1. ME 4447/6405 Microprocessor Control of Manufacturing Systems and
Introduction to Mechatronics
Instructor: Professor Charles Ume
Lecture #11

2. Homework Set #1 Solution
1. Write an assembly language program to clear the usable internal RAM in the
MC9S12C32 [$0800 to $0DFF]
Solution: Clear Locations $ $0DFF.
ORG $1000
LDX #$0800
LOOP CLR $00,X
INX
CPX #$0E00
BNE LOOP
SWI
END
(NOTE: You should hand-assemble this program.)

3. Homework Set #1 Solution
2. Write a program to add even/odd numbers located in addresses $0000 through $00FE.
Solution: (for adding odd numbers)
ORG $1000
LDY #$ CD 0000
LDX #$ CE 0000
IND LDAA #$FF FF
LDAB $00,X E6 00
BITB #$01 100A C5 01
BEQ EVEN 100C
ABY 100E 19 ED
EVEN INX
STX $ E 2000
EORA $ B8 2001
BNE IND ED
SWI F
END OR
DECA 4A
If you use
this, then
IND should
be moved
down to the
next line
1st Time
$2001:
Acc A:

4. Homework Set #1 Solution
2. Write a program to add even/odd numbers located in addresses $0800 through $0900. Solution: (for adding odd numbers)
ORG $1000
LDY #$ CD 0000
LDX #$ CE 0800
IND LDAB $00,X E6 00
BITB #$ C5 01
BEQ EVEN 100A
ABY 100C 19 ED
EVEN INX E 08
CPX #$ F 8E 0901
BNE IND ED
SWI F
END

5. Homework Set #2 Solution
Write an assembly language program to find the largest signed number in a list of numbers
stored in address $0020 thru $0029, and store it in $ Repeat for unsigned number.
FIRST EQU $0020
LAST EQU $002A
ANSWR EQU $2000
ORG $1000
LDX #FIRST
LDAA FIRST
STAA ANSWR
INX Goes with DECB. If DECB is removed, then
LDAB #$0A LDAB must be removed
NUB LDAA $00,X
CMPA ANSWR (A - ANSWR)
BLE NEXT; (signed)*
*(Branches if 2's complement number in register A is less than or OR equal to the 2's complement number represented by ANSWR) BLS NEXT; (unsigned)** **(Branch if lower or same, i.e, Branch if register A was lower or same as ANSWR) STAA ANSWR
NEXT INX
CPX #LAST OR DECB
BNE NUB
SWI
END

6. Homework Set #2 Solution Continued
Signed Example:
#$3F (positive)
#$F4 (negative)
Convert Numbers to Decimal:
#$F4 = %
1's Comp #$0B = %
2's Comp #$0C = %
#$F4 = -1210
#$3F = %
#$3F = 6310
#$3F is larger than #$F4 since 6310 is larger than -1210
Un-Signed Example:
#$3F (positive)
#$F4 (positive)
Convert Numbers to Decimal:
#$F4 = %
#$F4 = 24410
#$3F = %
#$3F = 6310
#$F4 is larger than #$3F since is larger than 6310

7. Homework Set #3 Solution
1. Write a subroutine to save the first 5 odd (8-bit) numbers pointed to by the x-register (passed in) onto the stack.
Solution:
SUB1 LDAA #$05
PULY
NEXT BRSET $00,X #$01 ODD
INX
BRA NEXT
ODD LDAB $00,X
PSHB
DECA
BNE NEXT
PSHY
RTS
Note: For subroutine to execute in main program, a BSR or JSR command must be used in main program.

8. Homework Set #3 solution
2. Write a program to output a square wave thru port S pin The output can be observed on the scope, and the period T of the wave should be measured. More than one period wave should be generated. The machine cycle time of the MC9S12C32 should be estimated. Draw the square wave.

9. Homework Solution Cont’d
ORG $2000
FCB $XX
FCB $04
ORG $1000
LDAA #$04
STAA $024A
LDAB #$04
CYCLE LDAA $2001 3
STAA $0248 3
EORA #$04 1
STAA $2001 3
LDAA $2000 3
NEXT DECA 1
BNE NEXT 3/1
DECB 1
BEQ EXIT 3/1
JMP CYCLE 3
EXIT SWI
END
Address
Opcode
Postbyte
Operand
1002 7A 024A
1005 C6 04
1007 B
100A 7A 0248
100D
100F 7A 2001
1012 B FD
101B
101E 3F
Configures PS2
as output.
Helps us to output two
periods.
Outputs signal from PS2
Switch $2001(toggle)
Delay
(Note: BNE and BEQ take 3 machine cycles to execute if the branch is taken and one machine cycle if it is not taken)
T/2=Machine Cycle Time*[ COUNT*(1)+(COUNT-1)* ]
(Note: Count = #$XX)

10. ASCII Character Codes

11. Examples: ASCII Character Codes
ASCII – American Standard Code for Information Interchange
ASCII assigns a hexadecimal ‘code’ to individual characters
Examples:
Character ASCII
‘A $41
‘E $45
‘e $65 (Note: Different codes for lower and upper case)
‘1 $31
BS $08 (Note: BS is Backspace. )
A microcontroller must send these codes to a display terminal in order for the terminal to display these characters.

12. ASCII Character Codes (continued)
Hex to ASCII Conversion table from Programming Reference Guide Page 58

13. An array of characters is called a string
ASCII Character Codes
An array of characters is called a string
Example:
character array String ASCII Representation
‘H ‘e ‘l ‘l ‘o “Hello” $48 $65 $6C $6C $6F
(Note:
Character ASCII
‘H $48
‘e $65
‘l $6C
‘o $6F )

14. Assembly Directives

15. Assembly Directive Types
Assembly Control
ORG, END
Symbol Definition
EQU
Data Definition/Storage Allocation
FCC, FCB, FDB, RMB, ZMB, BSZ, FILL
Listing Control
PAGE, OPT

16. Assembly Control

17. ORG and END
ORG : Store translated machine language instructions in sequence starting at given address for any mnemonic instructions that follow
END: Stop translating mnemonics instructions until another ORG is encountered
(Note: These were already discussed in Lecture 7)

18. Symbol Definition

19. EQU EQU lets you refer to a number or an address as a variable name.
Example:
VALA EQU $10 *LABEL VALA USED TO REFER TO $10
ORG $1000
LDAA #VALA *LOAD HEX NUMBER $10 IN ACCUMULATOR A
LDAB VALA *LOAD CONTENT OF MEMORY LOCATION $10
*IN ACC. B
SWI
END
Same As:
LDAA #$10 *LOAD HEX NUMBER $10 IN ACCUMULATOR A
LDAB $10 *LOAD CONTENT OF MEMORY
*LOCATION $10 IN ACC. B

20. Data Definition/Storage Allocation

21. FCC FCC – Form Constant Character string
FCC stores ASCII characters into consecutive bytes of memory.
Any printable ASCII characters can be contained in the string.
String is specified between two identical delimiters, which can be any printable ASCII character.
First non-blank character after the string is used as a delimiter.

22. FCC Cont’d Result Example: ORG $0400 FCC “12345” Address Prebyte
Opcode
Operand
$0400 31 32
$0401 33
$0402
$0403 34
$0404 35

23. FCB FCB – Form Constant Byte FCB has one or more operands.
Value of each operand is truncated to eight bits, and is stored in single byte of object program.
Operand may be a numeric constant, character constant, a symbol or an expression.
Multiple operands are separated by commas, and are stored in successive memory
bytes.

24. FCB Cont’d Result Example: VALA EQU $10 ORG $0400
FCB $34,’A, $28AC, $0A ,VALA
Address
Prebyte
Opcode
Operand
$0400 34 41
$0401 AC
$0402
$0403 0A
$0404 10

25. FDB FDB – Form Constant Double Byte
FDB stores a double (two byte) word.
May have one or more operands separated by commas.
Operand may be a numeric constant, a character constant, a symbol, or an expression.

26. Note: ASCII value for & is $26
FDB Cont’d
Result
Example:
ORG $0400
FDB $1234,’&,’G
Address
Prebyte
Opcode
Operand
$0400 12
$0401 34 00
$0402
$0403 26
$0404 00
$0405 47
Note: ASCII value for & is $26
ASCII value for G is $47

27. RMB Result RMB – Reserve Memory Byte
RMB saves a place in memory for a number.
Example:
ORG $0400
XVAR RMB *TWO MEMORY
*LOCATIONS $0400
*and $0401 ARE
*RESERVED FOR XVAR
ORG $1000
LDD #$FFAA
STD XVAR
SWI
END
Result
Address
Prebyte
Opcode
Operand
$0400 FF AA
$0401

28. Question a student asked:
What happens if you change the previous example to "XVAR RMB 3" instead of "XVAR RMB 2" ? What happens to the 3rd reserved byte when a 2 byte number is stored in XVAR?
Answer: Remains unchanged
Modified Program:
ORG $0400
XVAR RMB 3
ORG $1000
LDD #$FFAA
STD XVAR
SWI
END

29. ZMB, BSZ ZMB – Zero Memory Byte and BSZ – Block Storage of Zero
These directives fill a given number of memory locations with zero.
Causes assembler to allocate a block of memory bytes, and each memory byte is assigned a value of zero.
Both directives do the same thing.
Number of bytes allocated is given in the operand field.
Result
Example:
ORG $0400
ZMB #$02
BSZ #$02
Address
Prebyte
Opcode
Operand
$00 00
$01 00
$02 00
$03 00

30. FILL Fill given number of memory locations with any
number. (Note: Fill uses one byte. If two bytes are specified, then it will truncate it and use LS Byte.)
Example:
ORG $0400
FILL #$FF, #$02
Result
Address
Prebyte
Opcode
Operand
$00 FF FF
$01

31. Question a student asked:
What happens when the previous example is changed to "FILL #$9ABC, #$02" instead of "FILL #$FF,#$02"? What happens if you fill memory with a 2 byte number?
Answer: FILL will just use the LS Byte
Modified Example:
ORG $0400
FILL #$9ABC,#$02
END
(Note: There is no ”go 1000” on the screen since these are just assembly
directives and not a program)

32. Listing Control

33. PAGE PAGE The PAGE directive causes a page break in the list file.
If no source listing is being produced, the PAGE directive will have no effect.
The directive is not printed on the source listing.

34. OPT OPT Allows for various options in assembly of a program, including
generating a listing and counting instruction cycles.
Options:
nol-no output listing (default)
l-do an output listing
noc-no cycle number count (default)
c-turn on cycle count using zero initial value
contc-turn cycle count on, begin with last value
cre-create a cross reference table (default anyway) RMB
s-create a symbol table (default anyway) EQU
Example:
OPT l – Print source listing from this point

35. Homework 3 Solution Solution:
Write a subroutine to save the first 5 odd (8-bit) numbers pointed to by the x-register (passed in) onto the stack.
Solution:
SUB1 LDAA #$05
PULY
NEXT BRSET $00,X #$01 ODD
INX
BRA NEXT
ODD LDAB $00,X
PSHB
DECA
BNE NEXT
PSHY
RTS
Note: For subroutine to execute in main program, a BSR or JSR command must be used in main program.

36. Homework 4 solution
Write a program to output a square wave thru port S pin 2. The output
can be observed on the scope, and the period T of the wave should
be measured. More than one period wave should be generated. The
machine cycle time of the MC9S12C32 should be estimated. Draw
the square wave.

37. Homework 4 Solution Cont’d
ORG $2000
FCB $XX
FCB $04
ORG $1000
LDAA #$04
STAA $024A
LDAB #$04
CYCLE LDAA $2001 3
STAA $0248 3
EORA #$04 1
STAA $2001 3
LDAA $2000 3
NEXT DECA 1
BNE NEXT 3/1
DECB 1
BEQ EXIT 3/1
JMP CYCLE 3
EXIT SWI
END
Address
Opcode
Postbyte
Operand
1002 7A 024A
1005 C6 04
1007 B
100A 7A 0248
100D
100F 7A 2001
1012 B FD
101B
101E 3F
Configures PS2
as output.
Helps us to output two
periods.
Outputs signal from PS2
Switch $2001(toggle)
Delay
(Note: BNE and BEQ take 3 machine cycles to execute if the branch is taken and one machine cycle if it is not taken)
T/2=Machine Cycle Time*[ COUNT*(1)+(COUNT-1)* ]
(Note: Count = #$XX)

38. QUESTIONS???