Created
April 26, 2015 21:55
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| var newFFTPitch = { | |
| calc: function (fft, samplerate, fftSize) { | |
| var fftLen = fft.length; | |
| var i; | |
| var max = fft[0]; | |
| var curVal; | |
| for (i=1; i < fftLen; i += 1) { | |
| curVal = fft[i]; | |
| if (curVal > max) max = curVal; | |
| } | |
| // now find the index of 1/3 | |
| var leastIndex = -1; | |
| var minVal = max / 3; | |
| var leastVal = max; | |
| for (i = 0; i < fftLen; i += 1) { | |
| curVal = fft[i]; | |
| if (curVal > minVal && curVal < leastVal) { | |
| leastVal = curVal; | |
| leastIndex = i; | |
| } | |
| } | |
| var prevIndex = leastIndex - 1; | |
| var nextIndex = leastIndex + 1; | |
| var prevVal = fft[prevIndex]; | |
| var nextVal = fft[nextIndex]; | |
| /* | |
| // https://ccrma.stanford.edu/~jos/parshl/Peak_Detection_Steps_3.html#sec:peakdet | |
| explanation uses alpha for prevVal, beta for leastVal, gamma for nextVal | |
| then gives 1/2 * ( (alpha - gamma) / (alpha - 2*beta + gamma) ) | |
| for the parabole peak location, which is the position it would have in the array | |
| the frequency would be (sampleFreq*p)/ fftsize ? | |
| */ | |
| var p = 1/2 * ( (prevVal-nextVal) / (prevVal - (2*leastVal) + nextVal)); | |
| var freq = (samplerate * (p + leastIndex)) / fftSize; | |
| console.log('freq: ' + freq); | |
| return freq; | |
| } |
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