Here are the answers to the assignment questions: 1. No overflow occurs when adding 00100110 + 01011010 in two's complement. The sum is 10001000. 2. See textbook 1 problem 2-1.c for the solution. 3. See textbook 1 problem 2-11.c for the solution. 4. See textbook 1 problem 2-19.c for the solution. 5. The decimal equivalent of the hexadecimal number 1A16 is 2610.

1. NUMBER SYSTEM

2. Positional Number Systems
- Each digit position has an associated
weight
- The value of the number i a weighted
h l f h b is i h d
sum of the digits
- Th di it i position I h weight ri ,
The digit in iti has i ht
Where r is the radix (base)
( )

3. O
Octal and Hexadecimal Numbers
- Radix 8 and 16
- Useful for representing multi-bit numbers
p g
-Conversion from binary is done
by separating the bits into groups of
three or four (R-L) and replace each group
with the corresponding octal or hexadecimal
number

4. - If the number contains digits to the right of the
binary point, we group the part after the binary
point(L-R)
- To convert from Octal and Hexadecimal, we
replace each digit with the corresponding 3 or
4 bits string
string.
- Conversion from Decimal to Binary,Octal &
y,
Hexadecimal is achieved by dividing by the
radix.

5. Fractions
Binary to decimal
10.1011 >
10 1011 =>
1 x 2-4 = 0.0625
1 x 2-3 = 0.125
0 x 2-22 = 0.0
0 0
1 x 2-1 = 0.5
0 x 20 = 0.0
1 x 21 = 2.0
2 0
2.6875

6. Fractions

7. Fractions
(0.375)10
0.375 X 2 = 0.750 0
0.750 X 2 = 1.500 1
1.500
1 500 X 2 = 1 000 1
1.000
000 X 0 = 0 0
(0.375)10= (0.011)2

8. • Octal Addition (&
Subtraction)
• 11
456
123
601

9. Hexadecimal Addition (&
Subtraction)
A C 5 A 9
E D 6 9 4
D

10. Representation of negative numbers
ep ese tat o o egat ve u be s
- Singed-magnitude system: the number consists
Singed magnitude
of magnitude and symbol. The MSB is for the sign
0 means positive &1 means negative
- There are two representation for 0 1000 ( 0) &
0, (-0)
0000 (+0).
- For n-bit integer the range is –(2n-1 – 1) to +(2n-1-1)
+18 = 00010010
-18 = 10010010
18

11. - One complement : The MSB is for the sign.
- Boolean complement all bits to negate
+18 = 00010010
-18 = 11101101
- Two representations of zero: 0000 (+0)
1111 (-0)
- The range is –(2n-1 – 1) to +(2n-1-1).
(2n (2n 1).

12. - Two’s complement: MSB is for the sign
sign.
- The range is –(2n-1) to (2n-1-1)
(2 -1).
- 3 = 00000011
Boolean complement gives 11111100
B l l i
Add 1 to LSB +1
11111101

13. - Only one representation for 0
0.
0 = 00000000
Bitwise not 11111111
Add 1 t LSB
to +1
Result 1 00000000
Overflow is ignored, so:
-0=0

14. -T ’
Two’s C
Complement Additi
l t Addition
Overflow: An addition overflows if the
signs of the addends are the same and the
sign of the sum i diff
i f h is different f
from
the addends’ sign
addends sign.

15. 1001 1100
+ 0101 + 0100
1110 = -2 10000 = 0
(a) ( 7) + (+5)
(-7) (b) ( 4) + (+4)
(-4)
0011 1100
+ 0100 + 1111
0111 = 7 11011 = -5
(c) (+3) + (+4) (d) (-4) + (-1)

16. 0101 1001
+ 0100 + 1010
1001 = Overflow 0011 = Overflow
(e) (+5) + (+4) (f) (-7) + (-6)

17. Subtraction l
S bt ti rules:
Perform a bit-by-bit complement of the
subtrahend and add the complemented
subtrahend to the minuend with an initial
carry in of 1 instead of 0.

18. 0010 0101
+ 1001 + 1110
1011 = -5 10011 = 3
(a) M = 2 = 0010 (b) M = 5 = 0101
S = 7 = 0111 S = 2 = 0010
-S = 1001
S -S = 1110
S

19. 1011 0101
+ 1110 + 0010
11001 = -7 0111 = 7
(c)
( ) M = -5 = 1011
5 (d) M = 5 = 0101
S = 2 = 0010 S = -2 = 1110
-S = 1110 -S = 0010

20. 0111 1010
+ 0111 + 1100
1110 = Overflow 10110 = Overflow
(e) M = 7 = 0111 (f) M = -6 = 1010
S = -7 = 1001
7 S = 4 = 0100
-S = 0111 -S = 1100

21. The Byte Nibble, and Word
Byte, Nibble
• 1 byte = 8 bits
• 1 nibble = 4 bits
• 1 word = size depends on d t pathway
d i d d data th
size.
– Word size in a simple system may be one
byte (8 bits)
– Word size in a PC is eight bytes (64 bits)

22. -Codes are group of special symbols used t
C d f i l b l d to
represent numbers, letters or words.
1- Binary codes for decimal numbers (BCD)
y ( )
- Binary Coded Decimal (BCD) is another way
to present decimal numbers in binary form.
- BCD is widely used and combines features of
both decimal and binary systems.

23. - Each digit of a decimal is represented by its
four bit
four-bit binary equivalent (1 to 9)
- To represent the decimal number 10 we need
n mber e
eight bits (0001 0000)

24. • To convert the number 87410 to BCD
8 7 4 (decimal)
1000 0111 0100 (BCD)
• Each digit always uses four bits.
ac d g t a ways ou b ts.
• The BCD value can never be greater than 9
• Reverse the process to convert BCD to decimal.

25. • BCD i not a number system.
is t b t
• BCD is a decimal number with each digit
encoded to its binary equivalent.
equivalent
• A BCD number is not the same as a straight
binary number.
number
• The primary advantage of BCD is the relative
ease of converting to and from decimal.

26. Decimal Binary Octal Hexadecimal BCD
0 0 0 0 0
1 1 1 1 0001
2 10 2 2 0010
3 11 3 3 0011
4 100 4 4 0100
5 101 5 5 0101
6 110 6 6 0110
7 111 7 7 0111
8 1000 10 8 1000
9 1001 11 9 1001
10 1010 12 A 0001 0000
11 1011 13 B 0001 0001
12 1100 14 C 0001 0010
13 1101 15 D 0001 0011
14 1110 16 E 0001 0100
15 1111 17 F 0001 0101

27. 2
2- American Standard Code for Information
Interchange (ASCII)
- It is a seven bit code. It has 27 possible code
groups.
- Represents characters and functions found on
a computer keyboard.
- Examples of use are: to transfer information
between computers, between computers and
printers, and f i
i d for internal storage.
l

28. 3-
3 Gray code
Two successive values differ in only one bit

31. • 4- Codes for detecting and correcting
errors
• Error means corruption of data
data.
• - Parity Bit
• - Hamming C d
i Code

32. Parity bit:
It is an extra bit that is attached to
a code group that is being transferred from one
location to another. It is made either 0
or 1 Depending on the number of 1 that
1. di h b f 1s h
are contained in the code group.
Even parity:
The value of the parity bit is chosen so that
the total number of 1s including the parity
1s,
bit, is an even number;
1 1000011

33. Odd parity:
The value of the parity bit is chosen
so th t th t t l number of 1 i l di th
that the total b f 1s, including the
parity bit, is an odd number;
1 1000001

34. Assignment (1)
1- Indicate whether or not overflow occurs
when adding the following 8-bits
8 bits
two’s complement numbers 00100110
+ 01011010
(2.12 d textbok2)
2- 2-1.c (textbook1)
3- 2-11.c
4 2 19.c
4- 2-19.c
5- 2-24