1.6 - Proof of the First Shifting Theorem
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Introduction to Laplace Transform and the First Shifting Theorem
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Today we are discussing the First Shifting Theorem, which is crucial for handling exponential terms in the Laplace Transform. Can anyone explain why we might need this theorem?
Is it because in engineering we often deal with systems that have exponential growth or decay?
Exactly! By applying the theorem, we can shift the Laplace Transform to simplify our calculations. The basic idea is that multiplying by e^(at) shifts the variable s to s-a.
Can you show us the general formula for the theorem?
Certainly! If ℒ{f(t)} = F(s), then ℒ{e^(at)f(t)} = F(s-a). Remember, this means that whenever we see something multiplied by an exponential, we shift the transform.
Proof of the Theorem
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Let’s look at the proof. The Laplace Transform definition is an integral representation. Who can recall that?
It’s the integral of e^(-st)f(t) from 0 to infinity!
Right! Now, substituting our function, we get ℒ{e^(at)f(t)} = ∫_0^∞ e^(-st)e^(at)f(t) dt. Can someone simplify that?
We can combine the exponents to see e^(-(s-a)t).
Exactly! This integral is now the Laplace Transform of f(t) evaluated at (s-a), confirming our theorem.
So that’s why it’s called the First Shifting Theorem!
Applications of the Theorem
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Let’s explore where we might apply this theorem. Can anyone provide a scenario where this theorem is useful?
I think it would help with solving ODEs that have exponential forcing functions.
Correct! It’s also useful in control system design. Any other applications?
How about in electrical engineering with signals that contain exponential terms?
Absolutely! Understanding how to manipulate these transforms allows us to analyze systems effectively.
Examples of Using the First Shifting Theorem
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Let’s apply what we've learned. I’ll start with the function f(t) = sin(bt). What would its Laplace Transform be?
It’s F(s) = b / (s^2 + b^2)!
Great! Now, what about ℒ{e^(at) sin(bt)} using the First Shifting Theorem?
So we replace s with (s-a), right? It becomes (s-a)/(s-a)^2 + b^2.
Perfect! That shows the practical use of our theorem. Always remember to apply these shifts correctly!
Common Mistakes and Recap
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Before we wrap up, what are some mistakes students might make when applying this theorem?
Confusing the signs while shifting!
Exactly! It’s important to remember that for e^(at), we shift left. Now, who can summarize the key points of today's lesson?
Multiplying by e^(at) shifts the transform; the proof relies on understanding integration, and it helps solve different ODEs!
Great summary! Keep practicing these concepts in your exercises!
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
This section explores the First Shifting Theorem, which states that multiplying a function by an exponential term results in a horizontal shift in the Laplace domain. The theorem is proven mathematically, and its applications are highlighted, including solving ordinary differential equations and modeling systems in engineering.
Detailed
Proof of the First Shifting Theorem
The First Shifting Theorem is a fundamental property of the Laplace Transform that simplifies solving linear differential equations in engineering and applied mathematics. This theorem states that if the Laplace Transform of a function f(t) is given by ℒ{f(t)} = F(s), then multiplying f(t) by an exponential term e^(at) results in a shift in the Laplace domain: ℒ{e^(at)f(t)} = F(s - a). This means that the exponential multiplication in the time domain translates to a shift of the complex frequency variable s in the Laplace region.
Proof of the Theorem
The proof starts with the definition of the Laplace Transform as follows:
ℒ{e^(at)f(t)} = ∫_0^∞ e^(-st)e^(at)f(t) dt
Simplifying the exponent gives:
ℒ{e^(at)f(t)} = ∫_0^∞ e^{-(s-a)t}f(t) dt.
This integral is recognized as the Laplace Transform of f(t) evaluated at (s-a), hence proving that:
ℒ{e^(at)f(t)} = F(s-a).
Applications
The First Shifting Theorem is crucial in various engineering fields, such as:
- Solving ordinary differential equations (ODEs) involving exponential forcing functions.
- Designing and analyzing control systems, especially where damping or growth is modeled.
- Handling electrical engineering problems with exponential inputs, particularly in circuit analysis.
- Analyzing mechanical systems under exponentially varying forces.
Understanding this theorem not only aids in solving mathematical problems efficiently but also streamlines the process of modeling complex real-world systems.
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Definition of the Laplace Transform
Chapter 1 of 4
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Chapter Content
Let’s begin with the definition of the Laplace Transform:
$$
ℒ{e^{at}f(t)} = \int_0^{\infty} e^{-st} e^{at}f(t) dt
$$
Detailed Explanation
The Laplace Transform is a mathematical operation that transforms a time-domain function into a function in the s-domain (Laplace domain). Here, we are starting by expressing the Laplace Transform of a function multiplied by an exponential term. The notation ℒ represents the Laplace Transform, while the variables t and s denote time and the complex frequency, respectively. The integral definition shows how we calculate the transform by integrating over the time from 0 to infinity, where e^{-st} is the exponential decay component and e^{at} is the multiplicative factor related to growth rates.
Examples & Analogies
Consider a car traveling at a speed that changes over time. The Laplace Transform helps us understand how the distance changes as a function of time (the time-domain function), while the exponential ‘growth’ factor can be thought of as an increasing speed due to acceleration. Just like we want to calculate the total distance traveled over time, the Laplace Transform helps us transition from the time view to a frequency view.
Simplifying the Exponent
Chapter 2 of 4
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Chapter Content
Simplify the exponent:
$$
= \int_0^{\infty} e^{-(s-a)t}f(t) dt
$$
Detailed Explanation
In this step, we simplify the expression by combining the exponents in the integrand. The original expression has two exponential terms: one is e^{-st} and the other is e^{at}. When combined, we get a single exponential term e^{-(s-a)t}. This represents the effect of shifting the variable s by a, allowing us to see how the function f(t) behaves when weighted by the exponential growth or decay.
Examples & Analogies
Imagine you're adjusting the speed of a moving object while keeping track of the time. By combining the adjustments into one measure (like combining acceleration into one effective speed), we make it easier to handle and analyze, leading us into clearer insights about its behavior over time.
Final Result of the Laplace Transform
Chapter 3 of 4
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Chapter Content
= ℒ{f(t)} evaluated at (s−a)
$$
= F(s-a)
$$
Detailed Explanation
This part highlights that the integral we evaluated can be viewed as the Laplace Transform of the original function f(t), now evaluated at the corrected variable (s − a). Here F(s) represents the Laplace Transform of f(t). Therefore, we can conclude that the Laplace Transform of the function multiplied by an exponential factor leads to a shift in the s-domain.
Examples & Analogies
Think of it like changing the perspective of viewing the performance of an athlete. If we shift our point of view (the variable s), we are adjusting to see how extra factors like fatigue (the exponential factor) modify their performance metrics, leading to more accurate assessments.
Conclusion of the Proof
Chapter 4 of 4
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Chapter Content
Hence,
$$
ℒ{e^{at}f(t)}= F(s−a)
$$
This completes the proof.
Detailed Explanation
With this statement, we wrap up the proof of the First Shifting Theorem. It confirms that applying the theorem transforms a time-domain function multiplied by an exponential factor into a shifted version in the s-domain. The proof is important in demonstrating how complex systems can be analyzed simply by understanding this relationship.
Examples & Analogies
Think of a music playlist that has different tempos. By shifting the speed of the music (like shifting s), you can find new rhythms that match the original song’s vibe, allowing for an entirely different listening experience without changing the song itself.
Key Concepts
-
Laplace Transform: A technique for transforming differential equations into algebraic equations through variable substitution.
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Shift Theorem: A principle that allows the manipulation of Laplace Transforms by multiplying functions by exponentials, shifting the variable in the process.
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Proof Structure: The theorem's proof revolves around integration and substitution, yielding a significant simplification.
Examples & Applications
Using the First Shifting Theorem, the Laplace Transform of f(t) = sin(bt) results in ℒ{e^(at)sin(bt)} = b / ((s-a)^2 + b^2).
Given f(t) = t, its Laplace Transform is ℒ{e^(2t) t} = 1 / ((s-2)^2).
To find the Laplace Transform of e^(-3t)cos(4t), we leverage the theorem to yield ℒ{e^(-3t)cos(4t)} = 1 / ((s + 3)^2 + 16).
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
When an exponential's near, shift 's' without fear!
Stories
Imagine a train ('s') pulling into a station ('a') as time stretches out; every time the exponential 'e^at' shows up, the train shifts left into a new reverse track!
Memory Tools
A-shift when e-a-lens bright; keep F(s-a) in your sight!
Acronyms
SAME
Shift
Apply
Multiply
Evaluate!
Flash Cards
Glossary
- Laplace Transform
A mathematical transformation that converts a function of time f(t) into a function of a complex variable s.
- First Shifting Theorem
States that multiplying a time-domain function by e^(at) results in a Laplace transform that shifts the variable from s to (s-a).
- Exponential Function
A mathematical function of the form e^(at), where e is the base of natural logarithms and a is a constant.
- ODE (Ordinary Differential Equation)
A differential equation containing one or more functions of one independent variable and its derivatives.
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