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Chapter 4: Problem 3

Show that the Fourier transform of the "Gaussian" \(f(x)=\exp \left(-\left(x-x_{0}\right)^{2} / a^{2}\right) x_{0}, a\) real, is also a Gaussian: $$ \hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^{2}} e^{-i k x_{0}} $$

Short Answer

Expert verified
The Fourier transform of the Gaussian \(f(x)=\exp\left(-\frac{(x-x_{0})^2}{a^{2}}\right)\) is also a Gaussian: \[\hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^2} e^{-i k x_{0}}.\]

Step by step solution

01

Write the Definition of Fourier Transform

Recall that the Fourier transform \(\hat{F}(k)\) of a function \(f(x)\) is given by \[\hat{F}(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx.\]
02

Substitute the Gaussian Function

Substitute the given Gaussian function \(f(x) = \exp\left(-\frac{(x-x_0)^2}{a^2}\right)\) into the integral: \[\hat{F}(k) = \int_{-\infty}^{\infty} \exp\left(-\frac{(x-x_0)^2}{a^2}\right) e^{-i k x} \, dx.\]
03

Complete the Square in the Exponent

Complete the square in the exponent to combine the Gaussian and exponential terms. Let \(u = x - x_0\), then the integral becomes: \[\hat{F}(k) = \int_{-\infty}^{\infty} \exp\left(-\frac{u^2}{a^2}\right) e^{-i k (u + x_0)} \, du.\]
04

Simplify the Exponent

Combine the exponents: \[\hat{F}(k) = e^{-i k x_0} \int_{-\infty}^{\infty} \exp\left(-\frac{u^2}{a^2} - i k u\right) \, du.\]
05

Use a Change of Variables

Use the substitution \(v = u - \frac{(i a^2 k)}{2}\) to shift and simplify the integral, so: \[\hat{F}(k) = e^{-i k x_0} \int_{-\infty}^{\infty} \exp\left(-\frac{v^2}{a^2}\right) \, dv \exp\left(-\frac{(i a k)^2}{4}\right).\]
06

Evaluate the Gaussian Integral

The remaining integral is a standard Gaussian integral which evaluates to \(a \sqrt{\pi}\): \[\int_{-\infty}^{\infty} \exp\left(-\frac{v^2}{a^2}\right) \, dv = a \sqrt{\pi}.\]
07

Combine the Results

Combine all the factors to obtain the final result: \[\hat{F}(k) = a \sqrt{\pi} \exp\left(-\frac{(k a / 2)^2}{1}\right) \exp\left(-i k x_{0}\right).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

headline of the respective core concept
The Fourier transform is a key tool in many fields such as physics, engineering, and signal processing. It transforms a function of time (or space) into a function of frequency. Mathematically, given a function \( f(x) \), its Fourier transform \( \hat{F}(k) \) is defined as: \[ \hat{F}(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx. \] This formulation shows us how every point in the time domain can be transformed into the frequency domain by integrating over the entire range of the function multiplied by a complex exponential. The resulting function \( \hat{F}(k) \) describes the original function in terms of its frequency components.
headline of the respective core concept
The Gaussian integral is fundamental in probability, statistics, and quantum mechanics. The standard Gaussian function is of the form \( g(x) = \exp(-x^2) \). Its integral over all space is: \[ \int_{-\infty}^{\infty} \exp(-x^2) \ dx = \sqrt{\pi}. \] In the given exercise, we deal with a shifted and scaled Gaussian function \( f(x) = \exp\left(-\frac{(x - x_0)^2}{a^2}\right) \). Through the process of completing the square and clever substitutions, we transform and solve it using properties of Gaussian integrals, where the integral part evaluates to \( a \sqrt{\pi} \).
headline of the respective core concept
Complex exponentiation is a method to handle oscillatory behaviors in functions. It involves expressions like \( e^{-i k x} \), where \( i \) is the imaginary unit, and \( k x \) is a product of a frequency and a variable. It links directly to Euler's formula: \[ e^{i \theta} = \cos(\theta) + i\sin(\theta). \] This formula allows us to represent sinusoidal functions as exponentials, making them easier to manipulate analytically, especially in the Fourier transform. In the given problem, complex exponentials interplay with the Gaussian function to translate time-domain functions into the frequency domain.
headline of the respective core concept
The change of variables is a powerful technique in integral calculus to simplify the integration process. It involves substituting a new variable that makes the integral easier to solve. In the given problem, we used such a substitution: \( u = x - x_0 \) transforms the function to center around zero. Next, we utilize \( v = u - \frac{(i a^2 k)}{2} \), which shifts and simplifies the Gaussian in the exponential. This results in an integral of a standard Gaussian form, which greatly simplifies the evaluation and provides the necessary insight to solve the problem effectively.

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Most popular questions from this chapter

Given the wave equation $$ \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}=0 $$ and the boundary conditions $$ \begin{array}{ll} \phi(x, t=0)=0, & \frac{\partial \phi}{\partial t}(x, t=0)=0 \\ \phi(x=0, t)=0, & \phi(x=\ell, t)=f(t) \end{array} $$ (a) show that the Laplace transform of the solution is given by $$ \hat{\Phi}(x, s)=\frac{\hat{F}(s) \sinh s x}{\sinh s \ell} $$ where \(\hat{F}(s)\) is the Laplace transform of \(f(t)\) (b) Call the solution of the problem when \(f(t)=1\) (so that \(\hat{F}(s)=1 / s)\). to be \(\phi_{s}(x, t) .\) Show that the general solution is given by $$ \phi(x, t)=\int_{0}^{t} \frac{\partial \phi_{s}}{\partial t^{\prime}}\left(x, t^{\prime}\right) f\left(t-t^{\prime}\right) d t^{\prime} $$

(a) Show that the inverse Laplace transform of \(\hat{F}_{1}(s)=\log s\) is given by \(f_{1}(x)=-1 / x\) (b) Do the same for \(\hat{F}_{2}(s)=\log (s+1)\) to obtain \(f_{2}(x)=-e^{-x} / x\). (c) Find the inverse Laplace transform \(\hat{F}(s)=\log ((s+1) / s)\) to obtain \(f(x)=\left(1-e^{-x}\right) / x\), by subtracting the results of parts (a) and (b). (d) Show that we can get this result directly, by encircling both the \(s=0\) and \(s=-1\) branch points and using the polar representations \(s+1=\) \(r_{1} e^{i \theta_{1}}, s=r_{2} e^{i \theta_{2}},-\pi \leq \theta_{i}<\pi, i=1,2\)

Use Laplace transform methods to solve the ODE $$ \frac{d^{2} y}{d t^{2}}-k^{2} y=f(t), \quad k>0, \quad y(0)=y_{0}, \quad y^{\prime}(0)=y_{0}^{\prime} $$ (a) Let \(f(t)=e^{-k_{0 t}}, k_{0} \neq k, k_{0}>0\), so that the Laplace transform of \(f(t)\) is \(\hat{F}(s)=1 /\left(s+k_{0}\right)\), and find $$ \begin{aligned} y(t)=& y_{0} \cosh k t+\frac{y_{0}^{\prime}}{k} \sinh k t+\frac{e^{-k_{0} t}}{k_{0}^{2}-k^{2}} \\ &-\frac{\cosh k t}{k_{0}^{2}-k^{2}}+\frac{\left(k_{0} / k\right)}{k_{0}^{2}-k^{2}} \sinh k t \end{aligned} $$ (b) Suppose \(f(t)\) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section \(4.5\) ) to find $$ y(t)=y_{0} \cosh k t+\frac{y_{0}^{\prime}}{k} \sinh k t+\int_{0}^{t} f\left(t^{\prime}\right) \frac{\sinh k\left(t-t^{\prime}\right)}{k} d t^{\prime} $$ (c) Let \(f(t)=e^{-k_{0} t}\) in \((\mathrm{b})\) to obtain the result in part (a). What happens when \(k_{0}=k ?\)

Use Laplace transform methods to solve the ODE $$ L \frac{d y}{d t}+R y=f(t), \quad y(0)=y_{0}, \quad \text { constants } L, R>0 $$ (a) Let \(f(t)=\sin \omega_{0} t, \omega_{0}>0\), so that the Laplace transform of \(f(t)\) is \(\hat{F}(s)=\omega_{0} /\left(s^{2}+\omega_{0}^{2}\right) .\) Find $$ \begin{aligned} y(t)=& y_{0} e^{-\frac{\varepsilon}{L} t}+\frac{\omega_{0}}{L\left((R / L)^{2}+\omega_{0}^{2}\right)} e^{-\frac{k}{L} t}+\frac{R}{L^{2}} \frac{\sin \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \\ &-\frac{\omega_{0}}{L} \frac{\cos \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \end{aligned} $$ (b) Suppose \(f(t)\) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section \(4.5\) ) to find $$ y(t)=y_{0} e^{-\frac{k}{L} t}+\frac{1}{L} \int_{0}^{t} f\left(t^{\prime}\right) e^{-\frac{k}{L}\left(t-t^{\prime}\right)} d t^{\prime} $$ (c) Let \(f(t)=\sin \omega_{0} t\) in (b) to obtain the result of part (a), and thereby verify your answer. This is an example of an "L,R circuit" with impressed voltage \(f(t)\) arising in basic electric circuit theory.

Show that the inverse Laplace transform of the function $$ \hat{F}(s)=\frac{1}{\sqrt{s^{2}+\omega^{2}}} $$ \(\omega>0\), is given by $$ f(x)=\frac{1}{\pi} \int_{-\omega}^{\omega} \frac{e^{i x r}}{\sqrt{\omega^{2}-r^{2}}} d r=\frac{2}{\pi} \int_{0}^{1} \frac{\cos (\omega x \rho)}{\sqrt{1-\rho^{2}}} d \rho $$ (The latter integral is a representation of \(J_{0}(\omega x)\), the Bessel function of order zero.) Hint: Deform the contour around the branch points \(s=\pm i \omega\), then show that the large contour at infinity and small contours encircling \(\pm i \omega\) are vanishingly small. It is convenient to use the polar representations \(s+i \omega_{1}=r_{1} e^{i \theta_{1}}\) and \(s-i \omega_{2}=r_{2} e^{i \theta_{2}}\), where \(-3 \pi / 2<\theta_{i} \leq \pi / 2, i=1,2\), and \(\left(s^{2}+\omega^{2}\right)^{1 / 2}=\sqrt{r_{1} r_{2}} e^{i\left(\theta_{1}+\theta_{2}\right) / 2} .\) The contributions on both sides of the cut add to give the result.
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