Erin's question via email about a Fourier Transform

The theory is the same as in the text, but everything is nicely organized and it is more powerful.In summary, the Fourier Transform of the given function is equal to the expression $\frac{1 - \cos{ \left( 2\,\omega \right) }}{\omega ^2}$. This can be derived using distribution theory, which simplifies the calculations and allows for a more powerful understanding of Fourier transforms.
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Find the Fourier Transform of $\displaystyle \begin{align*} f \left( t \right) = \begin{cases} 1 - \frac{t}{2} \textrm{ if } 0 \leq t \leq 2 \\ 1 + \frac{t}{2} \textrm{ if } -2 \leq t < 0 \\ 0 \textrm{ elsewhere } \end{cases} \end{align*}$

$\displaystyle \begin{align*} F \left( \omega \right) &= \mathcal{F} \left\{ f \left( t \right) \right\} \\
&= \int_{-\infty}^{\infty}{ f\left( t \right) \mathrm{e}^{-\mathrm{j}\,\omega \, t}\,\mathrm{d}t } \\
&= \int_{-\infty}^{-2}{ 0\,\mathrm{d}t } + \int_{-2}^0{ \left( 1 + \frac{t}{2} \right) \mathrm{e}^{-\mathrm{j}\,\omega\,t}\,\mathrm{d}t } + \int_0^2{ \left( 1 - \frac{t}{2} \right) \mathrm{e}^{-\mathrm{j}\,\omega\,t}\,\mathrm{d}t } + \int_2^{\infty}{0\,\mathrm{d}t} \\
&= 0 + \left[ \left( 1 + \frac{t}{2} \right) \left( \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{-j\,\omega} \right) \right]_{-2}^0 - \int_{-2}^0{ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{-2\,\mathrm{j}\,\omega}\,\mathrm{d}t } + \left[ \left( 1 - \frac{t}{2} \right) \left( \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{-\mathrm{j}\,\omega} \right) \right]_0^2 - \int_0^2{ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{2\,\mathrm{j}\,\omega}\,\mathrm{d}t } + 0 \\
&= \frac{1}{-\mathrm{j}\,\omega} - \left[ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{2\,\mathrm{j}^2\,\omega^2} \right] _{-2}^0 - \left( \frac{1}{-\mathrm{j}\,\omega} \right) - \left[ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{-2\,\mathrm{j}^2\,\omega^2} \right] _0^2 \\
&= \left[ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{2\,\omega^2} \right] _{-2}^0 - \left[ \frac{\mathrm{e}^{-\mathrm{j}\,\omega\,t}}{2\,\omega^2} \right] _0^2 \\ &= \frac{1}{2\,\omega^2} \left\{ \left[ \mathrm{e}^{-\mathrm{j}\,\omega\,t } \right] _{-2}^0 - \left[ \mathrm{e}^{-\mathrm{j}\,\omega\,t} \right] _0^2 \right\} \\
&= \frac{1}{2\,\omega^2} \left[ \left( 1 - \mathrm{e}^{2\,\mathrm{j}\,\omega } \right) - \left( \mathrm{e}^{-2\,\mathrm{j}\,\omega\,t} - 1 \right) \right] \\ &= \frac{1}{2\,\omega^2} \left[ 2 - \left( \mathrm{e}^{2\,\mathrm{j}\,\omega} + \mathrm{e}^{-2\,\mathrm{j}\,\omega} \right) \right] \\ &= \frac{1}{\omega^2} \left[ 1 - \left( \frac{\mathrm{e}^{2\,\mathrm{j}\,\omega} + \mathrm{e}^{-2\,\mathrm{j}\,\omega}}{2} \right) \right] \\ &= \frac{1 - \cos{ \left( 2\,\omega \right) }}{\omega ^2} \end{align*}$
 
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What is a Fourier Transform?

A Fourier Transform is a mathematical operation that decomposes a function or signal into its constituent frequencies. It is used to analyze and interpret complex signals, such as sound waves and electrical signals, in terms of their frequency components.

What is the purpose of a Fourier Transform?

The purpose of a Fourier Transform is to simplify complex signals into their individual frequency components, allowing for easier analysis and interpretation. It is also used in many applications, such as signal processing, data compression, and image analysis.

How is a Fourier Transform performed?

A Fourier Transform is typically performed using a mathematical algorithm, such as the Fast Fourier Transform (FFT). This involves breaking down a signal into smaller parts and then combining them to create a representation of the signal in terms of its frequency components.

What are some real-world applications of a Fourier Transform?

A Fourier Transform has many practical applications, including audio and image processing, data compression, and signal analysis. It is also used in physics, engineering, and other fields to study and understand complex systems.

Is a Fourier Transform reversible?

Yes, a Fourier Transform is reversible. This means that it is possible to reconstruct the original signal from its frequency components using the inverse Fourier Transform. However, some information may be lost in the transformation process, so the reconstructed signal may not be identical to the original.

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