082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

18.2 Linearization using first-order Taylor polynomials 511
reason, it is not possible to use the principle of superposition when analysing the
network. For example, if the effect of the pressure at S 1
for a given flow rate was
calculatedseparatelyforeachoftheinputstothesystem--thereservoirsR 1
,R 2
,R 3
--
then the effect of all the reservoirs could not be obtained by adding the individual
effects. It is not possible to obtain a linear model for this system except under very
restrictive conditions and so the analysis of water networks isvery complicated.
Having demonstrated the valueoflinear models itisworthanalysing howand when
a non-linear system can be linearized. The first thing to note is that many systems may
containamixtureoflinearandnon-linearcomponentsandsoitisonlynecessarytolinearizecertainpartsofthesystem.AsystemofthistypehasbeenstudiedinEngineering
application 10.6. Thereforelinearization involves deciding which components ofasystemarenon-linear,decidingwhetheritwouldbevalidtolinearizethecomponentsand,
ifso, then obtaining linearized models of the components.
Consider again Figure 18.1. Imagine it illustrates a component characteristic. The
actual component characteristic is unimportant for the purposes of this discussion. For
instance, it could be the pressure/flow relationship of a valve or the voltage/current
relationship of an electronic device. The main factor in deciding whether a valid linear
model can be obtained is the range of values over which the component is required to
operate. If an operating point Q were chosen and deviations from this operating point
were small then it is clear from Figure 18.1 that a linear model -- corresponding to the
tangent tothe curve atpointQ-- would be an appropriate model.
Obtaining a linear model is relatively straightforward. It consists of calculating the
first-order Taylor polynomial centred around the operating point Q.This isgiven by
p 1
(x) =y(a) +y ′ (a)(x −a)
Then p 1
(x)isusedasthelinearizedmodelofthecomponentwithcharacteristicy(x).It
isvalidprovidedthatitisonlyusedforvaluesofxsuchthat |x−a|issufficientlysmall.
Asstated, p 1
(x)isalsotheequationofthetangenttothecurveatpointQ.Therangeof
values for which the model is valid depends on the curvature of the characteristic and
the accuracy required.
Engineeringapplication18.3
Powerdissipationinaresistor
The power dissipated in a resistor varies with the current. Derive a linear model for
thispowervariationvalidforanoperatingpointof0.5A.Theresistorhasaresistance
of 10 .
Solution
For a resistor
where
P=I 2 R
P = power dissipated (W)
I = current (A)
R = resistance ().
➔