Hi Bill,

Your program works OK but measure method is not enough precise.

The problem is that you make loop in 5 machine cycles (2.5 us for 24 MHz oscillator) when you wait the start and end of the measure period. So theoretically you can get error in 5 us for measure 10 kHz (100 us period) or 5 percents. You got (10140-9896)*100/10000 = 2.5 percents. Good result! What did you wish else?

Well to solve problem you can use several periods of the signal to reduce measure error like in AN426.PDF.

Be careful, I have just now seen AN426 and found out logical mistake there.

Instead of using this fragment

ORG 00000H
SJMP START ;RESET VECTOR
ORG 0000BH ;TIMER/COUNTER INTERRUPT ROUTINE
TIMER0: INC R3 ;INCREMENT R3 (3rd BYTE OF TIMER)
MOV A,#008H ;TEST FOR TIME OUT (R3 > 08)
SUBB A,R3 ;IF R3 > 08, CARRY IS SET
RETI

you can use following (for several periods)

TIMER0: DJNZ R3,NOEND
CLR TR0
NOEND: RETI

Don't forget remainder in TH0 & TL0.

I hope it helps you a little. Please email me the result.

Bye,

George