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11 Low level programming and Interfacing


Learn

Binary to hex and hex to binary conversion.

Bitwise operators

Simple interfacing

11.1 Binary to hex and hex to binary conversion.

A computer may be programmed to control external devices, for example, turning on or off

lights, pumps, motors or to monitor sensors such as smoe detectors or burglar alarms etc.

!hese types of control systems usually re"uire the computer to process incoming signals and

output control signals to external devices almost instantaneously. Because of its immediate

response such a system is called a real#time control system. $rogramming a real#time control

system will usually involve a substantial amount of low level programming where data are

manipulated at bit level. !he best way to visuali%e a low level operation is to express the data

in binary form. &nfortunately ' cannot represent data in binary form. But data can, however,

be represented in hex format. !he conversions between these two number systems are

relatively easy. !he conversion methods can be found in most textboos on digital electronics

but are repeated here as a reminder.

!o convert a binary number to a hex number, you must first organise the binary number in a

group of ( bits starting from the rightmost bit. )ext, for each group, write its hex e"uivalent

using the 1* symbols+ to - and A to where A to represent values from 1 to 1/.

or example, consider an 0#bit binary number 11111.

Step1+ rganise it into groups of ( bits starting from the rightmost bit, thus, giving

1 1 1 1 1

Step 2 + 3rite the hex e"uivalent for each group.

1 1 1 1 1

4ence the hex e"uivalent of binary number 11111 is xBA.

!o convert a hex number to it binary e"uivalent, write the (#bit binary e"uivalent of each hex

digit.

or example, to convert hex number x56 to binary, write the (#bit binary e"uivalent for hex

digits 5 and 6.

5 6

!he binary e"uivalent of x56 is 111111.

175

AB

111111

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8our turn

1 'onvert the following binary numbers to their hexadecimal e"uivalent+

9a: 111111 9b: 1111 9c: 1111 9d: 1111

2 'onvert the following hexadecimal numbers to their binary e"uivalent.9a: x(/ 9b: x* 9c: x-6 9d: xA 9e: x;

xx

11.2 Bitwise operators

' comes with a set of operators that allow us to manipulate data bitwise. !hese operators is a

great help in low level programming which deals mainly with data bitwise !hese bitwise

operators are described below.

11.2.1 The ~ operator (!T"#omplement$

!he < operator performs the one=s complement function. It inverts the bits of the data so that

a 1=s becomes a and a becomes a 1. Assuming that you have an 0#bit data represented by

the following binary pattern 1111 and you wish to invert it, this what you have to do +

'onvert it into hex format+ 1111 x'A 9' cannot represent data in binary

format:

3rite xA' and num_2> x*5, then result of num_1?

num_2will yield x2(. !he operation is illustrated below.

num_1> xA' > 1 1 1 1

num_2> x*5 > 1 1 1 1 1

####################################################

num_1 & num_2 > 1 1 > x2(

11.2. The ) operator (!*$

!he @ operator wors on two 9or more: bits and returns a 1 if at least one bit is 1 and returns a

if all the bits are %ero. 'onsider again the two variables num_1> xA' and num_2> x*5.

ing these two variables, we have

170

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num_1> xA' > 1 1 1 1

num_2> x*5 > 1 1 1 1 1

####################################################

num_1 | num_2 > 1 1 1 1 1 1 1 > x;

11.2.+ The , operator (-!*$

!he expression returns a if both the bits are or both the bits are 1. therwise it returns a

1. !he operation on numC1 and numC2 are shown below.

num_1> xA' > 1 1 1 1

num_2> x*5 > 1 1 1 1 1

####################################################

num_1 ^ num_2 > 1 1 1 1 1 > x'B

11.2. The // (shift left$ and 00 (shift right$ operators

!he DD operator shifts the data left by the number of bit positions specified by the operand

immediately following it.

or example, the statement

num_1 1 1 1 1

num_1 > 4;

means Eshift the value of num_1right by ( bit positionsF.

17-

Lost bits H shifted out

illed up with

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num_1> 1 1 1 1

num_1 >>4 > 1 1

Below is a program example that illustrates the use of these bitwise operators.

/* Program 11-1- Using bitwise operators */

#in!u"e

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num_1and num_2are declared as type intwhich occupy 2 bytes each. 4ence num_1>

xA' and num_2> x*5.

8our turn

1 ive the one=s complement of the following 9int: numbers+9a: 1 1111 9b: 11 11 11 11 9c: 11 11

9d: x2*5 9e: x*5af 9f: x12 9g: xcb

9h: 71 9i: (/72 9J: 12( 9: 0

2 !he following variables are declared+int num1 ( 3420 num2 ( )=0 num3 ( )a=;

ind the values of the following expressions+

%a num2

%b num1 | num2

% num1| num2| num3

%" num2 num3

%e num1 ^ num2 ^ num3

% num1 & num2 & num3

%g num1 >2

%i %num2 >1 %num2

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!he computer communicates with the outside world through its input#output 9IK: ports. Let

us imagine that the computer you are woring on has an 0#bit output port with address x17.

urther assume that this port is used to control a set of L;6s as shown in ig 11.1.

utput port + x17

Bit b5 b* b/ b( b7 b2 b1 b

ig 11.1

Bit b5 is the most significant bit and b is the least significant bit. An L;6 is turned on if the

bit to which it is connected is asserted high 9at logic 1: and is turned off if it is pulled low 9at

logic :. or example to turn on L;6s connected to bits b and b7, the binary pattern

11 9>x-: has to be sent to the output port x17. !urbo ' has the following

function to output data to an output port.

outportb%port_a""0 out"ata;

!his function outputs a byte 9hence the b after the word outport: of out"atato a hardware

port with address port_a"". Its definition is found in dos.h header file. In this example

port_a""> x17 and out"ata> x-. !he program to turn on L;6s at b7 and b is

shown below.

/* Program 11-2 @urn on AB:s at b3 an" b) */

#in!u"e

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!he delay()function, whose prototype is also found in the dos.hheader file, suspends

program execution for the interval of time 9in ms: specified in the parenthesis.

8our turn

1 3ith reference to ig 11.1, give the binary patterns and their hex e"uivalents re"uired tobe sent to the output port to turn on L;6s connected to bits

9a: b5 and b*,

9b: b5, b/, b7 and b1,

9c: b*, b( and b2,

9d: b5, b7, and b,

2 Assume that L;6s at b5, b*, b2 and b are now on, what is the value9 in hex: re"uired to

turn L;6s at b5 and b off while L;6s at b2 and b* remain on.

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11.+ a3ing L4's flash

;xamine $rogram 11#2 again. If the last three statements are put in a loop, then the L;6s at

b7 and b will be turned on and off repeatedly at a rate of about 1 second. 3e would have

effectively send a train of pulses with a fre"uency of / 4% to these two bits and cause the

L;6s to flash. See ig 11.2

ig 11.2

/* Program 11-3 Ea?ing AB:s at b3 an" b) !as$*/

#in!u"e

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main%'

int port_a"" ( )13);int out"ata ( ))C; /* binar pattern D ))))1))1 */

w$i!e%1'

outportb%port_a""0 out"ata; /*on AB:s at b3 an" b) */"e!a %1))); /* or 1))) ms %1 s*/outportb%port_a""0 ))); /* t$en turn t$em o */"e!a %1)));

!he last delay91: eeps the L;6s off for 1 second.

8our turn

1 3rite a program to mae L;6s at b2 and b/ flash at a rate of / 4%.

2 odify your program so that the L;6s at b1 and b/ flash alternately. !hat is, when b2is on b/ is off and when b2 is off, b/ is on.

94int+ &se two binary data patterns to be sent to the output port.:

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11. a3ing the lights move

3e now want to turn the L;6s on one at a time starting with the L;6 at b. !his will give

the illusion that the light is moving from right to left. 3e will have to start with data pattern

1 9>x1: and send it out to the output port. !o turn on the next L;6, the pattern is

shifted one bit position to the left before it is sent out to the output port . !he shifting process

is repeated until all the L;6s have been lit up. !he shifting binary patterns to be sent to the

output port are shown below.

b5 b* b/ b( b7 b2 b1 b L;6 on

1 b

1 b1 1 b2

1 b7

1 b(

1 b/

1 b*

1 b5

$rogram 11#( below EmovesF the lights from is left to right.

1((

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/* Program 11-4 EoFing !ig$ts */#in!u"e

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9d: 3hich L;6s are turned on when the loop is entered for the second time N

9e: 3hen will the program brea out from the loop N

9f: 6escribe, briefly, what the program does.

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11.5 &nother interfacing example 6 driving a stepper motor

!his interfacing example uses the Bytronic stepper motor. !he Bytronic stepper motor comes

with five connections and are labelled as 6, $, ;, /O and 12O as shown in ig.11.7

ig 11.7 Bytronic stepper connections

!he meaning of the connections and the re"uired inputs to these connections are listed below

Label eaning e"uired inputs

6 6irection A logic applied to this input will cause the motor

to rotate in the clocwise direction

A logic 1 will cause the motor to rotate in the anti

clocwise direction

$ $ulse A rising edge of a pulse is re"uired to rotate the

motor one step. !his stepper motor re"uires (0

pulses to mae one revolution.

ne related feature is that the specified maximum

pull#in rate of the stepper motor is 7/ steps per

second 97/ pulsesKsec.:. $ull#in rate is the

maximum switching rate at which a motor can

move without losing a step. If the pulse rate exceed

this figure, the motor will not wor.

; ;arth 'onnected to the )6 or O terminal of the power

supply.

/O / volts !o be connected to the / O power supply.

12O 12 volts !o be connected to the 12 O power supply.

1(*

6 $ ; /O 12O

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8our turn

1 iven above that (0 pulses are re"uired to rotate the motor through one complete

revolution.

9a: 3hat is the step angle N

9b: If 7 pulses are applied to the stepper motor, find the angular displacement.

9c: 4ow many pulses are re"uired to give an angular displacement of 9i: 7o, 9ii: *o,

9iii: -o, 9iv: 10o, 9v: 25o, 9vi: 77o.

2 If a continous train of pulses with a fre"uency of (0 4% is applied to the $ connection,

what is the rotational speed 9in rpm: of the motor.

7 epeat 2 for a fre"uency of 1 4%.

( &sing the maximum pull#in rate of 7/ stepsKs,

9a: state the maximum fre"uency a pulse train that can be fed to this stepper motor

9b: calculate the minimum duration between each pulse

9c: show that the maximum speed is about (( revolutions per minute 9rpm:.

9d: state if the following pulse trains shown in ig 11.( are suitable to drive the motor. If

not, explain why. All duration shown are in milliseconds 9ms:.

$ulse waveform or

i

ii

iii

1(5

7 7periodic

2 7 periodic

27 periodic

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iv

v

ig 11.(

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11.7 #onnection to the comp8ter

Let us use bit and bit 1 of the output port to drive the stepper motor. Bit is connected to

the $ input and bit 1 is connected to the 6 input of the stepper motor. !he connection

between the computer and stepper is shown in ig 11./ below.

Bit b5 PP b7 b2 b1 b

ig 11./

!o drive the motor one step a pulse is re"uired to be sent to bit . Let us create a pulse withthe mar and space ratio of 1+1 and a width of ( ms, i.e, mar > 2 ms and space > 2ms. See

ig 11.*

ig 11.*

!o set b to 1, we output the following binary pattern 1 9or x1: and use the delay9:

function to eep it high for 2 ms. After that, we set it bac to by sending the binary pattern 9x: to the output port. !he statements are shown below+

1(0

6 $ ; /O 12O

/ O 12 O )6

!o data signal

gnd

$ower Supply

1 1 periodic

aperiodic/ 2 7

2 2

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outportb9outCadd, x1:Q

delay92:Q

outportb9outCadd, x:Q

delay92:Q KR not necessary if we only want to drive the motor one step RK

)ote also that the above instructions drives the stepper motor one step in the clocwise

direction as b1 is maintained at logic .

!he complete program to drive the stepper one step in the clocwise direction is shown

below.

/*Program 11- :riFing t$e stepper motor one step */#in!u"e #"eine E+GH ))1#"eine IP+,B )))

main%

'int port_a"" ( )13);int ount;

1(-

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or%ount ( ); ount

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revolutions. All you have to do is to add an outer loop to repeat the statements which cause

the motor to mae 1 revolution. 'ompare your program with the one below./*Program 11-K Eotor rotates 3 reFo!utions in t$e !o?wise "iretion */#in!u"e #"eine E+GH ))1#"eine IP+,B )))#"eine ,A,H9MIB )))#"eine ,_,AH9MIB ))2 /*))))))1) */

main%'

int port_a"" ( )13);int ount0 no_o_reF(3;

w$i!e%no_o_reF L()'

or%ount ( ); ount

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no_o_reF --;

!he A value is ed with the 'C'L3IS; value resulting in the data shown below

A > x1 > 1

'#'L3IS; > x2 > 1

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

> 1 1

!he resulting data gives a pulse to the stepper motor and at the same time set the direction to

anti clocwise9A'3:. Similarly the expression S$A'; @ 'C'L3IS; results in a data value

of x2 which eeps the direction A'3 because bit b1 is still at logic 1.

Below is another program example which mae the motor rotates / revolutions in the

clocwise dircection and then followed by 7 revolutions in the anti clocwise direction. !he

cycle repeats until the ;S' ey is pressed.

/Program 11-1) N = reFo!utions ,9 an" 3 reFo!utions +,9 */#in!u"e

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w$i!e%no_o_reF L() /* 3 ,,9 reFo!utions */'

or%ount ( ); ount

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int port_a"" ( )13);int ount0 no_o_reF0 m?0sp0time_";$ar ?e;

Foi" pu!se_motor %int m?_"ata0 int sp_"ata0 int time_"e!a;

w$i!e %?e L(BI,'

"o'no_o_reF(=;w$i!e%no_o_reF L() /* = ,9 reFo!utions */'

m? ( E+GH | ,AH9MIB;sp ( IP+,B | ,AH9MIB;time_" ( 2;or%ount ( ); ount

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2 !he mar and space duration in the pulseCmotor9: function is the same. odify thefunction so that the mar and space can have dissimilar duration.

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11.: as3ing o8t 8nwanted bits

Sometimes, we wish only to focus our attention on certain bits in a data and ignore the others.

3e can Tmas= out the unwanted bits by setting these bits to . !his is achieved by A)6ing

9with the ? operator: the unwanted bits with and the re"uired bits with 1. or example, to

chec b7, we A)6 the data with this binary pattern 1 9>x(:. !he masing

operation is shown below.

6ata b5 b* b/ b( b7 b2 b1 b

as 1

6ata ?

as

b7

!he masing operation shows that all the unwanted bits are set to while the re"uired bit

9b7: remains.

4ere is a program example that chec if b/ is at logic or 1. !he mas to chec b/ is

1 9>x2:.

/*Program 11-13 mas?ing operation */#in!u"e

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94int + 8ou need two masing patterns:

1/*

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7 ;xamine $rogram 11#1( and answer the following "uestions.

/*Program 11-14 */#in!u"e

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Another of version of $rogram 11#1( which mae use of the for loop is shown below+

/*Program 11-1= +not$er Fersion o Program 11-. */#in!u"e

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3hen the switch is pressed, a logic 1 9/ O: is applied to the bit of the port to which this

switch is connected, otherwise a logic 9 O: is applied to it.

!o read the status of the switch, we use the function

inportb%port_a"";

which reads a byte from a hardware port with address port_a"". !he prototype is found in

the "os$header file.

!he program below checs if switch S3 are pressed.

/*Program 11-1 ,$e? swit$ I9) */#in!u"e#in!u"e

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!he algorithm for the pump control system is shown in the flow chart in ig 11.1 below.

ig 11.1

1*1

n $1ff $2

ead input

L1reached N

ff $1n $2

L7reached N

L7reached N

Sound Alarmff $1 ? $2

ead input

Start

eset N

L2reached N

8es

)o

8es

8es

8es

8es

)o

)o

)o

)o

9L1 may be faulty so

chec L7:

9$1 may not have turned

off, so chec L7:

93ait for the esetey 9: to be hit:

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Before we write the program, let us summarise the values that will be obtained at the input

and the values re"uired at the output.

At the input port x11, the input assignments are as follows+

bit device Oalues and meanings Action

b Sensor L1 H 9open contacts: Li"uid has not reachedthis level

1 H 9close contacts: Li"uid has reached

level

#

n $2, ff $1

b1 Sensor L2 H 9open contacts: Li"uid has fallen below

this level

1 H 9close contacts: Li"uid has not fallen

below this level

#

n $1, ff $2

b2 Sensor L7 H 9open contacts: Li"uid has not reached

this level

1 H 9close contacts: Li"uid has reached

level

#

ff $1 ? $2

n Alarm

b7#b5 )ot used

At the output port x17, the bit assignment are as follows+

bit device Oalues and meanings

b $ump $1 H !urn off pump

1 H !urn on pump

b1 $ump $2 H !urn off pump

1 H !urn on pump

b2 Alarm

Indicator

H ff annunciation

1 H n annunciation

b7#b5 )ot used

As the addresses of the IK ports are constants, we will define them to be so. 4ence

Udefine I)CA66 x11

Udefine &!CA66 x17

3e define as constants also the masing patterns to test bit b,b1and b2 as follows+

Udefine ASCL1 x1 KR1RK

Udefine ASCL2 x2 KR1RK

Udefine ASCL7 x( KR1RK

or output control, we have

Udefine )C$1 x1 KR1RK

Udefine )C$2 x2 KR1RK

Udefine )CAL x( KR1RK

!he program for the control system is shown in $rogram 11#11.

1*2

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/* Program 11-1K Pump ,ontro! Istem */#"eine M_+:: )11)#"eine U@_+:: )13)#"eine E+IH_A1 ))1 /*)))))))1*/#"eine E+IH_A2 ))2 /*))))))1)*/

#"eine E+IH_A3 ))4 /*)))))1))*/#"eine _P1 ))1 /*)))))))1*/#"eine _P2 ))2 /*))))))1)*/#"eine _+AE ))4 /*)))))1))*/#in!u"e

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print%5Istem ai!ure7n7n8;print%5,$e? pumps an" sensors7n8;print%59$en au!ts are retiie"0 press G to reset sstem7n8;Geset%;a!arm_A3();!eFe! ( );

/* en" o w$i!e%1 */ /* en" o main% */

/* User "eine" untion */

int Geset%'

$ar ?e;

w$i!e %?e L( TG && ?e L(r ?e(get$%;

!he above program is by no mean the Tbest= program. 8ou may write a more compact

program maing use of functions and pointers. !ry writing it.

8our turn

1 odify the program to print out the time at which the pumps are turned on. !he display

may loo as follows+

2 &se graphics, display the control system on the computer. &se a colour code to indicate ifa pump is on or off.

xx

Pump ,ontro! Istem-------------------

Itatus N orma!

11N3= Pump 1N Pump 2N RR12N=) Pump 1N RR Pump 2N

11_lowlevel

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