Page 460 - Mechanical Engineers' Handbook (Volume 2)

2 Laplace Transforms 451

complete solution of a differential equation require the evaluation of the integration constants
by use of the initial conditions. In the case of the Laplace transform method, initial conditions
are automatically included in the Laplace transform of the differential equation. If all initial
conditions are zero, then the Laplace transform of the differential equation is obtained simply
2
2
2
by replacing d/dt with s, d /dt with s , and so on.
Consider the differential equation
dy
2
5 3y ƒ(t) (15)
dt 2
with ˙y(0) 1, y(0) 0. By taking the Laplace transform of Eq. (15), we get
L 5 dy 3y
2
dt 2 L[ƒ(t)]
Now using Theorems T1 and T2 (page 449) yields
2
5[sY(s) sy(0) y(0)] 3Y(s) F(s)
Thus
F(s) 5
Y(s) (16)
2
5s 3
For a given ƒ(t), say ƒ(t) u (t), the unit step function is given as
s
1
F(s) (17)
s
Substituting Eq. (17) in Eq. (16) gives
5s 1
Y(s)
s(5s 3)
2
Then y(t) can be found by computing the inverse Laplace transform of Y(s). That is,
y(t) L [Y(s)] L 5s 1
1
1
2
s(5s 3)
This will be discussed in Section 2.4.

2.3 Transfer Function
The transfer function of a LTI system is defined to be the ratio of the Laplace transform of
the output to the Laplace transform of the input under the assumption that all initial con-
ditions are zero.
For the system described by the LTI differential equation (1),
L[output] L[y(t)] Y(s)

L[input] L[u(t)] U(s)
bs b s m 1 bs b
m
m m 1 1 0 (18)
n
as a n 1 s n 1 as a 0
n
1
It should be noted that the transfer function depends only on the system parameters
characterized by a ’s and b ’s in Eq. (18). The highest power of s in the denominator of the
i
i