boldsymbol{u(t)}$

##### Finding F

$\boldsymbol{F} = e^{\boldsymbol{A} \Delta t}$

we've already solved this for $$A = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right]$$

$\boldsymbol{F} = \left[ \begin{matrix} 1 & \Delta t\\ 0 & 1\\ \end{matrix} \right]$

##### Finding G

$\boldsymbol{F} = \int _{0}^{ \Delta t}e^{\boldsymbol{A}t}dt \boldsymbol{B}$

The general formula for $$\int _{0}^{ \Delta t}e^{\boldsymbol{A}t}dt$$ is the power series:

$\int _{0}^{ \Delta t}e^{\boldsymbol{A}t}dt= \Delta t \left( I+ \frac{\boldsymbol{A} \Delta t}{2!}+ \frac{ \left( \boldsymbol{A} \Delta t \right) ^{2}}{3!}+ \frac{ \left( \boldsymbol{A} \Delta t \right) ^{3}}{4!}+ \ldots \right)$

$\boldsymbol{A}^{2} = \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix} \right]$

The higher powers of $$\boldsymbol{A}$$ are also equal to 0.

$\int _{0}^{ \Delta t}e^{\boldsymbol{A}t}dt= \Delta t \left( \left[ \begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix} \right] + \left[ \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right] \frac{ \Delta t}{2} \right) = \left[ \begin{matrix} \Delta t & \frac{1}{2}\Delta t^{2}\\ 0 & \Delta t\\ \end{matrix} \right]$

$\boldsymbol{G} = \int _{0}^{ \Delta t}e^{\boldsymbol{A}t}dt\boldsymbol{B} = \left[ \begin{matrix} \Delta t & \frac{1}{2}\Delta t^{2}\\ 0 & \Delta t\\ \end{matrix} \right] \left[ \begin{matrix} 0\\ 1\\ \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{2}\Delta t^{2}\\ \Delta t\\ \end{matrix} \right]$

Now, we can write down the state extrapolation equation:

$\left[ \begin{matrix} p_{n+1}\\ \dot{p}_{n+1}\\ \end{matrix} \right] = \left[ \begin{matrix} 1 & \Delta t\\ 0 & 1\\ \end{matrix} \right] \left[ \begin{matrix} p_{n}\\ \dot{p}_{n}\\ \end{matrix} \right] + \left[ \begin{matrix} \frac{1}{2}\Delta t^{2}\\ \Delta t\\ \end{matrix} \right] a$

#### Example continued: mass-spring-damper system

Recall, that the state-space representation of the mass-spring-damper system:

$\left[ \begin{matrix} \dot{p}\\ \dot{v}\\ \end{matrix} \right] = \left[ \begin{matrix} 0 & 1\\ -\frac{k}{m} & \frac{c}{m}\\ \end{matrix} \right] \left[ \begin{matrix} p\\ v\\ \end{matrix} \right] + \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right] F$

when:

$\boldsymbol{A} = \left[ \begin{matrix} 0 & 1\\ -\frac{k}{m} & \frac{c}{m}\\ \end{matrix} \right]$

$\boldsymbol{B} = \left[ \begin{matrix} 0\\ \frac{1}{m}\\ \end{matrix} \right]$

In this example computing the matrix exponential is not so easy, since the high powers of A are not zero.

Solution of this differential equation is beyond the scope of this tutorial. You can use a computer software to solve the equation. If you want to solve it by yourself, you can find a useful guide on the following pages:

You can also find a solution of the second order differential equation of the mass-spring-damper system in the following presentation (slides 29-34):