Question about modems

[Brent Hilpert]

On 2019-Nov-14, at 10:23 PM, Jim Brain via cctalk wrote:
> If you look at the values received by an 8N1 connection from a sender using the different settings, you get:
> 
> 	AT
> 	at
> 	At
> 	aT
> 7E1
> 	E174
> 	41D4
> 	E1D4
> 	4174
> 7O1
> 	61F4
> 	C154
> 	6154
> 	C1F4
> 7M1
> 	E1F4
> 	C1D4
> 	E1D4
> 	C1F4
> 7S1
> 	6174
> 	4154
> 	6154
> 	4174
> 8N1
> 	6174
> 	4154
> 	6154
> 	4174
> 
> Obviously, still trying to find the magic boolean logic equation to tease out the parity, but you could brute force it with these values and only aT would cause you issues requiring looking at CR (7E1 would send 8d, while 7S1/8N1 would send 0d.


(If it is of any consequence at this point)
If those vertical groups of 4 are intended to correspond to the first group of AT/at/At/aT, then you have the case bit (0x20) inverted, uppercase are bit 0x20 OFF (lower value), lower case are 0x20 ON (higher value).

If it's of any interest, here's some (untested) code, using your finding that the case must be consistent:

===============================================

/* Return:	-1	the two characters are not ASCII A/a and T/t
 *		-2	the case is not consistent between the two characters
 *		 0	SPACE parity
 *		 1	ODD parity
 *		 2	EVEN parity
 *		 3	MARK parity
 */
int
AssessATParity( chA, chT )
	int	chA, chT;
 {	int	pp;

	if( (chA&0x5F)!='A' || (chT&0x5F)!='T' )	// check characters are A/a and T/t
		return( -1 );

	if( ((chA^chT)&0x20) != 0 )			// check case is consistent
		return( -2 );

	pp = ((chA&0x80)>>6) | ((chT&0x80)>>7);		// extract and combine the two parity bits into two-bit value (0..3)

	if( pp==0 || pp==3 )				// check for Space or Mark parity
		return( pp );

	return( (chA&0x20) ? pp : (3-pp) );		// Odd or Even parity, but pp value (1 or 2) must be swapped depending on case
 }

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