Laplace Transform
Laplace transform converts a time domain function to s-domain
function by integration from zero to infinity
of the time domain function, multiplied by e-st.
The Laplace transform is used to quickly find solutions for differential
equations and integrals.
Derivation in the time domain is transformed to multiplication by
s in the s-domain.
Integration in the time domain is transformed to division by s in
the s-domain.
Laplace transform
function
The Laplace transform is defined with the L{} operator:

Inverse Laplace transform
The inverse Laplace transform can be calculated directly.
Usually the inverse transform is given from the transforms table.
Laplace transform table
| Function name |
Time domain function |
Laplace transform |
|
f (t) |
F(s) = L{f (t)} |
| Constant |
1 |
 |
| Linear |
t |
 |
| Power |
t n |
 |
| Power |
t a |
Γ(a+1) · s -(a+1) |
| Exponent |
e at |
 |
| Sine |
sin at |
 |
| Cosine |
cos at |
 |
| Hyperbolic sine |
sinh at |
 |
| Hyperbolic cosine |
cosh at |
 |
| Growing sine |
t sin at |
 |
| Growing cosine |
t cos at |
 |
| Decaying sine |
e -at sin ωt |
 |
| Decaying cosine |
e -at cos ωt |
 |
| Delta function |
δ(t) |
1 |
| Delayed delta |
δ(t-a) |
e-as |
Laplace transform properties
Laplace transform examples
Example #1
Find the transform of f(t):
f (t) = 3t + 2t2
Solution:
F(s) = 3/s2 + 4/s2
Example #2
Find the inverse transform of F(s):
F(s) = 3 / (s2 +
s - 6)
Solution:
In order to find the inverse transform, we need to change the s
domain function to a simpler form:
F(s) = 3 / (s2 +
s - 6) = 3 / [(s-2)(s+3)] = a / (s-2)
+ b / (s+3)
[a(s+3) + b(s-2)] / [(s-2)(s+3)] = 3
/ [(s-2)(s+3)]
a(s+3) + b(s-2) = 3
To find a and b, we get 2 equations - one of the s coefficients
and second of the rest:
(a+b)s + 3a-2b =
3
a+b = 0 , 3a-2b = 3
a = 3 , b = -3
F(s) = 3 / (s-2) - 3 / (s+3)
Now F(s) can be transformed easily by using the transforms table
for exponent function:
f (t) = 3e2t
- 3e-3t
See also