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 4 of 5.
Uday said:
8 years ago
Why not study hard? @Venkatesh.
Rama Krishna Sambari said:
8 years ago
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.
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.
Pavankumar said:
8 years ago
Good explanation. @Vinsian.
Manish said:
8 years ago
1010011 =83. How?
Dimpie said:
8 years ago
Well explained, Thank you @Vinsan.
Ankit said:
8 years ago
Thanks for the explanation @Vinsan.
Mohini said:
7 years ago
1010011 first convert to octal or hexadecimal eg. In hexadecimal, it is 53 then 53 into decimal 5*16^1 + 3*16^0=80+3=83 decimal number.
Ashok said:
7 years ago
@Vinsan.
Excellent explanation, Thanks.
Excellent explanation, Thanks.
Karthikeyan said:
7 years ago
64 32 16 8 4 2 1
1 0 1 0 0 1 1 = 83.
1 0 1 0 0 1 1 = 83.
Kahindi said:
6 years ago
Nice explaination, thanks @Vinsan.
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