Laplace Transform of Constant: Super Easy Explanation!

The Laplace Transform, a fundamental concept in signal processing, finds extensive application in control systems analysis. Electrical engineers at institutions like MIT frequently leverage it. A key question that arises is, what is the laplace transform of a constant? It's a deceptively simple derivation and crucial for understanding more complex transformations with tools such as MATLAB. Let’s dive into an explanation.

Image taken from the YouTube channel Killian O'Brien , from the video titled Laplace transform of the constant function .

Laplace Transform of a Constant: A Simple Guide

This article focuses on understanding "what is the Laplace transform of a constant." We will break down the definition, the mathematical derivation, and provide examples to make the concept clear.

Defining the Laplace Transform

The Laplace transform is a mathematical tool that converts a function of time, f(t), into a function of complex frequency, F(s). It's particularly useful for solving differential equations, especially those involving initial conditions.

The Core Formula

The fundamental formula for the Laplace transform is:

F(s) = ∫₀^∞ e^(-st) f(t) dt

Where:

• F(s) is the Laplace transform of f(t).
• f(t) is the function of time.
• s is a complex frequency variable.
• The integral is evaluated from 0 to infinity.

Laplace Transform of a Constant Function

Now, let's consider the case where f(t) is a constant, say k. This means f(t) = k for all t.

Derivation

To find the Laplace transform of k, we substitute k for f(t) in the Laplace transform formula:

F(s) = ∫₀^∞ e^(-st) k dt

We can take the constant k outside the integral:

F(s) = k ∫₀^∞ e^(-st) dt

Now we need to evaluate the integral. The antiderivative of e^(-st) with respect to t is -e^(-st)/s. Therefore:

F(s) = k [-e^(-st)/s]₀^∞

Evaluating the Limits

Now we evaluate the antiderivative at the limits of integration (infinity and 0):

F(s) = k [lim (t→∞) (-e^(-st)/s) - (-e^(-s*0)/s)]

Assuming Re(s) > 0 (the real part of s is greater than zero) to ensure convergence, as t approaches infinity, e^(-st) approaches 0. Also, e^(0) = 1. Thus:

F(s) = k [0 - (-1/s)]

F(s) = k/s

Therefore, the Laplace transform of a constant k is k/s.

Examples

Here are a few examples to solidify your understanding:

• Example 1: f(t) = 5

  The Laplace transform of 5 is 5/s.

• Example 2: f(t) = -2

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  The Laplace transform of -2 is -2/s.

• Example 3: f(t) = π

  The Laplace transform of π is π/s.

Practical Applications

While seemingly simple, understanding the Laplace transform of a constant is crucial because:

1. Foundation for More Complex Functions: It's a building block for finding the Laplace transform of more complex time-domain functions.
2. System Analysis: It arises frequently when analyzing linear time-invariant (LTI) systems, particularly in control systems and electrical engineering. For example, it appears as a step input function.
3. Solving Differential Equations: Constants often appear in the differential equations that describe physical systems, and knowing their Laplace transform simplifies the solution process.

Table of Laplace Transforms (Partial)

Here is a small table illustrating the transform of the constant and some additional functions for context:

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