When we have a data source oscillating at a frequency of say 4KHz , this implies that the Nyquist sampling rate would be at least 8KHz. We know that the inverse of the signal frequency is the time it takes for a period to complete, i.e 1/4KHz = 250µs, but, what if we take the inverse of the Nyquist frequency, we get this 1/8000 = 125µs, which is half the period of the original signal. what does that give us?, initialy I was thinking it is the time when each of the sampling pulse is sent, but it doesn't make any sense that only two sampling points would be gotten in a cycle of the source signal, I need help please.
Electronic – Sample and Hold
multiplexer
Best Answer
As you're noticing, Nyquist really speaks to the minimum sampling frequency for a given Bandwidth. You often get people who broadly apply the 2X+ factor and take it no further. But as you noticed there are other effects, what about the phase alignment in sampling? (which will affect amplitude), what about aperture size (the time period of sampling). etc.
What is important is that you need to understand what it is about your waveform that is the salient features. Say for 8Khz signal you really care about the shape of the waveform, which could very well mean that you have to go up to the 3rd or 4th harmonic. In this case then you might want to over sample by 4X2.1 = 8.4X (I use 2.1 as an example simply because Nyquist says greater than but not equal to.
Or conversely is the 8Khz a carrier with a modulation at 100Hz, then you could down convert with a mixer and sample at ~ 250Hz or you can sample at 16KHz and do the down conversion in the digital domain ending up with a 250 Hz data rate.