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40 Chapter Two
equation(s) on which well-established algebraic operations are
applied. We illustrate this by converting the equation of motion of
the mass–spring–damper system [Eq. (2.10)] from the time domain
to the Laplace domain:
⎡ 2 ⎤
t
t
[( )] = m dx () + c dx () + kx () t ⎥ (2.19)
⎢
ft
⎣ dt 2 dt ⎦
Instead of calculating the Laplace transform of the function on the left
side of Eq. (2.19) by solving the integral from Eq. (2.11), we can use the
interesting properties described in Eqs. (2.12) to (2.18). Thanks to
the linearity theorem [Eqs. (2.12) and (2.13)], the Laplace transform of
the equation of motion can be written as the sum of the Laplace trans-
forms of the individual subsystems (inertia, damper, and spring). By
introducing the Laplace transform F(s) of the force signal f(t) Eq. (2.19)
can thus be rewritten as
⎡ 2 ⎤ ⎡ dx t () ⎤
dx t ()
Fs () = m ⎢ ⎥ + ⎥ + x t)] (2.20)
c ⎢
k [(
⎣ dt 2 ⎦ ⎣ dt ⎦
Because the remaining functions to be transformed are just differ-
entials of x(t), we can now apply the differentiation theorem [Eq. (2.14)]
and introduce the Laplace transform X(s) of the position signal x(t) to
rewrite Eq. (2.20) as
+
F s) = m s X s) − sx( − − 0 )] c sX s) − x( − + kX s() (2.21)
x( − +
2
0
0
)
)]
(
[
[
(
(
where x(0-) is the initial position x of the mass and x(0− is the initial
)
0
speed v of the mass. By separating the dynamic terms from the initial
0
conditions [Eq. (2.21)] can be rewritten as
+
=
+
2
Fs() [ ms + cs k X s() − mv − ( ms c x (2.22)
)
]
0 0
When we assume the mass to be initially at rest (v = 0) and define
0
the position x such that x is zero, we can isolate the Laplace trans-
0
form of the system from those of the signals:
Fs()
= ms + cs + k (2.23)
2
Xs()
This is a transfer function in the Laplace domain for the transfer of a
position signal X(s) into a force signal F(s).
Remember that the external force applied to the mass–spring–
damper system is actually the input, whereas the position of the mass
is the output. In the time domain, the position x(t) in Eq. (2.19) can