"1250025000hz" - 1 õppematerjal

Mikrokontrollerid ja praktiline robootika

Counts read by register at 500 Hz =  x1 500 Hz Fs Counts read by proceessor at 500.01 Hz =  x2 500.01Hz Fs Fs The difference is 50 counts, then x1  x2  50   500 Hz 500.01Hz Or the sample clock frequency: Fs  1250025000Hz  1.25GHz In this case, the clock frequency must be 1.25GHz in order to achieve 50 count differences for a resolution of 0.0001 MHz at the input frequency of 5 MHz.