Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Inverse Laplace Transform: Properties and Examples, Lecture notes of Signals and Systems

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

2021/2022

Uploaded on 09/12/2022

marphy
marphy 🇺🇸

4.3

(30)

284 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
The Inverse Laplace Transform
1. If L{f(t)}=F(s), then the inverse Laplace transform of F(s) is
L1{F(s)}=f(t).(1)
The inverse transform L1is a linear operator:
L1{F(s) + G(s)}=L1{F(s)}+L1{G(s)},(2)
and
L1{cF (s)}=cL1{F(s)},(3)
for any constant c.
2.Example: The inverse Laplace transform of
U(s) = 1
s3+6
s2+ 4,
is
u(t) = L1{U(s)}
=1
2L12
s3+ 3L12
s2+ 4
=s2
2+ 3 sin 2t. (4)
3.Example: Suppose you want to find the inverse Laplace transform x(t) of
X(s) = 1
(s+ 1)4+s3
(s3)2+ 6.
Just use the shift property (paragraph 11 from the previous set of notes):
x(t) = L11
(s+ 1)4+L1s3
(s3)2+ 6
=ett3
6+e3tcos 6t.
4.Example: Let y(t) be the inverse Laplace transform of
Y(s) = e3ss
s2+ 4.
pf3
pf4
pf5

Partial preview of the text

Download Inverse Laplace Transform: Properties and Examples and more Lecture notes Signals and Systems in PDF only on Docsity!

The Inverse Laplace Transform

  1. 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.

  1. 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

  • 3 sin 2t. (4)
  1. 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

  • e^3 t^ cos

6 t.

  1. 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).

  1. 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)

  1. 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)

  1. 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).

  2. 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)

  1. 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).