Fundamentals of Electrical and electronic engineering Binary code,grey code,octal code Module - 1.2.ppt

This ppt is about the fundamental of Electrical and electronic engineering

1. Binary Codes
Coding is the process of altering the characteristics
of information to make it more suitable for
intended application
Representation of data using 0’s and 1’s is called
binary code
Binary codes are classified as
Numeric codes
Eg: 8421,XS-3,Gray code
Alphanumeric codes
EBCDIC, ASCII

2. Binary codes may be
weighted
non weighted
Positively weighted –Assigned weights are
positive
Eg: 8421,2421,5211 etc
Negatively weighted-Some of the assigned
weights are negative
Eg:642-3,631-1 etc

3. Non weighted code
Does not obey positional weighting
principle
Eg: Excess 3 code(Xs-3), Gray code

4. BCD code( Binary Coded
Decimal)
The digits of a decimal number are encoded into
groups of binary digits
It is a weighted code
The weights attached are 8,4,2,1
Advantages
Ease of conversion
Disadvantage
Less efficient than pure binary
Invalid are : 1010,1011,1100,1101,1110,1111Decimal digit0 1 2 3 4
BCD 0000 0001 0010 0011 0100
Decimal digit5 6 7 8 9
BCD 0101 0110 0111 1000 1001

5. Excess-3 (XS-3) code
It is a non weighted code
The binary code word of it is corresponding to
8421 +3
Eg: 5= 0101 +0011 = 1000
The invalid states are:
0000,0001,0010,1101, 1110 &1111

6. Gray code
This is a non weighted code
It is not sutable for arithmetic operations
It is cyclic
Successive code differ in one bit position only
Good for error detection.DecimalBinaryGray CodeDecimalBinaryGray code
0 0000 0000 8 1000 1100
1 0001 0001 9 1001 1101
2 0010 0011 10 1010 1111
3 0011 0010 11 1011 1110
4 0100 0110 12 1100 1010
5 0101 0111 13 1101 1011
6 0110 0101 14 1110 1001
7 0111 0100 15 1111 1000

7. Binary-to-Gray Code Conversion
Retain most significant bit.
From left to right, add each adjacent pair of binary
code bits to get the next Gray code bit, discarding
carries.
Example: Convert binary number 10110 to Gray code.
(10110)
2 = 11101
Gray10110Binary

1 Gray 1+0110Binary

1 1 Gray 10+110Binary

11 1 Gray 101+10Binary

111 0 Gray 1011+0Binary

1110 1Gray

8. Gray-to-Binary Conversion
Retain most significant bit.
From left to right, add each binary code bit
generated to the Gray code bit in the next
position, discarding carries.
Example: Convert Gray code 11011 to binary.11011Gray

1 Binary 1 1011Gray
+
1 0 Binary 11 011Gray
+
10 0 Binary 110 11Gray
+
100 1 Binary 1101 1Gray
+
1001 0Binary
(11011)
Gray= (10010)
2

9. Alphanumeric Codes
Apart from numbers, computers also handle textual data.
Character set frequently used includes:
alphabets: ‘A’ .. ‘Z’, and ‘a’ .. ‘z’
digits: ‘0’ .. ‘9’
special symbols:‘$’, ‘.’, ‘,’, ‘@’, ‘*’, …
non-printable:SOH, NULL, BELL, …
Usually, these characters can be represented using 7 or 8
bits.
Used for transmitting data between computers and i/o
devices
Eg: ASCII,EBCDIC

10. ASCII (American standard Code for
Information Interchange)
ASCII: 7-bit, plus a parity bitfor error
detection (odd/even parity).CharacterASCII Code
0 0110000
1 0110001
. . . . . .
9 0111001
: 0111010
A 1000001
B 1000010
. . . . . .
Z 1011010
[ 1011011
\ 1011100

11. ASCII
ASCII table:MSBs
LSBs000001010011100101110111
0000NULDLESP 0 @ P ` p
0001SOHDC1 ! 1 A Q a q
0010STXDC2“ 2 B R b r
0011ETXDC3# 3 C S c s
0100EOTDC4$ 4 D T d t
0101ENQNAK % 5 E U e u
0110ACKSYN & 6 F V f v
0111BELETB ‘ 7 G W g w
1000BSCAN ( 8 H X h x
1001HTEM ) 9 I Y i y
1010LFSUB * : J Z j z
1011VTESC + ; K [ k {
1100FFFS , < L \ l |
1101CRGS - = M ] m }
1110 O RS . > N ^ n ~
1111SIUS / ? O _ oDEL

12. EBCDIC(Extended Binary Coded Decimal
Interchange code)
It is an 8 bit alphanumeric code
It can be used to encode all the symbols and
control characters
Because 2
8
= 256 bit patterns are available
A
I
R
C1
C9
D9
0
9
+
-
F0
F9
4E
6D

13. Error detecting and correcting code
When data transmitted through a channel, due
to noise they may be get corrupted
Error detection and correcting codes are used
to detect and correct errors
Encoding may be done for error correction
and error detection.
Eg: Parity code, Hamming code

14. Error Detecting Code (Parity)
The simplest technique for detecting errors
Implemented by adding an extra bit to each
word being transmitted
Odd Parity –Set to a 0 or a 1 such that the total
number of 1 bits in the word including the parity
bit is an odd number
Even Parity -Set to a 0 or a 1 such that the total
number of 1 bits in the word including the parity
bit is an even number

15. 7 bit Hamming Code (Error correcting)
Three parity bits are added to transmit four
data bits
Parity is added at bit positions 2
0
(1),2
1
(2) and(
2
2
(4)
The word formed is
P1P2D3P4D5D6D7
D is data bits
P is parity bits

16. Hamming Code(2)
P1 is set to 0 or 1, so that it establishes even
parity over the bits 1,3,5,7(p1,d3,d5,d7)
P2 is set to 0 or 1, so that it establishes even
parity over the bits 2,3,6,7(p2,d3,d6,d7)
P4 is set to 0 or 1, so that it establishes even
parity over the bits 4,5,6,7(p4,d5,d6,d7)

17. Hamming code(3)
Eg:
Encode 0011 to hamming code
P1 P2D3P4D5D6D7
0 0 1 1
P1(P1,D3,D5,D7)-P1 001 1
P2(P2,D3,D6,D7)-P2 011 0
P4(P4,D5,D6,D7)-P4 011 0
Final code is 1000011

18. Decoding of Hamming code
At the receiving end the hamming code is
decoded
Bits 1,3,5,7 bits 2,3,6,7 and bits 4,5,6,7 are
checked
If an error
A three bit binary number is generated out of
the parity checks and error bit is complimented

19. Received word 1001001
(1,3,5,7)-1001 no error –put 0 in 1’s position
(2,3,6,7)-0001 error –put 1 in 2’s position
(4,5,6,7)-1001 no error –put 0 in 4’s position
Error word is 010(2)
Compliment 2
nd
bit
Correct code is 1101001

20. Binary Arithmetic Operations (1/6)
ADDITION
Like decimal numbers, two numbers can be
added by adding each pair of digits together
with carry propagation. (11011)2
+ (10011)2
(101110)2
27
10
+ 19
10
----------------
46
10

21. Binary Arithmetic Operations (2/6)
Digit addition table:BINARY DECIMAL
0 + 0 + 0 = 0 00 + 0 + 0 = 0 0
0 + 1 + 0 = 0 10 + 1 + 0 = 0 1
1 + 0 + 0 = 0 10 + 2 + 0 = 0 2
1 + 1 + 0 = 1 0 …
0 + 0 + 1 = 0 11 + 8 + 0 = 0 9
0 + 1 + 1 = 1 01 + 9 + 0 = 1 0
1 + 0 + 1 = 1 0 …
1 + 1 + 1 = 1 19 + 9 + 1 = 1 9 0
1
1
0

1
Carry
inCarry
out (11011)2
+ (10011)2
(101110)2 1
1
1
1

1 1
0
0
1

0 0
1
0
1

0 0
1
1
0

1 1
0
0
1

0

22. Binary Arithmetic Operations (3/6)
SUBTRACTION
Two numbers can be subtracted by subtracting
each pair of digits together with borrowing,
where needed. (11001)2
- (10011)2
(00110)2 (627)10
- (537)10
(090)10

23. Binary Arithmetic Operations (4/6)
Digit subtraction table:BINARY DECIMAL
0 - 0 - 0 = 0 00 - 0 - 0 = 0 0
0 - 1 - 0 = 1 10 - 1 - 0 = 1 9
1 - 0 - 0 = 0 10 - 2 - 0 = 1 8
1 - 1 - 0 = 0 0 …
0 - 0 - 1 = 1 10 - 9 - 1 = 1 0
0 - 1 - 1 = 1 01 - 0 - 1 = 0 0
1 - 0 - 1 = 0 0 …
1 - 1 - 1 = 1 19 - 9 - 1 = 1 9
Borro
w (11001)2
- (10011)2
(00110)2 0
1
1
0

0 0
0
1
1

1 1
0
0
1

1 1
1
0
0

0 0
1
1
0

0 0
0
0
0

0

24. Binary Arithmetic Operations (5/6)
MULTIPLICATION
To multiply two numbers, take each digit of the multiplier
and multiply it with the multiplicand. This produces a
number of partial productswhich are then added. (11001)2 (214)10Multiplicand
x (10101)2 x (152)10Multiplier
(11001)2 (428)10
(11001)2 (1070)10 Partial
+(11001)2 +(214)10 products
(1000001101)2 (32528)10Result

25. Negative Numbers Representation
Unsigned numbers: only non-negative values.
Signed numbers: include all values (positive and negative).
Till now, we have only considered how unsigned (non-
negative) numbers can be represented.
There are three common ways of representing signed
numbers
Sign-and-Magnitude
1s Complement
2s Complement

26. Negative Numbers: Sign-and-Magnitude
(1/4)
Negative numbers are usually written by writing a
minus sign in front.
Example:
-(12)
10, -(1100)
2
In sign-and-magnitude representation, this sign is
usually represented by a bit:
0for +
1for -

27. Negative Numbers: Sign-and-Magnitude
(2/4)
Example: an 8-bit number can have 1-bit sign
and 7-bit magnitude.
sign
magnitude

28. Negative Numbers: Sign-and-Magnitude
(3/4)
Largest Positive Number: 0 1111111 +(127)
10
Largest Negative Number: 1 1111111 -(127)
10
Zeroes: 0 0000000 +(0)
10
1 0000000 -(0)
10
Range: -(127)
10to +(127)
10
Question: For an n-bit sign-and-magnitude
representation, what is the range of values that
can be represented?

29. Negative Numbers:Sign-and-Magnitude
(4/4)
To negate a number, just invert the sign bit.
Examples:
-(0 0100001)
sm = (1 0100001)
sm
-(1 0000101)
sm= (0 0000101)
sm

30. 1s and 2s Complement
Two other ways of representing signed numbers for
binary numbers are:
1s-complement
2s-complement
They are preferred over the simple sign-and-
magnitude representation.

31. 1s Complement (1/2)
Essential technique: invertall the bits.
Examples: 1s complement of (00000001)
1s= (11111110)
1s
1s complement of (01111111)
1s = (10000000)
1s
Largest Positive Number: 0 1111111 +(127)
10
Largest Negative Number: 1 0000000 -(127)
10
Zeroes: 0 0000000
1 1111111
Range: -(127)
10to +(127)
10
The most significant bit still represents the sign:
0 = +ve; 1 = -ve.

32. 1s Complement (2/2)
Examples (assuming 8-bit binary numbers):
(14)
10= (00001110)
2
-(14)
10= -(00001110)
2= (11110001)
1s
-(80)
10= -( ? )
2= ( ? )
1s

33. 1’s compliment subtraction
Represent the number in true binary form
Add 1’s compliment of subtrahend to the minuend
If a carry, bring the carry around and add it to the
LSB
Look at the sign bit , If it is 0 result is positive and in
true binary
If MSB is 1, the result is negative and is in 1’s
compliment form
Take 1’s compliment to get the magnitude in binary

34. 46-14
[phone]
-14 11110001
46 00101110
-14 + 11110001
----------------------
+32 1 00100000 Add carry to LSB
Example

35. 2s Complement (2/4)
Essential technique: invertall the bits and add
1.
Examples:
2s complement of
(00000001)
2s= (11111110)
1s(invert)
= (11111111)
2s(add 1)
2s complement of
(01111110)
2s= (10000001)
1s(invert)
= (10000010)
2s(add 1)

36. 2s Complement (3/4)
Largest Positive Number: 0 1111111
+(127)
10
Largest Negative Number: 1 0000000 -(128)
10
(special Case)
Zero: 0 0000000
Range: -(128)
10to +(127)
10
The most significant bit still represents the sign:
 0 = +ve; 1 = -ve.

37. 2s Complement (4/4)
Examples (assuming 8-bit binary numbers):
(14)
10= (00001110)
2
-(14)
10= -(00001110)
2= (11110010)
2s
-(80)
10= -( ? )
2= ( ? )
2s

38. 2’ compliment subtraction
Add the 2’s compliment of subtrahend to the
minuend
If there is a carry ignore it
If MSB is 0, result is positive and in true binary
form
If MSB is a 1 result is negative and is in 2’s
compliment form
Take 2’s compliment

39. Example
46-14
[phone]
-1411110010
[phone]
-14 [phone]
----------------------
+32 100100000(Ignore carry)