Hi there, yet another question about DFT/IDFT processing. Here's what I do (assuming 80% overlap): 1) Take let's say 4096 source samples. 2) Perform windowing (using Hann). 3) DFT the result. 4) Here should be some spectral processing, but now it is not. 5) IDFT the result. 6) Perform windowing again (could this be omitted?) 7) Multiple each resulting sample by 2.0 / 4096. I don't remember why, but it is caused by FFT algorithm (to be same for DFT and IDFT). 8) Multiple each resulting sample by 1.0 / 5.0 (80% overlapping). 9) Add the result to our temporary output buffer, which is also 4096 samples long and accumulates the result. First 20% (about 700 samples I guess) of it is final, the rest is moved, so that output overlapping can be performed. Similarly input temp is moved by 20% to the left and incoming data cover the last 20%. I somehow copied this code from Stephan Bernsee pitch transposition source. This should obviously do nothing except the introductory fade-in caused by windowing. It really does nothing (when I tried some equalization, it worked too), but I have a few questions: 1) Why does it work? Originally I thought each window should have some predefined overlap size, which equals to the sample with 0.5 value, so that when I apply windowing and overlapping to two successive blocks, they would generate the original signal. Which means: window[i] + window[i + overlapsize - fftsize] = 1 for every i But this code applies the overlap-adding by any constant, so on the first sight it seems, that if I choose really bad overlap size, it would generate waves in the output amplitude (e.g. overlap = 10%). Or am I missing something? 2) If the code is correct, how should I ensure the same signal power on the result? Without windowing it would be correct, since each target sample would be generate from five source samples (the same ones) divided by 5. But with the windowing each sample might be created from different ratios. I tried to approximate by : (1.0 - OverlapSize) / WindowSum (which is from ((1.0 - OverlapSize) / fftsize) * (fftsize / WindowSum) The result is not so bad, but still 20% lower in amplitude. Thanks guys again! dmnc

# DFT output signal power

Started by ●November 19, 2008

Reply by ●November 20, 20082008-11-20