Question

\(\Delta T = [\Delta T_1, \Delta T_2, \ldots, \Delta T_{19}, \Delta T_{20}]\), and \(\Delta T_u = [\Delta T_{u1}, \Delta T_{u2}, \ldots, \Delta T_{u19}, \Delta T_{u20}]\) denote the magnetic data observed at heights 0 km and 5 km, respectively, along a profile of length 100 km. What is the maximum attenuation at a height of 5 km?
[Use wave number in radian/km].

• A. \(e^{-\pi/2}\)
• B. \(e^{-\pi}\)
• C. \(e^{-2\pi}\)
• D. \(e^{-3\pi}\)

Hint:

Upward continuation attenuates each Fourier component by \(e^{-kh}\); maximum attenuation occurs at the Nyquist wavenumber set by the 20-sample, 100 km profile spacing.

Answer 1

The Correct Option is B

Instead of working with wavenumber directly, work with the shortest wavelength the profile can carry, then convert.

With 20 samples spread over a 100 km profile, the sample spacing is \(\Delta x = 100/20 = 5\) km. By the sampling theorem, the shortest wavelength that can be represented without aliasing is twice the sample spacing:
\[\lambda_{min} = 2\Delta x = 10\ \text{km}\]

The corresponding (largest) angular wavenumber is
\[k_{max}=\frac{2\pi}{\lambda_{min}}=\frac{2\pi}{10}=\frac{\pi}{5}\ \text{rad/km}\]

which is identical to the Nyquist wavenumber used above. This shortest-wavelength component is the one that loses the most amplitude on upward continuation, since higher spatial frequencies decay fastest with height. Evaluating the standard upward-continuation attenuation \(e^{-kh}\) at \(k=k_{max}=\pi/5\) rad/km and \(h=5\) km:
\[e^{-(\pi/5)\times5}=e^{-\pi}\]

This confirms, via the wavelength route, that the maximum attenuation at 5 km height is \(\boxed{e^{-\pi}}\), option (B).