I have heard anecdotaly that sampling complex signals need not follow Nyquist sampling rates but can actually be gotten away with half Nyquist sampling rates. I am wondering if there is any truth to this?

From Nyquist, we know that to unambiguously sample a signal, we need to sample at least higher than double the bandwidth of that signal. (I am defining bandwidth here as they do in the wiki link, aka, the occupancy of the positive frequency). In other words, if my signal exists from -B to B, I need to sample at least > 2*B to satisfy nyquist. If I mixed this signal up to fc, and wished to do bandpass sampling, I would need to sample at least > 4*B.

This is all great for real signals.

My question is, is there any truth to the statement that a complex baseband signal (aka, one that only exists on one side of the frequency spectrum) need not be sampled at a rate of at least > 2*B, but can in fact be adequately sampled at a rate of at least > B?

(I tend to think that if this is the case this is simply semantics, because you still have to take two samples (one real and one imaginary) per sample time in order to completely represent the rotating phasor, thereby strictly still following Nyquist...)

What are your thoughts?


Your understanding is correct. If you sample at rate $f_s$, then with real samples only, you can unambiguously represent frequency content in the region $[0, \frac{f_s}{2})$ (although the caveat that allows bandpass sampling still applies). No additional information can be held in the other half of the spectrum when the samples are real, because real signals exhibit conjugate symmetry in the frequency domain; if your signal is real and you know its spectrum from $0$ to $\frac{f_s}{2}$, then you can trivially conclude what the other half of its spectrum is.

There is no such restriction for complex signals, so a complex signal sampled at rate $f_s$ can unambiguously contain content from $-\frac{f_s}{2}$ to $\frac{f_s}{2}$ (for a total bandwidth of $f_s$). As you noted, however, there's not an inherent efficiency improvement to be made here, as each complex sample contains two components (real and imaginary), so while you require half as many samples, each requires twice the amount of data storage, which cancels out any immediate benefit. Complex signals are often used in signal processing, however, where you have problems that map well to that structure (such as in quadrature communications systems).

  • $\begingroup$ Right, and just to be clear, we can then say that we can sample at a rate of at least > B for each branch, real and imag, but we would still have taken 2 samples in that sampling time so technically speaking it is still 2 samples/ sample time. I suppose from say, an A/D perspective, we only need to run 2 A/Ds at sample rate at least > B , with one of them for the real, and the other for the imag. $\endgroup$ – Spacey Nov 20 '11 at 20:05

There is also a simple approach to explain this: If you real baseband signal has a spectrum from -B to +B you sample with 2B, so you make sure that the spectral repetitions of the spectrum don't overlap. A overlap would mean that you get aliasing and cannot reconstruct the original spectrum.

Now with a complex signal, the spectrum ranges, as mentioned by Jason, from 0 to B. (Theoretically it can also have spectrum at negative frequencies, but for most of practical cases it will range from 0 to B.) If you sample with rate B, since there is are no parts at negative frequencies in original spectrum, the repetitions of spectrum will not overlap --> unambiguous reconstruction is possible!


Since sines and cosines can be expressed as sums of complex exponentials, you can always think of any signal as complex. For real valued signals it just works out that the imaginary parts cancel.

If the sines and cosines in a real-valued signal use frequencies from 0 to B, the corresponding sum of complex exponentials will include frequencies from -B to B. The Nyquist signaling rate of 2B is really just the full width of frequencies, taking both positive and negative frequencies into account.

In general, for any complex valued signal, the full width of frequencies equals the maximum rate at which chosen complex values can appear. Choosing only real values is a special case.


I'd say it's a qualified 'No', in the sense that the number of individual real samples hasn't properly been clarified, along with the purpose for choosing the signal digitisation rate.

First, all real world signals are Real, rather than complex. That is, any time we are faced with a complex representation, we actually have two (real) data points, which should be factored in to the 'Nyquist' limit.

The second issue is 'negative frequencies', as perceived from baseband. Almost all sampling teaching is from a baseband perspective, so the frequencies tend to be 0..B, which is then sampled at fs. The negative frequencies are sort of ignored (using the complex conjugate identity).

It is possible to think of the baseband signal as if it's being modulated at zero frequency, however starting the carrier modulation at the nominal fs/2 point can be illuminating, as we then see the two sidebands, and the (mathematical) complex term from the carrier. The previously negative frequency has shifted. And we may no longer have the complex conjugate identity.

If the complex conjugate identity has been eliminated we no longer have the frequency folding, and we have a simple wrap around aliasing.

So if we have an HF real signal being sampled to provide demodulation of the complex representation, without folding, we in some sense end up with an fs/4 bandwidth (+/-B). For every 4 data samples (0, 90, 180, 270 deg) we output two values which represent the in-phase (0 - 180) and quadrature (90 - 270) components of the overall complex sample.

In a fully complex world, if the signal is complex, the sampling frequency is complex, resulting in twice the terms. It depends on what mathematical features you need out of the sampled signal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.