18.03 Class 27, April 17, 2006
Laplace Transform II: inverse transform, tderivative rule,
use in solving ODEs; partial fractions: coverup method; sderivative
rule.
Definition:
F(s)
=
L[f(t)]
=
integral_{0}^infty f(t) e^{st} dt
,
Re(s) >> 0
Rules:
L
is linear:
L[af(t)+bg(t)]
=
aF(s) + bG(s)
F(s)
essentially determines
f(t)
sshift:
L[ e^{at}f(t)
=
F(sa)
tderivative:
L[f'(t)]
=
s F(s)  f(0+)
if we omit the singularity
of
f'(t)
at
t = 0 .
Computations:
L[1]
=
1/s
L[e^{as}]
=
1/(sa)
L[cos(omega t)]
=
s/(s^2+omega^2)
L[sin(omega t)]
=
omega/(s^2+omega^2)
L[delta(ta)]
=
e^{as}
[1] The tderivative
rule:
Compute:
L[f'(t)]
=
integral_{0}^infty f'(t) e^{st} dt
u
=
e^{st}
du
=
s e^{st} dt
dv
=
f'(t) dt
v
=
f(t)
...
=
e^{st} f(t) _{0}^infty + s integral f(t) e^{st}
dt
We continue to assume that
f(t)
doesn't grow too fast with
t (so that
the integral defining
F(s)
converges for
Re(s)
sufficiently large).
This means that for
s
sufficiently large, the evaluation of the first
term at
infinity
becomes zero.
Since we are always assuming rest
initial conditions, the evaluation at
zero is also zero. Thus
...
=
s F(s)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentNow, what is
f'(t) ? If
f(t)
has discontinuities, we must mean the
generalized derivative.
There is one discontinuity in
f(t)
that we
can't
just wish away:
f(0) = 0 , while we had better let
f(0+)
be whatever
it wants to be.
We have to expect a discontinuity at
t = 0 .
Just to keep the notation in bounds, let's suppose that
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Staff
 Fractions, Derivative, Laplace

Click to edit the document details