In this post we shall discuss about various
number systems, conversion of numbers from one base to the other, complements
and binary codes. In addition to this, the students will also be familiarized
with subtracting two numbers using complements.
Activity Outcomes:
The students will be able to represent
any number in any base. They will also learn how to convert a number from one
base to any other base. The students will be acquainted with the subtraction of
numbers using complements.
Instructor Note:
Read
the lab notes carefully and thoroughly. All the examples have been solved in an
easy to understand manner. You should not find it difficult to learn the topics
given in these lab notes. Try to solve all the exercises yourself that are
given at the end of notes. These exercises will certainly help you in achieving
the objectives of this lab.
Number System
A number
system is a set of numbers
together with one or more operations
(e.g. add, subtract). Commonly used number systems are decimal, binary, octal
and hexadecimal number systems. All these number systems are also called
weighted number systems.
Decimal
Number System or Base (or Radix) 10 System
Before digital
computers, the only known number system was the decimal number system.
It has a total of
ten digits: {0,1,2,….,9}-
A decimal number
such as 7392 represents a quantity equal to 7 thousands plus 3 hundreds plus 9
tens plus 2 units. The thousands, hundreds etc are powers of 10 (in decimal
number system representing the position of the coefficients. To be more exact,
7392 should be written as 103 x 7 + 102 x 3 + 101
x 9 +100 x 2
In general if a
decimal number is represented as ……a4 a3 a2 a1
a0 . a-1 a-2 a-3 a-4………..
So this decimal
number can be represented by multiplying coefficients of integral part by
increasing powers of 10 (from right to left) and multiplying coefficients of
fractional part by decreasing powers of 10 (from left to right) and then adding
all these terms as
……..+103
x a3 + 102x a2+101 x a1 +100
x a0 +10-1 x a-1 +10-2x a-2
+ 10-3 x a-3 + 10-4 x a-4 +……..
Example
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9375
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Binary Number System (base-2)
Binary number system is a different number system. This
number system has only two possible values i.e. 0 and 1. The decimal equivalent
of a binary number can be found by multiplying coefficients of integral part by increasing
powers of 2 (from right to left) and multiplying coefficients of fractional
part by decreasing powers of 2 (from left to right) and then adding all these
terms as
……..+23 x a3 + 22x
a2+21 x a1 +20 x a0 +2-1
x a-1 +2-2x a-2 + 2-3 x a-3
+ 2-4 x a-4 +…
where ……a4
a3 a2 a1 a0 . a-1 a-2
a-3 a-4……….. are coefficients of binary number.
Example
Example
Write decimal
equivalent of binary number (101011.1011)2
Octal Number System (or Base-8 System)
This number system has only eight possible values i.e. 0,
1, 2, 3, 4 ,5 ,6 and 7. The decimal equivalent of an octal number can be found
by multiplying
coefficients of integral part by increasing powers of 8 (from right to left)
and multiplying coefficients of fractional part by decreasing powers of 8 (from
left to right) and then adding all these terms as
……..+83 x a3 + 82x
a2+81 x a1 +80 x a0 +8-1
x a-1 +8-2x a-2 + 8-3 x a-3
+ 8-4 x a-4 +…
where ……a4
a3 a2 a1 a0 . a-1 a-2
a-3 a-4……….. are coefficients of octal number.
Example: Convert the octal number (375)8
into its equivalent decimal number.
Example
Write decimal
equivalent of octal number (630.45)8
Hexadecimal Number System (or Base-16
System)
This number system has sixteen possible values i.e. 0, 1,
2, 3, 4 ,5 ,6, 7, 8, 9, A, B, C, D, E and F. The decimal equivalent of a
hexadecimal number can be found by multiplying coefficients of integral part by increasing
powers of 16 (from right to left) and multiplying coefficients of fractional
part by decreasing powers of 16 (from left to right) and then adding all these
terms as
……..+163 x a3 + 162x
a2+161 x a1 +160 x a0 +16-1
x a-1 +16-2x a-2 + 16-3 x a-3
+ 16-4 x a-4 +…
where ……a4
a3 a2 a1 a0 . a-1 a-2
a-3 a-4……….. are coefficients of hexadecimal number.
Example
Example
Write decimal
equivalent of hexadecimal number (2C5.FB)16
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Binary Addition (cont.)
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Q: How to verify?
A: Convert to decimal
45
+ 39
-----------
84
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1 0
1 1 0
1
1
0 0 1
1 1
1
0 1 0
1 0 0
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Carries
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Augund
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Addend
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Sum
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1
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1
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1
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1
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1
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ubtractionBinary
S
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Minuend
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Subtrahend
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Converting Octal to Binary
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Since 8 can be written as 23,
therefore each octal digit corresponds to three binary digits. The
conversion from octal to binary is easily accomplished by writing each
octal digit into a group of three bits.
Example:
(673.124)8 = (110 111
011 . 001 010 100)2
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6
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7
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3
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1
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2
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4
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First convert the octal
number to binary number using the same procedure as described above. Then
make groups of 4 bits starting from right to left for integral part and
from left to right for fractional part. In the process, complete the
unformed group by placing extra zeros (on the left of integral part’s
incomplete group and on the right of fractional part’s incomplete group).
Finally write the hexadecimal equivalent of each group of four bits.
Example:
(673.124)8 =
(110 111 011 . 001 010
100)2
(First converting Octal number to Binary)
= (110 111 011 . 001
010 100)2
= (0001 1011
1011 . 0010 1010 0000)2 (Making groups of four bits)
= (1 B B . 2 A )16 (Writing
hexadecimal equivalent of every group)
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6
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7
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3
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1
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2
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4
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B
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B
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0
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2
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A
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1
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Since 16 can be written as 24,
therefore each hexadecimal digit corresponds to four binary digits. The
conversion from hexadecimal to binary is easily accomplished by writing
each hexadecimal digit into a group of four bits.
Example:
(30DF.A14)16 = (0011 0000
1101 1111 . 1010 0001
0100)2
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3
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0
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D
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F
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A
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1
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4
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First convert the hexadecimal
number to binary number using the same procedure as described above. Then
make groups of 3 bits starting from right to left for integral part and
from left to right for fractional part. In the process, complete the
unformed group by placing extra zeros (on the left of integral part’s
incomplete group and on the right of fractional part’s incomplete group).
Finally write the octal equivalent of each group of three bits.
Example:
(1
B B . 2 A 0 )16 =
(0001 1011 1011 . 0010 1010
0000)2 (First
converting hexadecimal number to Binary)
= (000110111011 . 00101010 0000)2
= ( 110 111 011 . 001 010 100)2 (Making groups of three
bits)
=(6 7 3 . 1 2 4) 8 (Writing octal
equivalent of every three bit group)
= (673.124)8 (Writing hexadecimal
equivalent of every group)
(110 111
011 . 001 010 100)2
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1
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A
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0
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6
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7
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3
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1
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2
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4
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B
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B
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2
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9
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9
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9
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10
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9
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9
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9
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10
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Complements are used in digital computers for simplifying the
subtraction operation and for logical manipulation (i.e. finding
complements of Boolean functions).
For every base ‘r’ systems, we have two complements:
(i)
r’s
Complement
(ii)
(r – 1)’s
Complement
Hence for base 10, we have 10’s complement and 9’s complement.
Similarly for base 2 system, we have 2’s complement and 1’s complement.
r’s Complement:
If we have a positive number N in any base ‘r’ with an integer part
of ‘n’ digits, then
rn – N if N ≠ 0
r’s Complement of N =
0 if N = 0
Example:
Find 10’s complement of
(i)
(52520)10 (ii)
(0.3267)10
(iii) (25.639)10
Solution:
(i)
Subtract the
first non-zero digit (from right) from 10 and subtract all remaining digits
from 9. So, we get
– 5 2 5 2 0
4 7 4 8 0
(ii)
–
0 . 3 2 6 7
0 .
6 7 3 3
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Complements are used in digital computers for simplifying the
subtraction operation and for logical manipulation (i.e. finding
complements of Boolean functions).
For every base ‘r’ systems, we have two complements:
(iii)
r’s
Complement
(iv)
(r – 1)’s
Complement
Hence for base 10, we have 10’s complement and 9’s complement.
Similarly for base 2 system, we have 2’s complement and 1’s complement.
r’s Complement:
If we have a positive number N in any base ‘r’ with an integer part
of ‘n’ digits, then
rn
– N if N ≠ 0
r’s Complement of N =
0 if N = 0
Example:
Find 10’s complement of
(ii)
(52520)10 (ii)
(0.3267)10
(iii) (25.639)10
Solution:
(ii)
Subtract the
first non-zero digit (from right) from 10 and subtract all remaining digits
from 9. So, we get
– 5 2 5 2 0
4 7 4 8 0
(ii)
–
0 . 3 2 6 7
0
. 6 7 3 3
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9
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9
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10
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9
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Note:
In the above formula
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(iii)
– 2 5 . 6 3 9
7 4 . 3 6 1
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10
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9
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9
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9
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9
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9
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9
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9
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10
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:Example
Find 2’s complement of
(i)
(10110)2 (ii)
(0.0110)2
(iii) (1011.101)2
Solution:
(i)
Subtract the
first non-zero bit (from right) from 2 and subtract all remaining bits from
1. So, we get
– 1 0 1 1 0 0
0 1 0 1
0 0
(ii)
–
0 . 0 1 1 0
0 .
1 0
1 0
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2
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1
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1
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1
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2
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1
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1
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(iii)
– 1 0 1 1 . 1 0 1
0 1 0 0 . 0 1 1
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2
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1
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1
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1
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1
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1
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1
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1
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Note:
In the above formula
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(r-1)’s Complement:
If we have a positive number N in any base ‘r’ with an integer part
of ‘n’ digits and a fraction part of ‘m’ digits, then
rn – N – r-m if N ≠ 0
(r-1)’s Complement of N =
0 if N = 0
Example:
Find 9’s complement of
(i)
(52520)10 (ii)
(0.3267)10
(iii) (25.639)10
Solution:
(i)
Subtract each
and every significant digit from 9. So, we get
– 5 2 5 2 0
4 7 4 7 9
(ii)
–
0 . 3 2 6 7
0 .
6 7 3 2
(iii)
– 2 5 . 6 3 9
7 4 . 3 6 0
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9
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9
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9
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9
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9
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9
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9
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9
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9
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9
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9
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9
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9
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9
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Note:
In the above formula
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:Example
Find 1’s complement of
(i)
(10110)2 (ii)
(0.0110)2
(iii) (1011.101)2
Solution:
(i)
Subtract each
and every significant bit from 1. So, we get
– 1 0 1 1 0 0
0 1 0 0 1 1
(ii)
–
0 . 0 1 1 0
0 .
1 0
0 1
(iii)
– 1 0 1 1 . 1 0 1
0 1 0 0 . 0 1 0
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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Note:
In the above formula
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Subtraction using r’s Complement:
The
subtraction of two positive numbers M and N (i.e. M – N), both of base ‘r’
may be done as follows:
1. Add
the minuend M to the r’s complement of the subtrahend N
i.e. M – N = M + (r’s complement of N)
2. Inspect
the result obtained in step (1) for an end carry.
(i)
If an end carry
occurs, discard it.
(ii)
If an end carry
does not occur, again take the r’s complement of the result obtained in
step (1) and place a negative sign
in front of the answer.
Note:
Make
number of digits same for both the numbers by adding additional zeros to
the left of integral part and to the right of fractional part.
Example:
Use 2’s complement to perform M
– N with the given binary numbers:
(i)
M = 1010100 N = 1000100
Solution:
M
– N = M + (2’s Complement of N)
= 1010100 + (2’s Complement of
1000100)
= 1010100 + 0111100
= 1
0 1 0
1 0 0
+ 0
1 1 1
1 0 0
1 0 0
1 0 0
0 0
Since
end carry exists, therefore we discard it. Hence the answer is 0010000 or
10000.
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End Carry
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Verification:
84
-
68
16
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(ii)
M = 1000100 N = 1010100
Solution:
M
– N = M + (2’s Complement of N)
= 1000100 + (2’s Complement of 1010100)
= 1000100 + 0101100
= 1
0 0 0
1 0 0
+ 0
1 0 1
1 0 0
1 1
1 0 0
0 0
Since
no end carry exists, so final answer = - (2’s Complement of 1 1
1 0 0
0 0) = - 0 0 1 0 0 0 0
= - 1 0 0 0 0
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Verification:
68
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84
- 16
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Subtraction using (r-1)’s Complement:
The
subtraction of two positive numbers M and N (i.e. M – N), both of base ‘r’
may be done as follows:
1. Add
the minuend M to the (r-1)’s complement of the subtrahend N
i.e. M – N = M + ((r-1)’s complement of N)
2. Inspect
the result obtained in step (1) for an end carry.
(iii)
If an end carry
occurs, add 1 to the least significant digit in order to get the final
answer.
(iv)
If an end carry
does not occur, again take the (r-1)’s complement of the result obtained in
step (1) and place a negative sign
in front of the answer.
Note:
Make
number of digits same for both the numbers by adding additional zeros to
the left of integral part and to the right of fractional part.
Example:
Use 1’s complement to perform M
– N with the given binary numbers:
(i)
M = 1010100 N = 1000100
Solution:
M
– N = M + (1’s Complement of N)
= 1010100 + (1’s Complement of
1000100)
= 1010100 + 0111011
= 1
0 1 0
1 0 0
+ 0
1 1 1 0 1 1
1 0 0 0 1 1 1 1
1 +
0 0 0
0 1 0
0
Since
end carry exists, therefore we have added it to the above result in order
to get the final answer which is 0010000 or 10000.
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(ii)
M = 1000100 N = 1010100
Solution:
M
– N = M + (1’s Complement of N)
= 1000100 + (1’s Complement of 1010100)
= 1000100 + 0101011
= 1
0 0 0
1 0 0
+ 0
1 0 1
0 1 1
1 1
0 1 1
1 1
Since
no end carry exists, so final answer = - (1’s Complement of 1 1
0 1 1
1 1) = - 0 0 1 0 0 0 0
= - 1 0 0 0 0
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Comparison between 1’s Complement and 2’s
Complement:
The
1’s complement has advantage over 2’s complement because digital components
has to change only 0’s to 1’s and 1’s to 0’s in case of 1’s complement. But
in case of 2’s complement, we have to take 1’s complement and then have to
add 1 to get the 2’s complement or we have to leave all the least
significant 0’s and the first 1 unchanged and then change all 0’s into 1’s
and 1’s into 0’s.
The
2’s complement has advantage over 1’s complement while subtracting two
binary numbers using complements. In 2’s complement, we have to perform
addition for just one time, whereas in 1’s complement, we have to perform
addition twice, when an end carry exists.
The
digital components prefer using 2’s complement because when we subtract two
same numbers using 1’s complement and 2’s complement separately, we have to
remember two representations of 0 in case of 1’s complement; one with
positive zero and the other with negative zero.
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Binary Codes:
Electronic
digital systems (computers) use signals that have two distinct values and
circuit elements that have two stable states. A binary number of ‘n’ digits
may be represented by ‘n’ binary circuit elements, each having an output
signal equivalent to a 0 or 1. A bit is a binary digit To represent a group
of 2n distinct elements in a binary code requires a minimum of
‘n’ bits. For example, a group of four (22) distinct quantities
can be represented by a 2 bit code, with each quantity assigned one of the
bit combinations: 00, 01, 10, 11.
Similarly a group of 8 (23) elements requires a 3-bit code.
Decimal Codes:
Binary
codes for decimal digits require a minimum of four bits (because to
represent digits from 0 to 9 we need at least four bit code). Numerous
different codes can be obtained by arranging four or more bits in the
distinct possible combinations. A few possibilities are shown in the
following table:
In
the above table, Excess-3 and 2 of 5 are the only un-weighted codes. Excess-3
was used in some old computers. Its code assignment is obtained from the
corresponding value of BCD after adding 3 in it.
The
Excess-3 , the 2 4 2 1 and the 8 4 -2 -1 are self-complementary codes, i.e.
the 9’s complement of the decimal number is easily obtained by changing 1’s
into 0’s and vice versa. For example, the decimal 395 is represented in 2 4
2 1 code by 0011 1111 1011. Its 9’s complement 604 is represented by 1100
0000 0100 which is easily obtained by replacing 0’s into 1’s and 1’s into
0’s of the original number.
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Q. No. 4. Convert following Binary Numbers into Decimal
—
(11010)2 = ( )10
—
(110101.11)2 =
( )10
Q. No. 5. Convert the following Octal Numbers
into the base indicated
—
127.4)8= (
)10
—
(673.12)8= ( )2
—
(407.06)8= ( )16
—
—
—
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Q. No. 6. Convert the following Hexadecimal
Numbers into the base indicated
(B65F)16 = (
)10
(3A6.C)16
= (
)2
(40A.EC)16 = ( )8
Q. No. 7. Convert the following Decimal numbers
into the base indicated:
—
(153.513)10 = (
)8
—
(41.6875)10 =
( )2
(106.16)10 = ( )16
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Q. No. 8. Express
the following numbers in decimal:
a. (10110.0101)2
b. (26.24)8
c. (1010.1010)2
Q. No. 9.Obtain the 1's and 2's
complements of the following binary numbers:
(100000)2
(11011010)2
(1000101)2
Q. No. 10. Perform the addition (59F)16
+ (E46)16:
Q.
No. 11.Convert the following numbers from the given base to the other three
bases listed in the table:
Hexadecimal
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Octal
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Binary
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Decimal
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110011001.101
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472.3
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ADF9.C
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5324.25
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Q.
No. 12 Using 1’s and 2’s complement, perform M – N with the following
binary numbers. Also check the result by performing subtraction directly.
(ii)
M
= 11010 N = 1101
(iii)
M
= 11010 N = 10000
(iv)
M
= 10010 N = 10011
Q.
No. 13 For the weighted codes (a) 3 3 2 1 (b) 4 4 3 -2 for the decimal
digits, determine all possible tables so that the 9’s complement of each
decimal digit is obtained by changing 1’s to 0’s and 0’s to 1’s.
Q.
No. 14 Represent the decimal number 8620 (a) in BCD (b) in Excess-3
(c)
in 2 4 2 1 code
Q.
No. 15 Obtain a binary code to represent all base-6 digits so that the 5’s
complement is obtained by replacing 1’s by 0’s and 0’s by 1’s in the bits
of the code.
Hint: complements of each other, 2 and 3
are complement. In 5’s complement, 0 and 5 are complements of each other, 1 and 4 are
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