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Today we will explore the Laplace Transform with some examples. Let's start with the Laplace Transform of a constant function. Can anyone tell me what the Laplace Transform of f(t) = 1 is?
Is it one divided by s?
Exactly! We express it mathematically as: β{1} = β«β^β e^(-st) * 1 dt, which evaluates to 1/s for s > 0. This is a critical result because it simplifies the constant function into an algebraic form.
Why does the condition s > 0 matter?
Great question! The condition ensures that the integral converges. If s were less than or equal to zero, the integral wouldn't yield a finite result. Remember: 's must always be positive for convergence!'
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Next, let's apply the Laplace Transform to an exponential function, such as f(t) = e^(at). What do you think the result will be?
Would it be something like 1/(s - a)?
Correct! The precise calculation involves the integral β{e^(at)} = β«β^β e^(-st) * e^(at) dt, which simplifies to 1/(s - a) for s > a. Can anyone remind us why s must be greater than a?
To ensure the integral converges, right?
Exactly! Always remember: 's must be larger than the exponential growth rate for convergence!'
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Finally, let's consider polynomial functions. For f(t) = t^n, can someone tell me how we might find the Laplace Transform?
Is there a formula for that?
Yes, there is! The result is given by β{t^n} = n!/s^(n + 1), with s > 0. Each polynomial can be transformed into a neat formula! Why do we care about this form?
Because it makes it easier to solve differential equations?
Absolutely! Remember the key principle: 'Laplace Transform simplifies the solving process!'
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
The section provides examples demonstrating how to apply the Laplace Transform to specific functions, such as a constant, an exponential function, and a polynomial function, highlighting how the transform simplifies these mathematical expressions.
In this section, we explore practical applications of the Laplace Transform through specific examples. Each example illustrates the process of transforming a time-domain function into the s-domain, showcasing key mathematical principles involved in the Laplace Transform.
$$\mathcal{L}\{1\} = \int_{0}^{\infty} e^{-st} \cdot 1 \, dt = \frac{1}{s}, \quad s > 0$$
demonstrates the Laplace Transform's ability to simplify functions into algebraic forms. The simplicity of this transform can be an effective strategy in solving initial value problems.
$$\mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \frac{1}{s - a}, \quad s > a$$
This example highlights the need for the variable s to satisfy specific conditions relative to the parameter a for the existence of the transform.
$$\mathcal{L}\{t^{n}\} = \int_{0}^{\infty} e^{-st} t^{n} \, dt = \frac{n!}{s^{n + 1}}, \quad s > 0$$
This case demonstrates how the Laplace Transform efficiently manages polynomial expressions, further underscoring its utility in solving differential equations.
Through these examples, we illustrate the convergence conditions for the Laplace Transform's validity and its relevance in engineering applications.
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Example 1: Laplace Transform of a Constant
π(π‘) = 1
β{1} = β«β^β πβ»Λ£π‘ β
1 ππ‘ = [ ] = , for π > 0
βπ
π
In this example, we compute the Laplace Transform of a constant function, which is 1. The integral we need to evaluate is from 0 to infinity, where we multiply our constant function by the exponential decay term e^(-st). The resulting integral simplifies to 1/s, as we evaluate it from 0 to infinity under the condition that s is greater than 0. This reflects that the total contribution from the constant value over time results in a finite value in the s-domain.
Think of a steady light bulb that gives off a constant amount of light. If this light bulb is turned on for an indefinite amount of time, the total 'light output' in our analysis can be represented using the Laplace Transform, showing how that output translates into a measurable effect (brightness) in a different scenario (s-domain), where it equals 1/s.
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Example 2: Laplace Transform of an Exponential Function
π(π‘) = π^{ππ‘}
β{π^{ππ‘}} = β«β^β πβ»Λ£π‘ β
π^{ππ‘} ππ‘ = β«β^β π^{-(π βπ)π‘} ππ‘ = , for π > π
π βπ
In this example, we consider an exponential function, f(t) = e^(at). To compute its Laplace Transform, we set up the integral similarly, but now we have the exponential terms combined. This allows us to factor out e^(-(s-a)t), leading to a simpler form to integrate. When we evaluate this integral, we find that it converges to 1/(s-a) for s greater than a. This property of exponential functions shows how they decay and are transformed into manageable algebraic fractions.
Imagine a plant that grows exponentially at a certain rate a. If you want to measure its growth over time and summarize how much it contributes at different points, the Laplace Transform helps portray that continuous growth in a simplified manner, converting it into a ratio (1/(s-a)), which is easier to analyze than tracking the actual growth curves in real-time.
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Example 3: Laplace Transform of π‘^n
π(π‘) = π‘^n, π β β
β{π‘^π}= β«β^β πβ»Λ£π‘ π‘^n ππ‘ = , for π > 0
π!
In this example, we calculate the Laplace Transform of a polynomial function, t^n, where n is a non-negative integer. The integral we evaluate involves multiplying the polynomial by the exponential decay term. Upon integration, this yields a result of n!/s^(n+1), demonstrating a pattern where higher powers lead to factorial growth in the numerator, which is important for understanding how different types of polynomial systems behave in the s-domain.
Think about measuring the effect of time on a project that involves multiple stages or tasks. If the time required for each stage is represented by different powers of time, the Laplace Transform helps in summarizing and transforming that complex structure into a straightforward equation (n!/s^(n+1)), enabling us to easily compute and predict the overall effect on the project's completion.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Laplace Transform: A tool that converts time-domain functions into the s-domain, facilitating simpler algebraic manipulation.
Existence Conditions: Functions must be piecewise continuous and of exponential order for the Laplace Transform to exist.
Transform of Constant Function: The Laplace Transform of f(t) = 1 is 1/s for s > 0.
Transform of Exponential Function: The Laplace Transform of f(t) = e^(at) is 1/(s - a) for s > a.
Transform of Polynomial Functions: The Laplace Transform of f(t) = t^n is n!/s^(n + 1) for s > 0.
See how the concepts apply in real-world scenarios to understand their practical implications.
β{1} = 1/s, for s > 0; demonstrates the transformation of a constant.
β{e^(at)} = 1/(s - a), for s > a; shows how the exponential function converts based on growth rate.
β{t^n} = n!/s^(n + 1), for s > 0; illustrates how polynomial functions are simplified in transformation.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
For every constant one you see, itβs one over s, easy as can be!
Imagine transforming a constant into an algebraic gem, itβs as simple as placing it within a fraction, one over s again!
In Laplace, when you see 'e', just remember 's must be greater than the growth spree'.
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Review the Definitions for terms.
Term: Laplace Transform
Definition:
An integral transform used to convert a function of time (t-domain) into a function of a complex variable (s-domain).
Term: Convergence
Definition:
The condition where an integral achieves a finite value; critical for the existence of the Laplace Transform.
Term: Exponential Order
Definition:
A condition for functions implying there exist constants M, a, and T such that |f(t)| β€ M e^(at) for all t > T.