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84) A vibroseis source sweeps acoustic signal in the frequency range 10 Hz – 100Hz. The maximum sampling interval to correctly recover the recorded signal will be _______ milli seconds.

(Thanks to Chandra Sekhar, ANU)

Solution:

Given the frequency range (f) = 10 Hz -100 Hz
Sampling interval(t_{s}) = ?

The sampling frequency (f_{s}) = 2 * f_{m}

Maximum frequency is given (f_m) = 100 Hz


Therefore The sampling frequency (f_{s}) = 2 * f_{m}

                                                                 = 2*100

                                                                 =200 Hz

The Sampling interval(t_{s}) = \frac{1}{f_{s}}

                                              = \frac{1}{200}

                                              = 0.005 sec

Sampling interval(t_{s}) = 5 milli seconds


Referance: https://community.sw.siemens.com/s/article/digital-signal-processing-sampling-rates-bandwidth-spectral-lines-and-more

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