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

660 Chapter 21 The Laplace transform
G(s). Therefore, forEquation (21.5),
G(s) = 1
1 +s
The concept of a transferfunction isvery usefulinengineering. Itprovides a simple
algebraic relationship between the input and the output. In other words, it allows the
analysisofdynamicsystemsbasedonthedifferentialequationtoproceedinarelatively
straightforward manner. Earlier we noted that it was necessary to assume zero initial
conditions in order to form the transfer function. Without such an assumption, the relationship
between the input and the output would have been more complicated. What
is more, the relationship would vary depending on how much energy is stored in the
system at t = 0. By assuming zero initial conditions, the transfer function depends
purely on the system characteristics. Such an approach, whilst very convenient, does
have its limitations. The transfer function approach is useful for analysing the effect of
aninputtothesystem.However,ifonerequirestheeffectofthecombinationofasystem
inputand initial conditions,thenitisnecessarytocarryoutafullsolution ofthedifferential
equation as we did in Section 21.10. In practice there are many cases where the
simplified treatment provided by the transfer function is perfectly acceptable. An alternativeapproachisprovidedbystate-spacemodels,whichweexaminedinSection20.4.
Theseprovideanaturalwayofhandlinginitialconditions.Inordertosolveastate-space
model wherethereareinitialconditions,allthatisnecessaryistouseanon-zeroinitial
valueofthestatevector.RecallthatinSection20.4weonlysetupthestate-spacemodels
and did not solve them. However, the method for the standard solution of a state-space
model can befound inmany advanced textbooks on control and systems engineering.
Example21.28 Findthetransferfunctionsofthefollowingequationsassumingthat f (t)representsthe
inputandx(t) represents the output:
(a)
(b)
dx(t)
dt
d 2 x(t)
dt 2
−4x(t) =3f(t), x(0) =0
+3 dx(t)
dt
−x(t) = f(t),
dx(0)
dt
=0, x(0)=0
Solution (a) Taking Laplace transforms ofthe differential equation gives
sX(s) −x(0) −4X(s) =3F(s)
(s −4)X(s) =3F(s) asx(0) =0
X(s)
F(s) =G(s) = 3
s −4
(b) Taking Laplace transforms ofthe differential equation gives
s 2 X(s) −sx(0) − dx(0) +3(sX(s) −x(0)) −X(s) =F(s)
dt
(s 2 +3s −1)X(s) =F(s) as dx(0) =0 and x(0)=0
X(s)
F(s) =G(s) = 1
s 2 +3s−1
dt
Whencreatingamathematicalmodelofanengineeringsystemitisoftenconvenientto
thinkofthevariableswithinthesystemassignalsandelementsofthesystemasmeans