On Fri, Apr 9, 2010 at 01:25, Krishna S <krishna2...@yahoo.com> wrote:
> 1. I set the interpolation rate to 6. Does that mean baseband signal is > fed to DAC at the rate of 100MHz/6 = 16.6666666666 Msps? Not quite. It means that the FPGA is interpolating and creating 6 samples on the DAC bus for every sample that is sent to it over the GbE. Since the DAC is "consuming" samples at 100 Msps, the FPGA is consuming samples at 100 Msps/6 from the GbE. Thus, you need to generate (and being able to sustain) a sample stream on the host at 100 Msps/6, and by the Nyquist limit, it can contain up to ~16.666 MHz of spectral content. > 2. I want actually feed 15.36 Msps to DAC which is not possible as the > interpolation factor becomes non-integer. So do you suggest me to use > ‘resampler’ to convert 16.666666 Msps to 15.36Msps? The DAC always runs at 100 Msps. You need to understand this to know what needs to happen upstream. If what you really mean is that you have a host PC generated sample stream at 15.36 Msps and need to transmit it with the USRP2, then yes, you'd set the USRP2 FPGA interpolation to 6, then fractionally resample from 15.36 Msps to 16 Msps on the host. Depending on what the last DSP processing stage in your application is, you may be able to "fold" this resampling into the prior stage. For example, if the last step in your signal processing chain is a spectral shaping filter, like a root-raised-cosine filter, then you can reimplement this using a polyphase resampler and use the output filter taps there. This would combine the filter and resampling operation in one block and eliminate the need for a very high CPU resampling block. Where does the figure of 15.36 Msps come from? Johnathan _______________________________________________ Discuss-gnuradio mailing list Discuss-gnuradio@gnu.org http://lists.gnu.org/mailman/listinfo/discuss-gnuradio
