Fundamentals of Computer Engineering 1
Exercise 7
Fundamentals of
Computer Engineering 1
Exercise 7
Faculty of Engineering
Department of Electrical Engineering and Information Technology
Institute for Computer Engineering- Prof. Dr. A. Hunger
Fundamentals of Computer Engineering 1
Kevin Moheb, M. Sc.
Tel: 0203 - 379 - 3577
Kevin.moheb@uni-due.de
Slide 1
Fundamentals of Computer Engineering 1
Exercise 7
Objectives
• Learn how to represent signed numbers in
binary systems:
- Sign-Magnitude Representation
- 1s complement
- 2s complement
• Learn what Floating Points are and how tod eal
with them
Slide 2
1
Fundamentals of Computer Engineering 1
Exercise 7
Intro
What is 3710 as binary number ?
Æ100101
What is –3710 as binary number ?
Æ???
New representation needed
Slide 3
Fundamentals of Computer Engineering 1
Exercise 7
Sign-Magnitude Representation
Most Significant Bit is sign bit
0 Æ positive
1 Æ negative
?
all
t
a
-3710 y=bi-ts1001012
n
t
m-a 0100101
en
=
i
c
w
fi
Ho
uf ion
s
=In- m0000100101
= ??
at
Æ or
inf
Slide 4
2
Fundamentals of Computer Engineering 1
Exercise 7
Excursion: Representations
For every sensible conversion to a
signed binary system we need to know
• the representation system
• the number of bits
(=data word length)
Slide 5
Fundamentals of Computer Engineering 1
Exercise 7
Sign-Magnitude – Representation
Proper question:
What is –3710 as a 8 bit SignMagnitude number ?
-3710 = - 0010 01012 = 1010 01012,SM
Slide 6
3
Fundamentals of Computer Engineering 1
Exercise 7
Sign-Magnitude: Range
Sign
Highest
Smallest
2n-1-1
Magnitude
0
1111….111
-(2n-1-1) 1
1111…111
1 bit
n-1 bits
Range of
4 bit SM numbers: [-7, 7]
8 bit SM numbers: [-127,+127]
etc
Slide 7
Fundamentals of Computer Engineering 1
Exercise 7
1s complement
What is –3710 as 8 bit 1s complement?
-3710 = - 0010 01012
Invert each single bit
= 1101 1010
MSB indicates the sign but is not the
same as SM‘s sign bit!
Æ Magnitude cannot be read directly
Slide 8
4
Fundamentals of Computer Engineering 1
Exercise 7
1s complement: Range
MSB
Highest
2n-1-1
Smallest
0
111…1111
-(2n-1-1) 1
000…0000
1
Range of
4 bit 1s compl. :
8 bit 1s compl. :
etc
n-1
[-7, 7]
[-127,+127]
Slide 9
Fundamentals of Computer Engineering 1
Exercise 7
2s complement
What is –3710 as 8 bit 2s complement?
-3710 = - 0010 01012
Invert each single bit
= 1101 1010 + 1
= 1101 1011
MSB indicates the sign but is not the
same as SM‘s sign bit!
Slide 10
5
Fundamentals of Computer Engineering 1
Exercise 7
2s complement: Range
MSB
Highest
2n-1-1
0
111…1111
Smallest
-2n-1
1
000…0000
1
n-1
Range of
4 bit 2s compl. :
8 bit 2s compl. :
etc
[-8, 7]
[-128,+127]
Slide 11
Fundamentals of Computer Engineering 1
Exercise 7
Signed Number = Negative Number?
A signed number CAN be negative,
but it can also bear the sign „+“ !
-1
-2
-3
+0
0000
+1
0001
0010
1101
-4
1100
-5
1011
-6
1111
1110
1010
-7
+
0011
+3
0 100 = + 4
0100
+4
1 100 = - 4
0101
1001
+2
+5
Both are 2s
Complement
numbers !
-
0110
1000
-8
0111
+6
+7
Slide 12
6
Fundamentals of Computer Engineering 1
Exercise 7
Interpret Signed Binary Numbers
Which decimals could N= 101 10002 correspond to ?
7 bit SM:
101 1000 = - 24
7 bit 1s c.:
101 1000 = - 010 0111 = - 39
7 bit 2s c.:
101 1000 = - 010 1000 = - 40
All representations
2537
bit 2s compl.: with more then 7 bit
101 1000
1000 =
= 0..0101
0..0101 1000
1000 =
= 88
88
101
2530 zeros
Slide 13
Fundamentals of Computer Engineering 1
Exercise 7
Choosing a proper system (1)
Which representation system is sensible
for a voltage measurement device that
has a range from - 2 V to 2 V with a
sensitivity of 1 mV ?
Slide 14
7
Fundamentals of Computer Engineering 1
Exercise 7
Choosing a proper system (2)
Numbers needed:
-2000 mV to 2000 mV
Æ2n-1-1 >= 2000
Æn-1
>= ld 2001
Æn-1
>= 10.9966
Æn
= 12
Range: [-2047,+2047] (SM, 1s Compl.)
Slide 15
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Format (1)
Problem:
General Format needed that can store
0.00000234 - very small!!
as well as
129,802,323,443,123 - very big!!
Slide 16
8
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Format (2)
Solution:
Use Exponential Form:
0.00000234
= 2.34 • 10-6
129,802,323,443,123
= 1.29802323443123 • 1014
Slide 17
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Format (3)
Similar in binary:
0.001001101
= 1.001101 • 2-3
1101 1100 001 = 1.101 1100 0011 • 211
Slide 18
9
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Format (4)
In general :
Every number can be written as
± M • BE
M Æ Mantissa
B Æ Base
E Æ Exponent
Floating Point Format!!
Information to be stored:
how to store the sign, M, B and E ?
ÎMany choices
Slide 19
Fundamentals of Computer Engineering 1
Exercise 7
IEEE Floating Point Numbers (1)
NIEEE floating point = ± M • 2E
Sign
Exponent
0
0111 1100
1 bit
8 bit
Mantissa
0100
0000
0000
0000
0000
000
23 bit
IEEE Single Precision Floating Point Format (32 bit)
Slide 20
10
Fundamentals of Computer Engineering 1
Exercise 7
IEEE Floating Point Numbers (2)
- the sign Sign bit:
0
Æ positive
1
Æ negative
Slide 21
Fundamentals of Computer Engineering 1
Exercise 7
IEEE Floating Point Numbers (4)
- the mantissa Trick to save one bit:
Mantissa is always M = 1.m 1<=M<102 for definition
... float the point ...
ÆJust store the fraction m
Hidden Bit
ÆStore 24 bit in 23 “slots“
Slide 22
11
Fundamentals of Computer Engineering 1
Exercise 7
IEEE Floating Point Numbers (3)
- the exponent Negative Exponents ?
Æ Offset of 127 (Excess-127 representation)
Example:
Exponent is E = - 4
Store Characteristic
C = E + 127 = 123
E = C - 127
8 bit Characteristic Æ RangeC = [0,255]
Æ
RangeE = RangeC - 127 = [-127,128]
Slide 23
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Example (1)
Example : Conversion of IEEE single precision floating point number to decimal number.
Sign
0
Characteristic
Mantissa
00110 11110
0010 0000
0000 0000 0000 0000
11 1100
0100 0000 0000 0000 0000 000
Sign : 0 Æ Positive
Characteristic : 0 1 1 1 1 1 0 02
= 124 = - 3 + 127 Æ Exponent = - 3
Mantissa ( hidden bit ) : 1 . 0 1 0 0
0000
0000
0000
0000
000
-2
-1
-3
Number :001...1
0
00
1
01
0
1
1
0 0
0
1
010
00 0
010
000
000
0
000
000
000
00 0
000
000
000
00 0
000
000
000
00 0
0000
0000
0000
000 xx 2
2-0
Decimal number : 0 . 0 0 1 0 12 = 0.1562510
Slide 24
12
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Example (2)
Example: Conversion of decimal number to IEEE single precision floating point number.
- 9.8281210 Æ IEEE 32-bit floating point
9.8281210 = 1 0 0 1 . 1 1 0 1
0 011
1111
1111
1 0 1 02
1233
.011111 01
01011 1
000011 0
x0
= 10
.. .0000
01 1
00 0
11 00 11001111111111 1111111111111111 11111111
x x2
x222
Characteristic : 127 + 3 = 13010 = 1 0 0 0 0 0 1 02
Sign
Characteristic
1
1000 0010
Sign bit : 1 ( negative )
Mantissa
0011
1010
0111 1111 1111
010
Slide 25
Fundamentals of Computer Engineering 1
Exercise 7
Floating Point Example
Sign
Characteristic
1
1000 0010
Mantissa
0011
1010
0111 1111 1111
010
1100
0001
0001
1101
0011
1111
1111
1 010
C
1
1
D
3
F
F
A
-9.8281210 in IEEE 32 Bit Single Precision Floating Point,
visualized Hexadecimal
-9.8281210 Æ C1 1D 3F FA
Slide 26
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