6.3 - Proof of the Theorem
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Introduction to Laplace Transform of Integrals
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Today, we’re discussing the proof of a theorem that simplifies the analysis of differential equations through the Laplace Transform. Can someone remind me what the Laplace Transform does?
It transforms a function from the time domain into the s-domain.
Exactly! Now, when we integrate a function, we can actually simplify this process through a theorem we will prove today. What do you think this could help with?
It could help solve differential equations more easily!
Great thought! Solving integro-differential equations will be one of the key applications.
Understanding the Theorem
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The theorem states that if L{f(t)} = F(s), then the Laplace Transform of the integral from 0 to t of f(t) can be expressed as F(s)/s. Why is dividing by s important here?
It shows how integrating affects the Laplace Transform.
Exactly! It provides a direct correlation between operations in the time as compared to the s-domain.
Is this applicable in real-world problems, like electrical circuits?
Absolutely! This theorem greatly influences how we analyze systems with memory, like capacitors.
Proof of the Theorem
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Let's take a closer look at the proof. We denote g(t) as the integral of f from 0 to t. Can anyone identify the integral we're referring to?
It's g(t) = ∫f(τ)dτ from 0 to t!
Correct! Now, by applying the Laplace Transform, we work through changing the order of integration. Who remembers why we can do that?
That's Fubini's Theorem, right? It allows us to switch the order under certain conditions.
Very well! When we evaluate the integral through this method, we derive that the Laplace Transform can be simplified significantly, showcasing the utility of this theorem.
So this really does show that integrating makes it easier to manipulate functions!
Absolutely! Recapping our discussion, we’ve reinforced that the transformation relates integration in time to algebraic manipulation in the s-domain.
Application Insights
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Before we conclude, let's discuss how this theorem is applied in engineering. What practical issues can it help?
It can simplify the analysis of circuits or systems with memory like capacitors!
Exactly! It allows us to analyze charge accumulation effectively. This makes integral transforms a fascinating area to study.
Can we use this for signals processing too?
Yes! Integration plays a vital role there. Understanding these principles thoroughly will enhance your analytical skills.
Introduction & Overview
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Quick Overview
Standard
The proof of the theorem shows that the Laplace Transform of an integral can be computed by dividing the Laplace Transform of the original function by the variable s. This property aids in solving complex differential and integral equations, making it crucial for applications in engineering and mathematics.
Detailed
Proof of the Theorem
In this section, we explore the proof of a significant theorem in engineering mathematics, specifically concerning the Laplace Transform of integrals. The theorem states that if L{f(t)} = F(s), then the Laplace Transform of the integral of f(t) from 0 to t is given by:
$$L{\int_0^t f(\tau) d\tau} = \frac{F(s)}{s}$$
To understand this, we define the function g(t) as the integral of f(t) from 0 to t. We then apply the Laplace Transform to g(t) and utilize Fubini's Theorem to interchange the order of integration. By evaluating the inner integral, we arrive at the formula, confirming the relationship between integration in the time domain and manipulation in the Laplace domain. This theorem is foundational for resolving integro-differential equations and analyzing systems with memory, making it particularly relevant in engineering disciplines.
Audio Book
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Defining the Function g(t)
Chapter 1 of 5
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Chapter Content
Let’s denote:
t
g(t)=∫f (τ)dτ
0
Detailed Explanation
In this chunk, we introduce the function g(t), defined as the integral of another function f(t) from 0 to t. Here, τ is a dummy variable for integration. This means g(t) accumulates the values of f(τ) over the interval from 0 to t, essentially summing up the outputs of f(τ) as τ progresses from 0 to t.
Examples & Analogies
Think of g(t) as tracking your savings over time. If f(τ) represents the amount you save each day, then g(t) shows the total amount saved by day t. Just like accumulating savings, g(t) aggregates the contributions from f(τ).
Taking the Laplace Transform of g(t)
Chapter 2 of 5
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Chapter Content
Now, taking Laplace Transform of g(t),
∞ t
( )
L{g(t)}=∫e−st ∫f(τ)dτ dt
0 0
Detailed Explanation
In this step, we are applying the Laplace Transform to the function g(t). This involves calculating the double integral: first integrating f(τ) from 0 to t, then applying the e^{-st} factor, which is crucial for transforming the function into the Laplace domain. The outer integral sums the contributions from g(t) over time, while the inner integral sums f(τ) from 0 to t.
Examples & Analogies
Imagine you're calculating your total savings with respect to days and considering the effect of inflation (represented by the e^{-st} factor). The inner integral is like checking how much you've saved during each specific period, while the outer integral considers all those periods together to find a total adjusted for currency value over time.
Interchanging the Order of Integration
Chapter 3 of 5
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Chapter Content
Interchanging the order of integration (using Fubini's Theorem),
∞ ∞
( )
¿∫f(τ) ∫e−stdt dτ
0 τ
Detailed Explanation
Here we apply Fubini's Theorem, which allows us to change the order of integration. By doing this, we first integrate with respect to t, then with respect to τ. This mathematical tool is useful when dealing with double integrals and simplifies our calculations significantly, allowing us to evaluate the inner integral independently.
Examples & Analogies
Think of rearranging a group project where team members can work on their tasks in any order. By switching the tasks (integrations) around, each member can focus on their part more effectively, allowing the overall project to be streamlined. Similarly, changing the order of integrals often makes complex problems easier to manage.
Evaluating the Inner Integral
Chapter 4 of 5
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Chapter Content
Evaluating the inner integral:
∞ ∞
[−1 ] 1
∫e−stdt= e−st
=
e−sτ
s s
τ τ
Detailed Explanation
In this chunk, we compute the inner integral, which involves integrating e^{-st} with respect to t. This gives us an expression that simplifies further calculations. The result is [−1/s] e^{-st}, evaluated from 0 to ∞, which leads to e^{-sτ}/s. This step is crucial as it translates the time-based integral into a simpler form.
Examples & Analogies
Imagine calculating how much money you'd save over time if you received compound interest. By integrating your savings, we can find how much that account grows over various periods. We evaluate this growth (inner integral) at the limits of time, similar to how we assess our savings at key points.
Final Steps and Conclusion of the Proof
Chapter 5 of 5
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Chapter Content
So,
∞ 1 1 ∞ F(s)
L{g(t)}=∫f(τ)⋅ e−sτdτ= ∫f(τ)e−sτdτ=
=s s s
0 0
Hence proved.
Detailed Explanation
This chunk wraps up the proof by integrating the simplified expression from the previous step. The final result shows that the Laplace Transform of g(t) yields a new expression that is related to the Laplace Transform of f(t). Specifically, we find that the Laplace Transform of the integral of f(τ) corresponds to the Laplace Transform of f(τ) divided by s. This conclusion is key to understanding the relationship between integration in the time domain and manipulation in the Laplace domain.
Examples & Analogies
When putting together pieces of a puzzle, finding how they fit (proving a theorem) is just like integrating functions. By bringing all the parts (functions) together and seeing how they relate under certain transformations (like Laplace), we achieve a clearer picture or understanding. The proof demonstrates how integral calculations connect with Laplace Transforms, paving the way for more complex problems.
Key Concepts
-
Laplace Transform of an Integral: The theorem L{∫f(τ)dτ} = F(s)/s connects integration in time to algebraic manipulation in the s-domain.
-
Fubini's Theorem: Allows interchanging the order of integration, essential for deriving the theorem.
-
Application in Engineering: The theorem facilitates solving integro-differential equations crucial in engineering applications.
Examples & Applications
Example 1 illustrates finding the Laplace Transform of ∫sin(aτ)dτ, demonstrating how to use the theorem in practice.
Example 2 shows calculations for L{∫e^(2τ)dτ}, applying the theorem to find results.
Memory Aids
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Rhymes
When functions fold under integral's sway, In Laplace they play without delay!
Stories
Once upon a time, a troublesome integral couldn't reveal its secrets until the wizard Laplace showed how to transform it into the simpler realm of s-domain. With its newfound powers, the integral became a straightforward equation to solve!
Memory Tools
Remember ‘TEM’ for theorem: Transformation, Evaluate, Manipulate to help remember the steps of proving.
Acronyms
Use ‘LIFT’ for Laplace Integral Function Transformation to recall the main purpose of the theorem.
Flash Cards
Glossary
- Laplace Transform
A mathematical transformation that converts a time function into a complex frequency domain.
- Fubini's Theorem
A theorem that allows the interchange of the order of integration in double integrals under certain conditions.
- Integral
A mathematical concept representing the area under the curve of a function.
- Differential Equation
An equation involving derivatives of a function or functions.
- Piecewise Continuous Function
A function that is continuous on intervals, except for a finite number of jump discontinuities.
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