Show work!! (Use separate sheets if necessary) Name_AdaiA BRuN Fill in the reque
ID: 3596169 • Letter: S
Question
Show work!! (Use separate sheets if necessary) Name_AdaiA BRuN Fill in the requested register contents and flag status' as if the instructions were executed on the Z80 microprocessor. B (Hex) B (Dec) S Z CY LDB. 62H- SRA B SRA B (Signed Number) 2. LD B, 84H SRA B SRA B LD B, 16H SLA B SLAB LD DE, COUNT I 4. Compute delay time both including and excluding the instruction outside the loop for: a) COUNT=0 b) COUNT= 1 DLOOP: DEC DE LDA, E 4 OR D4 JP NZ, DLOOP IO f=3MHz LD D, OCOUNT LD E, lCOUNT NOP 4 DEC Ey JP NZ, ILOOP 1o NOP 4 5. Compute ICOUNT so that inner loop delay is .1 msec. LOOP: LOOP: 6. Compute OCOUNT so that total delay is 20 msec. T 500 nsec DEC D JP NZ, OLOOP IO WWEllIExplanation / Answer
Z80 instruction definition and architecture is defined.
1)
LD B 62H
This is a load instrcuction and it will load the values into B which is 62 in hexadecimal.
So a value of 62 will be loaded into B.
B (Hex) will be 62
B (dec) will be 98
S - 0 (There is no sign change with operation)
Z - 0 (zero flag no change)
Carry flag (CY) - 0 (no change in carry)
SRA instrcution means arithemtic shift right. In this bit 7 remains unchanged and others bits are shited right by 1 and 0th bit is moved to carry flag.
Now B has 62 value which means 0110 0010
SRA B instruction result will be
B (hex) will be 31 (right shift means it will be divided by 2)
B (dec) will be 49 (again right shift)
S - 0
Z - 0 (zero flag is zero)
CY - 0 ( bit 0 moved to carry flag)
Again SRA B instruction is executed.
B has 31 value in hex which means in binary 0011 0001
Now SRA result will be
B(Hex) will be 18 (Right shift by 1)
B (decimal) will be 24
S flag will be 0
C - 0
CY - 1( as the last bit was 1 before the operation)
2)
Z80 instruction definition and architecture is defined.
1)
LD B 84H
This is a load instrcuction and it will load the values into B which is 84 in hexadecimal.
So a value of 84 will be loaded into B.
B (Hex) will be 84
B (dec) will be 132
S - 1 (MSB is set for this number)
Z - 0 (zero flag no change)
Carry flag (CY) - 0 (no change in carry)
SRA instrcution means arithemtic shift right. In this bit 7 remains unchanged and others bits are shited right by 1 and 0th bit is moved to carry flag.
Now B has 84 value which means 1000 0100
SRA B instruction result will be
B (hex) will be C2 (right shift means it will be divided by 2 but sign will remain as it is so it will be 1100 0010)
B (dec) will be 194 (again right shift)
S - 1
Z - 0 (zero flag is zero)
CY - 0
Again SRA B instruction is executed.
B has C2 value in hex which means in binary 1100 0010
Now SRA result will be
B(Hex) will be A1 (Right shift by 1 after shift 1010 0001)
B (decimal) will be 161
S flag will be 1
C - 0
CY - 0
3)
Z80 instruction definition and architecture is defined.
LD B 16H
This is a load instrcuction and it will load the values into B which is 16 in hexadecimal.
So a value of 16 will be loaded into B.
B (Hex) will be 16
B (dec) will be 22
S - 0 (MSB is set for this number)
Z - 0 (zero flag no change)
Carry flag (CY) - 0 (no change in carry)
SLA instrcution means arithemtic shift left. In this bit 7 remains unchanged and others bits are shited left by 1 and 7th bit is moved to carry flag.
Now B has 16 value which means 0001 0110
SLA B instruction result will be
B (hex) will be 2C (left shift means it will be multiplied by 2 but sign will remain as it is so it will be 0010 1100)
B (dec) will be 44 (again left shift)
S - 0
Z - 0 (zero flag is zero)
CY - 0 (bit 7th is moved to carry flag)
Again SLA B instruction is executed.
B has 2C value in hex which means in binary 0010 1100
Now SLA result will be
B(Hex) will be 58 (left shift by 1 after shift 0101 1000)
B (decimal) will be 88
S flag will be 0
Z - 0
CY - 0 ( as the 7th bit was 0 before the operation)
The numbers in the loop are coming from the instruction execution time and the carry flag criteria. This is calculated bsaed on the overflow.