Fair Record

8/4/2019 Fair Record

1/86

EMBEDDED SYSTEMS LAB

Expt. No: 1 BLOCK MOVE

Date: 12/07/11

AIM

To transfer data from memory block 1 to memory block 2.

TOOLS REQUIRED

Vision from Keil for an 8051 microcontroller family.

PROCEDURE

1. Create a project in Vision.

2. Set the device to AT89C51.

3. The source code is written in the text window and is saved as .asm file.

4. Add the source file into the Source Group.

5. Build target.

6. Start the debug session.

7. Enter the data values in the source address through memory window.

8. Run the program.

9. Observe the result in the destination address through memory window.

DEPARTMENT OF ECE 1 SNGCE, KADAYIRUPPU


2/86


ALGORITHM

No

Yes


IsCounter

= 0?

Initialize source blockPointer

Start

StopDecrement counter

Increment both pointers

Move the content from source todestination and clear the source pointer

Initialize destinationBlock pointer

Copy length of block


3/86


PROGRAM:

ORG 0000H

SJMP 30H

ORG 30H

MOV R0, #35H

MOV R1, #45H

MOV A,@R0

MOV R3, A

INC R0

BACK: MOV A,@R0

MOV @R1, A

MOV @R0,#00H

INC R0

INC R1

DJNZ R3, BACK



4/86


SJMP $

END

RESULT WINDOW

Input Window:

Output Window:

RESULT:

An assembly program to move a block data bytes to internal memory location

has been executed and the output has been verifed.

Expt. No: 2 EXCHANGE OF FIVE BYTES OF DATA

Date: 12/07/11

AIM

To exchange five bytes of data between two internal memory locations.

TOOLS REQUIRED

Vision from Keil for an 8051 microcontroller family

PROCEDURE




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5. Build target.



8. Run the program.


ALGORITHM



Start




6/86


No

Yes

PROGRAM

ORG 0000H

MOV R0, #20H


Is

Counter= 0?

StopDecrement counter


Exchange content ofSource and destination

Memory location


7/86


MOV R1, #30H

MOV R3, #05H

LOOP: MOV A,@R0

XCH A,@R1

MOV @R0, A

INC R0

INC R1

DJNZ R3, LOOP

HERE: SJMP HERE

END

RESULT WINDOW

Input Window

Output Window

RESULT



8/86


An assembly program to exchange a block five data bytes between two

internal memory location has been executed and the output has been verifed.

Expt. No: 3 MULTIPLICATION OF TWO 8 BIT NUMBERS

Date: 15/7/11

AIM:

To write a program to multiply two 8 bit numbers using 8051c.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.




9/86


ALGORITHM:

Yes

No

No

Yes


Start

Stop

Take the multiplier and multiplicandfrom the location

Mul=Mul+multiplicand using addcinstruction

Take the multiplier as count

Is carry?Increment the register

Is count=0

Store the result


10/86


PROGRAM

ORG 0000HMOV R0, #50H

CLR A

MOV R1, A

MOV A,@R0

MOV R3, A

INC R0

MOV A,@R0

MOV R4, A

CLR A

BACK: ADD A, R4

JNC CARRY

INC R1

CARRY: DJNZ R3, BACK

INC R0

INC R0

MOV @R0, A

DEC R0

MOV A, R1

MOV @R0, A

HERE: SJMP HERE

END



11/86


RESULT WINDOW:

Input Window:

Output Window:

RESULT

An assembly program to multiplication of two 8 bit numbers has been

executed and the output has been verifed.



12/86


Expt. No:4 SQUARE ROOT OF AN 8 BIT NUMBER

Date: 15/07/11

AIM:

To find the square root of a number.

TOOLS REQUIRED:


PROCEDURE





5. Build target.



8. Run the program.



13/86



ALGORITHM:

YES

Yes

NoNo

Yes


Start

Set count to 0

Copy number from memory location

Subtract the number with odd number

Increment count on each subtraction

Is result=0

Is result=negativeStore count value to memory

location

Write FF to memory location


14/86


PROGRAM:

ORG 0000H

SJMP 30H

ORG 30H

MOV R0,#50H

MOV A,@R0

MOV R1,#01H

MOV R2,#00H

BACK: SUBB A, R1

INC R2

JZ HERE

JC HERE1

INC R1

INC R1

SJMP BACK

HERE: INC R0

MOV A,R2

MOV @R0,A


Stop


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SJMP $

HERE1:INC R0

MOV @R0,#0FFH

END

RESULT WINDOW:

Input Window:

Output Window:

RESULT:

An assembly program to find the square root of a number has been executed

and the output has been verifed.



16/86


Expt. No:5 SQUARE OF AN 8 BIT NUMBER

Date: 15/07/11

AIM:

To write a program to find the square of a number.

TOOLS REQUIRED:


PROCEDURE





5. Build target.



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8. Run the program.


ALGORITHM:

YES

No

Yes


Start


Set count =number

add the number with next odd number

Decrement the count

Is count=0

Store the result

Stop

Increment by 2


18/86


PROGRAM:

ORG 0000H

MOV R0, #50H

MOV A,@R0

MOV R2, A

MOV R1, #01H

CLR A

LOOP: ADDC A,R1

INC R1

INC R1

DJNZ R2, LOOP

INC R0

MOV @R0, A

HERE: SJMP HERE



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END

RESULT WINDOW:

Input Window:

Output Window:

RESULT:

An assembly program to find the square number has been executed and theoutput has been verifed.



20/86


Expt. No: 6 WAVE FORM GENERATION

Date: 19/07/11

AIM:

To write a program to generate the following waveform

1. sine wave,

2. Square wave,3. Staircase wave.

TOOLS REQUIRED:




21/86


PROCEDURE:





5. Build target.


7. Open the Logic Analyzer window

8. Load Setup values into the respective port

9. Run the program.

10. Observe the result in the logic analyzer window.

ALGORITHM:

1. SINE WAVE

Angle Sin(angle) VOUT=5V+(5Sin(angle)) CORRESPONDINGDECIMAL TO DAC

VALUE=VOUT*25.6

0 0 5 128

30 0.5 7.5 192

60 0.866 9.33 238

90 1.0 10 255



22/86


120 0.866 9.33 238

150 0.5 7.5 192

180 0 5 128

210 -0.5 2.5 64

240 -0.866 .669 17270 -1.0 0 0

300 -.866 .669 17

330 -0.5 2.5 64

360 0 5 128

2. SQUARE WAVE

Delay Calculation:

Required frequency = 49.99Hz with 50% duty-cycle =>

ON/OFF period = 10.002ms

Delay calculation:

Effective clock frequency = 24MHz/12 = 2MHz => 0.5 S

Delay = 10.002ms => Count = (10.002 x 10 -3 ) / (0.5 x 10-6 ) = 20004

TH1:TL1 = 0xB1DCh

3. STAIRCASE

Step Size: 0xFF/5 = 0x33h/51d

Delay calculation:

Effective clock frequency = 24MHz/12 = 2MHz => 0.5 S

Delay1 = 10ms = 10/5x 10-3 S => Count = (2 x 10-3 ) / (0.5 x 10-6 ) = 20000

TH1:TL1 = E0C0h

Delay2 = .0005ms = 5 x 10-4 S => Count = (5 x 10-4) / (0.5 x 10-6 ) = 1000

TH0:TL0 = FC18h



23/86


1. SINE WAVE


Load P1 with the new level

Decrement count

No

Yes

Define the lookup table partof the code space

Start

Initialize the samples count

Count = 0?

Get the next level from table

Stop


24/86


2. SQUARE WAVE:


Start

Clear ACC and load P2

Call DELAY2:Load TL0, TH0

Start Timer0: set TR0

NO

YES

Clear TR0, TF0

Add #33h and load P2Configure TMOD register

Call DELAY1:Load TL1, TH1


Clear TR1, TF1TF0 =1?

TF1 =

1?

A =0xFF?

YES

NO

YES NO


25/86


3.STAIRCASE WAVE:


Start

Clear ACC and load P2.3

Call DELAY:Load TL1, TH1


NO

YES

Clear TR1, TF1

Toggle P2.3

Configure TMOD register

TF1 =1?


26/86


PROGRAM:

1.SINE WAVE

ORG 0000H

AGAIN: MOV DPTR, #TABLE

MOV R2, #0CH

BACK: CLR A

MOVC A,@A+DPTR

MOV P1, A

INC DPTR

DJNZ R2, BACK

SJMP AGAIN

ORG 300H

TABLE: DB 128,192,238,255,238,192

DB 128,64,17,0,17,64,128

END

2. SQUARE WAVE

ORG 0000H

MOV TMOD, #20H

AGAIN: MOV TL1, #0DCH

MOV TH1, #0B1H

SETB TR1

BACK :JNB TF1, BACK

CLR TR1



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CPL P2.3

CLR TF1

SJMP AGAIN

END

3.STAIRCASE WAVE:

ORG 0000H

MOV TMOD, #10H

AGAIN: CLR A

MOV P2, A

ACALL DELAY1

LOOP: ADD A, #33H

MOV P2, A

ACALL DELAY1

CJNE A, #0FFH, LOOP

SJMP AGAIN

DELAY1: MOV TL1,#0C0H

MOV TH1,#0E0H

SETB TR1

BACK: JNB TF1,BACK

CLR TR1

CLR TF1

RET

DELAY2:MOV TL0, #18H



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MOV TH0,#0FCH

SETB TR0

BACK1: JNB TF1, BACK1

CLR TR0

CLR TF0

RET

END

RESULT WINDOW:

SINE WAVE

SQUARE WAVE



29/86


30/86


Expt. No:7 SORTING OF NUMBERS

Date: 19/7/11

AIM:

To arrange a block of data in descending order.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.




31/86


ALGORITHM:

No

Yes

No


Yes

No


Initialize outer counter

Initialize inner counter

Compare two numbers

Exchange two numbers

Decrement inner counter

Decrement outer counter

Is counter=0?

Is counter=0?

Is carry flag

set?

Increment memorypointers

Initialize memory pointers

Start


32/86


No

PROGRAM:

ORG 0000H

MOV R0, #50H

MOV R1, #60H

MOV R3, #0AH

BACK1: MOV A,@R0

MOV A,@R0

MOV @R1, A

INC R0

INC R1

DJNZ R3, BACK1

LOOP: MOV R3, #0AH

MOV R4, #0AH

MOV R0, #60H

BACK: MOV A,@R0

MOV B, A

INC R0

MOV A,@R0

CJNE A, B, LOOP1

LOOP1: JC LOOP2

MOV A,@R0

DEC R0

XCH A,@R0

INC R0

MOV @R0, A

LOOP2: DJNZ R3,BACK

DJNZ R4, LOOP


Stop


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HERE: SJMP HERE

END

RESULT WINDOW:

Input Window:

Output Window:

RESULT:

An assembly program to sort 10 numbers has been executed and the output

has been verifed.



34/86


Expt. No: 8 PALINDROME

Date: 22/07/11

AIM:

To check the given data is palindrome or not.

TOOLS REQUIRED:


PROCEDURE





5. Build target.



8. Run the program.

9. Observe the result in the destination address through memory window



35/86


ALGORITHM

No

Yes

No

No

Yes


IsCounter

= 0?


Decrement counter


Move the content from source todestination



Reload thecounter

Compare the first bytes from 2 locations

Is equal?

IsCounter

= 0?

Start


36/86


Yes

PROGRAM:

ORG 0000H

MOV R3, #09H

MOV R0, #20H

MOV R7, #00H

MOV DPTR, #MYDATA

LOOP: CLR A

MOVC A,@A+DPTR

MOV @R0, A

INC R0

INC DPTR

DJNZ R3, LOOP

MOV R3, #09H

MOV R1, #40H

MOV R0, #28H

LOOP1: MOV A,@R0

MOV @R1, A

DEC R0

INC R1

DJNZ R3, LOOP1

MOV R3, #09H

MOV R0, #20H

MOV R1, #40H

CLR C


Store FF to memory location

Decrement counter

Store 11 to memory

Stop


37/86


LOOP3: CLR C

MOV A,@R1

SUBB A,@R0

CJNE A, #00H, LOOP4

INC R0

INC R1

DJNZ R3, LOOP3

MOV R6, #11H

MOV R1, #60H

MOV A, R6

MOV @R1, A

SJMP HERE

LOOP4: MOV R6, #0FFH

MOV R1, #30H

MOV A, R6

MOV @R1, A

HERE: SJMP HERE

ORG 80H

MYDATA: DB "AABBCBBAG"

END

RESULT WINDOW:

Input Window:

Output Window:



38/86


RESULT

An assembly program to check a given string is palindrome has been executed and

the output has been verified.



39/86


40/86


ALGORITHM

No

Yes


Take no. from memory location

Is count==no

Start

Stop

Cube=Square*number

Take it as count

Get first odd no.

Square =0

Square=square + odd no

Increment by 2 to obtain next odd


41/86


42/86


Output Window:

RESULT:

An assembly program to find the cube of a number has been executed and

the output has been verifed.



43/86


Expt. No: 10 COPY THE MESSAGE FROM

Date: 26/07/11 ROM TO RAM

AIM:

To copy the message WELCOME from ROM to RAM.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.




44/86


45/86


PROGRAM:

ORG 0000H

MOV R3,#07H

MOV R0,#20H

MOV DPTR,#MYDATA

LOOP:CLR A

MOVC A,@A+DPTR

MOV @R0,A

INC R0

INC DPTR

DJNZ R3,LOOP

SJMP $

ORG 30H

MYDATA: DB"WELCOME"

RESULT WINDOW:

Input Window:

Output Window:



46/86


RESULT:

An assembly program to copy the message WELCOME from ROM to RAM

has been executed and the output has been verifed

Expt. No: 11 ADDITION OF TWO 16-BIT NUMBERS

Date: 26/07/11

AIM:

To perform addition of two 16-bit numbers

TOOLS REQUIRED:


PROCEDURE:





5. Build target.





47/86


8. Run the program.


ALGORITHM:


Start

Stop

Add MSBs With carry and store itin next location

Add LSBs and store it in nextmemory location

Copy LSB of second no.andincrement pointer

Copy MSB of first no.andincrement pointer

Copy LSB first no and incrementpointer

Load DPTR with memory location

Copy MSB of second no.andincrement pointer

Is carry?Increment location and store carry


48/86


49/86


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51/86


Expt. No: 12 SUM OF 5 BCD NUMBERS

Date: 26/07/11

AIM:

To find sum of 5 BCD numbers.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.




52/86


ALGORITHM:

Yes

No

No


Load the limit frommemory location

Load numbers frommemory location

Clear Sum and Carry

Add Sum with the content ofthe memory location

Add 0110 to nibble that havevalue >1001

Increment carry

Increment memory location

Decrement limit

If limit=0

Display output

If carry=0

Decrement limit

Stop

Start


53/86


54/86


PROGRAM:

ORG 0000H

MOV R0, #50H

MOV R1, #60H

MOV R2, #5H

CLR A

MOV B, A

AGAIN:ADD A,@R0

DA A

JNC NEXT

INC B

NEXT: INC R0

DJNZ R2, AGAIN

MOV @R1, B

INC R1

MOV @R1, A

HERE: SJMP HERE

END

RESULT WINDOW:

Input Window:

Output Window:



55/86


RESULT:

An assembly program to find sum of BCD numbers was executed

successfully.



56/86


Expt. No: 13 CONVERSION PACKED BCD TO ASCII

Date: 30/07/11

AIM:

To implement packed BCD to ASCII conversion.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.

9. Observe the result in the destination address through memory window



57/86


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60/86


61/86


ALGORITHM:

YES

No

Yes


Copy count from memory location

Increment location


Unpack number and separate MSB & LSB

Add 30 with LSB & MSB

Store result in memory location

Decrement count

Stop

Is count=0?

Start


62/86


63/86


64/86


65/86


PROGRAM


Start

Enter the count

Copy the number frommemory location

Multiply it by 0FH and swapit to get the MSB

Copy the next number frommemory location

Multiply it by 0FH to get theLSB

Store the result in the memory

location

Increment the memorylocation and decrement thecount

Ifcount=0

Stop

Add the MSB and LSB to getthe BCD number


66/86


67/86


68/86


69/86


Expt. No: 16 CHECKSUM

Date: 2/08/11

AIM:

To write a program to find the checksum of given data HELLO.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.



8. Run the program.




70/86


PROGRAM:

No

Yes


STOP

Start

Exchange contents of source anddestination

Initialize destination block pointer

Initialize source block pointer

Input string

Store data

Twos compliment

Add hex value of string

Is counter=0


71/86


72/86


Input window:

Output Window:

RESULT:

An assembly program to find the checksum of given data HELLO has been

executed and the output has been verifed.



73/86


Expt. No: 17 TRANSPOSE OF A MATRIX

Date: 2/08/11

AIM:

To generate transpose of a matrix of arbitrary size.

TOOLS REQUIRED:


PROCEDURE:





5. Build target.


7. Run the program.

8. Observe the result in the memory window.



74/86


75/86


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77/86


Output Window:

RESULT:

An assembly program to generate transpose of a matrix hs been executed and the

results verified in the memory window.



78/86


79/86


ALGORITHM:

Yes

No

Yes

No

Yes

No Yes

No


Start

Load the memory location

Increment the location and store the result

Increment location and copy second no.

Copy first operand

Increment location

Load choice from that location

Is choice=3

Is choice=4

Is choice=2

Is choice=1

Stop

Div=A/B

Mul=A*B

Sub=A-B

Sum=A+B

1

1

1

1

1


80/86


81/86


MOV @R0,A

SJMP HERE

LOOP2: CJNE R3,#03H,LOOP3

MOV B, A

MOV A, R4

MUL AB

INC R0

MOV @R0, A

SJMP HERE

LOOP3: CJNE R3,#04H,LOOP4

MOV B, A

MOV A, R4

DIV AB

INC R0

MOV @R0, A

INC R0

MOV A, B

MOV @R0, A

SJMP HERE

LOOP4: MOV A, #0FFH

INC R0

MOV @R0, A

HERE: SJMP HERE

END

RESULT WINDOW:

Input Window:



82/86


Output Window:

RESULT:

An assembly program to impliment calculator has been executed and the

output has been verifed.

Expt. No:19 FIBONACCI SERIES



83/86


84/86


85/86


PROGRAM:

ORG 0000H

CLR A

MOV R3, 50H

MOV R0, #51H

MOV @R0, #00H

INC R0

MOV @R0, #01H

MOV R1, #01H

DEC R3

DEC R3

LOOP: ADD A, R1

INC R0

MOV @R0, A

MOV R1, A

DEC R0

MOV A,@R0

INC R0

DJNZ R3, LOOP

HERE: SJMP HERE

END

RESULT WINDOW:

Input window:



86/86


Output Window:

RESULT:

An assembly program to generate Fibonacci series of 10 elements has been

written, executed and the output has been verifed.

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