Laplace Transform Introduction

The Laplace transform is a powerful mathematical tool used to convert a function of a real variable \(t\) (often time) to a function of a complex variable \(s\) (complex frequency). This transformation can simplify solving differential equations and analyzing linear time-invariant systems in fields like electrical engineering and control systems.

The definition of the unilateral (one-sided) Laplace transform \(F(s)\) of a function \(f(t)\) is given by the integral:

\[F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt\]

Here, \(s\) is a complex variable, and \(\mathcal{L}\) denotes the Laplace transform operator.

Transforming \(f(t) = e^{-bt} u(t)\)

We need to find the Laplace transform of the function \(f(t) = e^{-bt} u(t)\). The term \(u(t)\) represents the unit step function, which is defined as:

• \(u(t) = 0\) for \(t < 0\)
• \(u(t) = 1\) for \(t \geq 0\)

Because of the unit step function \(u(t)\), the function \(f(t)\) is non-zero only for \(t \geq 0\). This simplifies the lower limit of the integral to \(0\).

Substitute \(f(t) = e^{-bt} u(t)\) into the Laplace transform definition:

\[F(s) = \int_{0}^{\infty} e^{-st} (e^{-bt} u(t)) dt\]

Since \(u(t) = 1\) for \(t \geq 0\), the expression becomes:

\[F(s) = \int_{0}^{\infty} e^{-st} e^{-bt} dt\]

Combine the exponential terms using the property \(e^A e^B = e^{A+B}\):

\[F(s) = \int_{0}^{\infty} e^{-(s+b)t} dt\]

Integral Evaluation for Laplace Transform

Now, we evaluate this definite integral. This is a standard integral of the form \(\int e^{ax} dx = \frac{1}{a} e^{ax}\). Here, \(a = -(s+b)\) and \(x = t\).

\[F(s) = \left[ \frac{e^{-(s+b)t}}{-(s+b)} \right]_{t=0}^{\infty}\]

Apply the limits of integration:

\[F(s) = \lim_{T \to \infty} \left( \frac{e^{-(s+b)T}}{-(s+b)} \right) - \left( \frac{e^{-(s+b) \cdot 0}}{-(s+b)} \right)\]

For the integral to converge, we typically require that the real part of \((s+b)\) be greater than zero (\(Re(s+b) > 0\)). Under this condition, as \(T \to \infty\), \(e^{-(s+b)T}\) approaches \(0\).

\[\lim_{T \to \infty} \frac{e^{-(s+b)T}}{-(s+b)} = 0\]

The second term simplifies as \(e^0 = 1\):

\[\frac{e^{-(s+b) \cdot 0}}{-(s+b)} = \frac{1}{-(s+b)}\]

So, combining these, we get:

\[F(s) = 0 - \left( \frac{1}{-(s+b)} \right)\]

\[F(s) = \frac{1}{s+b}\]

This result is a fundamental Laplace transform pair: \(\mathcal{L}\{e^{-bt} u(t)\} = \frac{1}{s+b}\).

Conclusion and Options Analysis

Comparing our derived Laplace transform with the given options:

• Option 1: \(F(s)= \frac{s}{s+b}\) - Incorrect.
• Option 2: \(F(s)= \frac{1}{s(s+b)}\) - Incorrect.
• Option 3: \(F(s)= \frac{1}{s+b}\) - Correct.
• Option 4: \(F(s)= \frac{s}{1+b}\) - Incorrect.

The Laplace transform of \(f(t) = e^{-bt} u(t)\) is indeed \(F(s)= \frac{1}{s+b}\).