Question

Evaluate the Fourier transform of $$g(x)=\left\\{\begin{array}{ll}b-b|x| / a & \text { if }|x|a\end{array}\right.$$

Short answer

The Fourier transform output \(G(f)\) is obtained by following the steps outlined above. It involves interpreting the function \(g(x)\), applying the correct Fourier transform integral, performing the integration, and simplifying the result, where possible. However, the precise solution depends on the values of \(b\), \(a\), and \(f\) and requires the integral computations.

Step-by-step solution

Understand the Function Definition

Observe the function \(g(x)\). It’s a piece-wise function that is defined differently over different intervals. For \(|x| < a\), \(g(x)\) = \(b - b|x|/a\) where \(b\) and \(a\) are constants. For \(|x| > a\), \(g(x)\) is zero.

Apply Fourier Transform Formula

Recall the general formula for Fourier transform: \(G(f) = \int_{-\infty}^{+\infty} g(x) e^{-2\pi i f x} dx\). The transformation involves an integral from negative to positive infinity. However, observing the function \(g(x)\), it is understood that the integral only needs to be taken over the interval where \(g(x)\) is non-zero, that is, for \(|x| < a\). Therefore, the expression becomes \(G(f) = \int_{-a}^{+a} g(x) e^{-2\pi i f x} dx\).

Substitute \(g(x)\) in the Integral

Substitute \(g(x) = b - b|x|/a\) into the Fourier transform equation: \(G(f) = \int_{-a}^{+a} (b-b|x|/a) e^{-2\pi i f x} dx\). This equation could be split into two, at \(x = 0\), leading to two integrals: \(G(f) = \int_{0}^{+a} (b-bx/a) e^{-2\pi i f x} dx + \int_{-a}^{0} (b+bx/a) e^{-2\pi i f x} dx\).

Evaluate the Integrals

Now the integrals can be evaluated by finding the anti-derivatives and using the Fundamental Theorem of Calculus. After performing the integration, you'll find the Fourier transform \(G(f)\), which is the output for different frequency values \(f\).

Simplify the Result

Finally, simplify the resultant expression for \(G(f)\), if possible. This is the Fourier transform of the original function \(g(x)\).

Key concepts

Understanding Piecewise Functions

Piecewise functions are a special type of function that are defined by different expressions over different intervals of their domain. It's like having multiple "mini functions" in one, each activated depending on the input value.
For the given problem, the function \(g(x)\) is piecewise:

• If \(|x| < a\), \(g(x) = b - \frac{b |x|}{a}\). This part means for values of \(x\) within \(-a\) to \(a\), \(g(x)\) is a linear function.
• For \(|x| > a\), \(g(x) = 0\). This indicates that outside this interval, the function becomes zero.

Piecewise functions are powerful in modeling scenarios where a rule or condition changes based on the situation, such as in signal processing, financial models, or physics. They allow functions to represent complex behaviors that can't be described with a single equation.

Basics of Integral Calculus

Integral calculus is a key mathematical concept used to find areas, volumes, central points, and other values that are basically "summing up" a function. It's like finding the total area under the curve represented by a function on a graph.
The Fourier transform involves using integral calculus to analyze frequency components of a function. Here, we apply the integral from the Fourier transform formula: \[ G(f) = \int_{-a}^{+a} (b - \frac{b |x|}{a}) e^{-2\pi i f x} dx \]This integral computes how much of each frequency component is present in the original piecewise function.
When calculating integrals, especially with piecewise functions, one often calculates separate integrals over each interval where the function has different expressions. Doing so ensures you properly "sum up" the function over each relevant part of its domain, which is essential for accurate results.

The Role of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus provides a foundational link between derivatives and integrals, showing that integration can be reversed by differentiation and vice versa.
This theorem has two main parts:

• The first part states that if a function is continuous over an interval, then it has an antiderivative over that interval.
• The second part states that if you have an antiderivative of a function, the definite integral of the function over an interval can be found by evaluating the antiderivative at the boundaries of that interval.

When implementing the theorem in our problem, it allows us to evaluate the integrals from the Fourier transform. By finding the antiderivative of each piece of our piecewise function, we can compute the area under the curve (including weighted parts due to \(e^{-2\pi i f x}\) in the Fourier transform) over the intervals \(-a\) to \(0\) and \(0\) to \(a\). This step is crucial for solving the problem and obtaining the Fourier transform of \(g(x)\).