Logarithm of a 0<X<1

Projects General Guidance
1

Hey there Arduino folks!
I'm trying to make some calculation in my sketch, (frequency to midi Note), it's an easy one, you can find the formula anywhere which is:
midiNote = round(12*log2(440/440))+69;

There's no logarithm of base 2 in Arduino, but you can easily write it:
float log2 (float x) {
return (log(x) / log(2));
}

This calculation works fine in Processing but not in Arduino, for any argument between 0 and 1 (which has to result a negative number) it just gaves me and infinite (or something similar, it's been a couple days since I last checked that).

What am I might be doing wrong??

Thanx!

2

Have you tried making log(2) a constant, and use that throughout your program?

const float LOG2 = 0.30102999566398119521373889472449;

Note: you may have to crop some decimal places there!

And then:

float log2 (float x) {
return (log(x) / LOG2);
}

That will not only make your program a lot faster (because log(2) doesn't ever need to be calculated), but it will ease the burden on the Arduino.

3 
round(12*log2(440/440))+69;

That will only ever produce the value 69. If you try, for example,

round(12*log2(220/440))+69;

The value of 220/440 is zero because it is an integer division. You should use this:

round(12*log2(frequency/440.))+69;

which forces the division to be done with floating point.

Pete

4

Natural log is the area under the curve 1/x from 1 to the number whose log you are trying to find. Always from 1. There is no natural log for less than 1.

5

There is no natural log for less than 1.

There certainly is. The log (natural or not) of a number is defined for any real number above zero. Numbers between zero and one have a negative log.

Pete

6

el_supremo; THAT was certainly THE POINT! :sweat_smile:
Thanx so much! (now it works nice!)

7

el_supremo:

There is no natural log for less than 1.

There certainly is. The log (natural or not) of a number is defined for any real number above zero. Numbers between zero and one have a negative log.

Pete

I don't remember doing that but I guess you could integrate left-wise, it's just a matter of ordering the limits. It's been a while since I did the math.