Discussion
The inverse Fourier transform of \(e^{-|k|}\) is :
Answer: A
The inverse transform is \(\frac{1}{\pi}\frac{1}{1+x^{2}}\).
With the convention \(f(x) = \frac{1}{2\pi}\int F(k) e^{ikx} dk\), split the integral at \(k = 0\):
\[ \begin{aligned} f(x) &= \frac{1}{2\pi}\Big[\int_{0}^{\infty} e^{-k(1-ix)}dk + \int_{0}^{\infty} e^{-k(1+ix)}dk\Big] \\ &= \frac{1}{2\pi}\Big[\frac{1}{1-ix} + \frac{1}{1+ix}\Big] \\ &= \frac{1}{2\pi}\cdot\frac{2}{1+x^{2}} = \frac{1}{\pi}\frac{1}{1+x^{2}} \end{aligned} \]
This is the Lorentzian (Cauchy) profile. A Gaussian \(e^{-x^{2}}\) is the transform of a Gaussian, not of \(e^{-|k|}\); \(e^{-|x|}\) simply copies the input; and \(1/(1+x)\) is not even an even function, while the transform of an even function must be even.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development