Halve the volume of both waves: for(int i = 0 i < data.Length i++)ĭata = (byte)(128 + (63 * Math.Sin(i/2.0) + 63 * Math.Sin(i))) īetter. What happened? We overflowed again the sum was larger than 256 at many points. Try that out and see how you get the same tone, an octave lower. (Every octave down halves the frequency every octave up doubles it.)
Music note c in hertx full#
and therefore the tone will be a full octave lower. If you do that, you'll get peaks half as often: at i = 4, 16, 28. We don't want integer rounding! The 2.0 tells the compiler that you want the result in floating point, not integers. So let's make it happen half as often: for(int i = 0 i < data.Length i++)ĭata = (byte)(128 + 127 * Math.Sin(i/2.0)) Suppose you wanted to make an octave below your sample. Remember, that's all a WAV file is: a big list of air pressure changes. It's equal to the sum of the pressures at any given point in time. If you have one vibrating source that is pumping pressure up and down 440 times a second, and another one that is pumping pressure up and down 880 times a second, the net is not the same as a vibration at 660 times a second.
A chord which is A440 and an octave above, A880 is not equivalent to 660 Hz. How do chords work? Are they the average of the pitches? 1760 Hz is two octaves above A 440, so this is a slightly flat A tone.
(The Hertz is the measure of frequency how many peaks per second). How many peaks are there per second? 11025/2π = 1755 Hz. How far apart in time are those? Each sample is 1/11025th of a second, so the peaks are about 2π/11025 = about 570 microseconds between each peak. So the peaks are whenever i is close to one of those. Where are the peaks? That is, where does the sample attain a value of 255, or close to it? The change goes from the smallest change, 1, to the largest change, 255. The size of the pressure change is interpreted as the volume. If those changes form a repeating pattern then the frequency at which the pattern repeats is interpreted by the cochlea in your ear as a particular tone. The human ear detects incredibly tiny changes in air pressure. Now we have smoothly varying data that goes between 1 and 255, so we are in the range of a byte.
Music note c in hertx code#
Let's rework your code so that the sample is in the right range. Wait a minute though, sine goes from -1 to 1, so the samples go from -256 to +256, and that is larger than the range of a byte, so something goofy is going on here. Each sample is a number between 0 and 255 which represents a small change in air pressure at a point in space at a given time. Let's take a look at your example: for(int i = 0 i < data.Length i++) This creates (somehow) a constant sound - but I don't understand completely how the code correlates with the result.
Sound Data Size = Number Of Channels * Bits Per Sample * Samplesīyte data = new byte Int samples = 11025 * seconds //Create x seconds of audio WaveFile file = new WaveFile(channels, bitsPerSample, 11025) WaveFile is custom class to create a wav file. If giving code examples, I am using C# and the code I am currently using to create wav files is as follows: int channels = 1 any other relevant information relating to this.how is the result of multiple notes being inverse FFT'd converted to an array of bytes, which make up the data in a wav file?.How is the length of time to play each note specified, when the contents of the wav file is a waveform?.How do chords work? Are they the average of the pitches?.If my understanding is correct, this pitch is in the frequency domain, and so needs the inverse fast fourier transform applying to it to generate the time-domain equivalent?
Music note c in hertx how to#
I am interested in how to take musical notes (e.g A, B, C#, etc) or chords (multiple notes at the same time) and write them to a wav file.įrom what I understand, each note has a specific frequency associated with it (for perfect pitch) - for example A4 (the A above middle C) is 440 Hz (complete list 2/3 of the way down This Page).
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