The Enigmatic Signal: Unraveling the Fourier Transform of ( f(t) )

Table of Contents

## The Mystery Posed

### Question 3

Let $f(t)$ be a signal with Fourier transform $F(s)$ . We’re presented with a trio of clues:

1. $f(t)$ is real.
2. $f(t) = 0$ for $t \leq 0$ .
3. $\mathcal{F}^{-1}\{\text{Re}\{ F(s) \}\} = |t| e^{-|t|}$ .

Our quest: Find $f(t)$ .

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## The Mathematical Prologue: Odd and Even Functions

Before we delve into the complex world of Fourier transforms and real functions, let’s briefly focus on a mathematical gem: every real function $f(t)$ can be decomposed into odd and even parts. Mathematically, this is represented as:

$f(t) = f_{\text{even}}(t) + f_{\text{odd}}(t)$

where,

$f_{\text{even}}(t) = \frac{f(t) + f(-t)}{2}$

$f_{\text{odd}}(t) = \frac{f(t) - f(-t)}{2}$

In essence, this decomposition acts as the mathematical cornerstone that guides our journey to decode $f(t)$ .

## Charting the Unknown

### The Initial Pathway

The clue that $f(t)$ is real becomes our starting point. Armed with the knowledge that every real function can be split into odd and even components, we’re equipped to delve deeper.

### The Schematic Blueprint

Our exploratory journey is best illustrated through this architectural framework:

$\begin{array}{cccccccccccc} & & f(t)& \\ & & \updownarrow{\text{Decompose}}& \\ & f_{\text{odd}}(t) & + & f_{\text{even}}(t) \\ & \downarrow \mathcal{F} & & \downarrow \mathcal{F} \\ & F_{\text{odd}}(s) \, (\text{Imaginary, Odd}) & + & F_{\text{even}}(s) \, (\text{Real, Even}) \\ & \downarrow \text{Re\{\}} & & \downarrow \text{Re\{\}} \\ & 0 & + & \text{Re}\{ F_{\text{even}}(s) \} \\ & \downarrow \mathcal{F}^{-1} & & \downarrow \mathcal{F}^{-1} \\ & 0 & + & \mathcal{F}^{-1}\{\text{Re}\{ F_{\text{even}}(s) \}\} \\ & & \downarrow & \\ & & \mathcal{F}^{-1}\{\text{Re}\{ F(s) \}\} = |t| e^{-|t|} & \end{array}$

### The Crucial Insight

Since $f(t) = 0$ for $t \leq 0$ :

$\begin{array}{ccc} f_{\text{odd}}(t) + f_{\text{even}}(t) & = & 0 \, (\text{for } t < 0) \end{array}$

We deduce $f_{\text{odd}}(t) = - f_{\text{even}}(t) = -|t| e^{-|t|}$ for $t < 0$ . ## The Final Revelation As $f_{\text{odd}}$ is an odd function:

$\begin{array}{ccc} f_{\text{odd}}(t) & = & \begin{cases} -|t| e^{-|t|} & \text{for } t < 0 \\ t e^{-t} & \text{for } t > 0 \end{cases} \end{array}$

And so, the final piece of the puzzle falls into place. We find:

$\begin{array}{ccc} f(t) & = & \begin{cases} 0 & \text{for } t \leq 0 \\ 2 t e^{-t} & \text{for } t > 0 \end{cases} \end{array}$

## Epilogue: The Signal Decoded

In a grand revelation, $f(t)$ stands before us, no longer an enigma. What began as a mathematical mystery ended as a scientific accomplishment, all thanks to the foundational understanding that every real function can be decomposed into its odd and even components. Our journey through this intellectual landscape has led us to this defining moment: $f(t)$ is not just a set of symbols and equations; it’s a signal decoded.

## The Saga Continues: Unlocking Further Mysteries of Fourier Transforms

### Question 4

Consider two functions $f(t)$ and $g(t)$ , illustrated in Figure below. Their Fourier transforms are denoted as $F(s)$ and $G(s)$ , respectively.

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1. Without performing any integration, what is the real part of $F(s)$ ?
2. Given the imaginary part of $F(s)$ as $\frac{\sin(2 \pi s) - 2 \pi s}{4 \pi^2 s^2}$ , what is $G(s)$ ?

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### The Intricacies of Real Parts in Fourier Transformations

The first question poses a new challenge, asking us to discern the real part of $F(s)$ without resorting to integration.

Drawing from the invaluable insights obtained from our previous exploration, we’re reminded that $f(t)$ is not a standalone entity but a blend of a scaled even function $0.5 \Lambda(t)$ and its corresponding odd counterpart $0.5 g(t)$ .

This is where the Fourier transform reveals its true prowess. The transformation process clarifies that only the even component contributes to the real part. Consequently, the real part of $F(s)$ elegantly emerges as $0.5 \text{sinc}^2(s)$ .

### Solving for $G(s)$

The second question pivots us to $G(s)$ , extending the challenge with the gift of a known variable: the imaginary part of $F(s)$ .

Building upon our earlier discussion, the task becomes remarkably straightforward. We deduce that $G(s)$ is essentially $2F(s)$ . Thus, $G(s)$ is precisely expressed as $\frac{\sin(2 \pi s) - 2 \pi s}{2 \pi^2 s^2}$ .