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You can prove that L{H(t)} = 1/s,
from which it follows that L{U
o
⋅H(t)} = U
o
/s,
where U
o
is a constant. Also, L
-1
{1/s}=H(t),
and L
-1
{ U
o
/s}= U
o
⋅H(t).
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{H(t-k)}=e
–ks
⋅L{H(t)} = e
–ks
⋅(1/s) = (1/s)⋅e
–ks
.
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
.
y
x
x
0
(x
_
x)
0
H(x
_
x)
0
x
0
y
x
1