How to adjust values to a new range

[Ioannis]

Since you have done it in PBP and if you don't mind, could you post the code you tested please?

Maybe there is a way to overcome the Long need using functions of 32 bit like DIV32?

Ioannis

P.S. Test it in Excel and I think there is a mistake on the B3 function. My output result on B13 is always equal to B11.

[Ioannis]

There is a mistake on B13, it should be MIN(B12;B7)

Ioannis

[HenrikOlsson]

As I turn the pot, the formula should convert from 0 all the way to 1023, to a new range of 100 to 235

OutputValue = 100 + (InputValue */ 34)

[Ioannis]

Many ways to skin a cat but Henriks is the best!

Fast and simple!

Ioannis

[Demon]

OutputValue = 100 + (InputValue */ 34)

That's awesome Henrik. It works like a charm.

Can you explain how you got 34? It's not jumping out at me.

"A simple way to think about '*/' is that it shifts the result 8 places to the right,
resulting in an automatic division by 256."

I'm just not sure how you went about figuring that out.

I'd like to convert 34 into a formula for future use when I have to use a different range (cause I'm sure to have that need - several Flight Sim variables use pots and have their own range of values).

[Demon]

There is a mistake on B13, it should be MIN(B12;B7)

Ioannis

Good eye.

I had corrected the formula in B13, but had somehow missed the text in C13 (it's just text to show what goes on in the real cells in B column).

[Demon]

Since you have done it in PBP and if you don't mind, could you post the code you tested please?

Maybe there is a way to overcome the Long need using functions of 32 bit like DIV32?

Ioannis

P.S. Test it in Excel and I think there is a mistake on the B3 function. My output result on B13 is always equal to B11.

I divided by 4 first, then ran it through the formula.

"EDIT: Hmmm, what if I divide my input by 4, and adjust my input range accordingly..."

[Demon]

Since you have done it in PBP and if you don't mind, could you post the code you tested please? ....

Sure thing:

ADCinput    var WORD
ADCcalc     var WORD
ADCinLow    con 0
ADCinHigh   con 255
ADCoutLow   con 100
ADCoutHigh  con 235
        ADCcalc = ADCinput / 4          ' reduce to 0-255 to avoid WORD overflow
        ADCcalc = (ADCcalc MAX ADCinlow) - adcinlow ' adjust 0 - 255 range
        ADCcalc = ADCcalc * (ADCouthigh - adcoutlow)'   to 100 - 235 range
        ADCcalc = ADCcalc / (adcinhigh - adcinlow)
        ADCcalc = ADCcalc + ADCoutlow
        ADCcontrast = ADCcalc MIN ADCouthigh

I just kept things simple and did things one step at a time, just like Jerson explained it. And women say men can't follow instructions.

I know I can shrink this, but I'm just researching all the different bits of code I will need first, and documenting with comments what the code does and why.

[HenrikOlsson]

This has been covered many many times on the forum so why not one more time...

Basically we want multiply the ADC value by 135/1023=0.132 so in comes the */ operator.... 0.132*256=34 (rounded up).

1023*34/256=135
512*34/256=68
128*34/256=17

When the value you want to multiply by is <1 you can get more precision with the ** operator (same thing but inherent division by 65536). In this case it doesn't matter though because the output value range is so small.

Do note that there's no expensive division going on here, all it does is discard the bottom 8 bits (in the case of */) of the multiplication.

/Henrik.

[Ioannis]

If for any reason the input may have a value over 1023 then there would be a problem with the simple equation. But since the that value comes from a specific 10 bits ADC module it defines the upper limit of 1023.

If you want to use it in other cases, a MIN/MAX would be needed.

In any case Henrik's solution is neat and fast.

Ioannis

[Demon]

This has been covered many many times on the forum so why not one more time...

You've explained it, using my own problem as example, and I still don't completely understand it. My eyes sort of gloss over when I've read these explanations on the forum. And it's not getting better as I get older.


I really do appreciate you holding my hand while you pound this into my skull.

[HenrikOlsson]

Allright...

[1] I believe you're with me for the 135/1023=0.132 part, right?

[2] IF the description for the */ operator WOULD have said "....resulting in an automatic division by 1000" then I think you'd see that the solultion WOULD be OutputValue = 100 + (InputValue */ 132), because 0.132*1000=132, correct?

[3] But since the */ operator "....resulting in an automatic division by 256" we must multiply 0.132 by 256 instead of 1000. Hence the value 34.

Makes sense? If not, which step?

[Demon]

I follow. I just could not have thought of this by myself.

[Ioannis]

Or maybe use a formula

X=135*256/1023

where X is the number you wan to use in the */

135 is the max scale value

1023 is the max ADC value

Ioannis

[amgen]

error posting