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Is there a way to produce a 20Hz squarewave on one of the pins on the BX-24?
I would like this to run all the time and affect my normal program as little as possible. 


Right now my main program is the only task running on it.  I am using com1 for serial data
in and out.

The 20Hz doesn't have to be very accurate (within 5%) and can vary a bit if it has to.  Of
course a perfect signal is best.

Any ideas?
Tom
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> ... to produce a 20Hz squarewave...

I can suggest two methods.

You can run another task that sleeps for half of the 20Hz period, 
toggles (inverts) any pin, and sleeps again, repeating endlessly.  Task 
switching will result in some irregularity with this method.

sub TaskB()
  Do
   call sleep(0.5/20.0)
   register.PortD = register.PortD xor bx1000_0000 'Red, pin 25
  Loop
end sub

You can program Timer1 or Timer2 to produce 20Hz, exactly, independent 
of firmware, on pins 26 or 27 (Timer1) or 25 (Timer2).  You'll need to 
set the timer's mode registers for Ck/1024 prescaling, the compare match 
mode (to toggle the pin), and an appropriate output compare value for 
half of the period.  If the timer counts at Ck/1024, half of a 20Hz 
period would be represented by a compare count of 
(7372800/1024/20/2)=180; if the output pin is toggled accordingly, the 
result is 20Hz.  Here is a Timer2 implementation:

Sub Put20HzOnPin25()
  Const T2mode As Byte = bx0001_1111 'flip 25, clear, 7200Hz
  Register.TCCR2 = 0 'stop Timer2
  Register.TCNT2 = 0	
  Register.OCR2 = 180 '20Hz=(7372800/1024/20/2)=180
  Register.TCCR2 = T2mode
End Sub
	Tom
	

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Thanks Tom, that was exactly what I was looking for!
Tom

-----Original Message-----
From:	Tom Becker [mailto:gtbecker@gtbe...]
Sent:	Wed 3/8/2006 12:42 PM
To:	basicx@basi...
Cc:	
Subject:	Re: [BasicX] 20 Hz squarewave

> ... to produce a 20Hz squarewave...

I can suggest two methods.

You can run another task that sleeps for half of the 20Hz period, 
toggles (inverts) any pin, and sleeps again, repeating endlessly.  Task 
switching will result in some irregularity with this method.

sub TaskB()
  Do
   call sleep(0.5/20.0)
   register.PortD = register.PortD xor bx1000_0000 'Red, pin 25
  Loop
end sub

You can program Timer1 or Timer2 to produce 20Hz, exactly, independent 
of firmware, on pins 26 or 27 (Timer1) or 25 (Timer2).  You'll need to 
set the timer's mode registers for Ck/1024 prescaling, the compare match 
mode (to toggle the pin), and an appropriate output compare value for 
half of the period.  If the timer counts at Ck/1024, half of a 20Hz 
period would be represented by a compare count of 
(7372800/1024/20/2)=180; if the output pin is toggled accordingly, the 
result is 20Hz.  Here is a Timer2 implementation:

Sub Put20HzOnPin25()
  Const T2mode As Byte = bx0001_1111 'flip 25, clear, 7200Hz
  Register.TCCR2 = 0 'stop Timer2
  Register.TCNT2 = 0      
  Register.OCR2 = 180 '20Hz=(7372800/1024/20/2)=180
  Register.TCCR2 = T2mode
End Sub
	Tom
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> > ... to produce a 20Hz squarewave...
> ... Task switching will result in some irregularity with this method.

I spent some time looking at a couple of techniques to generate a 20Hz
squarewave in software.  The simplest, looping around a Sleep(0.025),
produces a small rate error because 0.025 second isn't an integral
tick count; the resulting rate is pretty stable, but wrong.  Another
method slaves the rate to the RTC; that results in an exact short-term
_average_ rate; the resulting rate is jittery, but correct.

I ran both techniques, each in a task, and drove the same LED with
them, simultaneously.  Since each task XORs the LED bit, the LED
displays the beat frequency, the difference between the two
techniques.  The difference appears as a slow intensity change of the
LED, actually 20Hz PWM.  The intensity cycles in about three seconds,
implying that the rate difference is about 0.3Hz, or about 2% error,
if my math is right.
	Tom
	      dim TaskAStack(1 to 31) as byte
      dim TaskBStack(1 to 31) as byte

sub Main()
      register.DDRD = register.DDRD or bx1010_0000   'LEDs out
      register.PortD = register.PortD or bx1010_0000 'LEDs off
     CallTask "TaskA", TaskAStack
     CallTask "TaskB", TaskBStack
      Do
          call Sleep(-1)
      Loop
end sub

sub TaskA()
	const iFreq as integer = 20
      dim iThen as integer, iNow as integer
      Do
		do
			iNow = (cint(register.RTCTick mod 512) * 2 * iFreq) \ 512
		loop until iNow <> iThen 'average rate is exactly 40Hz
		iThen = iNow
         register.PortD = register.PortD xor bx1000_0000      'flip LED
      Loop
end sub

sub TaskB()
	Do
		call sleep(0.025)	'rate is approximately 40Hz
         register.PortD = register.PortD xor bx1000_0000      'flip LED
	Loop
end sub
	

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Another academically interesting exercize, I think:

The code below runs two instances of the same task, each producing a
slightly different rate, set by Main before each task is launched.  If
you play with the SPeriod values, you'll find the effect of the tick
increment.
	    dim TaskStack1(1 to 31) as byte
    dim TaskStack2(1 to 31) as byte
    dim sPeriod as single

sub Main()
       register.DDRD = register.DDRD or bx1010_0000   'LEDs out
       register.PortD = register.PortD or bx1010_0000 'LEDs off
	sPeriod = 0.025  'rate is approximately 40Hz
      CallTask "TaskB", TaskStack1
	sPeriod = 0.024  'rate is quicker
      CallTask "TaskB", TaskStack2
       Do
           call Sleep(-1)
       Loop
end sub

sub TaskB()
	dim sP as single
	sP = sPeriod
	 Do
		 call sleep(sP)
          register.PortD = register.PortD xor bx1000_0000      'flip LED
	 Loop
end sub
	

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Even simpler.  Main becomes the second instance by a Call, instead of a 
second CallTask, so TaskB is run as both a task and as a endless sub of 
Main.  Nothing practical here, just amusing code.
	     dim TaskStack1(30) as byte
     dim sPeriod as single

sub Main()
        register.DDRD = register.DDRD or bx1010_0000   'LEDs out
        register.PortD = register.PortD or bx1010_0000 'LEDs off
	sPeriod = 0.025  'rate is approximately 40Hz
       CallTask "TaskB", TaskStack1
	sPeriod = 0.024  'rate is quicker
       Call TaskB
end sub

sub TaskB()
	dim sP as single
	sP = sPeriod
	 Do
		 call sleep(sP)
           register.PortD = register.PortD xor bx1000_0000      'flip LED
	 Loop
end sub

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