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2.0 Mathematical Backgroundsummarize

T he systems discussed on this site are restricted to linear systems. These are systems that can be modeled using a linear ordinary differential equation with constant coefficients. Fortunately this covers a wide range of engineering problems.

Keep in mind that there are several “systems” to consider. One is the plant – this is the systems before we add a control system. The second is the controller or compensator – these too are restricted to be linear systems that can be modeled using a differential equation. The third is the closed-loop system. The closed-loop system consists of the plant, the controller and any other hardware needed to for the connection, such as sensors or actuators. The closed-loop system can also be modeled as a differential equation.

2.1 Modeling a System

The system model can be represented in several different formats. These are all just different ways of representing the same differential equation. Each model has its advantages and disadvantages. We are going to focus on two representations: equation form and transfer functions (which includes block diagrams).

Equation Form

Equation form is the traditional mathematical expression for a differential equation. It is given in Eqn 2.1.1,

x(n)+an-1x(n-1)+an-2x(n-2)+ ... +a1x(1) + +a0x     
     = bn-1y(n-1)+bn-2y(n-2)+ ... +b1y(1) + +b0y
2.1.1

where x(n) = dnx/dtn

A key concept is that y is the input to the differential equation and x is the output. In order to determine x, the output of the system, you need to solve the equation for x given an input y.

The following example demonstrate how a DC motor can be modeled using differential equations. img Modeling a DC Motor with Differential Equations.

Transfer Function Form

The transfer function is found by taking the Laplace transform of the differential equation and solving for the ratio of the output to the input. The transfer function, G(s), can then be written as the ratio of two polynomials.

G(s) =
X(s)_________Y(s)
=
Output(s)_________Input(s)
2.1.2

When the initial conditions are zero we have

ℒ(
dx_________dt
) = sX(s)
2.1.3
ℒ(
dy_________dt
) = sY(s)
2.1.4

Combining 2.1.1 through 2.1.4 gives:

G(s) =
bn-1 sn-1 + bn-2 sn-2 + ... + b1s + b0____________an sn + an-1 sn-1+ ... + a1s + a0
2.1.5

The input is Y(s) and the output is X(s). Note that both Y(s) and X(s) are in Laplace space.

One advantage of using the transfer function form is that it allows you to easily combine differential equation and manipulate them algebraically. This is especially useful when using block diagrams discussed later.

To solve for the time domain solution you need to specify the input, y(t), find its Laplace transform Y(s) and then take the inverse Laplace transform of X(s) = G(s)Y(s)

x(t) = ℒ-1(X(s)) = ℒ-1(G(s)Y(s))
2.1.6