Re: functions for audio

[Paul Kienzle]

On Mon, Jun 04, 2001 at 12:26:19AM -0700, Aaron Brick wrote:
> hello dave,
> 
> thanks a lot for your rapid reply! :)
> 
> so, to create a CD-quality one-second sample of middle A:
> 
>       fs=44100
>       freq=440
>       saveaudio ("file", sin ( (2*pi) * (freq/fs) * (1:fs) ), "au", 16)
> 
> and then:
> 
>       sox -r 44100 -t ul file.au file.wav
>       play file.wav
> 
> the samples come out super-quiet; i multiplied the sine function by about
> 100 to solve this.

Looking at the source for saveaudio, it is expecting numbers in the
range -128:127 or -32768:32767.  You should be multiplying by 127.
Alternatively, you could use my ausave function which uses an audio
range -1:1, and allows you to save a .wav file directly.  See
the octave project on sourceforge:

http://cvs.sf.net/cgi-bin/viewcvs.cgi/octave/octave/scripts/nonnumeric/audio/

> 
> now for a more complex sound which i'm stuck trying to create. i'm
> having conceptual difficulties translating my mathematical ideas into the
> sine ( constants * range ) form that octave uses.
> 
> i'm used to sound wave equations looking like y = A * sine ( F * x ), where
> A is an adjustment in amplitude, and F in frequency. i want a sound whose
> amplitude increases over time; normally i would define it by setting A =
> (1.01)^x. i don't know syntactically how to do this in octave.

(1.01).^x, where x is a vector.

To get a frequency varying like F(x), you will need to do something like: 

        x = 0:1/fs:tmax;
        y = cos(cumsum(F(x))/fs);

E.g.,  the following has a frequency which varies sinusoidally:

        function p=F(x), p = 5000*(sin(2*pi*x)+1); end
        tmax=2;
        fs = 1000;
        y = cos(cumsum(F(x))/fs);
        imagesc(flipud(stft(y)));

> 
> and do you know how to calculate nth roots? i can't find any mention of them
> in the manual.

a^(1/n)

> 
> many thanks,
> 
> aaron.
> 

Paul Kienzle
address@hidden



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