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## Re: Calculating Semitones in an octave

 From: Doug Stewart Subject: Re: Calculating Semitones in an octave Date: Wed, 29 Jan 2020 12:24:05 -0500

On Wed, Jan 29, 2020 at 12:04 PM Andrew Janke <address@hidden> wrote:

On 1/29/20 11:53 AM, harish3679 wrote:
> Hi guys,
>
> To find frequency using semitones value from A=440Hz
>
> We have, f = 2^(n/12)*440 //where n is the semitone value
>
> Now, to find semitone value for A=432Hz from 440Hz
>
> It is 432 = 2^(n/12)*440
>
> I came up with 2^n=0.92668

I think this is wrong it should be :
2^(n/12)=432/440
and take the log2 of that:
n/12=log2(432/440)
n=12*log2(432/440)
n = -0.31767

>
> Although have no clue how to calculate this in octave rather got an error
> saying
>
> "2^(n)=0.92668
> parse error:
>
>   invalid left hand side of assignment"
>
> Someone, please shed light on this part
>
>
> Regards,
> Harish
>

You can't write arbitrary equations in Octave and have them figure it
out. The "=" sign is an assignment operation, and you need to have a
single variable on the left hand side. So re-work your equation so that
n is by itself on the left hand side.

I think this means taking the base-2 logarithm of both sides, and ending
up with:

n = log2(0.92688)

And that you can run in Octave:

>> log2(0.92668)
ans =   -1.098568597196845e-01

Cheers,
Andrew

--
DAS 