Page 748 - The Mechatronics Handbook

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The Laplace Transformation
The one-sided Laplace transformation, , is an integral operator that converts a signal f(t) to a complex-
valued function F(s) in the following fashion:

∞
∆
[
ft()] ≡ Fs() = − ∫ ft()e – st dt (23.156)
0
We refer to the transformed function F(s) as the Laplace transform of the signal f(t). Picture the lower
−
limit 0 of the integral as a specific instant prior to but infinitesimally close to t = 0. It is customary to
use a lowercase symbol (f ) to represent a signal waveform and an uppercase symbol (F) to represent its
Laplace transform. (Although we speak here of time signals, there is nothing in Eq. (23.156) that requires
f(t) to be a function of time. The transformation can be applied to functions of any quantity t.)
We shall use the Laplace transformation to transform the signals of time-invariant linear systems. The
behavior of such a system for t ≥ 0 depends only on the input signal for t ≥ 0 and on the prior state of
−
the output variable (at t = 0 ). Hence, it does not matter that the Laplace transformation ignores f(t) for
−
t < 0 .
The process of finding the time function f(t) that corresponds to a particular Laplace transform F(s)
−1
is called inverse Laplace transformation, and is denoted by . We also call f(t) the inverse Laplace
−
transform of F(s). Since the one-sided Laplace transformation ignores t < 0 , F(s) contains no information
− −
about f(t) for t < 0 . Therefore, inverse Laplace transformation cannot reconstruct f(t) for t < 0 . We
−
shall treat all signals as if they are defined only for t ≥ 0 . Then there is a one-to-one relation between
f(t) and F(s).
To illustrate the Laplace transformation, we find the Laplace transform of the decaying exponential,
−
−αt
f(t) = e , t ≥ 0 . The transform is
∞
( –
s +a)t
∞
e
Fs() = 0 − ∫ e – αt – st dt = --------------------
e
s +
a)
( –
0 −
(23.157)
∞
( –
σ +a)t −jωt
e
1
= e -------------------------- = ------------ for Re s[] > – a
( – s + a) − s + a
0
We must require σ > −α, where σ is the real part of s, in order that the real-exponent factor converge
to zero at the upper limit. (The magnitude of the complex-exponent factor remains 1 for all t.) Therefore,
the Laplace transform of the decaying exponential is defined only for Re[s] > −α. This restriction on the
−
domain of F in the complex s plane is comparable to the restriction t ≥ 0 on the domain of f.
The significant features of the complex-frequency function 1/(s + α) are the existence of a single pole
and the location of that pole, s = −α [rad/s]. (The pole defines the left boundary of that region of the
complex s plane over which the transform 1/(s + α) is defined.) The significant features of the corre-
sponding time function are the fact of decay and the rate of decay, with the exponent −α [rad/s]. There
are clear parallels between the features of f(t) and F(s). We should think of the whole complex-valued
function F as representing the whole time waveform f.
As a second transformation example, let f(t) = δ(t), the unit impulse, essentially a unit-area pulse of
very short duration. It acts at t = 0, barely within the lower limit of the Laplace integral. It has value zero
− −
at t = 0 . (Because we use 0 as the lower limit of the defining integral, it does not matter whether the
impulse straddles t = 0 or begins to rise at t = 0.) The impulse is nonzero only for t ≈ 0 , where e – st ≈ . 1
Therefore, the Laplace transform is
∞ – st ∞
∆ s() = − ∫ d t()e dt ≈ − ∫ d t() 1() dt = 1 (23.158)
0 0
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