Laplace of sin(2t)
Understand the Problem
The question is asking for the Laplace transform of the function sin(2t). The Laplace transform is a technique used to transform a time-domain function into a complex frequency domain representation.
Answer
The Laplace transform of $\sin(2t)$ is $\frac{2}{s^2 + 4}$.
Answer for screen readers
The Laplace transform of the function $\sin(2t)$ is given by:
$$ \mathcal{L}{\sin(2t)} = \frac{2}{s^2 + 4} $$
Steps to Solve
- Define the Laplace Transform
The Laplace transform of a function $f(t)$ is given by the formula:
$$ \mathcal{L}{f(t)} = \int_0^{\infty} e^{-st} f(t) , dt $$
where $s$ is a complex number.
- Substitute the Function
In this problem, we need to find the Laplace transform of $f(t) = \sin(2t)$.
We substitute this into the Laplace transform formula:
$$ \mathcal{L}{\sin(2t)} = \int_0^{\infty} e^{-st} \sin(2t) , dt $$
- Use Integration by Parts
To solve the integral, we will use integration by parts. Let:
- $u = \sin(2t)$, then $du = 2\cos(2t) dt$
- $dv = e^{-st} dt$, then $v = -\frac{1}{s} e^{-st}$
Using integration by parts, we have:
$$ \int u , dv = uv - \int v , du $$
- Evaluate the Integral
Apply the integration by parts formula:
$$ \int_0^{\infty} e^{-st} \sin(2t) , dt = \left[-\frac{1}{s} e^{-st} \sin(2t)\right]_0^{\infty} + \frac{2}{s} \int_0^{\infty} e^{-st} \cos(2t) , dt $$
The first term evaluates to 0 as $t \to \infty$, leaving us with:
$$ = \frac{2}{s} \int_0^{\infty} e^{-st} \cos(2t) , dt $$
- Finding the Remaining Integral
The integral of $e^{-st} \cos(2t)$ can also be computed and is given by:
$$ \int_0^{\infty} e^{-st} \cos(2t) , dt = \frac{s}{s^2 + 4} $$
- Combine Results and Simplify
Substituting back, we have:
$$ \mathcal{L}{\sin(2t)} = \frac{2}{s} \cdot \frac{s}{s^2 + 4} = \frac{2}{s^2 + 4} $$
The Laplace transform of the function $\sin(2t)$ is given by:
$$ \mathcal{L}{\sin(2t)} = \frac{2}{s^2 + 4} $$
More Information
The Laplace transform is widely used in engineering and physics, particularly for solving linear ordinary differential equations. The transform helps simplify the manipulation of functions in the frequency domain. The specific form of the transform for $\sin(kt)$ indicates how sinusoidal inputs can be represented in a more manageable mathematical form.
Tips
- Forgetting the limits of integration or assuming the exponential term converges without evaluation can lead to incorrect evaluations.
- Misapplying integration by parts or not correctly keeping track of terms when integrating.
