Question

Find the time delay for the delay subroutine shown below if the system has an ARM with a core clock frequency of 80 MHz: MOV

Using Assembly language
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Answer #1

Clock Frequency is 80MHz

80MHz/12 = 6.66MHz

1/6.66 = 0.150 us is the Machine Cycle.

MOV - 1 Machine Cycle

LDR - 1 Machine Cycle

NOP - 1 Machine Cycle

SUBS - 2 Machine Cycle

BNE - 1 Machine Cycle

SUBS - 2 Machine Cycle

BNE - 1 Machine Cycle

in the above code there are 2 loops. one is from here to here and other loop is from again to again.

so here: (1+2+1)*400=1600*0.15us = 240us

so again: (1+2+1)*200 = 800*0.15us = 120us

in this case here loop runs for 200 times so, 200*240 = 48000us

so the total delay is 0.15 + 48000 + 120 = 48120.15 us

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