1 Transfer functions and linear systems

Linear engineering systems are those that can be modelled by linear differential equations. We shall only consider those sytems that can be modelled by constant coefficient ordinary differential equations.

Consider a system modelled by the second order differential equation.

$a\frac{d^{2}y}{d{t}^2}+b\frac{dy}{dt}+cy=f\left(t\right)$

in which $a,b,c$ are given constants and $f\left(t\right)$ is a given function. In this context $f\left(t\right)$ is often called the input signal or forcing function and the solution $y\left(t\right)$ is often called the output signal .

We shall assume that the initial conditions are zero (in this case $y\left(0\right)=0,$ $y^{\prime }\left(0\right)=0$ ).

Now, taking the Laplace transform of the differential equation, gives:

$\left(a{s}^2+bs+c\right)Y\left(s\right)=F\left(s\right)$

in which we have used $y\left(0\right)=y^{\prime }\left(0\right)=0$ and where we have designated $\mathcal{L}\left\{y\left(t\right)\right\}=Y\left(s\right)$ and $\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ .

We define the transfer function of a system to be the ratio of the Laplace transform of the output signal to the Laplace transform of the input signal with the initial conditions as zero. The transfer function (a function of $s$ ), is denoted by $H\left(s\right)$ . In this case

$H\left(s\right)\equiv \frac{Y\left(s\right)}{F\left(s\right)}=\frac{1}{a{s}^2+bs+c}$

Now, in the special case in which the input signal is the delta function, $f\left(t\right)=\delta \left(t\right)$ , we have $F\left(s\right)=1$ and so,

$H\left(s\right)=Y\left(s\right)$

We call the solution to the differential equation in this special case the unit impulse response function and denote it by $h\left(t\right)u\left(t\right)$ (we include the step function $u\left(t\right)$ to emphasize its causality). So

$h\left(t\right)u\left(t\right)={\mathcal{L}}^{-1}\left\{H\left(s\right)\right\}\text{ when}f\left(t\right)=\delta \left(t\right)$

Now, keeping this in mind and returning to the general case in which the input signal $f\left(t\right)$ is not necessarily the impulse function $\delta \left(t\right)$ , we have:

$Y\left(s\right)=H\left(s\right)F\left(s\right)$

and so the solution for the output signal is, as usual, obtained by taking the inverse Laplace transform:

using the convolution theorem.

This approach provides yet another method of solving a linear system as Example 6 illustrates.

Example 6

Find the impulse response function $h\left(t\right)$ to a linear engineering system

modelled by the differential equation:

$\frac{d^{2}y}{d{t}^2}+4y={\text{e}}^{-t}y\left(0\right)=0y^{\prime }\left(0\right)=0$

and hence solve the system.

Solution

Here

$H\left(s\right)=\frac{1}{s^2+4}\left(=\frac{1}{a{s}^2+bs+c}\text{ with}a=1,b=0,c=4\right)$

This is obtained by replacing the forcing function ${\text{e}}^{-t}$ by the impulse function $\delta \left(t\right)$ and then taking the Laplace transform. Using this:

$h\left(t\right)={\mathcal{L}}^{-1}\left\{H\left(s\right)\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s^2+4}\right\}=\frac{1}{2}sin2t.u\left(t\right)$

Then the output $y\left(t\right)$ corresponding to the input ${\text{e}}^{-t}$ is given by the convolution of ${\text{e}}^{-t}$ and $h\left(t\right)$ . That is,

(Note: the last integral can be determined by integrating by parts (twice), or by use of a computer algebra system such as Matlab.)

Task!

Use the transfer function approach to solve

$\frac{dx}{dt}-4x=sintx\left(0\right)=0.$

Begin by finding the transfer function $H\left(s\right)$ :

Answer

Answer

Now complete the evaluation of this integral:

Answer