On Wed, May 27, 2009 at 11:35 PM, Karthik <karthik1...@gmail.com> wrote: > On Wed, May 27, 2009 at 9:18 PM, Firas Abbas <firasmail2...@yahoo.com> wrote: >> >> Hi, >> >>> On Wed, 5/27/09, Karthik <karthik1...@gmail.com> wrote: >>> >>> If you use usrp_c() source, then the real part of the complex values are >>> the I values and the imaginary parts are Q values. They are each 16 bits. >>> >>> Karthik >>> >> >> That is not correct. If you use usrp_c source, then I & Q each are 32 bits >> (total complex is 64bits). However, if you use usrp_s, then each are 16 bits >> (total complex is 32bits).. >> >> Best Regards, >> >> Firas >> > > Apologies for the wrong information. I don't mean to side track the > original query, but I have a couple of questions. > > I found this in the FAQ. > > "Regarding the bandwidth, we can sustain 32MB/sec across the USB. All > samples sent over the USB interface are in 16-bit signed integers in > IQ format, i.e. 16-bit I and 16-bit Q data (complex) which means 4 > bytes per complex sample. This resulting in a (32MByte per sec/4Byte) > 8Mega complex samples/sec across the USB." > > So I guess here we are assuming that we are using usrp_s() ? > > Also, since the ADC only has 12bit precision, are the extra bits used > to retain precision as we go through the CIC and Halfband filters? > > If I were to use the std_4rx_0tx.rbf and usrp_s() to receive 4 > channels. Then am I correct in saying that I need to have a > deinterleaver with 8 channels after the usrp_s() from which I will > club channel 0 and 1 into a new complex stream (similarly for channel > 2 and 3 etc) to get 4 complex streams? > > Thanks, > Karthik >
After snooping about a bit more, I found this on "The USRP under a 1.5X Magnifying Lens" , http://gnuradio.org/trac/attachment/wiki/UsrpFAQ/USRP_Documentation.pdf. "Q) ADC samples at 64MHz, and passes through both I and Q channels over the 24-bit RX bus. Internal to the FPGA, the CIC automatically decimates by a value of at least 4. The halfband decimating FIR filter internal to the FPGA decimates by a fixed value of 2. This gives a minimum decimation rate of 8, leaving 8Msps going over the USB of the USRP. Is this correct? A) Yes. Note that some FPGA builds don't contain the half-band. With 16-bit I & Q decim = 8 -> 8MS/sec -> 32MB/sec. With 8-bit I & Q decim = 4 -> 16MS/sec -> 32MB/sec " Does this mean that when we use usrp_c(), the data sent over USB is 16bit I and Q but the container used to hold it after it reaches the computer is 32 bit each. And when we use usrp_s(), the data sent is 8bit I and Q and the container used is 16bit? Karthik _______________________________________________ Discuss-gnuradio mailing list Discuss-gnuradio@gnu.org http://lists.gnu.org/mailman/listinfo/discuss-gnuradio
