"Complex sampling" can break Nyquist?

Question

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?

Answer 1

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).

Answer 2

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!

Answer 3

You are correct about the sampling rate, but this isn't a violation of Nyquist's bound.

Nyquist identified the maximum rate at which distinct values can appear in a signal. This is also the minimum sampling rate, since each distinct value needs to be sampled. It's simpler here to discuss the maximum signaling rate.

For a long complex-valued signal, the width of the range of frequencies allowed in its complex Fourier sum equals the maximum signaling rate. This is easy to prove if you assume the signals are periodic.

This bound holds regardless of whether the distinct signal values are real or complex. In the special case where all distinct values are real, there is frequency symmetry and we define the bandwidth to be just the width B of the range of positive Fourier frequencies. Thus the width of the full range used is 2B.

So I would say that the Nyquist bound doesn't actually change for complex signals, the definition of bandwidth does.

Answer 4

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.

Answer 5

A complex-valued signal $a \left( t \right) + j b \left( t \right)$ has half the one-sided bandwidth, $B$, of a real-valued signal $a \left(t \right)$ only if the imaginary parts of the complex values are the Hilbert transform of the real part, $b \left( t \right) = H \left[ a \left( t \right) \right]$.

This causes the complex-valued signal to be a single-sideband version of the real-valued one. (That happens when the real-valued signal is up-converted to center-frequency $B$, as $a \left( t \right) \cos \left( 2 \pi B t \right)$. The imaginary part is then $b \left( t \right) = -\sin \left( 2 \pi B t \right) a \left( t \right)$. Added together, they form a single-sideband signal, each sideband having bandwidth $B$).

In general, for any complex-valued signal, the real and imaginary parts are not so related, so the one-sided bandwidth of each is $B$. That is, the bandpass signal has a pass-band width of $2 B$ since it is not SSB in general. If then a given bandpass signal has bandwidth of $B$ only, each part (real, or in-phase, and imaginary, or quadrature-phase) at baseband will have a corresponding one-sided bandwidth of $B/2$ and can be sampled at it Nyquist rate $B$. (Since there are $2$ parts, the aggregate number of samples taken per second is still $2 B$).