keybd INPUT yes screen OUTPUT lf ~\n 1 nolist fmt \n\N yes junk 1 yes general 2 degrees accum Value in accumulator at entry to loop 200 cycle This contains the bus cycle time in seconds 5E-07 home This web computes the time required to execute a Motorola 68HC705J1a wait loop, given the value in the accumulator on entry. The system clock is assumed to be 4.0MHz, resulting in a 500 nanosecond cycle time. The timing loop, with cycles per instruction is: Await TAX ; 2 Olp TXA ; 2 TSTA ; 3 BNE Ilp ; 3 RTS ; 6 Ilp SUB #1 ; 2 NOP ; 2 NOP ; 2 NOP ; 2 NOP ; 2 BNE Ilp ; 3 DECX ; 3 BRA Olp ; 3 Time per execution of the outer loop is: 2+3+3+3+3=14 Time per execution of the inner loop is: 2+2+2+2+2+3=13 INSTRUCTIONS: 1) Place the initial accumulator value into the value field of cell accum. 2) Select Execute from the Special menu. 3) Read the answer in seconds in the value field of the home cell. 0.132061 indx Index of for loop for computing total bus cycles 0 innerCycles Number of bus cycles in one repetition of the inner loop 13 outerCycles Number of bus cycles attributable to outer loop code in one repetition of the outer loop 14 totalCycles This variable accumulates bus cycles during computation 264122