I can map the first Oscillator knobs properly to the C4, but using the OSC select switch to select the second or third oscillator, seems to start NRPN messages. AFAIK, NRPN messages are basically two 7-bit messages (msb+lsb). The manual says, that whenever NRPN is involved, the msb part is not used and should be treated with 0 value. Only the lsb part is used from the message, but here starts my problem: As long as i move only the osc select switch from left to right (osc1-osc3), my mapped vpots for each oscillator will work correct, but if i use other switches from the osc section in between the osc select switch, my mapping for osc 2&3 falls apart.
I attached the manual and the midi implementation pages start with 66-70. The part i dont understand is page 70, point 11(OSC select). What does those âbitsâ mean in terms of a raw midi message? How i generate a proper rule for that?
PS: If you own the V-Station VST plugin, you can use that for trying. The implementation is the same as for the hardware.
Well I canât pretend to completely understand what the bits all mean but the below should help get you started.
Here I take the incoming CC 12 and use the value to determine which oscillator to select (bits 0 and 1).
I put in rules to determine the other parameters but at this point, I left them all at 0. I determine the value and then bit shift them into position combining them with the original value (qq) make a full 7 bit (bits 0-6) value.
Here is an extract from the rules. There may be some other way that you want to do this but at least this gives you an idea of bit manipulation.
// bits 2 and 3
// set value between 0 and 2
uu=0
if uu>2 then log "Noise/Ring/Ext Out of range %uu%- ABORTING"
if uu>2 then exit rules, skip outgoing action
// shift bits
uu=uu<<2
qq=qq|uu
Ok, if i capture raw midi data from the OSC select switch it is as follows:
Osc1 select: B0 62 1A following by B0 06 00
Osc2 select: B0 62 1A following by B0 06 01
Osc3 select: B0 62 1A following by B0 06 02
This switch will stay like this, no matter what combination of the other switches i am doing. It will always output this. Lets assume, i dont touch this switch anymore and put it to Osc1 select.
Now if i want to use the Octave switch, it is as follows:
BUT this time, the only constant is B0 47, the last two digits will change, depending what other switches are used. So the next time, i want to use the Octave switch, it might change into something different like:
I would understand this, if i had changed the Osc select to Osc2 or Osc3, then this values would need to change. But remember, i did not touch that switch.
So this can only work, if i can make a rule that takes the B0 47 + the very last digit into account and ignore the first digit (because it will change) of the last two digits.
I guess that is what you mean with âshiftingâ.
This is NRPN LSB 26 (Hex 1A) with value of 0, 1 or 2.
In the manual LSB 24-31 says reserved for future use, so I guess they are using it now for Oscillator select.
This is CC 71 which uses packed parameter 4.
For binary the value bits bits are for 2C is
0010 1100
So bits 1 and 0 are 00 which means a value of -1
bits 3 and 2 are 11 which means a value of 2
bits 5 and 4 are 10 which means a value of 1
bit 6 is 0 which means Osc1>2 sync is off
For binary the value bits bits are for 6C is
0110 1100
So bits 1 and 0 are 00 which means a value of -1
bits 3 and 2 are 11 which means a value of 2
bits 5 and 4 are 10 which means a value of 1
bit 6 is 1 which means Osc1>2 sync is on
The below is a simple binary, decimal, hex lookup table so hopefully you can see what I mean when you look at it from a bit (binary) perspective.
Binary
Dec
Hex
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
8
8
1001
9
9
1010
10
A
1011
11
B
1100
12
C
1101
13
D
1110
14
E
1111
15
F
By breaking up a byte into 4 bit chunks the conversion is much easier.
For instance
2C = 0010 1100 which is decimal 44
the right digit is you multiply by 1
the next digit left you multiply by 2
then x4, x 8, x 16 , x 32, x 64, x128
Every digit from the right is 2x the previous.
In a nutshell, you need to only look at the digits you want to change and not the whole value.
Say you want to only change bits 0 and 1 then you would need rules like this where qq=the total incoming value
// mask out other bits (AND operation)
tt=qq&3
// you want to change the output from 3 to 2 but leave the other bits alone (OR Operation)
qq=qq|2
I hope this helps!
Steve Caldwell
Bome Customer Care
Also available for paid consulting services: bome@sniz.biz
This was the easiest part to understand. Yeah, it would also explain, why you cant find any âpacked parameter 11â in the manual pages, before that âreserved for future useâ quote. Unlike all other (1-10) packed parameters.
So this means if i want to use the C4 as a remote, that i need to emulate the Osc select switch each time, before i use any of the other switches of the Osc section that are mapped to the vpots of the C4. Is this correct?
Reminder: The Osc select switch, should be obsolete on the C4, because I can show all three Oscillators and all of their parameters on three displays of the C4.
It took me a while to hopefully understand this. So if the manual talks about âbitsâ on page 70, i need to use binarys, right?
Your quote here, explains the binary numbers, which seems that the bits (0-6) are read from right to left (wtf!), is this correct? And if this is correct, then please add this important fact to a explanation. You really cant expect that the average user is aware of this.
The resulting values (except bit 6) are for what? Only one Osc? All three Osc´s?
This was even harder to understand. This is not your fault and i am aware that the text editor cant show the formula. You for sure did the best to explain this . Only the rapid tables in the browser had helped me here to understand that, because there was a example:
Even that example, brings a result of 14 for me and not 13.
1101 = 1x8+1x4+0x2+1x2
is 14 for me.
Doing this with your â2C = 0010 1100 which is decimal 44â would be
(0010 = 32 )+ (1100 = 12) = 44.
Is this correct?
This is the hardest part for me to understand. I only understand the goals here and what operators to use for that.
// mask out other bits (AND operation)
tt=qq&3
so which bit is masked here? 3? What would that mean for my Octave switch? How i deal with binarys or bits and create rules in MTP? How do i tell MTP that the incoming hex value is treated like a binary?
A little rant here (not towards you ) : Why the hell are simple but important things handled in a way, that a normal person get headaches? Its really insane, the more I think about it. I guess i will need days to get those âpacked parametersâ working. Imagine coming home from a hard workday and then trying to do this in your freetime, without a single example in the manual. Its just gross. Its going from macro, to micro, to nano universe and in a not so well documented way.
To my understanding, i am pretty sure that a single CC + the value of it, could have handled all these âpacked parametersâ in one go. But no, i have to deal with binary numbers or bits. This is real pain.
I can only urge you people who are experts in this, to create proper documentations for these things, because generation Z will leave this $hit alone after 5min. trying to understand this. In a decade or two, people who where familiar with these kind of things will pass away and then is no one left who can explain this.
Where is the AI when I need it You know what i mean here I hope my post helps you to understand, why I am stuck here and how far my knowledge goes .
I really donât know. I can explain binary but not how the A-Station works.
Yes, when calculating you can see similar to decimal system and binary system.
Decimal 123 = ( 3 x 10^0 ) +( 2 x 10^1) + (1 x10 ^2)
Binary 0101 = ( 1 x 2^0 )+( 0x2^1)+ (1x 2^2) +( 0x 2^3)
It is (1x8) + (1x4) + (0x2) + (1x1) = 13
Yes
So â&â is not adding, it is ANDâing
1 AND 1 = 1
0 AND 1 = 0
0 AND 0 = 0
So if you AND 0101 1011 and 0011 you get
0101 1011
0000 0011
--------------
0000 0011
You are essentially masking out all bits but 0 and 1.
Yeah, this should be directed at the controller manufacturer. Iâm sure they have their reasons.
Binary numbering is how computers work. I doubt if everyone that understands binary will pass away because it is so prevalent in the computer industry. Most programmers know this stuff.
Steve Caldwell
Bome Customer Care
Also available for paid consulting services: bome@sniz.biz
I guess i will need a break from this here, because i am so confused now. I am not able to understand the code nor what to do with all this info. At least i can say i tried. Not being able to control a 4-way switch with midi, or need to know fundamental computer language for this, like it is some rocket science, is a bummer to me.
I will create non-working vpot dummys for these switch-cases and keep going on with parameters, that i understand, just to have some progress. Another option is creating rules for every possible case regarding the switch-values. At least they dont appear random to me. There is a pattern to this. Not elegant, but might be working. Unnecessary many rules, that even needs to be copied multiple times.
Ok, i see that and i know now, i will need that to solve my problem, but thats it.
All i can say is, that the actual generated code is getting worse since years, because the ânext generationâ of programmers are lacking of this fundamental knowledge already. This is what the web says at least.
Thank you Steve for all the help and patience with me so far .
,
. Only the rapid tables in the browser had helped me here to understand that, because there was a example:
) : Why the hell are simple but important things handled in a way, that a normal person get headaches? Its really insane, the more I think about it. I guess i will need days to get those âpacked parametersâ working. Imagine coming home from a hard workday and then trying to do this in your freetime, without a single example in the manual. Its just gross. Its going from macro, to micro, to nano universe and in a not so well documented way.
You know what i mean here