Download Inverse Laplace Transform: Properties and Examples and more Lecture notes Signals and Systems in PDF only on Docsity!
The Inverse Laplace Transform
- If L{f (t)} = F (s), then the inverse Laplace transform of F (s) is
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.
- Example: The inverse Laplace transform of
U (s) =
s^3
s^2 + 4
is
u(t) = L−^1 {U (s)}
=
L−^1
s^3
+ 3L−^1
s^2 + 4
s^2 2
- Example: Suppose you want to find the inverse Laplace transform x(t) of
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
+ L−^1
s − 3 (s − 3)^2 + 6
e−t^ t^3 6
6 t.
- Example: Let y(t) be the inverse Laplace transform of
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).
- Example: Let G(s) = s(s^2 + 4s + 5)−^1. The inverse transform of G(s) is
g(t) = L−^1
s s^2 + 4s + 5
= L−^1
s (s + 2)^2 + 1
= L−^1
s + 2 (s + 2)^2 + 1
− L−^1
(s + 2)^2 + 1
= e−^2 t^ cos t − 2 e−^2 t^ sin t. (5)
- There is usually more than one way to invert the Laplace transform. For example, let F (s) = (s^2 + 4s)−^1. You could compute the inverse transform of this function by completing the square:
f (t) = L−^1
s^2 + 4s
= L−^1
(s + 2)^2 − 4
L−^1
(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
− L−^1
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)
- Example: Find the inverse Laplace transform x(t) of the function
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
− L−^1
s 4(s^2 + 4)
(1 − cos 2t).
