Question

The following program is to have a time delay. Calculate and tell how much the time delay is. The CPU frequency is 24-MHZ.

     org $1000 ; starting address of the program

     ldx #80000     ; 2 E cycles

loop psha             ; 2 E cycles

     pula       ; 3 E cycles

     psha       ;

     pula       ;

     psha       ;

     pula       ;

     psha       ;

     pula       ;

     psha       ;

     pula       ;

     psha       ;

     pula       ;

     nop        ;     1 E cycle

     nop        ;

     dbne x,loop;     3 E cycles       

     swi

      end

Answer 1

 ldx #80000      
loop       psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           psha              ; 2 E cycles
           pula              ; 3 E cycles
           nop               ; 1 E cycle
           nop               ; 1 E cycle
           dbne x,loop       ; 3 E cycles
           dbne x,loop

To delay by 100 millisec, that is 24,000,000 * 100 / 1,000 EClocks, namely 2,400,000 instruction cycles.

The maximum register size available is 16-bits, so a loop counter value is chosen that is <= 65535.

Conveniently 60,000 is a factor of 2,400,000 being 60,000 * 40. So the inner loop is contrived to take 40 cycles.

Giving the required 40 cycles execution time.if you have interrupts, other processes, this hard coded method is not very accurate, and a timer interrupt would be better.

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