ME4447/6405 The George W. Woodruff School of Mechanical Engineering ME4447/6405 Microprocessor Control of Manufacturing Systems and Introduction to Mechatronics

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The George W. Woodruff School of Mechanical EngineeringThe George W. Woodruff School of Mechanical Engineering

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Microprocessor Control of Manufacturing Systemsand

Introduction to Mechatronics

Instructor: Professor Charles Ume

Lecture #9

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Reading Assignments

Reading assignments for this week and next week Read Chapters 5-8 in Basic Microprocessors and the 6800, by Ron

Bishop.

Chapter 5 Microcomputers-What Are They?Chapter 6 Programming ConceptsChapter 7 Addressing ModesChapter 8 M6800 Software

There will be questions and answers the rest of this week and next week based on your reading assignment.

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HC12S CPU can only understand instructions written in binary called Machine Language.

Writing programs in Machine Language is extremely difficult

Mnemonics are simple codes, usually alphabetic, that is representative of instruction it represents (example: LDAA [LoaD Accumulator A])

A program written using Mnemonic Instructions is called Assembly Language program

An Assembler can be used to translate Assembly Language program to Machine Language Program, and put it in S-Record Format.

Why use Assembly Language?

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Address: Common term for memory location. Always written in hexadecimal. $4000 is the 4000th

16 Memory Location (Note: “$” signifies hexadecimal)

Literal Value: A number used as data in program indicated by “#”. Can be represented in the following ways:

#$FF = hexadecimal number FF#%1011 = binary number 1011 (Note: “%” signifies binary)#123 = decimal number 123

A Literal Value can be stored in an addressExample: Literal Value #$FF can be stored in address $4000

Example 2: Literal Value #$FE0A is stored in address $2000 (Note: #$FE is stored in address $2000 #$0A is stored in address $2001)

Assembly Language Notations

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Directives: Instructions from the programmer to Assembler NOT to microcontroller

Example 1: ORG <address>

Store translated machine language instructions in sequence starting at given address for any mnemonic instructions that follow

Example 2: END

Stop translating mnemonics instructions until another ORG is encountered

(Note: More will be covered in later lectures)

Assembly Language Directives

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Assembly Language FormatLe

ft m

argi

n of

ass

embl

y pr

ogra

m

A Tab (8 white spaces) or Label(Note: A Label is another Assembly Directive and will be covered in later lectures)

Assembly Directive Or Mnemonic Instruction

(Note: Last three options are called Operands)

Data that the Assembly Directive uses

OrBlank if Mnemonic Instruction does not need Data

Or Offset Address used to modify Program Counter by a Mnemonic Instruction

OrData that Mnemonic Instruction uses

Or Address where the Data that Mnemonic Instruction will use is stored

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ORG $0800LDAA #$100A- -BNE Front 27 0ELDAB $2F,YSTAB $110C- -

Front INCX- -SWIEND

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In the previous slide, there were several options for the operand:• Blank if Mnemonic Instruction does not need Data• Offset Address used to modify Program Counter by a Mnemonic

Instruction• Data that Mnemonic instruction uses• Address were Data that Mnemonic instruction uses is stored

Which option a programmer uses is defined by the following addressing modes:

Addressing Modes

•Direct•Indexed•Relative

•Inherent•Immediate•Extended•Indexed Indirect

(Note: All instructions are not capable of all addressing modes.Example: BLE [Branch if Less than or Equal to Zero] is only capable of Relative addressing mode)

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Example: Programming Reference Guide Page 6

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Blank if Mnemonic Instruction does not need Data

If Mnemonic Instruction does not need data then it uses Inherent Addressing Mode

Example: Write a program to clear accumulator A. Start programming at address $1000

Solution:

ORG $1000

CLRA

SWI

END

CLRA [ CLeaR accumulator A] is an instruction using Inherent Addressing

(NOTE: SWI [SoftWare Interrupt] is a mnemonic instruction which tells the 9SC32 to store the content of cpu registers on the stack. Sets the I bit (the interrupt bit) on the CCR. Loads the program counter with the address stored in the SWI interrupt vector, and resume program execution at this location. If no address is stored in the SWI vector, the main program will stop execution at this point. Used in this course to return control to Mon12 Program)

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Data that Mnemonic instruction uses

The Mnemonic instruction is using Immediate Addressing mode if the operand is Data used by the instruction

Example: Write a program to load accumulator A with #$12. Start programming at address $1000

Solution:

ORG $1000

LDAA #$12

SWI

END

LDAA is an instruction using Immediate Addressing mode in this example

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Address were Data that Mnemonic instruction uses is stored

The following addressing modes apply if the operand is an Address containing Data used by Mnemonic instruction :

Direct•Data is contained in Memory locations $00 to $FF•Address is given as a single byte address between $00 to $FF •Instructions using Direct addressing has fastest access to memory

Example: LDAA $00 Loads accumulator A with Data value stored at memory location $00

Extended•Data is contained in Memory locations $0100 to $FFFF •Address is given as a two byte address between $0100 to $FFFF

Example: LDAA $2000Loads accumulator A with Data value stored at memory location $2000

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Example Problem 1

Example : Write a program to add the numbers 1010 and 1110.

Solution

ORG $1000

LDAA #$0A *Puts number $0A in acc. A

LDAB #$0B *Puts number $0B in acc. B

ABA *Adds acc. B to acc. A

STAA $00 *Stores results in address $00

SWI *Software interrupt

END

LDAB and LDAA use immediate addressing mode

STAA uses direct addressing mode

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Address were Data that Mnemonic instruction uses is stored (Continued)

Indexed:• Data is located within Memory locations $00 to $FFFF

Example: Store content of $2003 in Register A

LDX #$2000 LDAA $03,X

Loads accumulator A with Data value stored at memory location $2003

X + $03 = $2000 + $03 = $2003

(Note: LDX [ LoaD index register X])

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Why is Indexed Addressing Mode needed?

Example: Store Data Value #$20 into memory locations $2000 to $3000

Without Indexed Addressing Mode With Indexed Addressing Mode

ORG $1000LDAA #$20STAA $2000STAA $2001 . . .STAA $3000SWIEND

ORG $1000LDAA #$20

LDX #$2000LOOP STAA $00,X

INXCPX #$3001BNE LOOPSWIEND

Program on the Left is much longer than program on Right

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ORG $1000LDY #$1001LDAA #$20


INXDECYBNE LOOPSWIEND

Why is Indexed Addressing Mode needed?


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Why is Indexed Addressing Mode needed? (Continued)

ORG $1000LDAA #$20



Note: LDX [ LoaD accumulator A]STAA [ STore Accumulator A]INX [ INcrement X] CPX [ ComPare X]BNE [Branch if Not Equal] ( is using relative addressing in

conjunction with label “LOOP”)

LOOP, BNE LOOP, INX, and CPX #$3001 creates a loop.

Loop1: Data in accumulator A (#$20) is stored at $2000 + $00 Data in X is incremented #$2000 + #$0001 = #$2001 Data in X is compared to #$3001

Not equal so do another loop

Loop2: Data in Accumulator A (#$20) is stored at $2001 Data in X is incremented #$2001 + #$0001 = #$2002 Data in X is compared to #$3001

Not equal so do another loop

Etc…..

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Types of Indexed modes of Addressing

Indexed addressing may be implemented in multiple ways. HCS12 CPU uses X, Y, SP or PC as base index register for instruction. Offset is then added to base index register to form effective address.

Constant offset5-, 9- or 16-bit signed offset (-16 to +15, -256 to +255, -32768 to +32767)

5-Bit Constant Offset Indexed AddressingIndex mode uses 5-bit signed offset which is added to base index register (X, Y, SP or PC) to form effective address of memory location that will be affected by instruction. Offset ranges from -16 through +15. Majority of indexed instructions in real programming use offsets that fit in shortest 5-bit form of indexed addressing. Contents of base registers remain unchanged.

LDAA $00, X *load A with (X + $00)

The following three statements are equivalent:STAA -8,X Note: offset given in decimalSTAA -$08,X Note: offset given in hexSTAA $FFF8,X Note: offset given as 16-bit number

Let X contain #$3000. After program executed, content of A will be stored at address (#$3000 - #$08) = $2FF8

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9-Bit Constant Offset Indexed Addressing.

Uses 9-bit signed offset which is added to base index register (X, Y, SP or PC) to form effective address of memory location affected by instruction. Offset ranges from -256 through +255. Contents of base registers are not changed after instruction is executed. MSB (sign bit) of offset is included in instruction postbyte and remaining 8 bits are provided as extension byte after instruction postbyte in instruction flow.

LDAA $FF, X *Assume X contains $1000 prior to instruct. Execut.A6 E0 FF

LDAB -20, Y *Assume Y contains $2000 prior to instruct. Execut.E6 E9 EC

First instruction will load A with value from ($1000 + $FF) = $10FFSecond instruction will load B with value from ($2000 – 20) = $IFEC

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Accumulator Offset Indexed Addressing

In this indexed addressing mode, effective address is sum of values in base index register and unsigned offset in one of accumulator. Value in base index register is not changed. Indexed register can be X, Y, SP or PC and accumulator can be either 8-bit (A or B) or 16-bit (D)

– A, B, or D accumulator added to base index register to form address– Content of accumulator is unsigned offset

LDAA B, X

Instruction adds B to X to form address from which A will be loaded. B and X are not changed by this instruction.

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Types of Indexed modes of Addressing - Continued

•Auto Pre-/Post-Increment/Decrement•Base index register may be automatically incremented/decremented before or after instruction (Program Counter may not be used as base register)•No offset is available•May be incremented/decremented 1 to 8 timesPost-Increment (in ranges from 1 through 8):

LDX 2,SP+ *Index register X is loaded with contents of (Same as PULX) memory location in stack pointer, then stack

pointer is incremented twicePre-Increment (in ranges from 1 through 8):

LDX 2,+SP *Stack pointer is incremented twice, then Index register X is loaded with contents

of memory location in stack pointerPre-Decrement (in ranges from -8 through -1):

STAA 1,-X *Index register X is decremented, then Accumulator A is stored in memory location stored in Index Register X

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Indexed Addressing Mode Postbyte Encoding (xb)

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Indexed Addressing Mode Postbyte Encoding (xb) - Continued

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Why is Pre-/Post-Increment/Decrement Useful?


Without Post-Increment With Post-Increment

ORG $1000LDAA #$20



ORG $1000LDAA #$20

LDX #$2000LOOP STAA 1,X+

CPX #$3001BNE LOOPSWIEND

Program on the Left requires 1 more byte of program memory and takes 1 more cycle to execute per run through the loop than the program on the right. This may make a large difference when the program is large and complex or when dealing with values larger than 16-bits.

Note: “1” refers to the number of post increments, not an offset!

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•Indexed-Indirect [Differ Discussion of this for Now]•Like other indexed addressing modes, offset and base index register are added to form an effective address•Offset is either Accumulator D or a 16 bit constant offset•However, effective address is understood to contain address of memory location containing address to be acted upon.

Example: Clear the contents of the memory location pointed to by a pointer located in memory location $2000

ORG $2000PTR RMB 2 *Note: The user places the address of

the memory location to be cleared here ORG $1000

LDX $2000CLR [$00,X]SWIEND

(Note: CLR [CLeaR Memory Location])

•Useful for efficiently forming switch case statements and other program flow operations in high level programming languages

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Example: Hex number is stored in $2000

LDD #$2000 *Loads accumulator D with #$2000

STD $100A *Stores content of Register D in Addresses $100A and $100B

LDX #$1000 *Loads Register X with #$1000

LDAA [10,X] *#$0A is added to #$1000 in X-Register to form $100A. Next, CPU *goes to memory (address) locations $100A and $100B and fetch *address pointer $2000. Value stored in address $2000 is read and *loaded into accumulator A.


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As stated before the Assembler translates an assembly language program into a machine language program

Format of machine language program

Address where instruction is located

OperandOr

Blank if instruction does not use Operands

Instruction

Opcode

(Note: This format is for Lecture and Tests only !! The real format the assembler outputs is “S19” and will be shown to you in Lab)

Postbyte

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All Mnemonics and associated Op-codes can be found in Programming Reference Guide pages 6-19Example: Programming Reference Guide Page 12 (Note: LDAA outlined in red)

Postbyte and Opcode Reference

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Indexed Addressing Mode Postbyte Encoding (xb)

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Indexed Addressing Mode Postbyte Encoding (xb) - Continued

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Postbyte allows an op-code to be used for more than one instruction.

Determined from Tables 1, 3 or 4 in the programming reference guide

Postbyte

Instruction Opcode Postbyte

LDAA 0,X A6 00

LDAA $02,SP+ A6 B1

LDAA B,Y A6 ED

SUBB $1040,X E0 E2

SUBB D,Y E0 EE

SUBB -14,X E0 12Table 1 (Excerpt)

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The George W. Woodruff School of Mechanical EngineeringThe George W. Woodruff School of Mechanical Engineering(Programming Reference Guide)

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ORG $1000

LDAA #$0A

LDAB #$0B

ABA

STAA $00

SWI

END

Hand Assembling

Example: Assemble the following Program

$1000 86 0A

$1002

(Note: $1002 since $86 is now at $1000 and $0A is at $1001)

C6 0B

$1004 18 06

$1006 5A 00

$1008 3F

Address Opcode Postbyte Operand

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Example Problem 1 (revisited)

ORG $1000LDAB #$0A *Load acc. B with number 0ASTAB $1100 *Store acc. B in address $1100INCB *Increment acc. B by 1ADDB $1100 *Add memory location $1100

*to acc. BSTAB $1090 *Store acc. B in address $1090SWI *Software interruptEND

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Hand Assemble Example Problem 1 (Revisited)

ORG $1000

LDAB #$0A

STAB $1100

INCB

ADDB $1100

STAB $1090

SWI

END

Address Opcode Postbyte Operand

1000 86 0A

1002 7B 1100

1005 52

1006 FB 1100

1009 7B 1090

100B 3F

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Example Problem 2

Write a short assembly language program that stores the contentof Port T in memory location $3000 after waiting for 0.05 secondsfor the input data.

Solution

Recall: One machine cycle = 0.125 x 10-6 s (8 MHz Bus Clock)

We want the HCS12 to wait 0.05 s/0.125 x 10-6 s = 400,000 cycles

One good way to make the HCS12 wait is to create a loop.

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Wait Loop

LDY #$AD9C 2 cycles

LDD #$0000 2 cycles

LOOP ABA 2 cycles

CPX $2000 3 cycles

DEY 1 cycle

BNE LOOP 3 cycles

Assume the number of loops needed to wait is 2 bytes

(2 + 3 + 1 + 3)*X + 4 = 400,000 cycles

X = 44,44410 = $AD9C

Note: These instructions are included to increase the operation time

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Example 2 Solution

Solution*Remember that Port T is input upon reset

ORG $1000

LDY #$AD9C *Load Y with 44,44410

LDD #$0000 *Clear Accumulator DLOOP ABA *Add the contents of B to A

CPX $2000 *Compare X with the contents of*$2000

DEY *Decrement YBNE LOOP *Branch to LOOP if Y is not equal

*to zeroLDAB $0240 *Load acc. B with content of $0240STAB $3000 *Store content of acc. B in $3000SWI *Software InterruptEND

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Homework

Homework 1• Write an assembly language program to clear the internal RAM in the

MC9S12C32.

• Write a program to add even/odd numbers located in addresses $0800 through $0900.

Homework 2

• Write a program to find the largest signed number in a list of numbers stored in address $0A00 through $0BFF. Repeat for an unsigned number.

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QUESTIONS???

Documents

ME4447/6405 The George W. Woodruff School of Mechanical Engineering ME4447/6405 Microprocessor Control of Manufacturing Systems and Introduction to Mechatronics