Question

Let \(F\) denote the Fourier transform of the function $$ f(x)= \begin{cases}1, & 0 \leq x<1 \\ 0, & \text { otherwise }\end{cases} $$ Find the function \(g\) such that the Fourier transform \(G\) of \(g\) satisfies \(G(\omega)=\) \([F(\omega)]^{2} .\)

Short answer

The function \(g(x)\) is piecewise: \(x\) for \(0 \leq x < 1\), \(2-x\) for \(1 \leq x < 2\), and \(0\) otherwise.

Step-by-step solution

Determine the Fourier Transform of the Given Function

The Fourier transform of a function \( f(x) \) is given by the integral: \[ F(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i\omega x} \, dx \]Given \( f(x) = 1 \) for \( 0 \leq x < 1 \) and \( 0 \) otherwise, set up the integral:\[ F(\omega) = \int_{0}^{1} e^{-i \omega x} \, dx \]

Evaluate the Integral for the Fourier Transform

Calculate the integral determined in Step 1:\[ F(\omega) = \int_{0}^{1} e^{-i \omega x} \, dx = \left[ \frac{e^{-i \omega x}}{-i \omega} \right]_{0}^{1} = \frac{1 - e^{-i \omega}}{i \omega} \] Simplify this expression:\[ F(\omega) = \frac{1 - e^{-i \omega}}{i \omega} \]

Compute \([F(\omega)]^{2}\)

Square the transform \( F(\omega) \) to find \( G(\omega) \):\[ G(\omega) = \left( \frac{1 - e^{-i \omega}}{i \omega} \right)^2 = \frac{(1 - e^{-i \omega})^2}{(-\omega^2)} \] Simplify the numerator:\[ (1 - e^{-i \omega})^2 = 1 - 2e^{-i \omega} + e^{-2i\omega} \] Thus, the expression becomes:\[ G(\omega) = \frac{1 - 2e^{-i \omega} + e^{-2i\omega}}{-\omega^2} \]

Determine the Function g(x) from its Fourier Transform

The inverse Fourier transform \( g(x) \) can be found using:\[ g(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) e^{i \omega x} \, d\omega \] Using Fourier transform properties, recognize that squaring the Fourier transform \( F(\omega) \) leads to the convolution of \( f(x) \) with itself. Therefore, \[ g(x) = (f \ast f)(x) = \int_{-\infty}^{\infty} f(y)f(x-y) \, dy \] As \( f(x) = 1 \) from \( 0 \leq x < 1 \), this becomes:- \( g(x) = x \) for \( 0 \leq x < 1 \)- \( g(x) = 2-x \) for \( 1 \leq x < 2 \)- \( g(x) = 0 \) otherwise.

Key concepts

Fourier Series

Imagine breaking down a periodic function into a set of simpler sine and cosine functions. This is the core idea behind a Fourier series. By representing a function as a sum of sine and cosine terms, each multiplied by a coefficient, we can analyze its frequency components. These coefficients tell us how much of each sine or cosine wave is needed. It’s like mixing colors: different amounts create a unique shade.
\[ f(x) = a_0 + \sum_{n=1}^{\infty} \, [ a_n \cos(n\omega_0 x) + b_n \sin(n\omega_0 x) ] \]
Fourier series are incredibly useful in engineering and science. They allow us to handle signals, waves, and any repeating patterns. Although the original exercise dealt with Fourier transforms, using Fourier series helps in understanding the frequency aspects of functions in a more discrete setting.
In essence:

• Helps in breaking complex patterns into understandable waves.
• Widely used in signal processing and various fields.
• Makes analyzing periodic behaviors simpler.

Remember, whereas the Fourier series is about periodic functions, the Fourier transform expands this to any function, periodic or not.

Convolution

Convolution is an operation on two functions that yields a third. Imagine it as blending two functions to see their combined influence. It is particularly useful in signal processing, where it helps in understanding how a system modifies signals passing through it.
In mathematical terms, the convolution of two functions \( f \) and \( g \) is given by:
\[ (f \ast g)(x) = \int_{-\infty}^{\infty} f(y)g(x-y) \, dy \]
When you find the convolution of \( f(x) \) with itself, you are essentially measuring how the function overlaps with itself as one slides over the other. In the given exercise, this principle was applied to find the function \( g(x) \), which blended the original function \( f(x) \) with itself.
Key things to remember about convolution:

• It essentially integrates the product of two functions over a range.
• It can reveal properties like smoothing or blurring when applied to signals or images.
• In the context of the Fourier transform, convolution in the time domain becomes multiplication in the frequency domain, and vice versa.

Convolution might sound complex, but it reveals how functions interact and combine to form new patterns.

Complex Exponents

Complex exponents combine real and imaginary numbers, playing a pivotal role in oscillation and wave analysis. When you deal with expressions like \( e^{i\omega x} \), you tap into Euler's formula:
\[ e^{i \omega} = \cos(\omega) + i \sin(\omega) \]
This elegantly ties exponential functions to circular motions. It's like converting point coordinates into a rotation around a circle.
These exponents simplify calculations in waves since they thread through both the rise-fall behavior (sine, cosine) and growth-decay trends (exponentials). As seen in the Fourier transform exercise, complex exponents help reformulate integrals that describe frequency components of functions.
Looking at them is essential for understanding:

• How vibrations and oscillations are succinctly represented.
• The beauty and simplicity of rotating phasors in electrical engineering.
• Why they transform sinusoidal equations into exponential forms for ease of computation.

The duality of real-imaginary parts in complex exponents offers a stunningly simple way to work with rotations and waves.

Integral Calculus

Integral calculus deals with accumulation—be it areas, volumes, or total change. In the realm of the Fourier transform, integral calculus serves as a tool to switch functions from one domain (time/space) to another (frequency).
In the problem provided, you saw the application of integral calculus when calculating the Fourier transform. To compute \( F(\omega) \), the integral of the product of the function and a complex exponent was performed over a specific interval.
A few key notes on integral calculus, especially in this context, include:

• It helps in finding total areas under curves, important for signal analysis in transforms.
• The antithesis of differential calculus, it focuses on summing parts to find wholes.
• Central to processes that involve transforming functions or solving differential equations.

Understanding integrals is not just about areas but also about understanding changes and transitions, like moving from time-based information to frequency-based for deeper analysis.