Write the excess-3 equivalent codes of (6)10, (78)10 and (357)10, all in 16-bit format.0011001100111001, 0011001110101011, 0011011010001010Determine the Gray code equivalent of (10011)2 and the binary equivalent of the Gray code number110011.A 16-bit data word given by 1001100001110110 is to be transmitted by using a fourfold repetitioncode. If the data word is broken into four blocks of four bits each, then write the transmitted bitstream.1001100110011001100010001000100001110111011101110110011001100110
The Correct Answer and Explanation is :
Let’s go step by step to clarify each of your queries.
Excess-3 Code (XS-3):
Excess-3 (XS-3) is a binary-coded decimal (BCD) code where each decimal digit is encoded by adding 3 (0011) to the binary representation of the digit. You provided decimal values (6), (78), and (357) and asked for the 16-bit Excess-3 equivalent.
- (6)₁₀ in Excess-3 (16-bit format):
- First, we encode 6 using Excess-3 by adding 3 to it: (6 + 3 = 9), which in binary is
1001. - So, the Excess-3 code for 6 (in a 16-bit format) is
0000000000001001.
- (78)₁₀ in Excess-3 (16-bit format):
- The first digit, 7, in Excess-3: (7 + 3 = 10) (binary
1010). - The second digit, 8, in Excess-3: (8 + 3 = 11) (binary
1011). - So, the Excess-3 code for 78 (in a 16-bit format) is
0000101000101011.
- (357)₁₀ in Excess-3 (16-bit format):
- The first digit, 3, in Excess-3: (3 + 3 = 6) (binary
0110). - The second digit, 5, in Excess-3: (5 + 3 = 8) (binary
1000). - The third digit, 7, in Excess-3: (7 + 3 = 10) (binary
1010). - So, the Excess-3 code for 357 (in a 16-bit format) is
0000011000101000101010.
Gray Code:
Gray code is a binary numeral system where two successive values differ in only one bit. This encoding is used in various error correction applications because only one bit changes at a time.
- Gray code of (10011)₂:
- The first bit (most significant) is the same in Gray code as in the original binary number. So the first bit of the Gray code is 1.
- The subsequent bits of the Gray code are generated by XORing each pair of consecutive bits of the binary number:
- Gray code bit 2: (1 \oplus 0 = 1)
- Gray code bit 3: (0 \oplus 0 = 0)
- Gray code bit 4: (0 \oplus 1 = 1)
- Gray code bit 5: (1 \oplus 1 = 0)
- Therefore, the Gray code equivalent of (10011)₂ is 11010.
- Binary equivalent of the Gray code 110011:
- The first bit of the Gray code (most significant bit) is the same as the binary number: (1).
- To decode, XOR each subsequent Gray code bit with the previous binary bit:
- The first binary bit is 1.
- The second binary bit: (1 \oplus 1 = 0)
- The third binary bit: (0 \oplus 0 = 0)
- The fourth binary bit: (0 \oplus 0 = 0)
- The fifth binary bit: (0 \oplus 1 = 1)
- The sixth binary bit: (1 \oplus 1 = 0)
- So, the binary equivalent of Gray code 110011 is 100100.
Fourfold Repetition Code:
In a fourfold repetition code, each bit in the data is repeated four times to increase reliability. The given 16-bit data word is:
(1001100001110110)
To apply fourfold repetition, each 4-bit block is repeated four times. We break the 16-bit word into four 4-bit blocks:
- (1001 \rightarrow 1001 1001 1001 1001)
- (1000 \rightarrow 1000 1000 1000 1000)
- (1110 \rightarrow 1110 1110 1110 1110)
- (0110 \rightarrow 0110 0110 0110 0110)
Therefore, the transmitted bitstream is:
1001100110011001100010001000100001110111011101110110011001100110
Summary:
- Excess-3 Code: We converted each decimal number into its corresponding Excess-3 code and formatted it in 16 bits.
- Gray Code: We found the Gray code of a binary number and converted a Gray code back to binary.
- Fourfold Repetition: We applied a fourfold repetition code to each 4-bit block of the data word, resulting in a longer but more redundant transmission for error detection.
This explains the conversions and the steps in detail! Let me know if you have further questions.