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

Upward continuation of a potential field is a simple multiplication in the wavenumber domain: each spatial Fourier component of the field is attenuated according to
\[F_u(k) = F(k)\,e^{-kh}\]
where \(k\) is the angular wavenumber (in radian/km, as specified) of that component and \(h\) is the height gained (here 5 km). Since the attenuation factor \(e^{-kh}\) decreases as \(k\) grows, the STRONGEST (maximum) attenuation of the dataset happens for the shortest-wavelength (highest wavenumber) component that the sampled profile can actually carry - the Nyquist component.

The profile is 100 km long and is represented by 20 discrete samples \(\Delta T_1,\ldots,\Delta T_{20}\), so the station (sample) spacing is
\[\Delta x = \frac{100\ \text{km}}{20} = 5\ \text{km}\]

The highest (Nyquist) angular wavenumber that this sampling can resolve is
\[k_{max} = \frac{\pi}{\Delta x} = \frac{\pi}{5}\ \text{rad/km}\]

Putting this into the upward-continuation attenuation formula for \(h = 5\) km:
\[e^{-k_{max}h} = e^{-(\pi/5)(5)} = e^{-\pi}\]

So the maximum attenuation suffered by the data on being continued upward to 5 km is \(\boxed{e^{-\pi}}\), option (B).