On 11/11/2014 7:14 PM, robert bristow-johnson wrote:> On 11/11/14 4:29 PM, rickman wrote: >> On 11/11/2014 3:29 PM, robert bristow-johnson wrote: >>> On 11/11/14 1:53 PM, rickman wrote: >>>> On 11/11/2014 12:00 PM, Tim Wescott wrote: >>>>> On Tue, 11 Nov 2014 02:23:43 -0500, rickman wrote: >>>>> >>>>>> On 11/10/2014 8:37 AM, sharanbr wrote: >>>>>>> Hello All, >>>>>>> >>>>>>> I am reading DFT topic where quantization effect is described. >>>>>>> >>>>>>> One of the example in the book calculates the number of bits >>>>>>> needed to >>>>>>> compute the DFT of a 1024 point sequence with a SNR of 30 dB. >>>>>>> >>>>>>> Here b = 15 bits. My question is whether the size of the sample size >>>>>>> does not matter at all for computation of the DFT? >>>>>>> >>>>>>> If input is composed of 24 bits (for example), would number of bits >>>>>>> used for DFT internally would be independent of 24-bit size? >>>>>> >>>>>> I think you are saying you wish for the output spectrum to have 30 >>>>>> dB of >>>>>> SNR. To achieve that you need to consider the size of your FFT. My >>>>>> understanding is that you add about 1/2 bit of noise from roundoff >>>>>> error >>>>>> with each pass of the FFT. So a 1024 point FFT will have 10 passes >>>>>> and >>>>>> add 5 bits. To get 30 dB out (ENOB = 5) you would need to start with >>>>>> input data with 10 bits of significant data or 60 dB of SNR. You >>>>>> would >>>>>> need to do all calculations with at least 10 bits of precision. >>>>> >>>>> I think you're doing your computation backwards. >>>>> >>>>> Each bin of an N-point FFT acts like a bandpass filter, N points >>>>> long. So >>>>> for white noise at the input, the achievable SNR at each bin is (input >>>>> SNR) >>>>> + 3dB * N. >>>>> >>>>> So for a 30dB _input_ SNR you could expect (at best) a 60dB _output_ >>>>> SNR. >>>> >>>> You are talking about the averaging of the input noise. I am talking >>>> about the cumulative effects of round off error in each stage of the >>>> computation. >>> >>> >>> yup, but the only way doing the DFT with the simple dot-product can make >>> it better than the FFT is when your accumulator is much wider than the >>> original words (usually twice as wide) and you round (or dither and >>> round or noise-shape and round or all three) *once* at the end of the >>> summation. then the naive DFT will do better than the FFT. but i don't >>> think so otherwise. >> >> Why did you start talking about the DFT???? >> > > uhm, because those letters live in the Subject: field of this thread?? i > thought i was being on-topic with the listed subject of the conversation.Lol, yes, I see. I don't know how quickly the error grows with the DFT calculation. Why do you say the DFT produces less noise than the FFT? I have not seen an analysis of the DFT in that regard, but off the top of my head I would expect them to be about the same. The FFT noise is 1/2 log(n) bits from some paper I read many years ago. The DFT has more calculations total, but I believe each bin has about the same number of operations involved in producing it. It would depend on whether the DFT was done as a real or complex computation. -- Rick