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

21.11 Transfer functions 659
(c) x ′′ +x ′ −2x=1−2t,
x(0) =6,x ′ (0) = −11
(d) x ′′ −4x = 4(cos2t −1),
x(0) =1,x ′ (0) =0
(e) x ′ −2x−y ′ +2y=−2t 2 +7,
x ′
2 +x+3y′ +y=t 2 +6
x(0) =3,y(0) =6
(f) x ′ +x+y ′ +y=6e t ,
x ′ +2x−y ′ −y=2e −t
x(0) =2,y(0) =1
3 Using Laplace transformsfindthe particular
solution of
d 2 y
dt 2 −5dy dt −6y=14e−t
satisfyingy = 3 and dy
dt =8whent=0.
Solutions
1 (a) x(t)=3−2e −t
(b) 3 2 e−4t/3 + 1 2
1
(c)
4 + 15 4 e−2t
(d) y(t) = 7 8 + 41
8 e−2t
(e) y(t) = 1 +3.464e −0.5t sin0.866t
2 (a) 3sint+2t
(b) −3cos2t +2e −t
(c) t +6e −2t
(d)
e −2t
4 + e2t
4 − 1 2 cos2t+1
(e) x=t 2 +3,y=6−t
(f) x(t) = 6et
5 +2e−t − 6e−3t/2 ,
5
y(t) = 6e−3t/2
5
3 −2te −t + 8e−t
7
−2e −t + 9et
5
+ 13e6t
7
21.11 TRANSFERFUNCTIONS
It is possible to obtain a mathematical model of an engineering system that consists of
one or more differential equations. This approach was introduced in Section 20.4. We
have already seen that the solution of differential equations can be found by using the
Laplace transform. This leads naturally to the concept of a transfer function which will
bedeveloped inthissection.Consider the differential equation
dx(t)
+x(t) = f(t) x(0) =x
dt
0
(21.5)
andassumethatitmodelsasimpleengineeringsystem.Then f (t)representstheinputto
thesystemandx(t)representstheoutput,orresponseofthesystemtotheinput f (t).For
reasonsthatwillbeexplainedbelowitisnecessarytoassumethattheinitialconditions
associatedwiththedifferentialequationarezero.InEquation(21.5)thismeanswetake
x 0
tobe zero.Taking the Laplace transform of Equation (21.5) yields
sX(s)−x 0
+X(s) =F(s)
(1 +s)X(s) =F(s) assumingx 0
=0
so that
X(s)
F(s) = 1
1 +s
The function, X(s)/F(s), is called a transfer function. It is the ratio of the Laplace
transform of the output to the Laplace transform of the input. It is often denoted by