Conversion of
binary to ….
Decimal:
The steps to convert the binary numbers to the decimal:
- · Use the bit of binary number one-by-one (starting with the most significant (leftmost) bit. Begin with the value 0
- · Then , repeatedly double the prior value
- · Add the next bit to produce the next value
- · The next value is taken down as the prior value and multiply the 2 and also add the next bit.
- Eg: 102
- (0×2)+1=1( 1 = this value is then taken down to multiply 2 and adding the second bit,0 )
- (1×2)+0=2
- The operation is repeated until the last bit
Eg:
Prior value
|
×2and +
|
Next bit
|
Next value
|
0
|
1
|
1
|
|
1
|
0
|
2
|
|
2
|
1
|
5
|
|
5
|
1
|
11
|
|
11
|
0
|
22
|
|
22
|
1
|
45
|
|
45
|
0
|
90
|
10110102 =
90 10
Step 2:
Binary 1 0
1 1 0 1
0
Decimal 1×26 0×25
1×24 1×23
0×22 1×21 0×20
= 90
Eg 2:
Step 1:
11111012
Prior value
|
×2 and+
|
Next bit
|
Next value
|
0
|
1
|
1
|
|
1
|
1
|
3
|
|
3
|
1
|
7
|
|
7
|
1
|
15
|
|
15
|
1
|
31
|
|
31
|
0
|
62
|
|
62
|
1
|
125
|
11111012 = 125 10
Step 2:
Binary 1 1 1 1 1 0 1
Decimal 1×26
1×25 1×24 1×23 1×22 0×21 1×20
=125
:The conversion of binary fraction to decimal fraction=
Eg :
0.110102
Weight
|
...20
|
2-1
|
2-2
|
2-3
|
2-4
|
2-5
|
|
Value represented
|
… 1
|
0.5
|
0.25
|
0.125
|
0.0625
|
0.03125
|
|
Binary
|
…0
|
1
|
1
|
0
|
1
|
0
|
Converting the binary
fraction to decimal fraction:
Binary 1 1 0 1 0
Decimal 1×2-1 1×2-2 0×2-3 1×2-4 0×2-5
=0.5 + 0.25 +
0.0625
=0.8125 10
:: therefore , 0.110102
= 0.8125 10
Eg:
0.112
Weight
|
…20
|
2-1
|
2-2
|
2-3
|
Value represented
|
…1
|
0.5
|
0.25
|
0.125
|
Binary
|
…0
|
1
|
1
|
0
|
Converting binary
fraction to decimal fraction=
Binary 1 1
Decimal 1×2-1 1×2-2 =
0.5 + 0.25 = 0.75 10
:: therefore, 0.112
= 0.7510
Hexadecimal:
The following below is the Number System Conversion between hexadecimal and binary:
Hexadecimal
|
Binary
|
0
|
|
0001
|
|
2
|
0010
|
3
|
0011
|
4
|
0100
|
5
|
0101
|
6
|
0110
|
7
|
0111
|
8
|
1000
|
9
|
1001
|
A
|
1010
|
B
|
1011
|
C
|
1100
|
D
|
1101
|
E
|
1110
|
F
|
1111
|
To convert binary to hexadecimal:
·
- Grouped the binary into 4
For
example : 1100 00112 = 1100 | 00112
- · If the binary not enough to group into 4, then insert an extra 0 bit on the left side of binary (called padding )
For
example : 11000112 = 0110 |
00112 (0 is the extra 0 bits inserted
)
- · After grouping the binary , according the Number System Conversion ,convert the binary into hexadecimal
Eg :
1100 00112
First step :Grouping the binary into 4
1100 00112 = 1100
| 00112
Second step : According to the Number System Conversion…
1100 00112 = C316
Eg 2:
1011002
First step : Grouping the binary into 4
Since 1011002
is not enough to divide into 4
each ,therefore insert two extra
0 bits on the
left side of binary.
1011002 = 0010 |
11002Second step :According to the Number System Conversion…
1011002 =3C16
Eg 3:
1011.112
First step : Grouping the binary
into 4
Since 0.112
is a decimal point therefore insert 2 0 bits on the right side of the
decimal
point…
1011 .112 = 1011. |11002
Second step : According to the Number System Conversion between
hexadecimal and
the
binary…
1011.112 =B. C16
Octal:
It is easy to convert the binary into octal numeral system.
The steps to convert the binary into octal numeral system are:
- · Grouping the binary into 3 each
For
example: 1100112 = 110
| 0112
- · Likes the hexadecimal , if there isn't enough to group the binary into 3,then insert an extra 0 bit on the left side of the binary . This process is called padding.
For
example: 100112 = 010 | 0112
( 0 is the extra 0 bit
inserted)
- · Then according to the table as below:
Number
System conversion between octal and binary:
Octal
|
Binary
|
0
|
000
|
1
|
001
|
2
|
010
|
3
|
011
|
4
|
100
|
5
|
101
|
6
|
110
|
7
|
111
|
Eg:
110
0112
1.
Grouping
the binary into 3:
110 0112 =
110 | 0112
2.
According
to the table above:
110
= 6 ; 011
= 3;
So, 110
0112 = 638
Apart of that , there is also a way to convert binary to octal :
Eg 2:
1.
Grouping
the binary into 3 :
110 0112 =
110 | 0112
2.
Convert
the binary by using 2n system:
110 = (1×22) + (1×21) + (0 ×20) | 011= (0×22) + (1×21) + (1×20)
=4+2+0 | = 0 + 2 + 1
=6 | =3
Therefore,
110 0112 = 638
Eg 3:
10 0112
1.
Grouping
the binary into 3… but this binary is not enough to group into 3 so we have to
insert an extra 0 bit at the left side of this binary…
10 0112 = 010 | 0112 ( 0 is the extra 0 bit inserted )
2.
Then,
convert the binary into octal either according to the Number System Conversion
between octal and binary or by using the 2n system…
Method 1: According to Number
System Conversion between octal and binary
010 = 2 | 011 =3
Therefore,
Method 2 :By using 2n
system |
010 = (0×22) + ( 1×21) + (0×20) | 011= ( 0×22) + (1×21) + (1×20)
=2
| =2+1
=3
10 0112 = 238
Eg 4:
Eg 4:
100011.112
1. Grouping the binary into
3…
100011.112 = 100
| 011. |1102
2.By using 2n
system
100=
(1×22) + (0×21) + (0×20) | 011=(0×22) +(1×21) +(1×20) .| 110= (1×22)+(1×21)+(0×20)
= 4 +0 +0 | =0 + 2 +1 | =4 + 2 +0
= 4 | = 3 | =6
100011.112 = 43
.6 8
-Cai Ning
-Cai Ning
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