HP 50G, Руководство пользователя

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Page 16-13
Θ
Differentiation theorem for the n-th derivative.
Let f
(k)
o
= d
k
f/dx
k
|
t = 0
, and f
o
= f(0), then
L{d
n
f/dt
n
} = s
n
⋅
F(s) – s
n-1
⋅
f
o
−
…– s
⋅
f
(n-2)
o
– f
(n-1)
o
.
Θ
Linearity theorem. L{af(t)+bg(t)} = a
⋅
L{f(t)} + b
⋅
L{g(t)}.
Θ
Differentiation theorem for the image function. Let F(s) = L{f(t)}, then d
n
F/
ds
n
= L{(-t)
n
⋅
f(t)}.
Θ
Integration theorem. Let F(s) = L{f(t)}, then
Θ
Convolution theorem. Let F(s) = L{f(t)} and G(s) = L{g(t)}, then
Example 2 – As a follow up to Example 1, the acceleration a(t) is defined as
a(t) = d
2
r/dt
2
. If the initial velocity is v
o
= v(0) = dr/dt|
t=0
, then the Laplace
transform of the acceleration can be written as:
A(s) = L{a(t)} = L{d
2
r/dt
2
}= s
2
⋅
R(s) - s
⋅
r
o
– v
o
.
Example 3 – Let f(t) = e
–at
, using the calculator with ‘EXP(-a*X)’
`
LAP, you
get ‘1/(X+a)’, or F(s) = 1/(s+a). The third derivative of this expression can be
calculated by using:
‘X’
` ‚¿
‘X’
`‚¿
‘X’
` ‚¿ μ
The result is
‘-6/(X^4+4*a*X^3+6*a^2*X^2+4*a^3*X+a^4)’, or
d
3
F/ds
3
= -6/(s
4
+4
⋅
a
⋅
s
3
+6
⋅
a
2
⋅
s
2
+4
⋅
a
3
⋅
s+a
4
).
Now, use ‘(-X)^3*EXP(-a*X)’
`
LAP
μ
. The result is exactly the same.
{ }
). (
1
) (
0
s F
s
du u f
t
⋅ =
∫
L