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These topics are full of words I don't fully understand. So it's very possible I missed the point somewhere but I saw a feature on a microcontroller and wondered if it was applicable.

I have an ADC with several channels and they can be command to sample at a set phase offset from each other, say 90 degrees. Does sampling at a frequency and at 90* phase offset produce IQ data?

It feels like it does when I consider a sine wave but it's effect on more complex wave forms confuses me.

Questions which are reasonable:

Why not do it in analog? To save on parts

Why not do it digitally? To save on processor time.

foreverska
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Sampling at a 90 degree phase offset will only produce quadrature signals for frequencies that are narrowband and centered on the sampling rate. The phase offset between all frequencies within the Nyquist bandwidth of DC to $f_s/2$ will vary from 0 degrees at DC to 45 degrees at $f_s/2$. (You can intuitively verify this by considering very low frequency sinusoids sampled with such a clock, and what would happen as the frequency of the sinuosoid increased).

You can do quadrature sampling with a single ADC when the modulation signal is significantly oversampled and a given quadrature phase error is acceptable. In general given a fixed delay between samples $T$ in seconds, the phase between the samples for a signal at frequency $f$ in Hz is $-2\pi f T$ radians. The delay in the sampling clock can be chosen such that at the carrier frequency of the modulated signal, the phase is 90 degrees, however it will only be 90 degrees at this one frequency and consistent with a fixed delay, will vary in phase vs frequency (similar to the example above, and as given with the formula provided), while for quadrature demodulation we require the phase to be 90 degrees over all frequencies within the signal's bandwidth. The amount of phase variation (which leads to quadrature error) over the bandwidth of the signal is dependent on the amount of oversampling provided and what digital IF frequency the signal is centered at.

Dan Boschen
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  • I'd like to accept this answer but it would be interesting to know how narrowband we're talking. I found the Tayloe Quadrature Product Detector which seems really similar to my concept although uses 4 divisions. Is what you're talking about also inherent in this design and they've accepted that cost? – foreverska Sep 14 '22 at 03:28
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    Yes in the Tayloe Quadrature Product Detector, the carrier frequency is significantly wider than the signal bandwidth such that a 90 degree phase shift and 1/4 cycle delay are virtually identical. Ultimately some distortion will be introduced (and measured as EVM) but it may be of no concern; it depends on the waveform quality requirements if this is acceptable or not. To get a sense of the error, simply consider the Fourier Transform of a time delay (as $e^{j2\pi f T}$) and from that you can see the error from true quadrature versus a frequency offset (phase term is $2\pi f T$). – Dan Boschen Sep 14 '22 at 03:56
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    (I'd like to remove the first "No" in my answer as it is indeed feasible under such narrow band conditions as I describe) The No meant it is not the same as quadrature, but under narrow band condition the phase deviation from quadrature can well be under a limit of concern. – Dan Boschen Sep 14 '22 at 04:03
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I'd suggest writing down some equations to make the concepts clearer, since you don't seem to have a clear idea of what you want/need to do.

Some notes: since you want to demodulate, we need to assume the existence of a quadrature-modulated signal $$s(t) = s_I(t)\cos(2\pi f_c t)-s_Q(t)\sin(2\pi f_c t),$$ with a certain bandwidth and a certain carrier frequency.

This signal can be demodulated in multiple ways. One simple method is:

  • Bandpass-filter $s(t)$ to remove potential interference.
  • Multiply $s(t)$ by quadrature carriers to down-convert it.
  • After each of the two multipliers, low-pass filter the signals to recover $s_I(t)$ and $s_Q(t)$.
  • Sample the output of each LPF to obtain I/Q samples and continue processing in the discrete-time domain.

Try to write down the equations for each step. Then, think about the effect of delaying the sample clock of one of the ADCs. What effect does that have, and how does it help you?

MBaz
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  • The math is a bit beyond me. So I have to approach this intuitively. Let's say I have a 10Mhz CW signal and I'm sampling at 10Mhz. Assuming no jitter in either I will sample at the same point in the waveform every time. The effect is an unchanging ADC. Move the signal up by 1Khz and the sample point moves along the wave tracing a 1khz signal. Sample at a 90 degree offset and you get an out of phase 1Khz signal that does(?) the characteristic Q phase flop depending on if the signal is higher or lower than the sample rate. – foreverska Mar 03 '22 at 01:03
  • Look up "bandpass sampling" -- that may help you. – MBaz Mar 03 '22 at 02:14