# Computing Hardware and Software

```Computer Platforms
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Von-Neuman architecture
Early computers used to store the program and the data in separate areas of memory. This
meant that if you had 64Kb of program area you could only write programs that were a
maximum of 64Kb in length. The same problem occurred with data. If 64kb was allocated for
data and only 24Kb were needed then the extra memory was wasted.
Von-Neuman put forward the idea that both the program and the data could be stored in the
same memory area. This overcame the problem of having a maximum program length. Long
programs could be loaded but might restrict the amount of data. Short programs would allow
more data to be loaded i.e. the use of memory became more flexible. Became of this
flexibility Von-Neuman architecture became the normal model for all computers
Bit and Bytes
What is a Bit?
Bit stands for Binary Digit
A bit represents 2 states, which are usually represented as a 0 or a 1. Because a Bit can
represent 2 states its number base is 2. A bit is the smallest way data can be represented in a
computer.
What is a Nibble?
A Nibble is 4 Bits and can represent the binary numbers 0000 to 1111 and the hex numbers 0
to F (decimal 0 to 15).
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What is a Byte?
A Byte is 8 Bits and can represent the binary numbers 00000000 to 11111111 and the hex
numbers 00 to FF (0 to 255 decimal).
What is a Word?
A word is any collection of over 8 bits. Most words are prefixed with how many bits the word
contains. i.e. 16 bit Word, 32 bit Word. Words are usually used to describe the width of data
busses. Most words tend to be made up of 2 or more bytes.
Common sizes
When quoting bit and byte sizes in computing the following measures are used:
Kilobyte or Kilobit = 1042 bytes or bits or 210
Megabyte or Kilobit = 1048576 bytes or bit or 220
Gigabyte or Gigabit = 1073741824 bytes or bits or 230
Most people tend to round these numbers down to:
1,000, 1,000,000 and 1,000,000,000
Number Systems and Data Representation
Data Representation
All data is represented in a computer system in binary. The main types of data that needs to be
represented are
Alphabetic (upper and lower case)
Numeric (0 to 9)
Special characters (question marks, comma, etc.)
Control characters and codes
All these data types are represented using codes such as ASCII, Unicode and BCD.
American Standard code for Information Interchange (ASCII)
ASCII is a 7-bit code that is used to represent characters. Bit 8 is used as a parity check the
code was later extended to make an 8-bit code so that more characters could be represented.
The code consists of:
 Control characters from 0 to 32
 Symbols from 33 to 47and 58 to 64
 Numbers from 48 to 57
 Letters from 65 to 122
 Symbols from 123 to 126
 Delete. 127
Later the rest of the bits from 128 where used in the extended ASCII character set which uses
8 bits to represent extra characters. On of the problems with ASCII is that it can only be used
to represent 256 codes and as computing became more international extra codes were needed
to support other languages. Because of this reason ASCII is now being phased out and the
new UNICODE standard is being adopted.
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Unicode
This is the new International standard for representing characters Unicode is capable of
representing all the symbols, characters and controls for the world. www.unicode.org . The
main problem used to be that UNICODE uses 32 bits and many pieces of application software
will only support 8-bit ASCII coding, To overcome this there are 3 different Unicode
standards 8 bit 16 bit and 32 bit and most application software support these Unicode
standards.
Binary Coded Decimal BCD
Binary Coded Decimal is used to input and output characters. It is not used for performing
calculations. Its main use is in calculators.
You can have 4 bit or 8 bit BCD whatever system is used you must show all the bits. In the
examples below 25 cannot be written as 100101 it must be written as 00100101.
Advantages of BCD are the coding does not involve the conversion of numbers to binary.
Disadvantages Increases the length of the binary number and arithmetic is much more
difficult. The following examples show 4 bit BCD where 4 binary digits are used for each
number.
BCD Examples
Decimal
3
5
2
BCD
0011
0101
0010
Decimal
8
6
BCD
1000
0110
Decimal
7
1
4
BCD
0111
0001
0100
Bit masks are a way of representing data using individual bits in a byte or word. The data can
then be searched using and logic / or logic to find matches. The reason they are used is they
are very quick to search and a lot of data can be stored in a small amount of memory.
Suppose that 8 pieces of information needed to be stored about a client. Lets say:
1. Gender
2. Car Owner
3. House-owner
4. Married
5. Has Garage
6. Pension
7. Employed
8. Life Insurance.
Now these could be stored in 8 bits as follows
1. Gender
1 = Female 0 = Male
2. Car Owner
1 = Yes 0 = No
3. House-owner
1 = Yes 0 = No
4. Married
1 = Yes 0 = No
5. Has Garage
1 = Yes 0 = No
6. Pension
1 = Yes 0 = No
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7. Employed
1 = Yes 0 = No
8. Life Insurance.
1 = Yes 0 = No
Using the above a byte containing
11100011 would represent a female car owner with a pension, employed and has Life
Insurance .The decimal number 227 could represent this.
A bank of data can be quickly searched using AND. Looking at the above example if you
were looking for Employed people with a Life insurance then the search criteria would be
11000000. When an AND is performed on the data using this criteria a match will be found
11100011
Record
11000000
Search
AND
11000000
Match
10100011
Record
11000000
Search
AND
10000000
No match
More examples
Bit masks only use the binary numbers that have a single 1 in. So in our case you can use the
numbers 0001 0010 0100 and 1000. So with 4 bits you can represent 4 conditions. Here are
the 4 conditions we are going to use.
0010 Write
1000 Delete
Lets look at 4-person and give them some folder permissions.
Joe can read from the folder only.
Jill can read and write to the folder.
Now to make a bit mask
Joe
Joe can read from the folder only. To convert this information into a Bit Mask you compare
the information to the conditions.
0001
0001
Jill
Jill can read and write to the folder only. To convert this information into a Bit Mask you
compare the information to the conditions.
0001
Write
0010
0011
Fred
Fred can read, write and add to the folder. To convert this information into a Bit Mask you
compare the information to the conditions.
0001
Write
0010
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0100
0111
Freda
Freda can do all read, write, add and delete. To convert this information into a Bit Mask you
compare the information to the conditions.
0001
Write
0010
0100
Delete
1000
1111
Searching Data
Now lets assume that you wish to find who has permission to delete.
Delete
1000
1000
Now this mask is compared to each person in the database using AND
Person
AND Result
Joe
0001
1000
0000
Jill
0110
1000
0000
Fred
0111
1000
0000
Freda
1111
1000
1000
Match
Another Search
Who can add to a folder?
Person
AND Result
Joe
0001
0100
0000
Jill
0011
0100
0000
Fred
0111
0100
0100
Match
Freda
1111
0100
0100
Match
Lets try a more sophisticated search
Who can write AND add. This can be expressed as Person Mask AND (Write or Add) 0010
OR 0100 = 0110
Person
Result
Joe
0001
0110
0000
Jill
0011
0110
0010
Fred
0111
0110
0110
Match
Freda
1111
0110
0110
Match
Bit Maps
Bit maps are a method of storing graphic data. The most common file extension for a bit
mapped file is BMP. The easiest way to see a bit mapped file is to open Microsoft Paint.
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There are other bit mapped graphic formats such as gif. The biggest disadvantage of bit
mapped graphics is that the more detailed the picture the bigger the file size. The other
disadvantage is that if a bit mapped image is resized then quality is lost.
Number Systems
All number systems have a base, symbols and place values. By using these any number
system can be devised using a set of rules. When working in different number systems the
base is always indicated in subscript next to the number i.e. 237810
Three number systems
Decimal A number system based on 10
Binary
A number system based on 2
A number system based on 16
Binary base 2
We use 2 different symbols
01
When you have used the 2 symbols to indicate numbers larger than 1 the position of the
symbol is important. These positions are decided by raising the base 2 by its place value.
i.e.
23
22
21
20
(2 x 2 x 2) = 8
(2 x 2) = 4
(2 x 1) = 2
(2 x 0) = 1
So the decimal number 1310 is written as
23
22
8
4
1
1
21
2
0
20
1
1
When working in number systems this number is written as 11012
Denary or Decimal base 10
We use 10 different symbols
0123456789
To indicate numbers larger than 9 the position of the symbol is important. These positions are
decided by raising the base 10 by its place value.
i.e.
103
102
101
100
(10 x 10 x 10) = 1000 (10 x 10) = 100 (10 x 1) = 10
(10 x 0) = 1
So the number 2 thousand 3 hundred and seventy-eight (237810) is written as
103
102
101
100
1000
100
10
1
2
3
7
8
When working in number systems this number is written as 237810
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We use 16 different symbols
0123456789ABCDEF
When you have used the 16 symbols to indicate numbers larger than 15 the position of the
symbol is important. These positions are decided by the base of 16 being raised by the power
of its place value.
163
162
161
160
(16 x 16 x 16) = 4096
(16x16) = 256
(16 x 1) = 16
(16 x 0) = 1
So the decimal number 30010 is written as
163
162
4096
256
1
161
16
2
160
1
C
When working in number systems this number is written as 12C16
Converting Between Number Systems
Binary to Decimal
Example 1
Write the binary number under the relevant columns.
So the binary number 100110 would be converted by
1024 512
256
128
64
32
16
1
0
32
8
0
4
1
4
2
1
2
1
0
8
0
4
1
4
2
1
2
1
0
Answer = 32 + 4 + 2 = 3810
Example 2
So the binary number 110110 would be converted by
1024 512
256
128
64
32
16
1
1
32
16
Answer = 32+16 + 4+2 = 5410
Decimal to Binary
Example 1
To demonstrate converting 4710 to binary
1
2
Divide 47 by 2 and put the answer underneath and the
remainder alongside 23 remainder 1
remainder alongside 11 remainder 1
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23
23
11
Remain
1
1
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3
4
5
6
and the remainder alongside 5 remainder 1
and the remainder alongside 2 remainder 1
and the remainder alongside 1 remainder 0
and the remainder alongside 1 remainder 0
11
05
05
02
02
01
01
00
1
1
0
1
Read the columns upwards to get the answer is 4710 = 1011112
Example 2
To demonstrate converting 5410 to binary
1
2
3
4
5
6
45
27
28
13
13
06
06
03
03
01
01
00
Divide 54 by 2 and put the answer underneath and the
remainder alongside 27 remainder 0
remainder alongside 13 remainder 1
and the remainder alongside 6 remainder 1
and the remainder alongside 3 remainder 0
and the remainder alongside 1 remainder 1
and the remainder alongside 1 remainder 1
Remain
0
1
1
0
1
1
Read the columns upwards to get the answer is 5410 = 1101102
Converting Hex to Decimal method 1
This is easier to do the long way round if you do not have a calculator.
Step 1 Convert the hex number into groups of binary 3DEH
Hex
3
D
E
Binary
0011
1101
1110
So 3DEH = 1111011110 binary
Step 2 Convert binary to decimal
1024 512
256
128
1
1
1
512
256
128
64
1
64
32
0
16
1
16
8
1
8
4
1
4
2
1
2
1
0
Answer 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2 = 99010
Step 1 Convert the hex number into groups of binary 5AFH
Hex
5
A
F
Binary
0101
1010
1111
So 5AFH = 10110101111 binary
Step 2 Convert binary to decimal
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1024
1
1024
512
0
256
1
256
128
1
128
64
0
32
1
32
16
0
8
1
8
4
1
4
2
1
2
1
1
1
Answer 1024 + 256 + 128 + 32 + 8 + 4 + 2 +1= 145510
Converting Hex to Decimal method 2
Convert 4DDH to decimal
164
163
16 x 16 x 16 x 16 16 x 16 x16
162
16 x 16
4
4 x 16 x 16
1024
161
16
D
13 x 16
208
160
1
D
13
13
161
16
F
15 x 16
240
160
1
7
7
7
Answer 4DDH = 1024 + 208 + 13 = 124510
Convert D6F7H to decimal
164
163
16 x 16 x 16 x 16 16 x 16 x16
D
13 x 16 x 16 x16
53248
162
16 x 16
6
6 x 16 x 16
1536
Answer D6F7H = 53248 + 1536 + 240+7= 5503110
Example 1
To demonstrate converting 447710 to hex
1
2
3
4
Divide 4477 by 16 put the integer of the answer underneath
and the remainder alongside 279.8125 remainder 0.8125 x
16 = 13
Divide the answer 279 by 16 put the integer of the answer
underneath and the remainder alongside 17.4375 remainder
0.4375 x 16 = 7
Divide the answer 17 by 16 put the integer of the answer
underneath and the remainder alongside 1.0625 remainder
0.0625 x 16 = 1
Divide the answer 01 by 16 put the integer of the answer
underneath and the remainder alongside 0 remainder 1
4477
0279
Remain
13 in
hex
D
0279
0017
7
0017
0001
1
0001
0000
1
Remain
Read the columns upwards to get the answer is 447710 = 17DH
Example 2
To demonstrate converting 124510 to hex
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1
Divide 1245 by 16 put the integer of the answer underneath
and the remainder alongside 77.8125 remainder 0
.8125.9375 * 16 = 13
2 Divide the answer 77 by 16 put the integer of the answer
underneath and the remainder alongside 4.8125 remainder
0.8125 * 16 = 7
3 Divide the answer 4 by 16 put the integer of the answer
underneath and the remainder alongside 0.25 remainder 0.
25 = 4
Read the columns upwards to get the answer is 124510 = 4DDH
1245
0077
0077
0004
0004
0000
13 in
hex
D
13 in
hex
D
4
Converting a decimal number to BCD
Express 25 as a BCD
2
0010
5
0101
4
0100
7
1101
Express 347 as a BCD
3
0011
Using calculator
You can use the calculator supplied with Windows to perform conversions between Decimal,
1
Select, Programs, Accessories, Calculator
2 If needed
Select View
Scientific
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3 Click the
button to select
the mode of data
entry in this
example binary
4 Enter your
data in this
example 1101
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5 Click the
button for your
conversion in
this example
Dec
Representing real numbers in binary
Fixed Point
Fixed-point numbers are used to provide fast calculations. The big disadvantage of fixed-point
calculations is their inability to handle large numbers or small fractions. Addition or
subtraction is easy with fixed-point numbers but multiplication or division are difficult.
How do they work?
To create fixed-point number a set number of bits is divided into 2 parts, one part to represent
the integer (real component) and the other part to represent the fraction/decimal (rational
component).
Example
Suppose an 8-bit Integer number was made into a fixed-point number then 1 way of splitting
it would be
Sign bit
Real
Decimal
Rational
3
2
1
0
-1
2
2
2
2
2
2-2
8
4
2
1
0.5
0.25
1/2
&frac14;
0
1
1
1
1
.
1
1
1
1
1
1
1
.
1
1
+
8 + 4 + 2 + 1 = 15
.
0.5 + 0.25 = 0.75
8 + 4 + 2 + 1 = 15
.
0.5 + 0.25 = 0.75
So the maximum number that could be represented with this arrangement is from
+ 15.75 to – 15.75 to an accuracy of 0.25
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Accuracy
The more bits used the more accurate the number. Assume that 7 bits have been allocated to
the rational part then the accuracy would be to 1/128. This leads to problems when fixed point
numbers are used for calculations.
2-1
2-2
2-3
2-4
2-5
2-6
2-7
.5
.25
.125
.0625
0.03125
0.015625 0.0078125
&frac12;
&frac14;
1/8
1/16
1/32
1/64
1/128
Summary
Fixed point numbers can not represent very large or very small number
Fixed point numbers are not very accurate when used for calculations
Floating Point
Floating point numbers are used to represent real numbers to a high degree of accuracy.
Fixed-point numbers have accuracy problems as they can only use a set number of digits to
represent the decimal place. Floating point numbers overcome this by using scientific
notation. Scientific notation uses a mantissa, a base number and an exponent.
Here is the number 123.45678 in scientific notation 1.2345678 x 102
Here is the number 12345.678 in scientific notation 1.2345678 x 104
Here is the number 0.0012345678 in scientific notation 1.2345678 x 10-3
There are 2 types of floating point numbers single precision which use 32 bits and double
precision which use 64 bits. Only single precision will be explained below.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
How do they work
The decimal number is converted to binary (base 2) and stored in the computer in 3 parts:
 The sign
 The exponent
 The mantissa
When 32 bits (single precision) are used to store a number the bits are allocated as shown
below.
 The sign
 The exponent
 The mantissa
There is 1 sign bit which is used to store if the number is positive (0) or negative (1)
There are 7 exponent bits which can store a maximum number of 225 how this is used will be
explained later.
There 23 mantissa bits used to represent the fraction.
Example
For the purpose of this example how the decimal part of a real number is converted to binary
will be omitted.
Here is a real decimal number 3.6875 this would be represented as
The integer 3 in binary is 11. The decimal 0.6875 in binary is 1011
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So the binary number would be 11.1011
This is then normalised which means move the decimal point to the left of the first 1
Normalised = 1.11011 x 21
Now the sign bit is easy it is 0 as the number is positive.
The exponent is 127 + 1= 128 as the exponent is positive it is added to 127
The mantissa is 11011 the leading 1 is dropped, as it will always be there due to normalisation
To see how this number would be stored look at the first row.
For example 0.6875 in binary is 0.1011 when normalised is 1.011 x 2-1
The sign bit is 0
The exponent is worked out as 127 – 1 = 126 as a negative exponent is taken from 127
The mantissa is 011 the leading 1 is dropped, as it will always be there due to normalisation
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
To see how this number would be stored look at the second row
0 1 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Accuracy
There are two levels of accuracy with floating point numbers single and double precision.
Single precision numbers generally use 32 bits and double precision numbers use 64 bits. The
more bits used the larger number that can be represented. The word precision is a little
misleading, as it does not refer to the accuracy of the number just that double the number of
bits used.
You will come across single and double precision number in programming and the
importance is the amount of space needed to store the number and the range of the number
that can be stored. Numbers that fall outside these ranges will be subject to errors.
Storage: Single 32 bits = 4 bytes
Range: -3.402823E38 to -1.401298E-45 for negative values; 1.401298E-45 to 3.402823E38
for positive values
Storage: Double 64 bits = 8 bytes
Range: -1.79769313486232E308 to -4.94065645841247E-324 for negative values;
4.94065645841247E-324 to 1.79769313486232E308 for positive values
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Converting Floating Point to Binary
Well it works by using codes. So the first job is to calculate the code. Lets start with an easy
example such as 8.5
Step 1
Get the sign bit.
The sign bit is 0 as the number is positive.
Step 2
Convert the decimal to binary
Revision
27
26
25
24
23
22
21
20
2x2x2x2x
2x2x2
2x2x2x2x
2x2
2x2x2x2x
2
2x2x2x2
2x2x2
2x2
2
1
10000000 01000000 00100000 00010000 00001000 00000100 00000010 00000001
128
64
32
16
8
4
2
1
Now if we go the other way you get
2-1 2-2
2-3
2-4
2-5
&frac12;
&frac14;
1/8
1/16
1/32
2-6
2-7
1/64
1/128
.1 .01 .001 .0001 .00001 .000001 .0000001
0.5 0.25 0.125 0.0625 0.03125 0.015625 0.0078125
So 8.5 is 1000.1
Step 3
Get the exponent.
You have to move the decimal place 3 places to the left to end up with 1.0001. So the
exponent is 3. Now you have to add 127 to the 3 to make 130 which in binary is 10000010.
Step 4
Deal with the mantissa.
1.0001 as the first 1 is implicit it is left off so the mantissa becomes 0001
Step 5
Make up the number
Step 1 has the sign bit of 0
Step 3 has the exponent of 10000010
Step 4 has the mantissa of 0001 to which you have to add the remaining 0 to make the
number of 0 to 23.
So the final binary number for 8.5 decimal is:
01000001000010000000000000000000
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Example 2
Here is another number -15.078125
Step 1
Get the sign bit.
Negative is a 1
Step 2
Convert the decimal to binary
15.078125
15 is 8 + 4 + 2 + 1 which in binary is 1111
.078125 is 0.0625 + 0.015625, which in binary is .000101
So the complete binary number is 1111.000101
Step 3
Get the exponent.
1111.000101 so the exponent is 3 to make 1.111000101
Add 127 to make 130 which in binary is 10000010
Step 4
Deal with the mantissa.
1.111000101 is 11100010100000000000000
Step 5
Make up the number
11000001011100010100000000000000
To find out more
Do a search on www.google.co.uk with the criteria “floating point binary”. This will bring up
lots of hits relating to floating point numbers.
Summary
Floating point numbers are more accurate that fixed point
Floating point numbers can represent very small values
There are two floating-point numbers systems, single and double precision.
Binary Mathematics
Adding binary is the same as adding in decimal but you cannot have any number greater than
1. So 1 + 0 = 1, 0 + 1 = 1, 1 + 1 = 0 carry 1 and 1 + 1+ 1 = 1 carry 1.
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Example 1 Add 01111 to 01010 (15 + 10) the red number is the carry
Step 1
0 1 1 1 1
Step 2
0 1
0 1 0 1 0
0 1
1+0=1
1 + 1 = 10
1
1
0
1
1
0
1
0
1
1
Step 3
0
0
1
1
1
1
0
1
1
0
0
1
1
1
1
1
1
0
0
1
1
1
0
1 + 0 + 1 = 10
Step5
0
0
1
1
1+0+0=1
1
0
1
1
0
1
Example 2 Add 10111 to 01011 (23 + 11)
Step 1
1 0 1 1 1
0 1 0 1 1
1 + 1 = 10
0
1
Step 3
1
0
0
1
1
1
1
1
1
1
0
1
1
0
0
1
0
1
0
1
1
0
0
1
1
1
1
1
1
1
0
1 + 0 + 1 = 10
Step5
1 + 0 + 1 = 10
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1
1
0
0
1
Step 4
0
0
1
0
0
1
1
1
0
1
0
1
1
1
1
1
1
1
1
0
0
1
1
0
0
1
1
0
1
1
0
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
0
1
0
1
0
1
0
0
0
1
0
1 + 1 + 1 = 11
Step 6
Step 2
1 + 1 + 1 = 11
Step 4
1
0
1 + 1 + 0 = 10
Step 6
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Subtraction
Subtraction is carried out using the 2's complement method. This is a mechanical process of
inverting the negative number and adding 1 to the inverted number.
Important Point Make sure you keep the numbers the same length. You should always end
up with an extra number if you have carried out the process correctly.
Example 1 0111 - 0100 (7 - 4)
Invert the negative number
0
1
1
0
0
1
0
1
Now add the two numbers 0111 + 1100
Leave the spare 1 that
0 1
has fell off the end
1 1
1 0 0
Example 2 11010101 - 01100100 (213 - 100)
Invert the
0 1 1 0 0 1 0 0
negative
1 0 0 1 1 0 1 1
number
Add 1 to the inverted number
1
0
1
1
0
1
1
0
1
1
1
0
inverted
number
1
0
0
1
1
0
1
1
0
0
1
1
1
0
Now add the two numbers 11010101 - 10011100
Leave the spare 1 that
1 1 0 1 0 1 0 1
has fell off the end
1 0 0 1 1 1 0 0
1 0 1 1 1 0 0 0 1 Answer = 01110001 or 113
subtraction.
Multiplication and Division
Microprocessors can only add so this is how the do multiplication and subtraction is used to
carry out division. The process is a little more complicated than described below but the basic
principle is the same.
Multiplication
Multiplication 5 x 4 means 4 groups of 5 it can be shown as 5 + 5 + 5 +5 so the
microprocessor will perform 4 additions of 5 very quickly.
Division
Division is seeing how many times a number can be taken away from another number so 12 
3 can be show as
(12 - 3 = 9) (9 - 3 = 6) (6 - 3 = 3) (3 - 3 = 0) so the number 3 can be subtracted from the
number 12 a total of 4 times.
The microprocessor only carries out binary mathematics so all hexadecimal numbers are
converted to binary first. To carry out hexadecimal mathematics yourself it is best to convert
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1
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1
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the hexadecimal numbers to decimal and then carry out the calculation. Once the answer has
been obtained convert the answer from decimal to hexadecimal. What is important is that you
check your answers. Another way is to use binary groups of 4. This can be easy if you can
remember the binary number sequence 0 to 15.
Decimal Hex
Binary
0
0
0000
1
1
0001
2
2
0010
3
3
0011
4
4
0100
5
5
0101
6
6
0110
7
7
0111
8
8
1000
9
9
1001
10
A 1010
11
B 1011
12
C 1100
13
D 1101
14
E
1110
15
F
1111
Hex
Convert to decimal
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A4
A+5
10 + 5 = 15
15 = F
FF
+
+
5B
Convert to decimal
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FF
4+B
4 + 11 = 15
15 = F
Page 19 of 25
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Syllabuses
Unit 02: Computer Systems
Learning hours: 60
NQF level 3: BTEC National
Description of unit
This unit outlines the fundamental way in which a computer works, starting with simple logic
and progressing to a simple model of a microprocessor. This is followed by an overview of
computer hardware and peripherals to enable learners to understand the way in which
computer systems are constructed and how they work. The unit also covers a basic knowledge
of the purpose of operating systems, some elementary operating system processes and
elementary low-level programming.
The unit is intended as an introduction to a broad range of important computing concepts, and
consequently it is not necessary to deliver the content in great depth. Other units (in particular
Unit 23: Computer Hardware and Unit 19: Multimedia Technology) are intended to explore
some of the topics in this unit in much greater depth.
This unit presents opportunities to demonstrate key skills in application of number and
communication.
This unit presents opportunities to demonstrate key skills in application of number and
communication.
Summary of learning outcomes
To achieve this unit a learner must:
1 Describe the internal operations of a model microprocessor
2 Demonstrate a thorough knowledge of modern computer systems hardware
3 Write simple low level programs
4 Install and use a modern operating system.
1 Internal operations
Data representation: number system conversions, eg binary, denary, hexadecimal, floating
point numbers, ASCII, bit masks, graphic bitmaps, role of different number systems;
demonstration of the possibility of errors by inappropriate representation of decimal or other
numbers in various binary forms
Logic and fetch-execute cycle: concepts of logic and logic gates, simple arrays of logic gates,
truth tables, concepts of registers, buses, control unit, arithmetic and logic unit, memory,
clock and control signals in a model microprocessor and the fetch-decode-execute cycle,
without reference to performance enhancing hardware such as cache memory
2 Computer systems hardware
Computer selection and design: analyse user needs and select hardware matched to a machine
specification
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PC build: build and configure a PC to specification using appropriate safety and ESD
precautions as appropriate
Maintenance: maintain software and hardware using appropriate disk tools
Troubleshooting: locate and repair faults using appropriate faultfinding techniques
3 Low level programs
Machine and assembly language programming: design simple machine and/or assembly
language programs to use data transfer, arithmetic and jump and branch instructions.
4 Operating system
Operating system functions: overview of functions, eg user interface, machine and peripheral
management, etc; comparison between functions of different types of operating system
(personal computer, network, mainframe, etc).
Operating system installation: install operating system, eg Windows (modern versions),
Linux, AppleMacOS.
Computer operations: use of a proprietary operating system, generation of environment and
systems for a computer user (file/directory structures, tailoring of screen interface, backup
systems, etc).
In order to pass this unit, the evidence that the learner presents for assessment needs to
demonstrate that they can meet all of the learning outcomes for the unit. The criteria for a pass
grade describe the level of achievement required to pass this unit.
To achieve a pass grade the evidence must show that the learner is able to:
1.
2.
3.
4.
5.
6.
7.
State the basic concepts and principles involved in the representation of data (including
numeric, ASCII, bit masks and bit maps) within a computer system
Specify and select components, and build, conforming to safety standards, a PC to meet
a technical specification
Install and configure an operating system and software to meet a client specification
Convert numbers between binary, denary and hexadecimal notation
Add, subtract, multiply and divide numbers held in different formats including floating
point
Write very basic program functions (such as fetching and adding together two numbers,
and storing the result) in a low level language
Check the accuracy of own work and eliminate obvious errors.
To achieve a merit grade the evidence must show that the learner is able to:
1.
2.
3.
Understand the concepts and principles involved in the representation of data within a
computer system, the role played by each format and the possible errors caused by
inappropriate use of format
Describe, with the aid of diagrams, the hardware (including logic gates, registers, buses,
CU, ALU and memory) and its function within a model microprocessor
Describe the functions of an operating system within a modern computer system
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4.
Design and write programs in a low level programming language which include data
transfer, arithmetic, jump and branch instructions.
To achieve a distinction grade the evidence must show that the learner is able to:
1.
2.
3.
Describe in detail the function of the fetch- decode-execute cycle and its relationship to
the workings of a microprocessor, making detailed reference to the role played by
hardware and software in this relationship
Evaluate the work done in building and configuring a computer system to a given
specification including how well the finished system met the specification, what went
well and what went less well in the work and what improvements could be made to the
resulting system
Write coherently and comprehensively, including accurate drawings, which enhance the
text and show a good command of technical language.
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Higher Nationals
Unit 1: Computer Platforms
Learning hours: 60
NQF level 4: BTEC Higher National — H1
Description of unit
This unit is aimed at IT practitioners who need sufficient knowledge of computer architecture
to make rational and commercial decisions on the selection and specification of systems.
Learners will learn how to evaluate operating systems in order to create their own operating
environment. Many IT practitioners communicate with specialist technical support staff
during the specification and planning of systems implementation. This unit aims to give
learners the confidence to communicate with technical and non-technical specialists to justify
their recommendations.
It is expected that centres will use current personal computer and networking resources.
Learners should be encouraged to read current journals to investigate and evaluate new
hardware and software developments.
Summary of learning outcomes
To achieve this unit a learner must:
1 Investigate computer systems
2 Investigate operating systems
3 Design a computer system
1 Computer systems
Processor: description of components (Von-Neuman architecture), terminology (eg bits,
bytes, kilobytes etc), identification of factors affecting performance (eg millions of
instructions per second (MIPS), floating point operations per second (FLOPS), clock speed,
computed performance indexes, bus architectures, pipelining)
Backing store: identification of types (disc, CD, CD-R, CD-RW, DVD-RW, DVD-RAM etc),
performance factors (eg data transfer rate, seek times, capacity)
Peripherals: description of available peripherals (displays, printers etc), understanding of
performance factors (eg displays — performance, resolution, colour depth, video RAM,
refresh rate, interlacing, slot pitch, etc, printer — speed, resolution, image quality, software
requirements, postscript, PCL and associated printer control)
Computer selection: specification of requirements, performance of the selected system, costs,
user benefits
2 Operating systems
Operating system functions: overview of functions (eg user interface, machine and peripheral
management etc), comparison between functions of different types of operating system
(personal computer, network, mainframe etc)
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Computer operations: proprietary operating systems, creation of environment and systems for
a computer user (file/directory structures, tailoring of screen interface, backup systems etc),
user profile
3 Design a computer system
Selection: processor (eg speed, special characteristics), memory, storage devices, display,
peripherals, specialised components (eg sound cards, video cards, datalogging interfaces), bus
User needs: costs, productivity, particular requirements (eg power, display, special needs),
training needs
Test plan: software testing (eg: black box, white box), hardware testing methodologies,
documentation, health and safety issues (eg compliance)
User support planning: identifying user training needs, producing a training schedule,
functions of a help desk/help line/help software.
Security: physical and logical security measures, backup and recovery, hacking, encryption,
levels of access rights
Outcomes Assessment criteria for pass
To achieve each outcome a learner must demonstrate the ability to:
1 Investigate computer systems


Select machine components or sub-systems appropriate to given tasks
Evaluate the performance of the selected system
2 Investigate operating systems


Contrast the functions and features of different types of operating systems
Understand how to customise operating systems
3 Design a computer system


Investigate and identify the key components for a computer system for a particular
user
Specify a complete computer system to suit a given task
4 Test a computer system




Produce a plan that checks the main hardware and software components, using
standard techniques
Produce user documentation for your system
Produce a security policy for your system.
Demonstrate that the system meets health and safety requirements
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