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Examples and properties of the inverse Laplace transform, including its linearity, shift property, and use of partial fraction decomposition. It also demonstrates how to find the inverse Laplace transform of given functions.
Typology: Lecture notes
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The Inverse Laplace Transform
L−^1 {F (s)} = f (t). (1)
The inverse transform L−^1 is a linear operator:
L−^1 {F (s) + G(s)} = L−^1 {F (s)} + L−^1 {G(s)}, (2)
and L−^1 {cF (s)} = cL−^1 {F (s)}, (3)
for any constant c.
U (s) =
s^3
s^2 + 4
is
u(t) = L−^1 {U (s)}
=
s^3
s^2 + 4
s^2 2
X(s) =
(s + 1)^4
s − 3 (s − 3)^2 + 6
Just use the shift property (paragraph 11 from the previous set of notes):
x(t) = L−^1
(s + 1)^4
s − 3 (s − 3)^2 + 6
e−t^ t^3 6
6 t.
Y (s) =
e−^3 s^ s s^2 + 4
Don’t worry about the exponential term. Since the inverse transform of s/(s^2 +4) is cos 2t, we have by the switchig property (paragraph 12 from the previous notes):
y(t) = L−^1
e−^3 s^ s s^2 + 4
= H(t − 3) cos 2(t − 3).
g(t) = L−^1
s s^2 + 4s + 5
s (s + 2)^2 + 1
s + 2 (s + 2)^2 + 1
(s + 2)^2 + 1
= e−^2 t^ cos t − 2 e−^2 t^ sin t. (5)
f (t) = L−^1
s^2 + 4s
(s + 2)^2 − 4
(s + 2)^2 − 4
e−^2 t^ sinh 2t. (6)
You could also use the partial fraction decomposition (PFD) of F (s):
F (s) =
s(s + 4)
4 s
4(s + 4)
Therefore,
f (t) = L−^1 {F (s)}
= L−^1
4 s
4(s + 4)
e−^4 t
e−^2 t^ sinh 2t. (7)
that is, L−^1 {F (s)G(s)} = (f ∗ g)(t). (13)
Suppose that you want to find the inverse transform x(t) of X(s). If you can write X(s) as a product F (s)G(s) where f (t) and g(t) are known, then by the above result, x(t) = (f ∗ g)(t).
Example: Consider the previous example: Find the inverse transform q(s) of
Q(s) = 3 s (s^2 + 1)^2
Write Q(s) = F (s)G(s), where
F (s) =
s^2 + 1
and G(s) = s s^2 + 1
The inverse transforms are of F (s) and G(s) are f (t) = 3 sin t and g(t) = cos t. Therefore
q(s) = L−^1 {Q(s)} = L−^1 {F (s)G(s)} = (f ∗ g)(t)
= 3
∫ (^) t
0
sin (t − v) cos v dv. (14)
Even if you stop here, you at least have a fairly simple, compact expression for q(s). To do the integral (14), use the trigonometric identity
sin A cos B = sin (A + B) + sin (A − B) 2
With this, (14) becomes
q(s) =
∫ (^) t
0
sin t dv +
∫ (^) t
0
sin (t − 2 v) dv
t sin t. (15)
X(s) =
s(s^2 + 4)
If you want to use the convolution theorem, write X(s) as a product:
X(s) =
s
s^2 + 4
Since
L−^1
s
and
L−^1
s^2 + 4
sin 2t,
we have
x(t) =
∫ (^) t
0
sin 2v dv
(1 − cos 2t).
You could also use the PFD:
X(s) =
4 s
s 4(s^2 + 4)
Therefore,
x(t) = L−^1
4 s
s 4(s^2 + 4)
(1 − cos 2t).