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. 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? And, for D and F, the difference is 55 hertz.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). And then, C D makes 33 hertz, and note C and F, the difference is 88 hertz. Note A and note F is 220 minus 352, absolute value of which is 132 hertz. So, when A is paired with C, that makes a difference of 44 hertz. And, that's going to result in six different possible pairs, each of which will produce a beat frequency. So we have note A, C, D, and F all being played at the same time. ![]() If all four of these frequencies are played together, there's going to be a beat frequency for every pair of possible frequencies. And then, if both frequencies or beat frequencies will be present for all four. ![]() And notice this would be negative, but the absolute value bar just means we take the positive. And for part B, we're finding the difference in frequencies in D and F, which are 297 and 352, and this works out to 55 hertz. So, that's the difference between 264 hertz minus 220 hertz, which is a beat frequency of 44 hertz. So, for part A, the beat frequency will be the absolute difference between the frequency of the note A and the note C. The beat frequency, which is the frequency with which the amplitude is oscillating is going to be the absolute difference between the two frequencies. This is College Physics Answers with Shaun Dychko.
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