Laplace Transform Of E^(at)sin(bt) A Step-by-Step Guide

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Hey guys! Today, we're diving into the fascinating world of Laplace transforms, specifically focusing on finding the Laplace transform of the function f(t) = e^(at) sin(bt). This is a classic problem in mathematics and engineering, and understanding it can unlock a whole new level of problem-solving power. So, let's break it down step by step and make sure we've got a solid grasp on it.

Understanding Laplace Transforms

Before we jump into the specifics, let's quickly recap what Laplace transforms are all about. At its core, the Laplace transform is a mathematical tool that transforms a function of time, denoted as f(t), into a function of a complex variable s, denoted as F(s). Think of it as a way to switch from the time domain to the frequency domain, which can often simplify complex problems, especially those involving differential equations. The Laplace transform, makes the transformation easy and helps convert differential equations into algebraic equations, which are generally easier to solve. This transformation is defined by the following integral:

F(s) = ∫0^∞ e^(-st) f(t) dt

Where:

  • F(s) is the Laplace transform of f(t).
  • s is a complex variable (s = σ + jω, where σ and ω are real numbers, and j is the imaginary unit).
  • The integral is taken from 0 to infinity.

The beauty of the Laplace transform lies in its ability to turn differential equations into algebraic equations. This is a game-changer because algebraic equations are generally much easier to solve. Once you've solved the algebraic equation in the s-domain, you can use the inverse Laplace transform to get back to the solution in the time domain. This whole process is incredibly useful in various fields, including electrical engineering, control systems, and signal processing.

Why Use Laplace Transforms?

You might be wondering, why bother with Laplace transforms at all? Well, there are several compelling reasons:

  1. Solving Differential Equations: As mentioned earlier, Laplace transforms are fantastic for solving linear differential equations, especially those with constant coefficients. They convert the differential equation into an algebraic equation, which is much simpler to handle.
  2. Circuit Analysis: In electrical engineering, Laplace transforms are used extensively to analyze circuits, especially those with capacitors and inductors. They help determine the transient and steady-state responses of circuits.
  3. Control Systems: Control engineers use Laplace transforms to design and analyze control systems. The transfer function, a key concept in control systems, is defined in the s-domain using Laplace transforms.
  4. Signal Processing: Laplace transforms are also used in signal processing to analyze and manipulate signals. They can help in filtering, modulation, and demodulation of signals.
  5. Simplifying Complex Problems: In many situations, working in the s-domain can simplify the analysis of complex systems. The Laplace transform provides a different perspective that can reveal hidden structures and relationships.

In essence, the Laplace transform is a powerful tool that simplifies the analysis of dynamic systems. It provides a way to move from the complexities of the time domain to the more manageable frequency domain, making it an indispensable tool for engineers and mathematicians alike.

Finding the Laplace Transform of f(t) = e^(at) sin(bt)

Okay, now let's get to the heart of the matter: finding the Laplace transform of f(t) = e^(at) sin(bt). This particular function pops up quite often in various applications, so it's definitely worth knowing how to handle it. We'll use the definition of the Laplace transform and some clever integration techniques to arrive at the solution. So, buckle up, and let's get started!

The Definition and the Integral

As we discussed earlier, the Laplace transform of a function f(t) is defined as:

F(s) = ∫0^∞ e^(-st) f(t) dt

In our case, f(t) = e^(at) sin(bt). So, we need to evaluate the following integral:

F(s) = ∫0^∞ e^(-st) e^(at) sin(bt) dt

This integral might look a bit intimidating at first glance, but don't worry, we'll tackle it methodically. The first thing we can do is combine the exponential terms:

F(s) = ∫0^∞ e^((-s+a)t) sin(bt) dt

Now, we have an integral of the form ∫ e^(kt) sin(bt) dt, where k = -s + a. This type of integral can be solved using integration by parts. Integration by parts is a technique that allows us to integrate the product of two functions. The formula for integration by parts is:

∫ u dv = uv - ∫ v du

Applying Integration by Parts

To solve our integral, we'll apply integration by parts twice. Let's first identify our u and dv:

Let u = sin(bt), then du = b cos(bt) dt Let dv = e^((-s+a)t) dt, then v = -1/(s-a) e^((-s+a)t)

Applying the integration by parts formula, we get:

∫0^∞ e^((-s+a)t) sin(bt) dt = [-1/(s-a) e^((-s+a)t) sin(bt)]0^∞ - ∫0^∞ [-1/(s-a) e^((-s+a)t)] b cos(bt) dt

Let's simplify this a bit:

= [-1/(s-a) e^((-s+a)t) sin(bt)]0^∞ + b/(s-a) ∫0^∞ e^((-s+a)t) cos(bt) dt

Now, we need to evaluate the first term and solve the remaining integral. As t approaches infinity, the term e^((-s+a)t) approaches zero if the real part of (s - a) is positive, which is a crucial condition for the Laplace transform to converge. So, the first term becomes zero at the upper limit. At the lower limit (t = 0), sin(0) = 0, so the first term is zero overall. Now we are left with:

= b/(s-a) ∫0^∞ e^((-s+a)t) cos(bt) dt

We need to apply integration by parts again to solve this integral. Let's choose new u and dv:

Let u = cos(bt), then du = -b sin(bt) dt Let dv = e^((-s+a)t) dt, then v = -1/(s-a) e^((-s+a)t)

Applying integration by parts again, we have:

∫0^∞ e^((-s+a)t) cos(bt) dt = [-1/(s-a) e^((-s+a)t) cos(bt)]0^∞ - ∫0^∞ [-1/(s-a) e^((-s+a)t)] (-b sin(bt)) dt

Again, simplifying:

= [-1/(s-a) e^((-s+a)t) cos(bt)]0^∞ - b/(s-a) ∫0^∞ e^((-s+a)t) sin(bt) dt

The first term here, when evaluated from 0 to infinity, gives us:

= [0 - (-1/(s-a) * 1)] = 1/(s-a)

So, now we have:

∫0^∞ e^((-s+a)t) cos(bt) dt = 1/(s-a) - b/(s-a) ∫0^∞ e^((-s+a)t) sin(bt) dt

Putting It All Together

Let's plug this back into our original equation:

F(s) = b/(s-a) [1/(s-a) - b/(s-a) ∫0^∞ e^((-s+a)t) sin(bt) dt]

Notice that the integral on the right side is our original integral, F(s)! So, we can rewrite the equation as:

F(s) = b/(s-a)^2 - b2/(s-a)2 F(s)

Now, we can solve for F(s):

F(s) + b2/(s-a)2 F(s) = b/(s-a)^2

F(s) [1 + b2/(s-a)2] = b/(s-a)^2

F(s) [(s-a)^2 + b2]/(s-a)2 = b/(s-a)^2

Finally, we get:

F(s) = b / ((s - a)^2 + b^2)

And there you have it! The Laplace transform of f(t) = e^(at) sin(bt) is b / ((s - a)^2 + b^2). This is a crucial result to remember, and the process we went through demonstrates the power and elegance of Laplace transforms and integration by parts.

Conclusion: Mastering the Laplace Transform

So, guys, we've journeyed through the world of Laplace transforms and successfully found the Laplace transform of f(t) = e^(at) sin(bt). We started with the definition, tackled a tricky integral using integration by parts, and arrived at the solution: b / ((s - a)^2 + b^2). This exercise not only gives us a specific transform to add to our toolbox but also reinforces the fundamental concepts and techniques involved in Laplace transforms.

Key Takeaways

Let's recap the key takeaways from our adventure:

  • Laplace transforms are a powerful tool for converting functions from the time domain to the frequency domain.
  • They are particularly useful for solving linear differential equations and analyzing dynamic systems.
  • The Laplace transform of f(t) = e^(at) sin(bt) is b / ((s - a)^2 + b^2).
  • Integration by parts is a crucial technique for evaluating integrals that arise in Laplace transform calculations.
  • Understanding the conditions for convergence of the Laplace transform (like the real part of (s - a) being positive) is essential.

Further Exploration

If you're keen to delve deeper into Laplace transforms, there's a whole universe of fascinating topics to explore. You could investigate:

  • The Laplace transforms of other common functions, such as e^(at) cos(bt), t^n, and step functions.
  • The inverse Laplace transform and how to use it to get back to the time domain.
  • Applications of Laplace transforms in solving differential equations, circuit analysis, control systems, and signal processing.
  • The relationship between Laplace transforms and other transform techniques, such as Fourier transforms.

Practice Makes Perfect

The best way to truly master Laplace transforms is to practice, practice, practice! Work through examples, solve problems, and don't be afraid to make mistakes. Each mistake is a learning opportunity, and with enough effort, you'll become a Laplace transform pro in no time.

So, keep exploring, keep learning, and keep transforming! The world of mathematics and engineering is full of exciting challenges, and Laplace transforms are just one of the many tools that can help you conquer them. Keep up the great work, guys, and I'll catch you in the next mathematical adventure! This detailed explanation should help anyone understand and apply the Laplace transform effectively.