# Sampling rate versus number of sample of I and Q

I am acquiring data of an O-QPSK signal (chip rate : 2Mchip/s) with a SDR. I am undersampling the signal at 20 Mhz (carrier is at 2.8 GHz) so I respect the Shannon-Nyquist criterion that you need to sample 2 time the bandwidth of your signal, however when I take apart the I and Q phase, the signal look like it has been sampled at 10 Mhz. I observe this when I create my reference signal in matlab with a sampling rate of 20 Mhz for each phase I&Q. I end up with a signal x2 time the size of the real signal (2560 samples vs 5120 samples).

The chip period of my signal is 1 microsecond and I am sampling at a rate of a rate of 20 Mhz with 128 in my sampling period, you effectively end up with 2560 samples.

When you take each phase alone, each part of a chip will be 2 time the period chip which result in a signal of 2 microsecond, therefore is we don't divide the sampling rate by 2, we will end with 5120 sample.

This effect seem to be produce by the phase offset that the O-QPSK use. Can someone give an explanation about this phenomenon?

[EDIT]

The SDR I am using is the B200 Mini from Ettus Research.

• wait, why do you think the signal is undersampled? – Marcus Müller Jun 8 '16 at 20:34
• i.e. what's the bandwidth you're observing? most SDR frontends I'm aware of won't allow you to undersample without intrusive modification of the driver. – Marcus Müller Jun 8 '16 at 20:35
• are these 20 million complex equivalent baseband samples per second, or are are these 20 million real-valued samples from some superhet receiver? – Marcus Müller Jun 8 '16 at 20:36
• How exactly are you recording? Command lines / device configurations and the actual device you use might be helpful! – Marcus Müller Jun 8 '16 at 20:47
• @MarcusMüller undersampling, well the highest frequency of my signal was 2.4GHz + Bandwitdh. Sampling at 20Mhz don't respect the Nyquist theorem but it respect the Shannon-Nyquist theorem. I use the b200 mini USRP – MathieuL Jun 9 '16 at 13:34