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Problem 510

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 $$ (\mathrm{f} * \mathrm{~g})=\\{1 / \sqrt{(} 2 \pi)\\}^{\infty} \int_{-\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) \mathrm{d} \xi $$

Short Answer

Expert verified
In order to prove the convolution theorem for Fourier transforms, we first recall the definitions of the Fourier transform and convolution. Then, we substitute the given Fourier transforms F(k) and G(k) into their product and combine the integrals. Upon rearranging and comparing the resulting equation to the convolution definition, we can identify the similarity between the two expressions and conclude that h(x) = (f * g), thus proving the convolution theorem for Fourier transforms.
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Step 1: Recall the definition of the Fourier transform: $$ \mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} $$ #Step 2: Convolution Definition#

Recall the definition of convolution: $$ (\mathrm{f} * \mathrm{g})(\mathrm{x}) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) \, \mathrm{d}\xi $$ #Step 3: Starting with Fourier Transform Product#

Step 2: We know that F(k)G(k) = Φ{h(x)}. Let's write this out: $$ \mathrm{F}(\mathrm{k})\mathrm{G}(\mathrm{k}) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} $$ #Step 4: Substituting Fourier Transforms#

Now substitute the given Fourier transforms F(k) and G(k) in the equation from Step 3: $$ \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}\right)\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{g}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}\right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} $$ #Step 5: Combining Integrals#

Step 3: Now let's combine the product of the two integrals on the left side: $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) e^{-i \mathrm{k} \mathrm{x}} e^{-i \mathrm{k} \mathrm{\xi}} \, \mathrm{d}\xi \,\mathrm{d}\mathrm{x} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} $$ #Step 6: Rearrange and Compare to Convolution Definition#

Rewrite the left side of the equation from Step 5 using convolution definition: $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) e^{-i \mathrm{k} \mathrm{\xi}} \, \mathrm{d}\xi \right) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} $$ Now, we can identify the similarity between the left side of this equation and the convolution term defined in Step 2. Thus, we can conclude that h(x) = (f * g), which proves the convolution theorem for Fourier transforms.

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Most popular questions from this chapter

Chapter 17

Prove the following properties of Fourier transforms: a) The Fourier transform of \(\mathrm{f}(\mathrm{x})\) exists if \(\mathrm{f}\) is absolutely integrable over \((-\infty,+\infty)\), b) If \(\mathrm{f}(\mathrm{x})\) is real valued then $$ F(-k)=F^{*}(k) $$ where \(F^{*}(k)\) is the complex conjugate of \(F(k)\).

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 \(\mathrm{f}(\mathrm{t})\).

Chapter 17

a) Prove the attentuation property of Fourier transforms: $$ \Phi\left\\{\mathrm{f}(\mathrm{x}) \mathrm{e}^{\mathrm{ax}}\right\\}=\mathrm{F}(\mathrm{k}-\mathrm{ai}) $$ where $$ F(k)=\Phi\\{f(x)\\} $$ b) Prove the shifting property of Fourier transforms: $$ \Phi\\{\mathrm{f}(\mathrm{x}-\mathrm{a})\\}=\mathrm{e}^{\mathrm{ika}} \mathrm{F}(\mathrm{k}) $$ c) Prove the derivative properties of Fourier transforms: $$ \begin{aligned} &\Phi\left\\{f^{\prime}(x)\right\\}=-i k \Phi\\{f(x)\\} \\ &\Phi\left\\{f^{\prime}(x)\right\\}=-k^{2} \Phi\\{f(x)\\} \end{aligned} $$

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

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 \(\mathrm{b} \mathrm{rad} / \mathrm{s} ?\)

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