## Please, does someone know the inverse of this ?

Nov 29 2013 | 9:54 pm
I have tried and failed to make it work.

• Nov 29 2013 | 11:48 pm
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 30 2013 | 7:54 am
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
Danny
• Dec 02 2013 | 9:46 am
This will do it...
69.+12.*log10(\$f1/440.)/log10(2)
• Dec 02 2013 | 9:53 am
expr won't accept log2(\$f1)… so for log2(x) you can use: log10(\$f1)/log10(2)
• Dec 02 2013 | 11:28 am
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
Thanks again
Danny
• Dec 02 2013 | 8:09 pm
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.