Question

Problem 9: (10 Points) Find out the total time delay of the following code's execution: (The instruction cycle is 4 of the oscillator cycle, and assume the CPU is running on 40MHz crystal and TMP1 and TMP2 are user defined DM address) MOVLW MOVWF MOVLW MOVWF NOP NOP DECFSZ GOTO NOP NOP DECF BTFSS GOTO OXEE TMP2 OXAA TMP1 LOOP2 LOOP1 TMP1,F LOOP TMP2,F STATUS,Z LOOP2

Answer 1

Let y urdes Stand the a part by part The irst Instruchion, ovito mvesth value oxeF to The wosking neiste The hezt Statemert hove th value to TMPA "Total aGK tarregZgr these tao3netructiong Similarly, the hext too hdructo% untu tie loop label, hoe すhe value owe to mpl variable. "This taka 2rore Cycles Loopl lahelak yes In back inteixter this This takes I Cycle as TMPI ot enpty n Case TMP TMP-howん38 Value OxAA "Total operatiorys Irnik loop l noo tak a cycles "Ifǐ8 Ltopl i, exte uted "Therefore, tota cycle, bom th Block,以#x3+1 (fiv tte la사 gcle then value becomes o):17 (Converted DXAb her to cefrul) +1 FA Cycles. Then the nert tuo NopS take 2 Cycles

te Satus and charges ittz l take, I cycle to exeaute The pevfous twtncho Consume cycles; Given each initiuction takeatre-fourth tr 25% of ozgtation Cycle, each ydle thsuco ycle time( 146 p 0.00625 Micyo Secordk Total dklay to execule Ist Instructions E [8IX0 0065