Let's say a converter is sampling continuously at 48 kHz and the incoming waveform is a sine wave at 24 kHz... How does it faithfully recreate the waveform if the samples are taken at points that are 0?
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How does it faithfully recreate the waveform if the samples are taken a points that are 0?
Not at all. Things at $\frac{f_\text{sample}}2$ might just as well be constants. You can't tell [0 0 0 0] (sampled 24 kHz signal at zero-crossings) from [0 0 0 0] (sampled zero constant signal).
Nor can you tell [0.1 -0.1 0.1 -0.1] (sampled 24 kHz signal with amplitude 1 just a bit after each zero-crossing) from [0.1 -0.1 0.1 -0.1] (sampled 24 kHz signal of amplitude 0.1, sampled at its extrema!).
You need the sampling rate to be higher (not only just as high) than two times the highest signal frequency component for this to be unambiguous.
answered Aug 21, 2018 at 15:48
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$\begingroup$ That's what i was thinking also.. but after watching a video on why higher sample rates don't matter in audio production, i came away confused as the speaker seemed to suggest that samples not perfectly [1 -1 1 -1] can still be recreated perfectly at output. I'll leave link here for anyone how is interested (6:40): xiph.org/video/vid2.shtml $\endgroup$ Aug 21, 2018 at 16:02
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1$\begingroup$ Watch closely! 7:25: "if it differs even minutely, it's got frequency content at or beyond Nyquist…" So, having frequency components at Nyquist frequency introduces an error $\endgroup$ Aug 21, 2018 at 16:20
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That's why practical sampling operations that uses ADCs employ sampling frequencies about 10 percent higher (if not M times oversampling deliberately) than that's given by the theoretical limit of Nyquist rate which is twice the bandwidth. This is further supported by proper anti-aliasing filtering which would suppress enough at the stop band so that no significand content remains at the edges of the input signal's bandwidth of interest.
