1. HEXADECIMAL NUMBERS Code
WRITING one’s and zero’s can be error prone when dealing with large numbers. IBM came up with the following method of dealing with these numbers. BASE 16 with digits in base 10 are represented by the following notation in Hexadecimal.
016 = = = = 810
116 = = = = 910
216 = = 210 A16 = = 1010
316 = = 310 B16 = = 1110
416 = = 410 C16 = = 1210
516 = = 510 D16 = = 1310
616 = = 610 E16 = = 1410
716 = = 710 F16 = = 1510

2. EXAMPLE Convert A716 to binary
Base 16 Conversion
EXAMPLE Convert A716 to binary
Note: This is easy because of the relationship between the two bases
A = = 0111
Therefore 0A7H or $A7 =
Note leading 0 before A.
Since this is a number a different base than base 10, absence of the zero may confuse a compiler. Therefore it is customary to add the leading 0 to any hex character that begins with a letter of the alphabet (A,B,C,D,E,F) !! These are NUMBERS in base 16 !!

3. HEXADECIMAL TO DECIMAL
Similar Rules as to those in binary to decimal
Convert to binary then weighted multiplication -OR-
Leave in HEX and use HEX multiplication
EXAMPLE (repeated multiplication)
$F8E6H = F(16)3 + 8 (16)2 + E(16)1 + 6(16)0
= 15(16)3 + 8 (16)2 + 14(16)1 + 6(16)0
= 61,
= 63,718

4. DECIMAL TO HEXADECIMAL
Similar Rules as to those in decimal to binary Repeated HEX division
EXAMPLE (repeated division)
Divisors are for 16 binary digits (4 HEX) =
2479/16 = remainder 15 or F
154/16 = remainder 10 or A
9/ = remainder 9 or 9
Answer = 9AFH

5. DECIMAL TO HEXADECIMAL ALTERNATE
Same EXAMPLE (but weighted division)
Divisors are for 16 binary digits (4 HEX)
4096, 256, 16, 1 =
2479/256 = 9 remainder x256 =
175/16 = 10 or A remainder x 16 =
15/ = 15 or F
Answer = 9AFH

6. BINARY CODES 7421 6311 5421 5311 5211 DECIMAL 8421 BINARY 0 0000 0000
BCD
4 Bit BCD CODES
DECIMAL BINARY
....

7. MORE 4-BIT BCD CODES
Decimal /2/1 74/2/1
The / is subtract
weight

8. Decimal 2-out of-5 63210 Shift-Counter 86421 51111
5-BIT CODES
Decimal 2-out of Shift-Counter

9. Alphanumeric Codes OTHER CODES ASCII CODE 7 BITS EBCDIC CODE 8 BITS
UNICODE Bits

10. ASCII CODE Part 1

11. ASCII CODE CONTROL CODES

12. EBCDIC Part I

13. EBCDIC PART II

14. UNICODE 16 BIT CODE First Page 0000H - 00FFH Same as ASCII
Has not been standardized the remaining
0FFFFH minus 00FFH available for all remaining
countries and languages!

15. Representation of Negative Numbers
Number Systems
Representation of Negative Numbers
Three major schemes:
sign and magnitude
ones complement
twos complement
nines complement
tens complement
Assumptions:
we'll assume a 4 bit machine word
16 different values can be represented
roughly half are positive, half are negative

16. Number Systems Sign and Magnitude Representation
High order bit is sign: 0 = positive (or zero), 1 = negative
Three low order bits is the magnitude: 0 (000) thru 7 (111)
Number range for n bits = +/
Representations for 0 Two +/-
n-1

17. Number Systems Ones Complement
Subtraction implemented by addition & 1's complement
Still two representations of 0! This causes some problems
Some complexities in addition

18. Number Representations
Twos Complement
like 1's comp
except shifted
one position
clockwise
Only one representation for 0
One more negative number than positive number

19. 1 + 1 = 0 + carry 1 to next column RULES FOR SUBTRACTION 0 - 0 = 0
BINARY ARITHMETIC
RULES FOR ADDITION
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = carry 1 to next column
RULES FOR SUBTRACTION
0 - 0 = 0
0 - 1 = 1 and borrow from next column
1 - 0 = 1
1 - 1 = 0
Borrowing 1 from next column is equivalent to subtracting 1 from that column

20. EXAMPLE ADD 1010 to 1101 1010 + 1101 10111 SUBTRACT 1010 from 1101
Addition
EXAMPLE ADD 1010 to
SUBTRACT 1010 from 1101
borrow 4 subtract 2
nothing in 4 position
- 1010
0011

21. Number Representations
Addition and Subtraction of Numbers
Sign and Magnitude 4
+ 3 7
0100
0011
0111 -4
+ (-3) -7
1100
1011
1111
When signs is same,
result sign bit is the
same as the operands'
sign. Just add magnitudes
When signs differ,
operation is subtract,
sign of result depends
on sign of number with
the larger magnitude. This
is what we do in base 10. 4
- 3 1
0100
1011
0001 -4
+ 3 -1
1100
0011
1001

22. Sign and Magnitude Cumbersome addition/subtraction.
Number Systems
Sign and Magnitude
Cumbersome addition/subtraction.
Must compare magnitudes to determine the
sign of result.

23. /N = (2n - 1) - N Ones Complement
The symbol N, with a bar over it, is used to represent the complement. Also used to indicate inversion are /, ‘, or ! (CUPL and ABEL)
N is positive number, then N is its negative 1's complement. N is the precision, as many 1’s as required. Binary can be any precision, but BCD requires 4 bits.
Precision 4 bits
Therefore,
The general formula for finding
/N = (2n - 1) - N
To find 1's complement of 7
2 =
-1 =
1111
-7 =
1000
= -7 in 1's comp. 4

24. NOTE: Ones Complement ZERO Applying the general formula
/N = (2n - 1) - N
To find 1's complement of 0
2 =
-1 =
1111
-0 =
= -0 in 1's comp. 4
NOTE:
The complement of 0 is 1111, which means, there are
2 representations of 0.
Positive 0
Negative 0

25. SHORTCUT 1’s Complement
Shortcut method:
Replace all 0’s with 1’s, and all 1’s with 0’s
simply compute bit wise complement
7 in 1’s complement > 1000

26. Number Systems Addition and Subtraction of Numbers
Ones Complement Calculations 4
+ 3 7
0100
0011
0111 -4
+ (-3) -7
1011
1100
10111 1
1000
End around carry 4
- 3 1
0100
1100
10000 1
0001 -4
+ 3 -1
1011
0011
1110
End around carry

27. Number Systems Addition and Subtraction of Binary Numbers
Ones Complement Calculations
Why does end-around carry work?
Its equivalent to subtracting 2 and adding 1 n
For (M > N)
M - N = M + / N = M + (2n - N -1) = (M - N) + 2n - 1
For M + N < 2n-1
-M + (-N) = M + N = (2n - M - 1) + (2n - N - 1)
= 2n + [2n (M + N)] - 1
after end around carry the -1 is canceled if a carry, else not:
= 2n (M + N)
this is the correct form for representing -(M + N) in 1's complement!

28. Number Systems Two’s Complement Numbers
/N = 2n - N
Note: subtract number immediately. 4
2 =
7 =
1001 = represents -7
sub
Example: Twos complement of 7
2 =
-0 =
0000 = 0 4
sub
Example: Twos complement of 0
Only 4bits can represent number.
The most significant 1 is dropped!
Shortcut method Number 1:
Twos complement =1’s complement + 1
0111 -> > 1001 (representation of -7)
1001 -> > 0111 (representation of 7)

29. SHORTCUT Shortcut method Number 2:
Starting from the right, least significant bit, if 0 leave unchanged.
Continue from right to left, if 0, no change. When the first 1 is encountered,
we leave it unchanged, but complement each digit to the left.
710 = Apply method =
6810 = yields The lead 0 is important so that
the number to be converted
is positive! In base 10, the - sign
does that for us. With only on and
off (0,1) we need this extra bit
of information to describe the
signed number.
-6810 = yields =

30. Number Systems Addition and Subtraction of Binary Numbers
Twos Complement Calculations 4
+ 3 7
0100
0011
0111 -4
+ (-3) -7
1100
1101
11001 4
- 3 1
0100
1101
10001 -4
+ 3 -1
1100
0011
1111
Simpler addition scheme makes twos complement the most common
choice for integer number systems within digital systems

31. Number Systems Addition and Subtraction of Binary Numbers
Twos Complement Calculations
Why can the carry-out be ignored?
-M + N when N > M: n n
M* + N = ( M) + N = (N - M) n
Ignoring carry-out is just like subtracting 2
n-1
-M + -N where N + M < or = 2 n n
-M + (-N) = M* + N* = (2 - M) + (2 - N)
= (M + N) + 2 n n
After ignoring the carry, this is just the right twos compl.
representation for -(M + N)!

32. Number Systems Overflow Conditions
Add two positive numbers to get a negative number
or two negative numbers to get a positive number -1 +0 -1 +0 -2
1111 -2
0000 +1
1111
0000 +1
1110
0001
1110 -3 -3
0001 +2 +2
1101
1101
0010
0010 -4 -4
1100 +3
0011
1100 +3
0011 -5 -5
1011
1011
0100 +4
0100 +4
1010 -6
1010
0101 -6
0101 +5 +5
1001
1001
0110
0110 -7 +6 -7
1000 +6
0111
1000
0111
+1=6 (ok) -8 +7 -8 +7
+2=7 (ok)
+3=-8 (error)
= +7
4 bit representation
3 bits + sign in msb
range -8 to +7

33. This is true for both 1’s and 2’s complement addition.
OVERFLOW
If carry-in (penultimate) is added
to sign bit and there is a carry-out
then no overflow.
if carry-in differs from
carry-out then overflow
This is true for both 1’s and 2’s complement addition.

34. Number Systems Overflow Conditions Example is using 2’s Complement 5 3-8
carry -7 -2 7
carry
Overflow
Overflow
Note: the carry and penultimate carry are not the same!
This can be very easily implemented in hardware. XOR 5 2 7
carry -3 -5 -8
carry
No overflow
Overflow when carry in to sign does not equal carry out

35. BCD Addition BCD Number Representation
Decimal digits 0 thru 9 represented as 0000 thru 1001 in binary
Addition:
5 = 0101
3 = 0011
1000 = 8
5 = 0101
8 = 1000
1101 = 13!
Problem
when digit
sum exceeds 9
Solution: add 6 (0110) if sum exceeds 9.
5 = 0101
8 = 1000
1101
6 = 0110
= 1 3 in BCD
9 = 1001
7 = 0111
= 16 in binary
6 = 0110
= 1 6 in BCD

36. BCD Subtractin Must use 10’s complement
To find 10’s complement, first generate 9’s complement by subtracting
The number from 9.
If n = 9, 9-9 =0 9’s complement of 9 is 0 or 0000
If n = 8, 9-8 =1 9’s complement of 8 is 1 or 0001
If n = 7, 9-7 =2 9’s complement of 7 is 2 or 0010
If n = 6, 9-6 =3 9’s complement of 6 is 3 or 0011
If n = 5, 9-5 =4 9’s complement of 5 is 4 or 0100
If n = 4, 9-4 =5 9’s complement of 4 is 5 or 0101
If n = 3, 9-3 =6 9’s complement of 3 is 6 or 0110
If n = 2, 9-2 =7 9’s complement of 2 is 7 or 0111
If n = 1, 9-1 =8 9’s complement of 1 is 8 or 1000
If n = 0, 9-0 =9 9’s complement of 0 is 9 or 1001
Add 1 to 9’s complement to find 10’s complement.

37. Example of BCD Subtract
To subtract 2 from 5, find 10’s complement of 2 which is Bit
Representation.
The 1 carry, is normal form
and indicates answer is 3.
In binary, 1001 is exceeded!
Must add 6 to correct.
Known as BCD adjust.
0110
The 1 carry is normal form
answer is 3 base 10.

38. Addition Examples 4 BITS signed and unsigned

39. Addition Examples 4 BITS signed and unsigned
Note: overflow in all cases is determined by the precision, the carry, and in signed representation, the penultimate carry. This slide assumes a 4 bit precision. In base 10, the minimum sum is 0 and the maximum sum is 15 unsigned. A carry would indicate overflow. Base 10 max and min signed is -8 to +7, which is also true for 2’s and 10’s complement. The limit drops to + and - 7 for 1’s and 9’s complement because of the double representation of zero. Overflow in these cases is determined by the carry and the penultimate carry. If both are same, no overflow! If opposite, overflow.
Note in base 10 you 9’s and 10’s complement overflow conditions are recognized by the carry and penultimate carries. However, the range for single digit is -9 to +9 for 9’s complement, and -10 to + 9 for 10’s complement. If the arithmetic is using base 2, the overflow range will change to -7 to +7 in 9’s complement, and -8 to, +7 in 10’s complement.
A negative number is preceded by a 9 in base 10, so a -7 in 9’s complement is 92. In 10’s complement it is a 93. The next slide shows more examples. In a computer the carry will contain all the information when used with the penultimate carry to indicate over flow.

40. Excess-3 code is another important BCD Code.
To encode a decimal number, add 3 to each digit before converting to binary.
EXAMPLE (Note: 2 digits represented by 2 4 bit numbers)
to excess-3 = = =5
to excess-3 = = =12

41. EXAMPLES OF EXCESS ADDITION
CASE 1
In excess-3 addition, whenever we add two numbers whose sum is 9 or less, an excess-6 number is formed. To return to excess-3 we must subtract 3.
EXAMPLE
2 +5 = 7
1101 excess-6
1010 excess-3 equivalent of 7

42. EXAMPLES OF EXCESS ADDITION
CASE 2
In excess-3 addition, whenever we add two numbers whose sum is greater than 9, there will be a carry from one group to the next. When this happens, the group that produced the carry will revert to 8421 (BCD). To return to excess-3 we must subtract 3 from that group.
EXAMPLE
carry not carried to 10’s because of sum
excess-3 for 29
excess-3 for 39
first result
subtract less than 9, add 3
excess-3 for 68