Hey Harry! I have a feeling, that you should go with just one cycle of your waveform. So you would need a wavetable oscillator to load your waveform for one position in the full wavetable. That should solve your problems of pitch. I investigated one way of performing the full wavetable bit in csound and found one route, using a function table morfing opcode. Its part of the ftable read/access group of opcodes, just in case. Best wishes and good luck! Julien 050e010d0f12010401-0405-0d09-030f12011a0f0d- Such Is Life: Very Intensely Adorable; Free And Jubilating Amazement Revels, Dancing On - FLOWERS! ******** Find some music at ******** http://juliencoder.de/nama/music.html --------------------------------------------------------------- "If you live to be 100, I hope I live to be 95 and 37 days, so I can be sure, there's someone at your site, who loves you." (Not Winnie the Puh)

On 08/24/2012 10:35 PM, harryhaaren(a)gmail.com wrote: Hi Harry, You may mean the right thing and even have implemented it correctly but a halftone step over base_freq = base_freq * 2^(1/12) Even if you implemented it correctly, overtones or undertones can mislead ones perception quite easily. Looping can warp things back, too. -- I suggest to play the wavetable in a loop and check with a spectrum analyzer if you really have the fundamental. ciao, robin

On Fri, Aug 24, 2012 at 08:35:29PM +0000, harryhaaren(a)gmail.com wrote: A few things to consider. You need an interpolation method that doesn't create artefacts, in this case aliasing. You'll also want it to be fast and efficient. The easiest way in the case of wavetable synthesis is to upsample your waves by a factor of say 8, then use linear interpolation. Takes more memory, but memory is cheap these days. Aeolus does this for pipes that contain significant energy above 1/8 of the sample rate. Provided you have a *good* interpolation method, two things follow. * Your wavetable doesn't need exactly K cycles in N samples, K,N integer. If for example you know you have an integer number of cycles in say 123.4 samples, create a wavetable with 124 samples. On the first iteration, use samples 0...123, then when you reach 124 (beyond the end of the table), jump back 123.4 samples, so interpolate at 0.6, 1.6, 2.6 etc. Similar for all iterations through the loop, the third will be 0.2, 1.2, 2.2 etc. * If you are prepared to e.g. turn a C into a C# by interpolation, then you could as well interpolate to obtain the C. In other words, the wavetable doesn't need to be exactly C in the first place. You just need to know its exact pitch. Picking loop entry/exit points by looking for zero crossings can sometimes provide a reasonable result, but it's pure chance. First, the zero crossings could be everywhere between the samples. Second, the distance between two zero crossings doesn't need to be an exact number of cycles. The easiest way to obtain good loops of natural sounds is to have a playback engine that works as described above, and that allows you to modify the supposed lenght of the loop (the 123.4 in the example) in real time while listening to the result. The alternative is to compute the phase trajectory of the entire waveform, which will tell you the exact pitch with much more precision than looking at the waveform. Whatever the method, the result could be that you have a discon- tinuity between the start and end of the loop. There can be many causes for that. With recorded samples LF noise is one, or the harmonics could simply be not exactly harmonic. In such cases you need to test a different start and end points, until you find some that work well, or to 'bend' the waveform so the ends meet. Ciao, -- FA A world of exhaustive, reliable metadata would be an utopia. It's also a pipe-dream, founded on self-delusion, nerd hubris and hysterically inflated market opportunities. (Cory Doctorow)

On Sat, Aug 25, 2012 at 12:44:12AM +0100, Harry van Haaren wrote: Basically just multiply all samples indices by 8. For example if when using the original you would need the value at 3.4, and interpolate between samples 3 and 4, in the upsampled table that becomes 3.4 * 8 = 27.2 and you interpolate between samples 27 and 28. The run-time CPU load is just the same, you only need more memory. But since now you can interpolate witout artefacts, you don't need a perfect fit of K cycles in N samples, and that allows the waves to be shorter. Ciao, -- FA A world of exhaustive, reliable metadata would be an utopia. It's also a pipe-dream, founded on self-delusion, nerd hubris and hysterically inflated market opportunities. (Cory Doctorow)