Determine exactly how many instruction cycles the delay loop in the code below (between the comment lines Begin Delay Loop and End Delay Loop) takes as a function of the variables Count1 and Count2.
**Please Explain**
Answer :- CBLOCK in PIC assembly has the same meaning as the ENUM in C language. So the value held by variable Count1 and Count2 are 32 and 33 respectively(decimal value).
The MOVLW, MOVF and NOP instructions take 1 MC,
GOTO instruction takes 2 MC,
DECFSZ takes 1 MC if result is not zero, else it takes 2
MC.
Determine exactly how many instruction cycles the delay loop in the code below (between the comment...
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 Problem 9: (10 Points) Find out the total time delay of the following code's...