[hpsdr] Question about I and Q position in protocol

[Alberto I2PHD]

On 12/3/2012 10:40 PM, John Marvin wrote:

> /I have carefully followed the path and believe that bufl was the "left"
> sample from Hermes and bufr was the "right" sample from Hermes, and here
> the code is assigning the left sample to the imaginary component and the
> right sample to the real component.  The protocol document indicates
> that left is the I (in phase or real) sample and right is the Q
> (quadrature out of phase or imaginary) sample. So, am I missing
> something, or should the I and Q positions be switched in the protocol
> document?/

I am not familiar with PowerSDR, but if they coded it according to FlexRadio specs, it has
I and Q reversed with respect to the de-facto standard, i.e. I on left channel and Q on right.

Flex decided that "I" should be on the Right channel, and "Q" on the left, for reasons not known to me....
In my software, when I read a recorded file coming from Flex (easily recognized, and they
save the I/Q data as 32-bit floats, and they are the only ones to do this...) I automatically
activate the I/Q channel swap...


-- 
/*73 Alberto I2PHD*/
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