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Another important result, known as the second shift theorem for a shift to the
right, is that L
-1
{e
–as
⋅
F(s)}=f(t-a)
⋅
H(t-a), with F(s) = L{f(t)}.
In the calculator the Heaviside step function H(t) is simply referred to as ‘1’.
To check the transform in the calculator use:
1
`
LAP. The result is ‘1/X’,
i.e., L{1} = 1/s. Similarly, ‘U0’
`
LAP , produces the result ‘U0/X’, i.e.,
L{U
0
} = U
0
/s.
You can obtain Dirac’s delta function in the calculator by using:
1`
ILAP
The result is
‘
Delta(X)
’
.
This result is simply symbolic, i.e., you cannot find a numerical value for, say
‘
Delta(5)
’.
This result can be defined the Laplace transform for Dirac’s delta function,
because from L
-1
{1.0}=
δ
(t), it follows that L{
δ
(t)} = 1.0
Also, using the shift theorem for a shift to the right, L{f(t-a)}=e
–as
⋅
L{f(t)} =
e
–as
⋅
F(s), we can write L{
δ
(t-k)}=e
–ks
⋅
L{
δ
(t)} = e
–ks
⋅
1.0 = e
–ks
.
Applications of Laplace transform in the solution of linear ODEs
At the beginning of the section on Laplace transforms we indicated that you
could use these transforms to convert a linear ODE in the time domain into an
algebraic equation in the image domain. The resulting equation is then
solved for a function F(s) through algebraic methods, and the solution to the
ODE is found by using the inverse Laplace transform on F(s).
The theorems on derivatives of a function, i.e.,
L{df/dt} = s
⋅
F(s) - f
o
,
L{d
2
f/dt
2
} = s
2
⋅
F(s) - s
⋅
f
o
– (df/dt)
o
,
and, in general,
L{d
n
f/dt
n
} = s
n
⋅
F(s) – s
n-1
⋅
f
o
−
…– s
⋅
f
(n-2)
o
– f
(n-1)
o
,
are particularly useful in transforming an ODE into an algebraic equation.
