Question

Evaluate the Laplace transform of the step function \(F_{0} \mathcal{H}(t)\).

Short answer

The Laplace transform of the step function \(F_{0} \mathcal{H}(t)\) is \(\frac{F_{0}}{s}\).

Step-by-step solution

Recall the Definition of the Laplace Transform and Heaviside Step Function

The Laplace transform of a function f(t) is given by: $$ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt $$ where s is a complex variable with the real part greater than the abscissa of convergence of f(t). The Heaviside step function, denoted by \(\mathcal{H}(t)\), is defined as: $$ \mathcal{H}(t) = \begin{cases} 1, & \text{if } t > 0 \\ 0, & \text{if } t < 0 \end{cases} $$ Given that the exercise asks to calculate the Laplace transform of the step function \(F_{0} \mathcal{H}(t)\), where \(F_0\) is a constant, we will directly work with this function.

Apply the Definition of the Laplace Transform to the Step Function

We want to calculate the Laplace transform of \(F_{0}\mathcal{H}(t)\). Using the definition of the Laplace transform, we need to evaluate the following improper integral: $$ \mathcal{L}\{F_{0}\mathcal{H}(t)\} = \int_{0}^{\infty} e^{-st} (F_{0}\mathcal{H}(t)) dt $$ Since \(\mathcal{H}(t) = 1\) for \(t > 0\), the integral simplifies to: $$ \mathcal{L}\{F_{0}\mathcal{H}(t)\} = F_{0} \int_{0}^{\infty} e^{-st} dt $$

Evaluate the Integral

Now, we need to evaluate the integral: $$ \int_{0}^{\infty} e^{-st} dt $$ The integral can be found using integration by substitution. Let \(u = -st\), then \(du = -s dt\). So, $$ -\frac{1}{s} \int_{0}^{\infty} e^{u} du $$ Now, integrating \(e^{u}\) with respect to u, we have $$ -\frac{1}{s}\left[ e^{u} \right]_{0}^{\infty} $$ Substituting back for u, we get: $$ -\frac{1}{s}\left[ e^{-st} \right]_{0}^{\infty} $$ By evaluating the limits of the function, we have: $$ -\frac{1}{s}\left( \lim_{t \to \infty}e^{-st} - e^{0}\right) = -\frac{1}{s}\left( 0 - 1\right) = \frac{1}{s} $$

Multiply by the Constant

Now, we need to multiply our result by the constant \(F_{0}\): $$ F_{0} \cdot \frac{1}{s} = \frac{F_{0}}{s} $$

Final Answer

The Laplace transform of the step function \(F_{0} \mathcal{H}(t)\) is: $$ \mathcal{L}\{F_{0}\mathcal{H}(t)\} = \frac{F_{0}}{s} $$