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Increasing the pitch of audio using a varied value

开发者 https://www.devze.com 2023-02-28 04:09 出处:网络
Okay, this a bit of maths and DSP question. Let us say I have 20,000 samples which I want to resample at a different pitch. Twice the normal rate for example. Using an Interpolate cubic method found

Okay, this a bit of maths and DSP question.

Let us say I have 20,000 samples which I want to resample at a different pitch. Twice the normal rate for example. Using an Interpolate cubic method found here I would set my new array index values by multiplying the i variable in an iteration by the new pitch (in this case 2.0). This would also 开发者_高级运维set my new array of samples to total 10,000. As the interpolation is going double the speed it only needs half the amount of time to finish.

But what if I want my pitch to vary throughout the recording? Basically I would like it to slowly increase from a normal rate to 8 times faster (at the 10,000 sample mark) and then back to 1.0. It would be an arc. My questions are this:

How do I calculate how many samples would the final audio track be?

How to create an array of pitch values that would represent this increase from 1.0 to 8.0 back to 1.0

Mind you this is not for live audio output, but for transforming recorded sound. I mainly work in C, but I don't know if that is relevant.

I know this probably is complicated, so please feel free to ask for clarifications.


To represent an increase from 1.0 to 8.0 and back, you could use a function of this form:

f(x) = 1 + 7/2*(1 - cos(2*pi*x/y))

Where y is the number of samples in the resulting track.

It will start at 1 for x=0, increase to 8 for x=y/2, then decrease back to 1 for x=y.

Here's what it looks like for y=10:

Increasing the pitch of audio using a varied value

Now we need to find the value of y depending on z, the original number of samples (20,000 in this case but let's be general). For this we solve integral 1+7/2 (1-cos(2 pi x/y)) dx from 0 to y = z. The solution is y = 2*z/9 = z/4.5, nice and simple :)

Therefore, for an input with 20,000 samples, you'll get 4,444 samples in the output.

Finally, instead of multiplying the output index by the pitch value, you can access the original samples like this: output[i] = input[g(i)], where g is the integral of the above function f:

g(x) = (9*x)/2-(7*y*sin((2*pi*x)/y))/(4*pi)

For y=4444, it looks like this:

Increasing the pitch of audio using a varied value


In order not to end up with aliasing in the result, you will also need to low pass filter before or during interpolation using either a filter with a variable transition frequency lower than half the local sample rate, or with a fixed cutoff frequency more than 16X lower than the current sample rate (for an 8X peak pitch increase). This will require a more sophisticated interpolator than a cubic spline. For best results, you might want to try a variable width windowed sinc kernel interpolator.

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