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Problem 507

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 \(F(k)\).

Short Answer

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In summary, we proved the following properties of Fourier transforms: a) The Fourier transform of f(x) exists if f is absolutely integrable over \((-\infty,+\infty)\). We showed this by utilizing the definition of absolute integrability and the Fourier transform, and proved that \(|F(k)| < \infty\). b) If f(x) is real-valued, then \(F(-k) = F^{*}(k)\), where \(F^{*}(k)\) is the complex conjugate of \(F(k)\). We used the definitions of the Fourier transform and complex conjugate to show that the expressions for F(-k) and F^{*}(k) are equal when f(x) is real-valued.
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Step 1: Define absolute integrability

A function f(x) is absolutely integrable over an interval if: $$ \int_{-\infty}^{\infty} |f(x)| dx < \infty $$ Using this, we will prove that the Fourier transform of f(x) exists.

Step 2: Apply the definition of the Fourier transform

We will use the definition of the Fourier transform, which states that: $$ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx $$

Step 3: Apply absolute integrability to the Fourier transform

If f(x) is absolutely integrable, then the following condition holds: $$ \int_{-\infty}^{\infty} |f(x)| dx < \infty $$ We can rewrite the Fourier transform as follows: $$ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx = \int_{-\infty}^{\infty} f(x)(\cos(-2\pi k x)+i\sin(-2\pi k x)) dx $$ Now, we can split the integral into real and imaginary parts: $$ F(k) = \int_{-\infty}^{\infty} f(x) \cos(-2\pi k x) dx + i \int_{-\infty}^{\infty} f(x) \sin(-2\pi k x) dx $$ Since the absolute value of a complex number is less than or equal to the sum of the absolute values of its real and imaginary parts, we have: $$ |F(k)| \leq \int_{-\infty}^{\infty} |f(x)| |\cos(-2\pi k x)| dx + \int_{-\infty}^{\infty} |f(x)| |\sin(-2\pi k x)| dx $$ Both integrals on the right-hand side are absolutely integrable, since \(|\sin(-2\pi k x)| \leq 1\) and \(|\cos(-2\pi k x)| \leq 1\). Therefore, we can conclude that: $$ |F(k)| < \infty $$ Since the Fourier transform is finite, we have proven that the Fourier transform of f(x) exists if f is absolutely integrable over \((-\infty,+\infty)\). ##Part b: Real-valued functions and complex conjugate## Now we will prove that if f(x) is real-valued, then \(F(-k) = F^{*}(k)\).

Step 1: Rewrite the Fourier transform for F(-k)

We need to find the Fourier transform for F(-k), which is given by: $$ F(-k) = \int_{-\infty}^{\infty} f(x) e^{2\pi i k x} dx $$

Step 2: Use the conjugate of the Fourier transform

From the definition of the complex conjugate: $$ F^{*}(k) = \left(\int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx\right)^{*} $$ Now, we take the complex conjugate of the integral: $$ F^{*}(k) = \int_{-\infty}^{\infty} f(x)^{*} \left(e^{-2\pi i k x}\right)^{*} dx $$ Given that f(x) is real-valued, we have \(f(x)^{*}=f(x)\). Additionally, \((e^{-2\pi i k x})^{*} = e^{2\pi i k x}\), leading to: $$ F^{*}(k) = \int_{-\infty}^{\infty} f(x) e^{2\pi i k x} dx $$

Step 3: Compare F(-k) and F^{*}(k)

Comparing the expressions for F(-k) and F^{*}(k), we found that for a real-valued function f(x): $$ F(-k) = F^{*}(k) $$ This completes the proof of the property that if f(x) is real-valued, then the complex conjugate of \(F(k)\) at -k is equal to \(F(-k)\).

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Most popular questions from this chapter

Chapter 17

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.

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

a) Prove that if the functions \(\mathrm{g}(\mathrm{x})\) and \(\mathrm{F}(\mathrm{k})\) are absolutely integrable on \((-\infty,+\infty)\) and that the Fourier inversion integral for \(\mathrm{f}(\mathrm{x})\) is valid for all \(\mathrm{x}\) except possibly at a countably infinite number of points, then $$ \infty_{-\infty} F(k) g(-k) d k={ }^{\infty} \int_{-\infty} f(x) g(k) d x $$ where $$ \mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}, \mathrm{G}(\mathrm{k})=\Phi\\{\mathrm{g}(\mathrm{x})\\} $$ This is known as the second Parseval theorem of Fourier transform theory. b) From the above equation (1), prove the first Parseval theorem of Fourier transform theory, $$ \left.{ }^{\infty}\right|_{-\infty}|F(\mathrm{k})|^{2} \mathrm{dk}={ }^{\infty} \int_{-\infty}|\mathrm{f}(\mathrm{x})|^{2} \mathrm{~d} \mathrm{x} \text { . } $$

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

Define the Dirac delta function, \(\delta(\mathrm{x})\), and prove the sifting property of \(\delta(\mathrm{x})\) for all functions \(\mathrm{f}(\mathrm{x})\) which are continuous at \(\mathrm{x}=0\) $$ { }^{\infty} \int_{-\infty} \delta(\mathrm{x}) \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}(0) \text { . } $$

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