Introduction to Computer Science
Dr. Nagy Ramadan
E-mail: Nagyrdo@yahoo.com
Lecture - 5
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Outline

Signed-magnitude system.

Signed-complement system.

Excess System.

Examples
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Signed Binary Numbers
It is usual to represent the sign with a bit placed in the
leftmost position of the binary number.
Sign bit
The Most common notations are:

Signed-magnitude system.

Signed-complement system.

Excess System.
Sign bit 0 positive
Sign bit 1 negative
Sign bit 1 positive
Sign bit 0 negative
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Signed-magnitude system.
Ex1:
01001
11001


+9
–9
Ex2:
3-bit binary pattern
Bit Pattern
000 001 010 011 100 101 110 111
Signed-Magnitude
Decimal Value
+0 +1 +2
+3 -0
-1
-2
-3
Note:
For a n-bit binary pattern, the signed-magnitude decimal
range is –(2n-1-1)10, +(2n-1-1)10
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Signed-complement system

In this system, a negative number is indicated by its complement.

Since positive numbers always start with 0 (i.e. +) in the leftmost
position, the complement will always starts with 1 (i.e. -)

The signed-complement system can use either the 1’s complement
or the 2’s complement notations.

Changing the sign of the binary number in the 1’s complement system is
obtained by taking the 1’s complement of the binary number.

Changing the sign of the binary number in the 2’s complement system is
obtained by taking the 2’s complement of the binary number.
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EX:
Assuming the representation of the number 9 in binary
with 8-bits, we have the following cases:

Unsigned 9 or +9 has a the same representation in both
signed-magnitude and signed-complement systems
which is: 00001001

-9 has the signed-magnitude representation: 10001001

-9 has the signed-1’s complement representation: 11110110

-9 has the signed-2’s complement representation: 11110111
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EX
The signed-complement conversion table of a 3-bit binary
pattern is as follows:
Bit Pattern
000 001 010 011 100 101 110 111
Signed 1’s complement
decimal value
+0 +1 +2
+3 -3
-2
-1
-0
Signed 2’s complement
decimal value
+0 +1 +2
+3 -4
-3
-2
-1
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EX:
Obtain the decimal value of the binary number
(11111001.101)2 in case of:
A. Unsigned binary notation
B. Signed-magnitude notation
C. Signed-1’s complement notation
D. Signed-2’s complement notation
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Sol:
A. Unsigned binary notation
(11111001.101)2=1*27+1*26+1*25+1*24+1*23+1*20+1*2-1+1*2-3
=128 + 64 + 32 + 16 + 8 + 1 + 0.5 + 0.125=(249.625)10
B. Signed-magnitude notation
(s) (11111001.101)2= - (1*26+1*25+1*24+1*23+1*20+1*2-1+1*2-3)
= - (64 + 32 + 16 + 8 + 1 + 0.5 + 0.125)= - (121.625)10
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C. Signed-1’s complement notation
- 1’s complement of (11111001.101)2
= - (0000110.010) 2 = - (6.25)10
D. Signed-2’s complement notation
- 2’s complement of (11111001.101)2
= - (0000110.011) 2 = - (6.375)10
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Excess Notation

In this system, any binary number having 1 in the leftmost bit is
considered positive number.

All negative numbers have 0 in the leftmost bit.
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EX
The excess notation conversion table of a 3-bit
binary pattern is as follows:

Bit Pattern
000 001 010 011 100 101 110 111
Unsigned decimal
value
0
1
2
3
4
5
Excess notation
decimal value
-4
-3
-2
-1
0
+1 +2 +3
6
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The above table is called Excess Four Conversion Table, it is
obtained by subtracting 4 from the corresponding unsigned value.
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EX:
Convert each of the following excess eight notations to
its equivalent decimal form:
a) 1101
b) 0100
c) 0000
Sol:
Excess decimal value=unsigned decimal value – 8
a) (1101)  13 – 8 = + 5
b) (0100)  4 – 8 = - 4
c) (0000)  0 – 8 = - 8
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EX:
Convert each of the following decimal values to its
equivalent excess eight notations form:
a) 6
b) - 6
c) 0
Sol:
a) 6 + 8 = 14  (1110)
b) – 6 + 8 = 2  (0010)
c) 0 + 8 = 8  (1000)
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