Problem 3 - Advanced Engineering Mathematics

Question

Find the Fourier integral representation of the given function. $$ f(x)=\left\\{\begin{array}{lr} 0, & x<0 \\ x, & 03 \end{array}\right. $$

Step-by-Step Solution

Answer

The Fourier integral of \( f(x) \) is \[ \hat{f}(k) = \frac{3 e^{-6\pi ik}}{-2\pi ik} + \frac{e^{-6\pi ik} - 1}{(2\pi ik)^2}. \]

Step 1: Understanding the function

The given function \( f(x) \) is piecewise. It is defined as 0 for \( x < 0 \), linear \( x \) for \( 0 < x < 3 \), and 0 for \( x > 3 \). Thus, it is non-zero only in the interval \( (0, 3) \).

Step 2: Setting up the Fourier Transform

The Fourier transform \( \hat{f}(k) \) of a function \( f(x) \) is given by the integral \( \hat{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi ikx} \, dx \). For \( f(x) \), we only integrate over the interval where it is non-zero: \( \int_{0}^{3} x e^{-2\pi ikx} \, dx \).

Step 3: Evaluating the integral

To find \( \hat{f}(k) \), evaluate the integral \( \int_{0}^{3} x e^{-2\pi ikx} \, dx \) by parts. Let \( u = x \) and \( dv = e^{-2\pi ikx} \, dx \). Then \( du = dx \) and \( v = \frac{e^{-2\pi ikx}}{-2\pi ik} \). Apply integration by parts.

Step 4: Applying Integration by Parts

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Applying this gives:\[ \hat{f}(k) = \left[ \frac{x e^{-2\pi ikx}}{-2\pi ik} \right]_0^3 + \frac{1}{2\pi ik} \int_{0}^{3} e^{-2\pi ikx} \, dx. \]

Step 5: Simplifying the expression

Compute the boundary terms and the new integral. Boundary terms give:\[ \left[ \frac{x e^{-2\pi ikx}}{-2\pi ik} \right]_0^3 = \frac{3 e^{-6\pi ik}}{-2\pi ik} - \frac{0 e^{0}}{-2\pi ik} = \frac{3 e^{-6\pi ik}}{-2\pi ik}. \] The integral \( \int_{0}^{3} e^{-2\pi ikx} dx \) evaluates to \( \frac{e^{-2\pi ikx}}{-2\pi ik} \)_0^3\

Step 6: Finalizing the Fourier Transform

Evaluate the remaining integral:\[ \int_{0}^{3} e^{-2\pi ikx} \, dx = \left[ \frac{e^{-2\pi ikx}}{-2\pi ik} \right]_0^3 = \frac{e^{-6\pi ik} - 1}{-2\pi ik}. \] Substituting back gives: \[ \hat{f}(k) = \frac{3 e^{-6\pi ik}}{-2\pi ik} + \frac{1}{2\pi ik} \left( \frac{e^{-6\pi ik} - 1}{-2\pi ik} \right). \] Simplifying results in:

Step 7: Expression of Fourier Transform

Simplified, the Fourier transform becomes:\[ \hat{f}(k) = \frac{3 e^{-6\pi ik}}{-2\pi ik} + \frac{e^{-6\pi ik} - 1}{(2\pi ik)^2}. \] This is the Fourier transform representation of \( f(x) \).

Key Concepts

Fourier Transform

The Fourier Transform is a fundamental concept in mathematics and engineering. It transforms a function of time (or space) into a function of frequency. The process is essential for analyzing the frequencies present in a signal. Given a function \( f(x) \), its Fourier transform \( \hat{f}(k) \) is computed using the integral:

• \( \hat{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi ikx} \, dx \)

This equation essentially decomposes \( f(x) \) into its frequency components. In the exercise, the Fourier transform is adapted to evaluate only over the interval where the function is non-zero, i.e., \( (0, 3) \). The exponential term \( e^{-2\pi ikx} \) is crucial, as it represents complex sinusoids that help in analyzing the frequency components across the spectrum. This transformation allows us to visualize the "frequency content" of \( f(x) \), making it easier to analyze or manipulate signals in the frequency domain.

Piecewise Function

Piecewise functions are functions that have different expressions based on the value of the input variable. They are particularly useful when a function behaves differently over different intervals.

• For example, the function \( f(x) \) in the exercise is defined as:
• \( 0 \) for \( x < 0 \)
• \( x \) for \( 0 < x < 3 \)
• \( 0 \) for \( x > 3 \)

In this case, the function is non-zero only in the interval \( (0, 3) \). Outside this interval, the function stays at zero. This structure is practical when modeling real-world phenomena where a certain condition only applies for a specific range of inputs. Understanding piecewise functions is important in computing integrals like the Fourier Transform, as it requires focusing on the segments where the function is active. By identifying these intervals, computations can be simplified, reducing unnecessary calculations.

Integration by Parts

Integration by Parts is a valuable technique for solving integrals. It is especially useful when dealing with the product of two functions. The fundamental formula is:

• \( \int u \, dv = uv - \int v \, du \)

In this formula, \( u \) and \( dv \) are parts of the original integral that, when appropriately chosen, make the resulting calculations simpler. In the context of the exercise, integration by parts aids in tackling the integral \( \int_{0}^{3} x e^{-2\pi ikx} \, dx \). By choosing:

• \( u = x \)
• \( dv = e^{-2\pi ikx} \, dx \)

We find that \( du = dx \) and \( v = \frac{e^{-2\pi ikx}}{-2\pi ik} \). Applying the integration by parts formula declutters the original integral, turning it into a manageable form. This technique is not just about application but also about strategically selecting \( u \) and \( dv \) to simplify the problem.

Complex Exponential

Complex Exponentials are a key element in understanding the Fourier Transform. The term \( e^{-2\pi ikx} \) represents a complex sinusoid, effectively a combination of sine and cosine waves. This works because of Euler's Formula, which states:

• \( e^{ix} = \cos(x) + i \sin(x) \)

Here, \( i \) is the imaginary unit. Complex exponentials are particularly potent in Fourier analysis as they allow for efficient decomposition of functions into frequency domains. They enable the conversion of functions into their sinusoidal components, which simplifies processes like signal processing and audio analysis.
In the exercise, the complex exponential \( e^{-2\pi ikx} \) is integral in transforming the piecewise function into its frequency domain representation. This component not only aids in handling oscillatory functions but also facilitates the integration process by using its derivative properties. Understanding complex exponentials is crucial because they form the backbone of various transformations in both pure and applied mathematics.