>From: Nurkkala Eero.An (EXT-Offcode/Oulu) >Sent: 19 February, 2010 21:04 > >> Oops, I seemed to copy Q1.14 instead of Q14.1, but isn't it >still the same. >> When I added some debug messages it still seemed to be corret. >> >> Lowest number (-32768) is represented with 16th bit '1' and >the rest are >> zeros, right? That is 0x8000. >> -1 has all the bits set (0xffff) and 0 has all the bit >cleared (0x0000). >> Highest positive value has 16th bit cleared and the rest set >(0x7fff). >> >> Or did I interpret something wrong? >> >> Cheers, Ilkka > >I guess it's just fine, but let's see. Maybe I was lost in the >Q1.14: m + n + 1 ~ >binary: [sign bit, (m), (n)] where m is the integer portion, 0 >or 1, n is 14 bits.. >so if your input was [-32768... 32767] -> [-2,2] then, for example, >-32768 is in hex: 0x8000, but the 2nd most significant bit is >zero, which means >the integer portion (m) is not 1, which makes me doubt the >gain -32768 is >actually -1 (or 0), not -2. But then, as it's a two's >complement, it maybe just correct. > >So most likely it's just fine; I just had a thinko. Ok, now I got your point :) However, as far as I can understand the Q notation right, it basically defines how to interpret the value. That is, where is the decimal point. Thus, I would say that the code should work fine and based on quick tests, the gain control seems to behave ok too. Cheers, Ilkka-- To unsubscribe from this list: send the line "unsubscribe linux-omap" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
