Quick Math Laplace Transform
Enter expression in this form
Example:
- \( f(t) = \sin(bt) \) enter as sin(nt)
- \( f(t) = t^n \) enter as t^n
Laplace Transform
The Laplace Transform is a widely used integral transform in mathematics and engineering that converts a function of a real variable tt (usually time) into a function of a complex variable ss (complex frequency). This transform is particularly useful for analyzing linear time-invariant systems, solving differential equations, and transforming differential equations into algebraic equations.
Definition
The Laplace Transform \( \mathcal{L}\{f(t)\} \) of a function \( f(t) \) is defined by:
\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \]
where:
– \( t \) is a real number (usually representing time),
– \( s \) is a complex number (\( s = \sigma + i\omega \)).
Properties
– Linearity: \( \mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s) \)
– First Derivative: \( \mathcal{L}\{f'(t)\} = s F(s) – f(0) \)
– Second Derivative: \( \mathcal{L}\{f”(t)\} = s^2 F(s) – s f(0) – f'(0) \)
– Shifting: \( \mathcal{L}\{e^{at} f(t)\} = F(s-a) \)
Worked Examples
Example 1: Laplace Transform of \( f(t) = e^{at} \)
Problem: Find the Laplace Transform of \( f(t) = e^{at} \).
Solution:
\[ \mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \int_{0}^{\infty} e^{-(s-a)t} \, dt \]
\[ = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_{0}^{\infty} \]
\[ = \left. \frac{e^{-(s-a)t}}{-(s-a)} \right|_{0}^{\infty} = \frac{1}{s-a} \]
Answer: \( \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \).
Example 2: Laplace Transform of \( f(t) = \sin(bt) \)
Problem: Find the Laplace Transform of \( f(t) = \sin(bt) \).
Solution:
\[ \mathcal{L}\{\sin(bt)\} = \int_{0}^{\infty} e^{-st} \sin(bt) \, dt \]
Using the identity \( \sin(bt) = \frac{e^{ibt} – e^{-ibt}}{2i} \):
\[ \mathcal{L}\{\sin(bt)\} = \int_{0}^{\infty} e^{-st} \frac{e^{ibt} – e^{-ibt}}{2i} \, dt \]
\[ = \frac{1}{2i} \left( \int_{0}^{\infty} e^{-(s-ib)t} \, dt – \int_{0}^{\infty} e^{-(s+ib)t} \, dt \right) \]
\[ = \frac{1}{2i} \left( \frac{1}{s-ib} – \frac{1}{s+ib} \right) \]
\[ = \frac{1}{2i} \left( \frac{s+ib – (s-ib)}{(s-ib)(s+ib)} \right) \]
\[ = \frac{1}{2i} \left( \frac{2ib}{s^2 + b^2} \right) \]
\[ = \frac{b}{s^2 + b^2} \]
Answer: \( \mathcal{L}\{\sin(bt)\} = \frac{b}{s^2 + b^2} \).
Example 3: Laplace Transform of \( f(t) = t^n \)
Problem: Find the Laplace Transform of \( f(t) = t^n \) where \( n \) is a positive integer.
Solution:
\[ \mathcal{L}\{t^n\} = \int_{0}^{\infty} e^{-st} t^n \, dt \]
Using integration by parts:
Let \( u = t^n \) and \( dv = e^{-st} dt \). Then \( du = n t^{n-1} dt \) and \( v = \frac{e^{-st}}{-s} \).
\[ \mathcal{L}\{t^n\} = \left. \frac{t^n e^{-st}}{-s} \right|_{0}^{\infty} + \int_{0}^{\infty} \frac{n t^{n-1} e^{-st}}{s} \, dt \]
The first term evaluates to zero as \( t^n e^{-st} \) goes to zero at both limits.
\[ \mathcal{L}\{t^n\} = \frac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \, dt \]
Repeating the integration by parts \( n \) times:
\[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \]
Answer: \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \).
Note
The Laplace Transform is a powerful tool for transforming differential equations into algebraic equations, simplifying the process of solving them. It is widely used in engineering, physics, and control theory. Through the worked examples, we’ve seen how the Laplace Transform is applied to exponential, trigonometric, and polynomial functions.
