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davepusey said:From what i've just read in the datasheet the Clock Freq = Instruction Rate * 4
Am i correct. If I am that would mean that 22,050 instructions per secound * 4 = 88,200Hz clock freq.
Is all that correct? I hope so!
davepusey said:It takes to 2 intruction cycles to output each sample. So i figured I needed an excetion rate that is twice that of the sample rate.
Nigel Goodwin said:davepusey said:it would take 2 instruction cycles to output a single sample - any more will take considerably longer. You've got to jump back to the start of the loop, read the data from where ever it's stored, updating the address counter at the same time - loads more than 2 instructions!.
Not in my program (see attachment).