' bascom 2.0.7.3

' noel benoit f1tet ALGORITHME DE COMPRESSION TO SRAM 4 MBIT COMPRESSED  TO 20MBYTE DECOMPRESSED
 ' 25 08 2019

'$regfile = "m8515def.dat"
$regfile = "m168def.dat"
$crystal = 8000000
$hwstack = 100
$swstack = 100
$framesize = 40


 ' with a buffer switch adress 74HC373 8 BIT     LATCHED TO ALE SIGNAL

 ' the adc value onto 10 BITS IS ENCODED AS THE FOLLOWING
  ' on the fisrt 8 bits
  ' the 4 MSB ARE THE AVERAGE UPPER BYTES B5 TO B8 VALUE FROM 0 TO 64 TO 512  OF 8 EIGHT ADC SAMPLE DIVIDED / 8
  ' the 4 OTHER BIT ARE FOR INCREASE THE SPEED OF CHARGING/ DECHARGING CAPACITOR let 16 POSSIBILITY LET 8 BITS TO 0 OR TO 1
  ' WHEN 0 NO EXED CHARGING TIME WHEN 1 TIME MULTIPLIED PER 4
  ' the average value is outputed to adc port with relation Vaverage=vout*rc*T
  ' THAT CAN BE OUTPUTED TO ADC PORT AS OUTPUT THIS ALLOW TO INCREASE THE SPEED OF CHARGIN CAPACITOR DURING A CERTAIN TIME
  ' ON THE SECOND EIGHT SELECT IF THE OUTPUT IS UPPER OR UNDER THE AVERAGE VALUE   0 IS UNDER OR 1 IS UPPER DURING A FIXED TIME



  ' for resume there is 4 MAIN PARAMETER :
   'one average reference value TO PRECHARGE CAPACITOR WITH ADC port OUTPUT OR PWM OUTPUT WITH DIODES eight diodes
   ' one _ bit data to charge upper or discharge under the average value  i mean the second octet
   'other 4 bit that allow extra time if set one  under 16 possibility  THE FIRST 8 POSSIBILITY 0 OR 1  THE OTHER 8 0 OR 1
   'THE 8 ONE ARE EXTRA TIME IF SET ONE   AND THE SECOND 8 POSSIBILITY IS EXTRA TIME IF SET ONE


   Dim Value1 As Word                                       ' FIRST 10 BIT ADC VALEUR
   Dim Value2 As Word
   Dim Value3 As Word
   Dim Value4 As Word
   Dim Value5 As Word
   Dim Value6 As Word
   Dim Value7 As Word
   Dim Value8 As Word                                       ' EIGHT 10 BIT ADC VALEUR
   Dim Average As Dword                                     ' THE AVERAGE VALUE OF THE EIGHT MSB BITS LOWERED OF MAX 512 AND TRONCATED AT THE 4 LOWERS BYTES
    Dim Average1 As Dword                                   'AND DIVIDED BY 8 RESULT ONTO 4 BITS

   Dim Data2 As Byte                                        '' set upper or under the average value 0 or 1
   Dim Data1 As Byte                                        ' contain the 4 msb bit of average and the 4 bit coded of extra time cherging
   Dim Extand_charge As Byte


Config Adc = Single , Prescaler = 128 , Reference = Avcc



Start Adc




Dim W As Word , Channel1 As Byte


Channel1 = 1                                                'THE INCOMING SIGNAL IS CAPTURED IN ADC 1  EIGTH TIME AND A AVERAGE VALUE IS STORED IN DWORD AVERAGE
 Average = 0
Do
 Average = 0
 Value1 = Getadc(channel1)                                  ' SOME WAIT CAN BE INSERTED BEATWIN EACH ADC CAPTURE   SYNTAX WAIT TIME1
 Average = Average + Value1
 Value2 = Getadc(channel1)
 Average = Average + Value2
 Average = Average / 2                                      ' averaging the 2 first values   THIS METHOD INDUC A SMALL ERROR  ON THE FINAL RESULT
 Value3 = Getadc(channel1)
 Average = Average + Value3
 Average = Average / 2
 Value4 = Getadc(channel1)
 Average = Average + Value4
 Average = Average / 2
 Value5 = Getadc(channel1)
 Average = Average + Value5
 Average = Average / 2
 Value6 = Getadc(channel1)
 Average = Average + Value6
 Average = Average / 2
 Value7 = Getadc(channel1)
 Average = Average + Value7
 Average = Average / 2
 Value8 = Getadc(channel1)
 Average = Average + Value8
 Average = Average / 2
 Average1 = Average                                         ' STORING THE AVERAGE VALUE BEFORE MODIFICATIONS


 Rotate Average , Left                                      ' EXCLUDING THE MSB OF THE AVERAGE VALUE

Data2 = 0
If Value1 > Average1 Then Data2.0 = 1
If Value2 > Average1 Then Data2.1 = 1
If Value3 > Average1 Then Data2.2 = 1
If Value4 > Average1 Then Data2.3 = 1
If Value5 > Average1 Then Data2.4 = 1
If Value6 > Average1 Then Data2.5 = 1
If Value7 > Average1 Then Data2.6 = 1
If Value8 > Average1 Then Data2.7 = 1

Data1 = 0
Data1 = Average

'  THERE IS 512 BITS TO COMPENSATE WITH 16 POSSIBILYTIES LET 8 VALUES AT TIME1=512/3=170 BITS AND TIME2=512/3=170 BITS
' IF DATA2.0 POSITIF THE TIME1 AND TIME2  ARE ADDED  IF DATA2.0 ZERO THEN TIME 1 AND TIME2 ARE SUBSTRACTED FROM AVERAGE VALUE
' THIS ALLOW A DELTA OF 4 TIME1 MAX BETWEEN TWO CONCECUTIVE SIGNALS  IF T1=T2=170BITS   LET DELTA MAX 680BITS
' THE POSITIVE AND NEGATIVE TEST WITH AVERAGE VALUE MUST BE DONE TO COMPUTE THE 4BIT THAT CONTAIN THE 16 BITS VALUES
' for the restitusion signal there is tree time   let T+T1+T2=FINAL SIGNAL
'



If Data1 > Average1 Then Data2.0 = 1
If Data2 > Average1 Then Data2.1 = 1
If Data3 > Average1 Then Data2.2 = 1
If Data4 > Average1 Then Data2.3 = 1
If Data5 > Average1 Then Data2.4 = 1
If Data6 > Average1 Then Data2.5 = 1
If Data7 > Average1 Then Data2.6 = 1
If Data8 > Average1 Then Data2.7 = 1

Loop

End