8.4 The Laplace transform 193The inverse transform is written
f(t) = L-~[F(s)] (8.24)
F(s) is often written as f which is read 'f bar'.
The Laplace transforms of many different functions of time have been
determined and tabulated. Since these tables are readily available there is no
need to derive the required transform in any particular problem. However, a
few are derived here in order to show the method.The Laplace transform of the exponential function
If/(t) = exp (-at) then
oo oc
L[f(t)] = F(s) = f exp (-at) exp (-st) dt = f exp [-(s + a)t] dt
0 0
= [-1/(s + a){exp (-(s + a))t}]o
= 1/(s + a) (8.25)The Laplace transform of the unit step function
If f(t) = 1 for t -> 0 and f(t) = 0 for t < 0 then
oc
L[f(t)] = F(s) - f 1 exp (-st)dt = -(1/s)[exp (-st)]o
0- 1Is. (8.26)
It follows that, for a step function of amplitude A, the Laplace transform is
A/s.
The Laplace transform of the derivative of a function
The derivative of the function f(t) is d(f(t))/dt. Let this be denoted by f'(t).
Then
oG
L[f'(t)] = ff'(t) exp (-st) dt
0
Integrating by parts (fu dv = [uv]- fv du) we have, with u = exp (-st) and
dv = f'(t) dt (so that v = f(t)),
oo or
ff'(t) exp (-st) dt = [exp (-st)f(t)]o - ff(t){-s exp (-st)} dt
0 0
The upper limit of the first term on the right-hand side must be zero in order
that it tends to infinity as t tends to zero. Thus the right-hand side becomes
oo
-f(O) + sff(t) exp (-st) dt
0
and the second term is s times the Laplace transform of the original function of
t
.'. L[f'(t)] = sL[f(t)] - f(O) (8.27)