Vladislav D wrote on Sep 26th, 2012, 1:00pm:There is no need to use a window when you r doing simulation since you can precisely define all the input variables for fft.
This is correct for a Nyquist rate converter(more precisely, for evaluating a noise PSD that varies *less* steeply than the leakage sidelobes of the window), but not a noise-shaped one. The problem is not the leakage of the signal. By ensuring that the input is in one of the FFT bins(integer multiple of fs/nfft), signal leakage is avoided. But, the quantization noise, which is orders of magnitude higher out of band than in band will leak into the signal band. This can be easily seen by simulating a DSM in matlab and computing the PSD with and without a window.
http://www.ee.iitm.ac.in/~nagendra/shanthi_lectures/EE658/EE658_Lecture_43/EE658... has a description of this.
Vladislav D wrote on Sep 26th, 2012, 1:00pm:... that a simulator should save data points exactly at the particular moment of time. In the example above, every 62.5ns. You should specify it in the simulator's options.
The output is digital. So taking successive output samples will automatically ensure this. But the FFT must not be taken on the voltage values, but the quantized voltages so that the PSD is a true representation of the digital output.
RobG wrote on Sep 26th, 2012, 1:45pm:nrk1: In cadence AWD calculator I have Blackman available (not blackman harris), but no Hann. What do you mean by losing 9 bins? Can I still evaluate a two-tone test properly? With my hamming window they adjacent 3 bins appear to be averaged.
RobG: In the FFT interface, it is called cosine2, I think. Anyway, I prefer PSD instead of FFT-gives the plot from 0 to fs/2 and can also do averaging(FFT is taken of length "windowsize" and averaged if the record length is more than that). In that it is called Hanning. What I meant is, just like the Hann window spreads a sinusoid on a bin to two adjacent bins, blackmanharris will do it to 8 adjacent bins(4 on either side).
4096 is too few for the high OSR you have. You have 8 points in the signal band, and will lose two to dc offset and three each to the signal and its harmonics. You can hardly see the in-band noise. You can try putting the input signal just out of band so that you can see the in-band noise spectral density.