Digital Electronics - Number Systems and Codes - Discussion

As ASCII code is 7 bit.
So,
The first 7 bit is =1010011 which is equivalent to 83 equivalent ASCII= S.
2nd 7 bit is= 1010100 which is equivalent 84 equivalent ASCII= T.
3rd 7 bit is =1010101 ------------------- 85 equivalent ASCII= U.
4th 7 bit is= 1000100 ------------------- 68 equivalent ASCII= D.
5th 7 bit is = 1011001 ------------------- 89 equivalent ASCII= Y.
6th 7 bit is = 0100000 ------------------ 32 equivalent ASCII= space.
7th 7 bit is = 1001000 ------------------ 72 equivalent ASCII= H.
8th 7 bit is = 1000001 ------------------ 65 equivalent ASCII= A.
8th 7 bit is = 1010010 ------------------ 82 equivalent ASCII= R.
9th 7 bit is= 1000100 ------------------ 68 equivalent ASCII= D.

@All.

Correct explanation is available here:

First divide the code into 7 bits.
(1010011) (1010100) (1010101) (1000100) (1011011)
(0100000) (1001000) (1000001) (1010010) (1000100)

Then convert Binary data into Decimal form
1010011 = 1 * 2^6 + 0 * 2^5 + 1 * 2^4 + 0 * 2^3 + 0 * 2^2 + 1 * 2^1 + 1 * 2^0
= 64 + 0 + 16 + 0 + 0 + 2 + 1.
= 83.

Like this do all.........
we get decimal values.....
83, 84, 85, 68, 89, 32, 72, 65, 82, 68.

Characters of ASCII are.
32 is the space.
65-90 are (A-Z).
So, 65 = A.
66 = B.
67= C.Then 90=Z.
Now choose the correct form of letters then we get;
[STUDY HARD]
This is the correct way to understand easily.
Thank you.

Correct Explanation available here:

Step 1: Divide the code into 7-bits each

(1010011) (1010100) (1010101) (1000100) (1011001)
(0100000) (1001000) (1000001) (1010010) (1000100)

Step 2: Convert binary data to decimal individually

Step 3: (83) (84) (85) (68) (89) (32) (72) (65) (82) (68)

Step 4: Substitute characters of ASCII values

Example:-
65-A
66-B
67-C
....
....
90-Z
32-blank, etc.

i.e code converted into text as STUDY HARD

Note:- If given data is not consisting 7 multiples of bits then append 0's to MSB i.e left hand side of data.

Also you should remember that characters case i.e upper or lower.

Most of the people asking why to take 7bits particularly. The explanation is simple.

A-Z = 65-90
a-z = 97 - 122

2^0 = 1
2^1 = 2
2^2 s= 4
.
.
.
2^6 = 64
2^7 = 128
2^8 = 256

Here all the alphabets including upper and lower decimal values will not exceed 128 which is 2^7, which means the MSB is always 0 which is ignored here.

010100111010100101010110001001011001
01000001001000100000110100101000100
Add 0 to LHS to make it a multiplier of 4
now... 0101 0011=53..that is the hexadecimal of 'S'
Likewise 54 of 'T'
55 of 'U'
.
.
.
20 of space
Hence solved..

For ASCII

A = 65, B = 66, ............................, Z = 90.

a = 97, b = 98, ......................................., z = 122.

91 = [
92 = \
93 = ]
94 = ^
95 = _
96 = ` (above Tab key known as Back quote)

ASCII means 7-bit code so we have to split the given code from right to left as 7-bit like wise and after that assign the ASCII symbol based on the value eg:ASCII for A is 65.

We need take only 7bits because 2^7=128 which maximum ASCII value instead of that if we consider 8bits then 2^8 corresponds to 256 which is not the range of ASCII value.

ASCII code is a Alpha Numeric code, which is 7 bit code. So,divide each 7 bits apart and then follow the ASCII no logic.

A=65
B=67
.
.
.
so on
STUDY HARD will come.

Another way to do. First, we can observer question it consists of a 70. Each ASCII bit is 7. Total 70/7=10 characters. Only b option contains 10 characters.