Please, does someone know the inverse of this ?

Nov 29, 2013 at 1:54pm

Please, does someone know the inverse of this ?

I have tried and failed to make it work.


– Pasted Max Patch, click to expand. –


Nov 29, 2013 at 3:48pm

I don’t know how to make an inverse of lp.scampi.

And that expression seems like it’s mtof, right?

[mtof 0.]

is the inverse of:
[ftom 0.]

hope this helps.

Nov 29, 2013 at 11:54pm

cheers man,

I knew i could use ftom or sigmund for freq. to pitch/note conversion

It’s just that using object based programs like max, you rarely get to understand the function/maths properly.

I wanted to see what was under the hood, and try and get used to using the correct syntax for the expr object.

anyway, the correct equation was in the ftom maxhelp so thankyou :):)

I wasn’t a million miles away


Dec 2, 2013 at 1:46am

This will do it…


– Pasted Max Patch, click to expand. –
Dec 2, 2013 at 1:53am

expr won’t accept log2($f1)… so for log2(x) you can use: log10($f1)/log10(2)

Dec 2, 2013 at 3:28am

Hi Metamax, Cheers……. the answer was actually in a maxhelp file somewhere

They gave – expr (69. + (1./.057762265) *log($f1/440.))

and yours expr 69.+12.*log10(4f1/440.)/log10(2)

both work perfectly and do exactly the same frequency to midi note number conversion.

Sadly I wish i’d concentrated on maths more when i was in education now haha

Is it possible to describe to me, in laymans terms, whats actually happening & how both equations end with the same result.
recently i’ve been trying to understand what goes on behind the objects that i regularly use….. mtof, ftom etc. regarding both the maths and as importantly the syntax

eg. I succesfuly created the Split obect using [if $i1>=40 & $i1<=50 then $i1 else out2 $i1]

Thanks again


Dec 2, 2013 at 12:09pm

The inverse of [lp.scampi 1 0] is [lp.scampi 1 0] (or you could use [lp.scampf 1 0]). The ’round’ argument, however, makes the function non-invertible.

Interesting that the lp.scampi from Litter Power 1.8 is floating around.


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