If y(a) is a unique function which is continuous on [0, \(\infty\)] and also satisfy L[y(a)](b) = Y(b), then it is an Inverse Laplace transform of Y(b).
You can select a piecewise continuous function, if all other possible functions, y(a) are discontinuous, to be the inverse transform.
L^{-1}[Y(b)](a)
Definition of Inverse Laplace Transform
An integral defines the laplace transform Y(b) of a function y(a) defined on [o, \(\infty\)].
Also, the formula to determine y(a) if Y(b) is given, involves an integral.
Inverse Laplace transform table
To compute the inverse transform, we will use the table:
Function y(a) |
Transform Y(b) |
b |
1 |
\(\frac{1}{b}\) |
b>0 |
a |
\(\frac{1}{b^2}\) |
b>0 |
A^{i} , i = integer |
\(\frac{i!}{s^(i+1)}\) |
b>0 |
exp (ta), where t = constant |
\(\frac{1}{(b-t)}\) |
b>t |
cos (sa), s= constant |
\(\frac{b}{b^2+s^2}\) |
b>0 |
Sin (sa), s = constant |
\(\frac{t}{b^2+s^2}\) |
b>0 |
exp(ta)cos(sa) |
\(\frac{b-t}{(b-t)^2 +s^2}\) |
b>t |
exp(ta)sin(sa) |
\(\frac{s}{(b-t)^2 +s^2}\) |
b>t |
Example of Inverse Laplace
Example: Find the inverse transform of each of the following.
Y(b)= \(\frac{6}{b}\) -\(\frac{1}{b-8}\) – \(\frac{4}{b-3}\)
Solution:
Step 1: The first term is a constant as we can see from the denominator of the first term.
Step 2: Before taking the inverse transform, let’s take the factor 6 out, so the correct numerator is 6.
Step 3: The second term has an exponential t = 8.
Step 4: The numerator is perfect.
Step 5: The third term is also an exponential, t= 3.
Step 6: Now before taking the inverse transforms, we need to factor out 4 first.
Y(b) = 6 \(\frac{1}{b}\) -\(\frac{1}{b-8}\) – 4\(\frac{1}{b-3}\)
y(a) = 6(1) – e^{8a} +4 (e^{3t})
= 6 – e^{8a} +4 e^{3t}
Inverse Laplace Transform Calculator
The Inverse Laplace Transform Calculator helps in finding the Inverse Laplace Transform Calculator of the given function.
Remember, L^{-1}[Y(b)](a) is a function that y(a) that L(y(a) )= Y(b). So, generally, we use this property of linearity of Laplace transform to find the Inverse Laplace transform.