HP 50g User Manual page 493

y
You can prove that
from which it follows that
where U is a constant. Also, L
o
and
Also, using the shift theorem for a shift to the right, L{f(t-a)}=e
–as
e F(s), we can write L{H(t-k)}=e
Another important result, known as the second shift theorem for a shift to the
-1
right, is that L {e
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.,
Similarly, 'U0' ` LAP , produces the result 'U0/X', i.e., L{U
L{1} = 1/s.
= U /s.
0
You can obtain Dirac's delta function in the calculator by using: 1` ILAP
The result is
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
Also, using the shift theorem for a shift to the right, L{f(t-a)}=e
–as
e F(s), we can write L{ (t-k)}=e
(x _ x )
x
0
L{U
L
–as
F(s)}=f(t-a) H(t-a), with F(s) = L{f(t)}.
-1
{1.0}= (t), it follows that L{ (t)} = 1.0
y
0
1
x
L{H(t)} = 1/s,
H(t)} = U /s,
o o
-1
{1/s}=H(t),
-1
{ U /s}= U
o o
–ks
L{H(t)} = e
'Delta(X)' .
–ks –ks
L{ (t)} = e
H(x _ x )
0
x
x
0
H(t).
–as
–ks
(1/s) = (1/s) e
–as
–ks
1.0 = e .
L{f(t)} =
–ks
.
}
0
L{f(t)} =
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