• Red noise source was input to box 103 sp0.
• The input noise rms was set to 48 (via pnet --sigstat).
• 16 bit time domain sampling of polA was used.
• Decimations of 4,8,16, and 32 were used.
  
• hr_shift was kept constant at 10.
  
• This sent the 16 bit output of the digital low pass filter (dlpf) out to disc.
• the gain of each filter used was about .62.
• Data was taken for .1 seconds. This gave about 1e6 time samples.
  

• Page 1: 500 time samples of the 16 bits data for the various decimations. Black is digI and red is digQ. The vertical scale if 16 bit counts. The rms values for the entire 1e6 samples is printed in the upper left of each plot.
• Page 2: Histogram of the decimate by 4 data. The bottom plot is a blowup of the histogram.
•  Each bin is separated by 16 counts --> the bottom most 4 bits of the 16 bit number are always 0.
• Page 3: The measured rms for each decimation   divided by the sqrt(decimation).
  
• This is constant for the various decimations showing that the noise does increase by the sqrt(decimation).
  
• The top plot uses the entire 16 bits. The bottom plot divides by 16 to look at the 12 bits of actual data.
• Dividing by the sqrt(dec) gives the rms with no decimation. The bottom plot is 8. We started with an rms of 48. The filter gain is .62 so we would expect an rms/sqrt(dec) of 30 not 8.

Conclusions:

• The rms increases as the square root of the decimation.
• The algorithm for hr_shift to keep the same number of bits on the noise should be:
  
• hr_shift= 10 - log2(sqrt(decimation))
• The hi-res bits are in the upper most 12 bits of the 16 bit output.
•  The lower 4 bits are 0.
  
• This may be happening because the data is traversing a  12 bit bus somewhere in the fpga.
• When hi res spectra are taken, the upper most 12 bits are passed to the polyphase filter bank.
  
• The measured rms:8 (with no decimation) is smaller than the expected rms (48 measured*.62filter=30). I'm not sure where this discrepancy is coming from.
  
• The pshift parameter in the butterfly stages should work the same way since the butterfly  is a multiply add.
  
• The noise will grow as the sqrt(numAdds).
  
• Every two stages of the butterfly there should be a down shift.
• There are an odd number (13) of stages in an 8K butterfly.
  
• Since you want 1 shift every 2 stages, you can not maintain the same rms through the butterfly's:
• 7 shifts (0x1555) will decrease the rms by sqrt(2)
• 6 shifts (0x0aaa) will increase the rms by sqrt(2).
• You should increase(7),decrease(6) the input rms by sqrt(2) depending on the number of shifts used.
  

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