Floating point conversion, How I do

micman2 · 2009-05-13 10:47

Hi all,

i've a question:
I've a PID algorithmic ,


  encL := Enco.ReadEncoderLeft
  cur_posL := encL -  prevVL    
  prevVL := encL  
  
  cur_errorL := set_posL - cur_posL    

  PL := f.FMul(f.FFloat(cur_errorL),constKp)                            
  IL := f.FAdd(f.FFloat(IL),(f.FMul(f.FMul(f.FFloat(cur_errorL),cstdt),constKi)))
  eL := cur_errorL - pre_errorL
  DL := f.FDiv(f.FMul(f.FFloat(eL),constKd),cstdt)

  outputL := f.FAdd(f.FAdd(PL,IL),DL)

the result "outputL " is floating point algorithmic , The result is floating point ? how to convert in long ?

The next code limit -100 < outputL < 100 and send in PWM


  if outputL > 100
     outputL := 100

  if outputL < -100
     outputL := -100

  PWMg.setpwmDuty1(outputL,frequenzaPWM)

The constant is:

    constKp = 1.0
    constKi = 0.0
    constKd = 0.0

work but When I change in :

    constKp = 1.2     <------
    constKi = 0.0
    constKd = 0.0

1.2 (.2) (.1;.3;.4 ecc ecc ) send in over limit the "outputL"

Why?

stevenmess2004 · 2009-05-13 11:40

To turn a floating point number into an integer you need to use either f.FTrunc or f.FRound. Their both in the float32 object. You would use them like this

 outputL := f.FRound(f.FAdd(f.FAdd(PL,IL),DL))

and outputL will now be a normal integer

.

micman2 · 2009-05-13 11:53

Thanks you!

     cur_errorL := 5
     PL := f.FMul(f.FFloat(cur_errorL),f.FFloat(2))                             
     debugLout := f.FRound(PL)

debugLout = 10

The result is Ok ! ( I see with viewPort v4.1.504)

But when I change constKp with 1.8

constKp = 1.8

     cur_errorL := 5
     PL := f.FMul(f.FFloat(cur_errorL),f.FFloat(1.8))                             
     debugLout := f.FRound(PL)

The result is wrong
debugLout = 1 !!!!!! Why???

Post Edited (micman2) : 5/13/2009 12:01:48 PM GMT

stevenmess2004 · 2009-05-13 12:10

Could you please post the code thats producing the debugLout=1? It's a little hard to follow whats happening without it.

micman2 · 2009-05-13 13:04

Send source code

stevenmess2004 · 2009-05-14 07:59

micman2 said...

     cur_errorL := 5
     PL := f.FMul(f.FFloat(cur_errorL),f.FFloat(1.8))                             
     debugLout := f.FRound(PL)

The answer is quite simple, in fact, so simple that it took me awhile to see it

. You have entered cur_errorL as a floating point constant. However, you then try and convert it to a floating point as if it were an integer. Change the code above to this

     cur_errorL := 5
     PL := f.FMul(cur_errorL,1.8)                             
     debugLout := f.FRound(PL)

and you should have more luck.

micman2 · 2009-05-14 11:21

thanks you very mach!!!

Mike Green · 2009-05-14 13:20

You have to remember the ".0" otherwise the constant is an integer, not floating point:

cur_errorL := 5.0
PL := f.FMul(cur_errorL,1.8)
debugLout := f.FRound(PL)

Floating point conversion, How I do

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