Welcome to our YouTube channel! In this video, we dive into the fascinating world of engineering mathematics by exploring the Fourier Series of the Sawtooth Wave, explained in Hindi.
The Fourier Series is a fundamental concept in engineering math that allows us to represent complex periodic functions using a combination of sine and cosine functions. The Sawtooth Wave, characterized by its gradual rise and sudden drop, serves as an excellent example to demonstrate the principles of the Fourier Series.
Through this video, we aim to provide a comprehensive understanding of the Fourier Series of the Sawtooth Wave, particularly tailored for BTech students in Hindi. Our experienced instructor breaks down the mathematical concepts and derivations, making it accessible and easy to comprehend.
Topics covered in this video include:

Paragraph 1: Welcome to our YouTube channel! In this in-depth video, we explore the fascinating topic of the Fourier Series applied to the Sawtooth Wave in the context of Engineering Math for BTech students. The Fourier Series is a powerful mathematical tool that allows us to express complex periodic functions as a combination of simpler sine and cosine functions. In this tutorial, we delve into the principles and applications of the Fourier Series, with a focus on the Sawtooth Wave, explained in Hindi for better comprehension.

Paragraph 2: The Fourier Series plays a crucial role in engineering mathematics and signal analysis. By understanding its application to the Sawtooth Wave, BTech students can gain a solid foundation in periodic functions and their representation. Our experienced instructor breaks down the concepts step by step, providing clear explanations and demonstrations to ensure a thorough understanding of this important topic.

Paragraph 3: In this comprehensive video, we begin by introducing the Fourier Series and its significance in engineering mathematics. We explore how it enables us to analyze and manipulate periodic functions efficiently. Next, we define the Sawtooth Wave, known for its gradual rise and sudden drop pattern. We discuss its properties, mathematical representation, and practical applications in engineering.

Paragraph 4: Moving forward, we delve into the derivation of the Fourier Series formula for the Sawtooth Wave. Our instructor guides you through the mathematical steps, explaining each component in detail. We explore how the harmonics of the Sawtooth Wave contribute to its representation in the Fourier Series.

Paragraph 5: Graphical representations and visualizations are crucial for understanding complex mathematical concepts. In this video, we provide interactive visualizations to help you grasp the relationship between the Sawtooth Wave and its Fourier Series representation. These visuals enhance your learning experience and make it easier to comprehend the underlying principles.

Paragraph 6: Throughout the video, we emphasize the engineering applications of the Fourier Series and the Sawtooth Wave. We discuss how engineers and scientists utilize this mathematical tool in fields such as signal processing, telecommunications, audio engineering, and image processing. By understanding the Fourier Series and its application to the Sawtooth Wave, you’ll gain insights into practical engineering problems and solutions.

Paragraph 7: Additionally, we explore the mathematical properties of the Fourier Series, such as linearity, periodicity, and symmetry. These properties provide a deeper understanding of how the Fourier Series operates and can be leveraged in engineering applications.

Paragraph 8: Prerequisites for this tutorial include a basic understanding of calculus, trigonometry, and periodic functions. However, even if you are unfamiliar with these concepts, our instructor ensures that the explanations are accessible and easy to follow.

Paragraph 9: This video is specifically tailored to BTech students studying Engineering Math. We understand the challenges you face in comprehending complex mathematical topics, and our goal is to simplify the learning process. By explaining the Fourier Series of the Sawtooth Wave in Hindi, we eliminate language barriers and make the content more accessible to a wider audience.

Paragraph 10: In addition to the theoretical concepts, we provide practical examples and problem-solving techniques related to the Fourier Series and the Sawtooth Wave. These examples help reinforce your understanding and enable you to apply the concepts to solve engineering math problems effectively.

Paragraph 11: We encourage active participation and engagement throughout the video. Pause when necessary, attempt practice problems, and interact with the provided visuals. Learning is most effective when you actively involve yourself in the process.

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When solving f(x) = x over the interval 0 to 2π, we are essentially finding the values of x that satisfy the equation within that interval.

Since the function f(x) = x is a linear function, it passes through every point on a straight line with a slope of 1. In this case, the interval from 0 to 2π represents a full cycle of the unit circle.

Thus, the solution to the equation f(x) = x over the interval 0 to 2π is simply the set of all x values within that interval, which is [0, 2π]. In other words, x

Step 1: Determine the period. In this case, the function f(x) = x has a period of 2π. This means that the function repeats itself every 2π units.

Step 2: Calculate the Fourier coefficients. The Fourier coefficients for the function f(x) can be calculated using the formulas:

an = (1/π) * ∫[0 to 2π] f(x) * cos(n * x) dx
bn = (1/π) * ∫[0 to 2π] f(x) * sin(n * x) dx

Let’s calculate these coefficients step by step.

Step 3: Calculate the coefficient an. We need to evaluate the integral ∫[0 to 2π] f(x) * cos(n * x) dx. Since f(x) = x, the integral becomes:

an = (1/π) * ∫[0 to 2π] x * cos(n * x) dx

To solve this integral, we’ll use integration by parts. Let’s set u = x and dv = cos(n * x) dx.

Differentiating u gives du = dx, and integrating dv gives v = (1/n) * sin(n * x).

Applying integration by parts:

an = (1/π) * [u * v – ∫[0 to 2π] v * du]
= (1/π) * [x * (1/n) * sin(n * x) – (1/n) * ∫[0 to 2π] sin(n * x) dx]

Integrating sin(n * x) gives (-1/n) * cos(n * x), so the integral becomes:

an = (1/π) * [x * (1/n) * sin(n * x) + (1/n^2) * cos(n * x)] evaluated from 0 to 2π
= (1/π) * [(2π/n) * sin(2π * n) + (1/n^2) * cos(2π * n) – (0 * sin(0) + (1/n^2) * cos(0))]
= (1/π) * [(2π/n) * sin(2π * n) + (1/n^2)]

Simplifying further, we get:

an = (2/n) * sin(2π * n) + (1/π * n^2)

Step 4: Calculate the coefficient bn. Now, let’s calculate the coefficient bn. We need to evaluate the integral ∫[0 to 2π] f(x) * sin(n * x) dx. Since f(x) = x, the integral becomes:

bn = (1/π) * ∫[0 to 2π] x * sin(n * x) dx

We’ll use integration by parts again. Let’s set u = x and dv = sin(n * x) dx.

Differentiating u gives du = dx, and integrating dv gives v = (-1/n) * cos(n * x).

Applying integration by parts:

bn = (1/π) * [u * v – ∫[0 to 2π] v * du]
= (1/π) * [x * (-1/n) * cos(n * x) – (-1/n) * ∫[0 to 2π] cos(n * x) dx]

Integrating cos(n * x) gives (1/n) * sin(n * x), so the integral becomes:

bn = (1/π) * [x * (-1/n) * cos(n * x) + (1/n^2) * sin(n * x)] evaluated from 0 to 2π
= (1/π) * [(2π * (-1/n) * cos(2π * n) + (1/n^2) * sin(2π * n)) – (0 * cos(0) + (1/n^2) * sin(0))]
= (1/π) * [(-2π/n) * cos(2π * n) + (1/n^2) * sin(2π * n)]

Simplifying further, we get:

bn = (-2/n) * cos(2π * n) + (1/π * n^2) * sin(2π * n)

Step 5: Write the Fourier series. With the Fourier coefficients calculated, we can write the Fourier series representation of the function f(x) = x as:

f(x) = (a0/2) + Σ[(an * cos(n * x)) + (bn * sin(n * x))]

The series summation goes from n = 1 to infinity.

Substituting the calculated coefficients an and bn, the Fourier series becomes:

f(x) = (1/2) + Σ[((2/n) * sin(2π * n) + (1/π * n^2)) * cos(n * x) – ((2/n) * cos(2π * n) + (1/π * n^2)) * sin(n * x)]

This is the Fourier series representation of the function f(x) = x over the interval 0 to 2π.