From: ChrisAtUpw@.......

Date: Tue, 8 Feb 2000 09:36:57 EST

Hi there Barry, B> I have been running the 16 bit Burr Brown A/D at an oversampling rate of=20 30 readings per sample. Someone mentioned on a previous listing that=20 oversampling was the equivalent of a higher resolution A/D.=20 If you had a constant signal with no noise and a perfect A/D giving 16=20 bits, calling the converter would produce the same integer number n every=20 time. Averaging would produce the same integer. If you have a constant signal, but up to two bits of noise on the=20 converter, you will get a series of 16 bit readings of n-2, n-1, n, n+1, n+= 2=20 in random order. There will be more n readings than either n+1 or n-1 etc.=20 --- you get a statistical distribution and you are down to something like 14= =20 bit confidence accuracy on one reading. This is not good news. To halve the=20 average error on one single reading, you have to average four readings=20 together. For each factor of two, you need to average four times the previou= s=20 number --- to get 1/4 the average error, takes 16 readings. =20 If you add 16 off 16 bit readings together you get a 20 bit integer=20 number. You can a) use this number b) convert to 'real' numbers and divide b= y=20 16 c) stay with integers, div 16 and loose any extra accuracy of a size less= =20 than 1 bit and accept the 'rounding error' on the last bit d) stay with=20 integers, div 4 and use the 18 bit integer --- the last two or three bits of= =20 the 20 bit number will be noise that you don't want to know about anyway. =20 B> I agree with the consept that multiple readings will give one a more=20 accurate sample value but I wonder how I would get more than a count of =20= =B1=20 32768 (higher than 16 bit)?=20 The only way to represent an accuracy of less than 1 bit with a 16 bit=20 integer is to use a larger total number, 17, 18 bits etc.=20 If you have a couple of bits of noise on a 16 bit A/D, you can get more=20 than '16 bit accuracy' by taking and averaging enough readings 128, 512 ??=20 This can take a long time and you also have to ask yourself if your signal i= s=20 going to stay steady for this length of time. You need to have a filter on=20 your analogue input line which rejects any signal or noise of a frequency=20 greater than your sample time.=20 In the real world, it can be an advantage to average enough readings to=20 be reasonably certain of getting an accuracy within the last bit of the=20 converter, but beyond this, it starts to take a lot of time and effort and=20 you are usually better off using a higher accuracy A/D converter. When you have a lot of mains hum on the signal, which maybe changes from= =20 time to time, another technique which may be worth considering is to take an= d=20 average enough samples to exactly cover one mains cycle, but this does limit= =20 the sample rate to less than the mains frequency per second. If you have 60=20 Hz mains and a 20 mu Sec sample rate, you could set the coverter to sample=20 say eight channels in succession, 84 times. The average contribution to all=20 channels from mains hum over this period will be zero. In calculating the=20 sample rate, you have to take into account set-up times and settling times,=20 or you may also loose accuracy that way. Sampling eight channels with=20 suitable pauses 20 times equally spread out over a single cycle would=20 probably be better.=20 =20 Hope that this makes things clearer. Regards Chris _____________________________________________________________________ Public Seismic Network Mailing List (PSN-L)

Larry Cochrane <cochrane@..............>