Digital Electronics - Number Systems and Codes - Discussion
Discussion Forum : Number Systems and Codes - General Questions (Q.No. 6)
6.
Decode the following ASCII message.
10100111010100101010110001001011001
01000001001000100000110100101000100
Discussion:
47 comments Page 3 of 5.
VARUN said:
1 decade ago
1010011 = 83 S 1010100 = 84 T 1010101 = 85 U 1000100 = 68 D 1011001 = 89 Y.
0100000 = 32 1001000 = 72 H 1000001 = 65 A 1010010 = 82 R 1000100 = 68 D.
0100000 = 32 1001000 = 72 H 1000001 = 65 A 1010010 = 82 R 1000100 = 68 D.
Prem said:
10 years ago
I didn't understand this.
Bz Abhijeet said:
10 years ago
How could we know that?
A = 65 but why not a = 65.
Please explain.
A = 65 but why not a = 65.
Please explain.
Binu said:
10 years ago
What is the ASCII value of (91 - 98)?
Sukriti said:
9 years ago
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)
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)
Bhask said:
9 years ago
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.
A=65
B=67
.
.
.
so on
STUDY HARD will come.
Vipin said:
9 years ago
Your explanation is brilliant @Vinsan.
Venkatesh said:
9 years ago
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.
Y@seen said:
9 years ago
Good explanation @Vinsian.
Priya said:
8 years ago
We have to group 7 bits for each group. i.e., from least significant position to most significant position and then place them with equivalent ASCII code.
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