Page 751 - The Mechatronics Handbook

0066_Frame_C23 Page 59 Wednesday, January 9, 2002 1:56 PM

The two sides of Eq. (23.159) are identical functions of time. Therefore, the Laplace transforms of the
two sides of Eq. (23.159) are equal. Since the Laplace transformation is linear (properties 1 and 2 of
Table 23.12), and since the coefficients of the differential equation are constants, the Laplace transform
can be applied separately to the individual terms of each side. The result is
−
−
−
−
(
ms V 2 s() sv 2 0 ) v ˙ 2 0 )] + bsV 2 s() v 2 0 )] + kV 2 s() = bsV 1 s() v 1 0 )] + kV 1 s()
[
2
[
(
(
(
[
–
–
–
–
(23.160)
where the derivative properties of the Laplace transformation (properties 3 and 4 of Table 23.12) introduce the
−
−
prior values v 1 (0 ), v 2 (0 ), and v ˙ 2 0( − ) into the equation. According to Eq. (23.160), to fully determine
the transform V 2 (s) of the behavior v 2 (t), we must specify these prior values and also V 1 (s). It can be shown
that specifying the three prior values is equivalent to specifying the energy states of the spring and mass.
Let us assume that the independent source applies the constant velocity v 1 (t) = v c beginning at t = 0.
The corresponding transform, by item 2 of Table 23.11 and property 1 of Table 23.12, is V 1 (s) = v c /s.
Substitute the transform V 1 (s) into Eq. (23.160) and solve for
−
−
−
−
(
(
[
( bs + k)v c + msv ˙ 2 0 ) + bs v 2 0 ) v 1 0 )] + ms v 2 0 )
(
(
2
–
V 2 s() = --------------------------------------------------------------------------------------------------------------------------------------- (23.161)
sms + bs + k)
(
2
We could find the output signal waveform v 2 (t) as a function of the model parameters m, k, b, the
− − −
source-signal parameter v c , and the prior state information v 1 (0 ), v 2 (0 ), and v ˙ 2(0 ) , but the expression
for the solution would be messy. Instead, we complete the solution process for specific numbers: m =
− 2 − −
2 kg, b = 4 N · s/m, k = 10 N/m, v ˙ 2(0 ) = 0 m/s , v 1 (0 ) = 0 m/s, v 2 (0 ) = −1 m/s, and v c = 1 m/s. The
partial-fraction expansion of the transform and the inverse transform, both obtained by a commercial
computer program, are
2s +
2
V 2 s() = 1 ----------------------------- (23.162)
-- –
2
s ( s + 1) + 2 2
v 2 t() = 12e cos (), for t ≥ 0 (23.163)
−t
2t
–
We can take Laplace transforms of the system equations at any stage in their development. We can even
write the equations directly in terms of transformed variables if we wish. The process of eliminating variables
can be carried out as well in one notation as in another. For example, the operator G(p) in Eq. (23.154)
represents a ratio of polynomials in the time-derivative operator p. Therefore, Laplace transforming the
differential equation, Eq. (23.154), introduces the prior values of various derivatives of y 1 and y 2 . If the
prior values of all these derivatives are zero, then the Laplace-transformed equation is
Y 2 s() = Gs()Y 1 s() (23.164)
where the operator p in Eq. (23.154) is replaced by the complex-frequency variable s in Eq. (23.164). It
is appropriate, therefore, to define the transfer function directly in terms of Laplace-transformed signals:
Y 2 s()
Gs() = ------------ (23.165)
Y 1 s()
PV=0
where Y 1 (s) and Y 2 (s) are the Laplace transforms of the signals y 1 (t) and y 2 (t), and the notation PV = 0
−
means that the prior values (at t = 0 ) of y 1 (t) and y 2 (t) and the various derivatives mentioned above in
connection with Eq. (23.164) are set to zero. The frequency domain definition, Eq. (23.165), is equivalent
to the time domain definition, Eq. (23.155).
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