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Understanding Convolution
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Today, we're discussing the Convolution theorem, which is crucial in finding inverse Laplace transforms.
What exactly do we mean by convolution in this context?
Great question! Convolution combines two functions into a new function by integrating the product of one function and a time-reversed version of the other.
Can you give us a formula for convolution?
Absolutely! The convolution of functions \( f(t) \) and \( g(t) \) is defined as \( (f*g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau \).
What kind of problems can we solve using this theorem?
Convolution is particularly useful when handling products of Laplace transforms in differential equations and circuit analysis.
Now, letβs have a look at the first solved example to see how we can apply these concepts.
First Example Walkthrough
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For our first example, we want to find the inverse Laplace transform of \( \frac{1}{s(s+1)} \).
How do we start this problem?
Weβll use known transforms. We know that \( \mathcal{L}^{-1}\{\frac{1}{s}\} = 1 \) and \( \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t} \).
So we can use convolution here?
Exactly! We express \( (f * g)(t) \) as \( \int_0^t 1 \cdot e^{-(t - \tau)} d\tau \).
What do we get after integrating?
We find that \( (f*g)(t) = 1 - e^{-t} \). That's our final result.
To summarize, we utilized the convolution theorem to find the inverse Laplace transform, simplifying the initial product.
Second Example Exploration
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Now let's tackle our second example, where we find the inverse Laplace transform of \( \frac{1}{s^2(s+2)} \).
I noticed that itβs a bit more complicated compared to the first one.
You're correct! We'll break it down. Let \( F(s) = \frac{1}{s^2} \) and \( G(s) = \frac{1}{s+2} \).
What are the known inverses?
We have \( \mathcal{L}^{-1}\{\frac{1}{s^2}\} = t \) and \( \mathcal{L}^{-1}\{\frac{1}{s+2}\} = e^{-2t} \).
Does this mean we'll use convolution again?
Exactly! Weβll set up the integral \( \int_0^t \tau e^{-2(t - \tau)} d\tau \).
So how do we evaluate this integral?
We can apply integration by parts. This will ultimately give us the final expression.
To recap, we demonstrated how to apply the convolution theorem for more complex cases.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
The solved examples demonstrate practical applications of the Convolution Theorem. They provide insight into finding the inverse Laplace transform of various functions using convolution, simplifying complex operations in the field of differential equations and system analysis.
Detailed
Solved Examples
This section focuses on two solved examples that exemplify the application of the Convolution Theorem in calculating inverse Laplace Transforms.
Example 1
Problem Statement
Find the inverse Laplace transform of \( \frac{1}{s(s+1)} \).
Solution Breakdown
Starting with known parts:
- \( \mathcal{L}^{-1}\{\frac{1}{s}\} = 1 \)
- \( \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t} \)
Using convolution:
1. The convolution is given by:
\[(f * g)(t) = \int_0^t f(\tau)g(t - \tau) d\tau = \int_0^t 1 \cdot e^{-(t - \tau)} d\tau\]
2. Simplifying gives:
\[(f * g)(t) = e^{-t} \int_0^t e^{\tau} d\tau = e^{-t}(e^{t} - 1) = 1 - e^{-t}\]
Example 2
Problem Statement
Find the inverse Laplace transform of \( \frac{1}{s^2(s+2)} \).
Solution Breakdown
Let \( F(s) = \frac{1}{s^2} \), and \( G(s) = \frac{1}{s+2} \).
Using known inverses:
- \( \mathcal{L}^{-1}\{\frac{1}{s^2}\} = t \)
- \( \mathcal{L}^{-1}\{\frac{1}{s+2}\} = e^{-2t} \)
The convolution gets computed as follows:
1. \[(f * g)(t) = \int_0^t \tau e^{-2(t - \tau)} d\tau = e^{-2t} \int_0^t \tau e^{2\tau} d\tau\]
2. Applying integration by parts yields the final expression:
\[(f * g)(t) = (1 - e^{-2t}) + \frac{t}{4} e^{-2t}\]
Audio Book
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Example 1: Finding Inverse Laplace Transform
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Example 1:
Find ββ1{ \frac{1}{s(s+1)} }
Solution: We know,
β’ ββ1{ \frac{1}{s} } = 1
β’ ββ1{ \frac{1}{s+1} } = e^{-t}
So,
β^{-1}{ \frac{1}{s(s+1)} } = 1*e^{-t} = \int_0^t 1 \cdot e^{-(t-\tau)} d\tau
=t\int_0^t e^{-(t-\tau)} d\tau = e^{-t}\int_0^t e^{\tau} d\tau = e^{-t}(e^t - 1) = 1 - e^{-t}
Detailed Explanation
In this example, we want to find the inverse Laplace transform of the function 1/(s(s+1)). The solution begins by recognizing that we can break apart the inverse transform into simpler components: the first part corresponds to 1/s, which has an inverse transform of 1, and the second part corresponds to 1/(s+1), which has an inverse transform of exp(-t). We then utilize the property of convolution, which tells us that the inverse Laplace transform of the product of two transforms is the convolution of their respective time-domain functions. By computing the integral for the convolution, we arrive at the final result, simplifying the expression to 1 - exp(-t).
Examples & Analogies
Think of finding the inverse Laplace transform like breaking down a complex recipe into simpler parts. Instead of trying to prepare a complicated dish in one go, you might first gather all the ingredients (the parts of the Laplace transform), and then follow step-by-step instructions for each component before combining them at the end to create the final meal (the final function). This process of breaking down the recipe helps ensure that each part is correctly executed.
Example 2: Another Inverse Laplace Transform
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Example 2:
Find the inverse Laplace transform of \frac{1}{s^2(s+2)}
Solution: Let πΉ(s) = \frac{1}{s^2}, πΊ(s) = \frac{1}{s+2}.
We know:
β’ β^{-1}{ \frac{1}{s^2} } = t
β’ β^{-1}{ \frac{1}{s+2} } = e^{-2t}
So by convolution:
f* g = \int_0^t \tau \cdot e^{-2(t - \tau)} d\tau = e^{-2t}\int_0^t \tau \cdot e^{2\tau} d\tau.
Use integration by parts:
Let
β’ u = \tau β du = d\tau
β’ dv = e^{2\tau} d\tau β v = \frac{1}{2} e^{2\tau}.
Then, we have:
e^{-2t}[\tau \cdot \frac{1}{2} e^{2\tau}|_0^t - \int_0^t \frac{1}{2} e^{2\tau} d\tau] = e^{-2t}[\frac{1}{2}(te^{2t} - 0) - \frac{1}{2}(\frac{1}{2} e^{2t} - 1)]
= (1 - e^{-2t}) + \frac{1}{4}
Detailed Explanation
In the second example, we find the inverse Laplace transform of the function 1/(s^2(s+2)). We first define F(s) and G(s), which represent two simpler functions. Each of these functions has a known inverse Laplace transform (t and e^(-2t), respectively). Using the convolution theorem, we can compute the convolution of these two functions to get the final inverse transform. The integral involves integration by parts, allowing us to manage the computation effectively, resulting in a clear form that combines exponential decay with a linear term.
Examples & Analogies
Imagine trying to calculate the total distance traveled on a road trip that consists two different segments: one part of the trip has a steady speed (like the function \frac{1}{s^2}), and the other part involves driving a little slower due to traffic (like the function \frac{1}{s+2}). Just as you would calculate the distance for each segment separately before adding them up to arrive at the total distance traveled, here we similarly handle each part and combine effects through the convolution process, allowing us to achieve an understanding of the total outcome.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Convolution Theorem: The process of taking an inverse Laplace transform of the product of two Laplace transforms corresponds to the convolution of their time-domain functions.
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Properties of Convolution: Important properties like commutativity, associativity, and distributivity make convolution a flexible tool in solving various application problems.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example 1: The inverse Laplace transform of \( \frac{1}{s(s+1)} \) results in \( 1 - e^{-t} \).
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Example 2: For \( \frac{1}{s^2(s+2)} \), we find the inverse as \( (1 - e^{-2t}) + \frac{t}{4} e^{-2t} \).
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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To remember convolution's charm / Combine your functions, stay calm. / Integrate with g in reverse, / Time-shifted results, learn this verse.
π Fascinating Stories
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Imagine two rivers flowing towards a lake, each contributing water at different times; when mixed, they create a combined flow reflecting their individual strengths. This is like convolution in function analysis.
π§ Other Memory Gems
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Use 'C.I.D' for Convolution: Commutative, Integrative, Distributive.
π― Super Acronyms
Remember C.A.D. for Convolutionβs properties
- Commutative
- Associative
- Distributive.
Flash Cards
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Glossary of Terms
Review the Definitions for terms.
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Term: Convolution
Definition:
An operation that combines two functions to generate a third function by integrating the product of one function and a time-reversed version of another.
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Term: Inverse Laplace Transform
Definition:
The process of finding a time-domain function from its Laplace transform.
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Term: Laplace Transform
Definition:
An integral transform that converts a time-domain function into a complex frequency-domain function.