Unveiling the Mysteries of Fourier Transform: A Practical Series — Part 1

Fourier Transform of 4 e^{-j 4 \pi  s} \mathbf{sinc}^2(2s) + 5 e^{-j 8 \pi  s} \mathbf{sinc}^2(2s)

Welcome to the first installment of this practical series on Fourier Transform and its applications. As someone deeply engrossed in a course on this subject, I find it profoundly fascinating to explore the mathematical foundations and real-world applications of Fourier Transform. Let’s embark on this intellectual journey, where today we will delve into an intriguing problem that beckons us to make creative use of the \Lambda(\cdot) function.

## The Enigma: Transforming a Unique Signal

Our first question is an enticing one: What is the Fourier transform of a particular signal? The time-domain representation of this signal is illustrated in the figure below:

Rendered by QuickLaTeX.com

The signal in question is a sum of two scaled and shifted \Lambda functions:

    \[2\Lambda\left(\frac{x - 2}{2}\right) + 2.5\Lambda\left(\frac{x - 4}{2}\right)\]

### An Exploration into \Lambda

Before diving headlong into solving our central problem, let’s take a detour through the mathematical landscape of \Lambda(x). We begin by deriving the Fourier transform of the scaled function \Lambda\left(\frac{x}{c}\right).

The Fourier transform of this scaled function can be expressed as:

    \[\mathcal{F}\left(\Lambda\left(\frac{x}{c}\right)\right) = \int_{-c}^{c} \Lambda\left(\frac{x}{c}\right) e^{-i 2 \pi s x} dx \quad \text{(1)}\]

Through a substitution \frac{x}{c} = y, we can elegantly rewrite the integral:

    \[\mathcal{F}\left(\Lambda(y)\right) = c \mathbf{sinc}^2(cs) \quad \text{(2)}\]

### Generalizing the Fourier Transform

With the foundation laid, we introduce the shift relation in Fourier transform to broaden our discussion:

    \[g(t-b) \xleftrightarrow{} e^{-j 2 \pi b s} G(s)\]

From this, we can now generalize for a\Lambda\left(\frac{x-b}{c}\right):

    \[\mathcal{F}\left(a\Lambda\left(\frac{x-b}{c}\right)\right) = a c \, e^{-j 2 \pi b s} \mathbf{sinc}^2(cs) \quad \text{(3)}\]

### Solving the Enigma

Armed with these tools, we find the Fourier Transform of our signal as:

    \[\mathcal{F}\left(2\Lambda\left(\frac{x-2}{2}\right) + 2.5\Lambda\left(\frac{x-4}{2}\right)\right) = 4 e^{-j 4 \pi s} \mathbf{sinc}^2(2s) + 5 e^{-j 8 \pi s} \mathbf{sinc}^2(2s) \quad \text{(4)}\]

Below is a stunning visual representation of the Fourier Transform of this signal, displaying both its real and imaginary components:

Rendered by QuickLaTeX.com

Rendered by QuickLaTeX.com

## Conclusion: The Transformative Journey

Today, we’ve cracked the code of a complex signal using the Fourier Transform and unearthed the role of the \Lambda function in this process. As we journey deeper into this series, more mathematical enigmas await us. The Fourier Transform, it turns out, is not just a mathematical technique; it’s a lens through which we can view the world—distilled into its constituent frequencies.

Join me in the next part of this series, where we’ll explore more complex questions and applications of Fourier Transform. Until then, keep transforming!

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