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Problem 503

Develop the definition of the Fourier transform of a function \(\mathrm{f}(\mathrm{x})\) be extending the definition of the Fourier series of \(\mathrm{f}\) to the case where the discrete spectrum of Fourier coefficients becomes a continuous spectrum.

Short Answer

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In order to develop the Fourier Transform from the Fourier Series, we start by considering the limit as the period \(T\) goes to infinity, which leads to a continuous spectrum with angular frequency \(\omega\). Then, we replace the trigonometric functions with the exponential function using Euler's identity. This gives us the Fourier Transform definition: \[F(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i\omega x} dx\] And the inverse Fourier Transform: \[f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega x} d\omega\]
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Step 1: Review the Fourier Series Definition

The Fourier series represents a periodic function \(f(x)\) as a sum of sine and cosine functions. The general formula for a Fourier series is given by: \[f(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(\frac{2\pi nx}{T}) + b_n \sin(\frac{2\pi nx}{T})]\] where \(T\) is the period of function \(f(x)\), and the coefficients \(a_0\), \(a_n\), and \(b_n\) are defined as follows: \[a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(x) dx\] \[a_n = \frac{2}{T} \int_{-T/2}^{T/2} f(x) \cos(\frac{2\pi nx}{T}) dx\] \[b_n = \frac{2}{T} \int_{-T/2}^{T/2} f(x) \sin(\frac{2\pi nx}{T}) dx\]

Step 2: Introduce a continuous spectrum

Now, we will extend the definition of the discrete Fourier series to a continuous spectrum by considering the limit as the period \(T\) goes to infinity. In this case, the spacing between frequencies will become infinitesimally small and we can replace the discrete frequencies with a continuous variable, usually denoted as \(\omega\) (angular frequency). Instead of summing over discrete coefficients, we will integrate over the continuous variable \(\omega\). Also, note that \(\omega\) is defined as \(\omega = \frac{2\pi n}{T}\), which simplifies to \(n = \frac{T\omega}{2\pi}\).

Step 3: Develop the Fourier Transform Definition

Given the continuous spectrum, the Fourier series will be transformed into the Fourier transform. The formula for the Fourier transform is given by: \[F(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i\omega x} dx\] Where \(F(\omega)\) is the Fourier transform of the function \(f(x)\), and \(i\) is the imaginary unit (\(i^2 = -1\)). Note that we replaced the trigonometric functions with the exponential function using Euler's identity: \[e^{ix} = \cos(x) + i\sin(x)\] And the inverse Fourier transform is defined as: \[f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega x} d\omega\] The Fourier transform and its inverse allow us to move back and forth between the time domain (represented by the function \(f(x)\)) and the frequency domain (represented by its Fourier transform, \(F(\omega)\)).

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Most popular questions from this chapter

Chapter 17

Given the current pulse, $\mathrm{i}(\mathrm{t})=\mathrm{te}^{-\mathrm{bt} \text { . }}$ (a) find the total \(1 \Omega\) energy associated with this waveform; (b) what fraction of this energy is present in the frequency band from \(-\mathrm{b}\) to \(\mathrm{b} \mathrm{rad} / \mathrm{s} ?\)

Chapter 17

Find the Fourier transform, \(\mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}\) of the Gaussian probability function $\mathrm{f}(\mathrm{x})=\mathrm{Ne}^{(-\alpha \mathrm{x}) 2} \quad(\mathrm{~N}, \alpha=\( constant \))$ Show directly that \(\mathrm{f}(\mathrm{x})\) is retrievable from the inverse transform. I.e., show that $\mathrm{f}(\mathrm{x})=\left\\{1 / \sqrt{(2 \pi)\\}}^{\infty} \int_{-\infty} \mathrm{F}(\mathrm{k}) \mathrm{e}^{-\mathrm{ikx}} \mathrm{dx}=\Phi^{-1}[\Phi\\{\mathrm{f}(\mathrm{x})\\}]\right.$

Chapter 17

Use Fourier transform methods to find the time-domain response network having a system function $$ j 2 \omega /(1+2 j \omega) $$ if the unit is $$ \mathrm{V}(\mathrm{t})=\cos \mathrm{t} $$ (For a sinusoidal input cos \(t\), the Fourier transform is $$ \pi[\delta(\omega-)+\delta(\omega-1)]) $$

Chapter 17

Let $\mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}, \mathrm{G}(\mathrm{k})=\Phi\\{\mathrm{g}(\mathrm{x})\\}$ and suppose $\mathrm{F}(\mathrm{k}) \mathrm{G}(\mathrm{k})=\Phi\\{\mathrm{h}(\mathrm{x})\\}$. Prove the convolution theorem for Fourier transforms: If \(g(x)\) and \(F(k)\) are absolutely integrable on $(-\infty, \infty)\( and if the Fourier inversion integral for \)\mathrm{f}(\mathrm{x})$ is valid for all \(\mathrm{x}\) except possibly a countably infinite number of points, then $$ \mathrm{h}(\mathrm{x})=(\mathrm{f} * \mathrm{~g}) $$ where ( \(f * g\) ) is the convolution of \(f\) and \(g\) defined by $$ (\mathrm{f} * \mathrm{~g})=\\{1 / \sqrt{(} 2 \pi)\\}^{\infty} \int_{-\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) \mathrm{d} \xi $$

Chapter 17

a) If \(\mathrm{a}>0\) show that the Fourier transform of the function defined by $$ \begin{array}{cc} \mathrm{f}(\mathrm{t})=\mathrm{e}^{-a t} \text { coswdt } & \mathrm{t} \geq 0 \\\ & =0 & \mathrm{t}<0 \end{array} $$ is \((a+j w)^{2} /\left[(a+j w)^{2}+\omega^{2} d\right]\) Then find the total \(1 \Omega\) energy associated with the function $$ \begin{array}{rlrl} \quad \mathrm{f} & =\mathrm{e}^{-\mathrm{t}} \text { cost } & & \mathrm{t} \geq 0 \\ \text { and } & =0 & \mathrm{t} & <0 \end{array} $$ b) time domain integration. That is, find the total energy by integrating $$ \left.\mathrm{W}={ }^{\infty} \int_{0}[\mathrm{f}(\mathrm{t})\\}\right]^{2} \mathrm{dt} $$ c) frequency domain integration. That is, find the total energy by integrating $$ \mathrm{W}=(1 / 2 \pi)^{\infty} \int_{-\infty}|\mathrm{F}(\mathrm{w})|^{2} \mathrm{~d} \mathrm{w} $$ where \(F(\omega)\) is the Fourier transform of the function \(\mathrm{f}(\mathrm{t})\).

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