1.11 - Additional Exercise (Practice)
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Introduction to the First Shifting Theorem
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Today, we’re diving into the First Shifting Theorem for Laplace Transforms! Can anyone tell me what the theorem states?
Isn’t it that if we have ℒ{f(t)} = F(s), then ℒ{e^(at)f(t)} = F(s-a)?
Exactly! That horizontal shift in the Laplace domain is crucial when dealing with functions that include exponential factors. We can remember it using the acronym 'SHFT'—Shift Helps Find Transform!
How does this apply in real-world problems, especially in engineering?
Great question! This theorem helps in solving differential equations in control systems, electrical circuits, and mechanical vibrations. It makes handling exponential terms much simpler.
Can you give an example?
Sure! If we take f(t) = sin(bt), then by the theorem, ℒ{e^(at)sin(bt)} translates to F(s-a).
So we have to ensure that s > a for it to converge, right?
Yes, precisely! Understanding these conditions is vital. Let’s move on to some exercises to apply what we’ve learned.
Application Exercises
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Let's tackle the exercises from the section. The first one is to find the Laplace Transform of e^(4t)cos(3t). Any thoughts?
I think we should start with ℒ{cos(3t)} first, right? It’s s/(s^2 + 9).
Correct! Now, applying the theorem, what do we get?
Right, so it becomes ℒ{e^(4t)cos(3t)} = F(s-4), which means we replace s with s-4 in the original Laplace transform.
Excellent! You did that quickly and correctly. Now, let's try e^(-2t)t^2. Who wants to take a shot?
The Laplace transform of t^2 is 2/s^3! So replacing s with s+2 would give us 2/(s+2)^3.
Fantastic! Just a tiny detail—remember you are shifting to the left, what’s important? What inequality do we need to check?
We must ensure s+2 is greater than zero for convergence!
Exactly right! Convergence conditions are crucial. Let’s finish off with the last exercise!
Introduction & Overview
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Quick Overview
Standard
Users are encouraged to practice finding Laplace Transforms using the First Shifting Theorem. This section provides specific exercises to reinforce understanding of how to apply the theorem in given cases.
Detailed
The First Shifting Theorem is a crucial tool in engineering and applied mathematics, allowing us to find Laplace Transforms of functions multiplied by exponential factors. This section emphasizes the importance of practicing the theorem through exercises where learners are required to find the Laplace Transforms of various functions, showcasing the theorem's applications in problem-solving scenarios. Each exercise requires careful application of the theorem, ensuring that learners grasp both the conceptual and computational aspects of using this theorem effectively.
Audio Book
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Exercise 1: Laplace Transform of Exponential and Cosine
Chapter 1 of 3
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Chapter Content
Find the Laplace Transform of the following using the First Shifting Theorem:
1. 𝑒4𝑡 ⋅cos(3𝑡)
Detailed Explanation
To find the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{4𝑡} imes ext{cos}(3𝑡), we can apply the First Shifting Theorem. First, we note that the Laplace Transform of cos(3t) is known:
$$
ℒ{cos(3t)} = \frac{s}{s^2 + 9}
$$
Since we have an exponential term e^(4t), we can let 'a' be 4. According to the theorem, we will shift 's' to 's - 4'. Therefore:
$$
ℒ{ℓe^{4t} \cdot cos(3t)} = ℒ{cos(3t)} |{s=s-4} = ℒ{cos(3t)}|{s-4} = \frac{s-4}{(s-4)^2 + 9}
$$
Examples & Analogies
Think of it like tuning a radio. When you're trying to catch a specific station (the frequency of cos(3t)), sometimes you need to adjust your position slightly to account for interference, like the 4 from e^(4t). This adjustment represents shifting your frequency to get a clearer signal.
Exercise 2: Laplace Transform of Exponential and Quadratic
Chapter 2 of 3
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Chapter Content
- 𝑒−2𝑡 ⋅𝑡^2
Detailed Explanation
To solve for the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{-2𝑡} imes t^2, we first find the Laplace Transform of t². The Laplace Transform of t^n is:
$$
ℒ{t^n} = \frac{n!}{s^{n+1}} ext{ for } s > 0
$$
So, for n = 2:
$$
ℒ{t^2} = \frac{2!}{s^3} = \frac{2}{s^3}
$$
Next, applying the First Shifting Theorem with a = -2, we shift 's' to 's + 2':
$$
ℒ{ℓe^{-2t}t^2} = \frac{2}{(s + 2)^3}
$$
Examples & Analogies
Imagine you're pushing a toy car that slows down over time due to friction. The quadratic term represents the distance it covers, and the e^{-2t} is like the slowing force. We adjust our expectation of distance based on how quickly it slows down, which mirrors how we adjust 's' in the transform.
Exercise 3: Laplace Transform of Exponential and Sine
Chapter 3 of 3
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Chapter Content
- 𝑒5𝑡 ⋅sin(2𝑡)
Detailed Explanation
To find the Laplace Transform of the function 𝑓(𝑡) = 𝑒^{5𝑡} imes ext{sin}(2𝑡), we start by finding the Laplace Transform of sin(2t):
$$
ℒ{sin(2t)} = \frac{2}{s^2 + 4}
$$
Now, since we have an exponential e^ ext{5t}, and setting a = 5, we apply the First Shifting Theorem, which leads to shifting 's' to 's - 5':
$$
ℒ{e^{5t} \cdot sin(2t)} = \frac{2}{(s-5)^2 + 4}
$$
Examples & Analogies
Think of this process like adjusting the brightness on your smartphone screen. The sine function (sin(2t)) behaves predictively like the brightness levels, but when someone introduces a constant increase (e^{5t}), it's like someone adjusting the overall brightness. The transformation reflects this change in the 's' domain.
Key Concepts
-
First Shifting Theorem: A method used to find Laplace Transforms of functions multiplied by exponential terms.
-
Exponential Shift: The way the shifting property alters the parameter in the Laplace domain.
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Conditions for Application: Ensuring s > a for proper convergence when applying the theorem.
Examples & Applications
Example 1: ℒ{e^(4t)cos(3t)} = F(s-4) with F(s) = s/(s^2 + 9) results in s/(s^2 + 9) | s -> s-4.
Example 2: ℒ{e^(-2t)t^2} = 2/(s+2)^3 derived from ℒ{t^2} = 2/s^3.
Memory Aids
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Rhymes
If e is near, change the s with cheer, or else convergence will disappear!
Stories
Imagine a ship named 'Laplace' sailing from the point 's'. When it hits 'e^(2t)', it takes a joyful leap left, landing at 's-2' where the seas are calm and equations are solved!
Memory Tools
Remember: S.H.F.T for Shift Helps Find Transform when using the theorem!
Acronyms
S.H.F.T
= shift
= helps
= find
= transform.
Flash Cards
Glossary
- Laplace Transform
An integral transform that converts a function of time into a function of a complex variable, simplifying the analysis of linear systems.
- First Shifting Theorem
A theorem stating that ℒ{e^(at)f(t)} = F(s - a), which shows how multiplying by an exponential function shifts the Laplace transform.
- Exponential Function
A mathematical function of the form e^(at) where 'a' is a constant, often appearing in differential equations.
Reference links
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