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Chapter 2
1. 16
2. 8
3. 8
6. A = 15H + 13H = 28H (HEX) A = 0001_0101B + 0001_0011B = 0010_1000B (BINARY)
7. A and D
11. Low; high
12. Assembly Language uses less memory space then C and files are generally shorter.
13. Compiler
14. true
15. lst file
23. (a) MOV A, #55H - 2 bytes (b) MOV R3, #3 2 bytes (c) INC R2- 1 byte
(d) ADD A, #0 -2 bytes (e) MOV A, R1 -1 byte (f) MOV R3, A -1 byte
(g) ADD A, R2 -1 byte
37. (a) MOV A, #54H , ADD A, #0C4h CY=1
(b) MOV A, #00, ADD A, #0FFh CY=0
(c) MOV A, #250, ADD A, #05 CY=0

38. Add 25H and 70H and find the contents of the AC, CY, and P flags.
25h
70h
95h

AC=0
CY=0
P=0
42. 8 bits
43. bank 0
48. MOV SP, #52H stack starts at MOV A, #04 move 4 into acc MOV R0, #05H move value to r0 ADD A, R0 add and store to acc PUSH 0E0H push the value of acc.on the stack
PUSH 0 push value of r0 on the stack
SET PSW.4 select reg bank 2
POP 10H pop value at r0
0E0H pop the value.of 00
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Chapter 3
4. Short jump; 2 byte
5. long jump; 3 bytes
6. The LJMP instruction is 3-bytes in which the first byte is the opcode, and the second and third bytes represent the 16-bit address of the target location from 0000 to FFFFH. The SJMP is a 2-byte instruction where the first byte is the opcode and the second byte is the address location FFH
7. true
8. false
9. (c)LJMP

10. A short jump is a 2-byte instruction. Why?
The first byte is the opcode (e.g MOV, ADD, etc.) and the second byte is the relative address of the target location. 0-FFH MAX
11. true
14. The loop is executed 200x100 = 20,000 times
17. 3
18. 2
19. 2
20. 3
23. The reason is that the stack keeps track of where the CPU should return after completing the subroutine and must be balanced if PUSH and POP are used.
33. Find the time delay for the following delay subroutine, if the system has an 8051 with frequency of 11.0592 MHz.
DELAY: MOV R3, #200
HERE: NOP
NOP
NOP
DJNZ R3, HERE
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Chapter 4
8. input
11. MOV A, #0FFH ; A = FF hex
MOV P1, A ; make P1 an input port by writing all 1’s to it
BACK: MOV P1,A ; get data from P0 MOV P0,A ; send it to P0 MOV P2,A ; send it to P2 MOV P3, A ; send it to P3 SJMP BACK ; keep doing it
14. 0, because the chip has been reset.