2 Modelling linear systems by transfer functions

We have seen previously that an engineering system can be modelled by one or more differential equations. However, with the introduction of the transfer function we have an alternative model which we examine in this Section.

It will be helpful to develop a pictorial approach to system modelling. To begin, we can imagine a differential equation:

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

as being a model of the engineering system which transforms the input signal $f\left(t\right)$ into an output signal $y\left(t\right)$ (the solution of the differential equation). The system is characterised by the values of the coefficients $a,b,c$ . A different engineering system will be characterised by a different set of coefficients. These coefficients are independent of the input signal. Changing the input signal does not change the system. It is the system that changes the input signal into the output signal. This is easy to describe pictorially (Figure 21).

Figure 21 :

In a block diagram the system is represented by a rectangular box and the input and output signals represented by lines with an arrow to indicate the ’flow’.

After the Laplace transform of the differential equation is taken, the differential equation is transformed into

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

in which $H\left(s\right)$ is the transfer function. The latter characterises (in Laplace transform terms) the engineering system from which it was derived. The relation, connecting the Laplace transform of the output $Y\left(s\right)$ to the Laplace transform of the input $F\left(s\right)$ , can also be described schematically (Figure 22).

Figure 22 :

We can begin to model an engineering system directly in terms of transfer functions. In order to do this effectively we need to know how transfer functions are to be combined together. Before we do this we first extend our block diagrams to allow for ’interactions’. There are three basic components occurring in block diagrams which we now describe.

The first component , we have already met: the block relating the input to the output (Figure 23):

Figure 23 :

The second component is called a summing point (Figure 24):

Figure 24 :

Here we have shown two incoming signals $R\left(s\right),X\left(s\right)$ (but at a general summing point there may be many incoming signals) and one outgoing signal (there should never be more than one outgoing signal). The sign attached to the incoming signal defines whether the signal is adding to ( $+$ ) or subtracting from $\left(-\right)$ the summing point. The outgoing signal is then calculated in an obvious way, taking these signs into account.

The third component is a take-off point (Figure 25):

Figure 25 :

Here the value of the signal $Y\left(s\right)$ is found in such a way as not to affect the signal that is being transmitted. (This situation can never be precisely realised in practice, but using sensitive measuring devices it can be well approximated. As a simple example consider the problem of measuring the temperature of a certain volume of liquid. The act of putting a thermometer in the liquid will usually slightly affect the temperature we are trying to observe.) An example of a block diagram is the so-called negative feedback loop, shown in Figure 26 (we are using $G\left(s\right)$ to denote the transfer function):

Figure 26 :

Here, the output signal is tapped and subtracted from the input signal. Hence

$Y\left(s\right)=G\left(s\right)Y_1\left(s\right)$

because $Y_1\left(s\right)$ is the input signal to the system characterised by transfer function $G\left(s\right)$ . However, at the summing point $Y_1\left(s\right)=F\left(s\right)-Y\left(s\right)$ and so

$Y\left(s\right)=G\left(s\right)\left(F\left(s\right)-Y\left(s\right)\right)$

from which we easily obtain:

$Y\left(s\right)=\left[\frac{G\left(s\right)}{1+G\left(s\right)}\right]F\left(s\right)$

so that, in terms of input and output signals, the feedback loop is characterised by a transfer function

$\frac{G\left(s\right)}{1+G\left(s\right)}$ .

In some feedback loops the tapped signal $Y\left(s\right)$ may be modified in some way before feedback. Using the overall transfer function we can now picture the feedback loop in a simpler way (Figure 27):

Figure 27 :

Another type of block diagram occurs when the output from one system becomes the input to another system. For example consider the system of coupled differential equations:

in which $f\left(t\right)$ is a given input signal. transforms we have, as usual

$sX\left(s\right)+X\left(s\right)=F\left(s\right)3sY\left(s\right)-Y\left(s\right)=X\left(s\right)$

so the transfer function for the first equation ( $G_1\left(s\right)$ say) satisfies

$G_1\left(s\right)\equiv \frac{X\left(s\right)}{F\left(s\right)}=\frac{1}{s+1}$

whilst the transfer function for the second equation $G_2\left(s\right)$ satisfies

$G_2\left(s\right)\equiv \frac{Y\left(s\right)}{X\left(s\right)}=\frac{1}{3s-1}$

In pictorial terms this is shown in Figure 28:

Figure 28

So we have two transfer functions ‘in series’. To find how they combine we simply find an expression connecting the final output $Y\left(s\right)$ to the initial input $F\left(s\right)$ . Clearly

$X\left(s\right)=G_1\left(s\right)F\left(s\right)$ and so $Y\left(s\right)=G_2\left(s\right)X\left(s\right)=\left[G_2\left(s\right)G_1\left(s\right)\right]F\left(s\right)$

So transfer functions in series are simply multiplied together. In this case the overall transfer function $H\left(s\right)$ is:

$H\left(s\right)=G_1\left(s\right)G_2\left(s\right)=\frac{1}{\left(s+1\right)\left(3s-1\right)}$

Note that this result could be found directly from the differential equations used to model this system. If we differentiate the second differential equation of the original pair we get:

$3\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}=\frac{dx}{dt}$

Rearranging the first equation gives $\frac{dx}{dt}=f\left(t\right)-x$

$\text{Substituting gives:}3\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}=f\left(t\right)-x=f\left(t\right)-\left[3\frac{dy}{dt}-y\right]$

or

$3\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}-y=f\left(t\right)$

which, on taking Laplace transforms, gives the $s$ -relation $\left(3s^2+2s-1\right)Y\left(s\right)=F\left(s\right)$ implying a transfer function:

$H\left(s\right)=\frac{1}{3s^2+2s-1}=\frac{1}{\left(s+1\right)\left(3s-1\right)}$

as obtained above.