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From a hobbyist perspective, the sampling rate of an ADC must be at least twice of the input signal. As I was looking for an ADC to digitize a demodulated signal, I found that some ADCs like the AD4030-24 have 2Ms/s but 74MHz at −3 dB input bandwidth. How is this possible?

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3 Answers 3

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Consider the bandwidth specification as how fast a signal changes can be captured. And consider the sampling rate as how many times it can be captured per unit time.

Another way to think of it: You have your signal, run it through a low-pass filter (Fc = 74MHz), and run the output of that through a 2Ms/s ADC without a bandwidth limit.

From DC to around 74Mhz, there is no problem. Past that, the effective signal that gets sampled is now attenuated (presumably, 6dB/octave or 20dB/dec, for a first-order rolloff)

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  • \$\begingroup\$ So what you are saying is that this ADC operate in this frequency range (DC=>74MHz) and it is capable of sampling 2MHz per second step by step so 74÷2 = 37 seconds to fully sampl the full range specified? But because of Nyquist sampling theorem my bandwidth should be 37 MHz! \$\endgroup\$
    – Tintin
    Commented Jan 31 at 11:24
  • \$\begingroup\$ @Tintin your applying the theorem wrong. Also, "74÷2 = 37 seconds" is both mathematically and dimensionally incorrect. \$\endgroup\$
    – MOSFET
    Commented Jan 31 at 16:43
  • \$\begingroup\$ Can you corrected for me please? \$\endgroup\$
    – Tintin
    Commented Feb 1 at 11:59
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The formula is actually twice the bandwidth of the input signal, not twice the highest frequency. For example, a signal from 99 MHz to 101 MHz has a bandwidth of 2 MHz and thus requires 4 msps for Nyquist. In this case you would need an ADC with at least a 101 MHz bandwidth or else your signal would be attenuated by the ADC during sampling.

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Aliasing isn't just something annoying that happens; it's a part of the signal path like anything else.

The simplest case, and the most commonly presented, is when the signal of interest is baseband (zero to ±BW bandwidth), the sample rate is higher (Fs ≥ 2BW), and the Nyquist sampling theorem is applied.

In particular, aliasing can be used to advantage, say when frequency-conversion effects are desirable. For example, at a 1MHz sample rate, the aliasing sidebands repeat every 1MHz; if an RF pre-converter stage filters received signals into 1MHz bands, the ADC can sample that directly without additional conversion stages; the pre-stage provides image filtering.

Or suppose we have a quasi-periodic signal, from which we measure a timer trigger, measuring samples relative to it (how they are timed, doesn't matter, so long as we resolve the timing later in reconstruction). If we sample such that we reconstruct a time series of samples where each sample is a small increment with respect to the trigger event, then we have equivalent time sampling, effectively dividing a time-domain waveform at high frequency down to a much lower frequency, while preserving its harmonic content (i.e. the waveform, the (quasi-)periodic signal).

There is nothing wrong with either of these schemes, as, in the first example, the Nyquist reconstruction theorem is satisfied by the image filter (it doesn't matter which aliasing band it's placed in; baseband is merely a special case!), and in the latter, the passband is spread throughout the fundamental and harmonics, a narrow bandwidth at each point, the total equaling the acquisition bandwidth. (The tradeoff is, not being able to observe changes in the waveform from cycle to cycle -- hence, best suited for periodic or otherwise repetitive signals.)

We might also simply not care at all about reconstruction, and whatever aliasing happens, happens. Many oscilloscopes acquire and display data this way; often a sinc reconstruction filter is applied before visualization, but sometimes this gives hilarious results (ringing and overshoot where none is actually present). A control loop might prioritize response time over smoothness of response, and thus tolerate aliasing (the most common downside being, the control loop dominant pole ends up very close to the unit circle, i.e. prone to instability.)

Or we might be interested in the statistics, rather than the form of the signal itself; assuming we can avoid correlation, and thus spurious aliasing, a random sampling is as good as a full-bandwidth sampling for purposes of measuring the RMS for example. A power-line monitor could sample 50/60Hz mains at merely some ~Hz, if all it needs is a long-term (minutes to hours) representation of line voltage (or load current or whatever), to within only a modest accuracy (better the longer the acquisition window).

In any case, an ideal ADC has aliasing products from DC to light*, so the above can be done pretty much how ever you like; real devices however are band-limited in some manner or another, and eventually the aliasing bands fall away.

*Which, I suppose, as a corollary: an ideal, finite-impedance ADC has zero SNR, because the uncertainty of a truly band-unlimited signal is total, i.e. the Johnson noise integrated over all bandwidth is "yes". Clearly, it's pretty useful having an ADC of finite bandwidth, heh.

Physically what's happening is, consider an ADC of the type where the input pin is connected to a sampling capacitor via some analog switch resistance. The resulting RC time constant limits the bandwidth at which the sample node (capacitor) can follow the input. Related, there is a time constant due to the switch itself opening and closing (aperture), which is to say: if the input changes suddenly in the instant the switch is turning off, how much is the sampled value "smeared out" due to finite switch turn-off speed? This limit is also typically near the bandwidth; though they do have different causes and effects.

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  • \$\begingroup\$ "the most commonly presented, is when the signal of interest is baseband (zero to ±BW bandwidth), the sample rate is higher (Fs ≥ 2BW), and the Nyquist sampling theorem is applied" in my case => Baseband:0.5115, 1.023, 2.046, 2.5575, 5.115, 10.23 Subcarrier frequency:6.138, 15.345 so i need a 30 to 40 Ms/s ? \$\endgroup\$
    – Tintin
    Commented Jan 31 at 11:12
  • \$\begingroup\$ Those frequencies in MHz levels \$\endgroup\$
    – Tintin
    Commented Jan 31 at 11:18
  • \$\begingroup\$ I don't know what you're doing. You'll need at least enough bandwidth to sample the carriers plus sidebands, and depending on channel spacing, modulation, etc., you may find it easier to just sample the whole thing (>40MS/s). \$\endgroup\$ Commented Jan 31 at 11:26

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