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I have a frequency signal range from 25 Khz to 35 Khz. I know that the the minimum signal sampling to assure reconstruction of the signal from the samples is 70 Khz.

Would it be correct to assume that using the Nyquist's sampling criterion, the maximum data rate that the signal can carry for the 10 Khz bandwidth be 700 bits per second?

Do I multiply the samples per second (70) by bandwidth, which yields 700 or by 8 bits per sample?

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  • \$\begingroup\$ Minimum sample rate is 2*Fmax = 70kHz. Maximum baud rate (NOT data rate) is bandwidth/2 = 5kbaud. \$\endgroup\$ – markt Dec 13 '14 at 1:34
  • \$\begingroup\$ I forgot to edit the minimum sampling rate. Updated... I'm confused on how to correctly get the data rate. \$\endgroup\$ – Rick Dec 13 '14 at 1:40
  • \$\begingroup\$ Would I need to apply the following formula? D = 2 B log2K If so, I know B is the bandwidth, but not sure what K (signal levels) is. \$\endgroup\$ – Rick Dec 13 '14 at 1:47
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The minimum sampling rate to enable reconstruction of a bandlimited signal is related to the bandwidth of the signal, not the highest frequency of the signal. If a signal spans the frequency range from near DC to a maximum frequency, say Fmax, then the bandwidth of the signal is equal to the highest frequency and the minimum sampling rate is indeed 2Fmax. But if the signal starts at a frequency other than near DC, then the minimum sampling rate need only be twice the bandwidth. In your case, since the signal spans from 25 kHz to 35 kHz, the bandwidth is only 10 kHz and the minimum sampling rate is only twice that or 20 kHz. The maximum data rate that can be sent over a given bandwidth is determined by 2 factors: one, the bandwidth and two, the signal-to-noise ratio. The relationship is given by the Shannon formula: C = Blog(1+SNR) where C is the maximum bit rate, B is the channel bandwidth, SNR is the power signal-to-noise ratio, and the log is to base 2.

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    \$\begingroup\$ Your 20kHz sample rate proposition is only true if the data-carrying component of the signal has been shifted down so that sampling it at 2*B will recover the original signal - as per your edits. In this case, with no such frequency shift specified, the OP must sample at 2*Fmax or he'll get garbage. \$\endgroup\$ – markt Dec 13 '14 at 2:21
  • \$\begingroup\$ How can I calculate the Signal to Noise Ratio with the given values? \$\endgroup\$ – Rick Dec 13 '14 at 2:27
  • \$\begingroup\$ @markt, no, he's right. The signal can be reconstructed from samples at 20 kHz, however the reconstruction process will not just be a sinc filter and you need to have a priori knowledge of what band the signal occupies. \$\endgroup\$ – The Photon Dec 13 '14 at 16:23

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