Bandwidth with complex sampling

Question

On the transmit side, I have a 20 MHz carrier frequency carrying a signal with bandwidth of 40 MHz (so 0 Hz to 40 MHz, center at 20 MHz).

On the receive side, I have a dual channel ADC with each channel sampling at 40 MS/s (channel 1 samples at 40 MS/s, channel 2 samples at 40 MS/s). The hardware is setup to input I and Q into the two channels.

Am I able to properly sample the received signal, without aliasing? I saw that this was discussed before ("Complex sampling" can break Nyquist?), and the answer seemed to be "BW = fs because you get the negative frequencies", and I'm not 100% sure if the "negative frequencies" thing is a caveat, or I can just do the math as if I am getting 80 MS/s (which I technically am), so I drummed up this numerical example.

Answer 1

Complex sampling does not "break" Nyquist. IQ quadrature sampling produces twice as many bits per second of information (at the same sample rate for real or complex samples), and the 90 degree phase offset between the I and Q channel in those bits provides extra information about the spectrum.

One typical example to demonstrate aliasing is that the samples of DC and the samples of a sinusoid of a frequency at the sample rate look the same. A single channel of samples at 40 MHz would alias 0 Hz and a 40 MHz sinusoid together. Only if you strictly bandlimit the input spectrum to below Fs/2 in bandwidth would you prevent this aliasing. If you feed a strictly real data vector to an DFT, the result would be conjugate symmetric, half the FFT result would be redundant, and the upper half can be called the negative frequency image.

But with quadrature sampling (using a 40 MHz sampling clock plus a 90 degree phased shifted copy), a DC signal and a 40 MHz sinusoid would be clearly differentiated. I and Q would be the same for DC, but different for 40 MHz, e.g. if I is the same, the Q would be different, or vice versa. Therefore the spectrum between half the IQ sample rate and the sample rate will not be aliased with spectrum below, as is the case with single channel sampling. If you feed a DFT using the IQ data as a complex input vector, the lower and upper half of the result can be non-symmetric, thus represent independant information, thus twice the bandwidth (between 0 and Fs), compared to a strictly real DFT of data at the same sample rate.

Answer 2

To fully sample a signal of bandwidth F, Nyquist requires two samples per period.

One way to do this is to take real samples per period, to sample at 2F. This is the most common way the Nyquist sampling rate is expressed.

Another way to do this is to take a real sample and an imaginary sample every period. This is two independent samples per period. Stating it that way, it is clear that a complex sample is really TWO samples.

So you could say that Nyquist requires you to take 2 Real samples or 1 Complex sample per period. Both of these work out to two samples per period. Both of these give a full sampling of a signal band-limited to a bandwidth F=1/Period.