Transcript Sign bit
Introduction to Computer Science
Dr. Nagy Ramadan
E-mail: [email protected]
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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