Transcript represents 74

Computer Studies Today
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Chapter
1
» Numbers within computers can be
expressed by
– Sign-and-magnitude
representation
– Two’s complement representation
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Chapter
2
Sign-and-magnitude Representation
» The most significant bit represents
the sign of the number:
1 for negative and 0 for positive
» The remaining bits represent the
magnitude of the number.
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3
Bits for magnitude
Sign
bit
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Chapter
4
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Chapter
5
Binary Fraction in Sign-andmagnitude Representation
Binary fraction
0 0 0 1 1  1 1 0 = 11.112
1
0
1
2
=2 +2 +2 +2
Sign bit
= 2 + 1 + 0.5 + 0.25
= 3.7510
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Binary fraction
1 1 0 1 1 1  1 1 = 10111.112
= (2 + 2 + 2
0
1
2
+2 +2 +2 )
= (16 + 4 + 2 + 1
+ 0.5 + 0.25)
= 23.7510
4
Sign bit
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1
7
Two’s Complement Representation
» The most significant bit represents
a negative quantity.
» The magnitude of the number is the
sum of all digit values of the bits.
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8
Most significant bit
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Chapter
9
» With the two’s complement
representation, it is easy to change
a positive number to the
corresponding negative number
and vice versa.
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Chapter
10
» The steps are as follows:
– Change all 0s to 1s.
– Add 1 to the rightmost digit.
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Chapter
11
» Change 2510 to 2510.
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Chapter
12
» Change 2310 to 2310.
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Chapter
13
Binary Fraction in Two’s
Complement Representation
Binary fraction
0 0 1 0 1  1 1 0 = 101.1102
2
0
1
2
=2 +2 +2 +2
Most significant bit
= 5.7510
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Binary fraction
1 0 0 1  0 0 1 1 = 1001.00112
= 2 + 2 + 2
4
+2
= 6.812510
3
Most significant bit
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0
3
15
Addition and Subtraction in Two’s
Complement Representation
–)
61
74
61
+) –74
–13
–13
00111101
+) 10110110
11110011 = 13
(valid)
» Subtraction can be done by
complementation and addition.
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16
+)
61
74
00111101
+) 01001010
10000111 = 121 (invalid)
135
» The result is invalid because of an
overflow error.
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17
–61
+) –74
–135
11000011
+) 10110110
1 01111001 = 121 (invalid)
The extra bit is discarded
because this is an 8-bit code.
» The result is invalid because of an
overflow error.
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18
–61
–) –74
–61
+) 74
13
13
11000011
+) 01001010
1 00001101 = 13
(valid)
The extra bit is discarded
because this is an 8-bit code.
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19
Comparisons between Sign-andmagnitude and Two’s Complement
Representation
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Chapter
20
» A number is composed of a
fraction part called a mantissa and
a power called an exponent.
» It represents numbers in a wider
range.
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Sign of
mantissa
Sign of
exponent
Mantissa
15  14 13 12 11 10 9
8
7
6
5
Exponent
4
3
2
1
0
» Bit 0 to bit 7 represent the exponent
and bit 8 to bit 15 represent the
mantissa.
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Normalisation
» A standardised floating point
representation.
» The first bit of the mantissa must
be non-zero.
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23
» For example,
7.2510 = 111.012  2
0
= 0.111012  2
Mantissa
3
Exponent
0 1 1 1 0 1 0 0 0 0 0 0 0 0 1 1
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Chapter
24
Express 1001.0112 in floating-point
representation.
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Solution
1001.0112 = 0.10010112  2
4
 The floating-point representation is:
0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 0
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26
Express 0.0000011112 in floatingpoint representation.
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Solution
0.0000011112 = 0.11112  2
5
 The floating-point representation is:
0 1 1 1 1 0 0 0 1 0 0 0 0 1 0 1
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Convert 1110000100000110, which is
in floating-point representation, into a
decimal value.
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Solution
1110000100000110
= 0.11000012  2
= 0110000.12
6
= 48.510
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Chapter
30
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Chapter
31
» It means dropping off any extra
digits which cannot be
accommodate.
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Truncate 123.88 to 4 significant
figures.
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Solution
123.88
123.8
This causes a truncation error.
The error is 123.88  123.8 = 0.08
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Chapter
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Truncate 0.10101012 to 6 bits.
Solution
0.10101012
0.1010102
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Chapter
35
» It means shortening a number by
deleting one or more digits in the
least significant positions.
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Chapter
36
» Increasing by one the remaining
far right digit if the truncated
digit is greater than or equal to
half of the number base.
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Chapter
37
Round 123.88 to 4 significant figures.
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Chapter
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Solution
123.88
123.9
This causes a rounding error.
The error is 123.9  123.88 = 0.02
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Chapter
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Round 0.10101012 to 6 bits.
Solution
0.10101012
0.1010112
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Chapter
40
» Since the fixed-point and floatingpoint representations of a number
have their own ranges, this causes
– An overflow error if the number
exceeds the range that can be
represented.
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– An underflow error if the
number is too small to be
represented.
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