For the above given signal each bit is separated by a vertical line. If total time is given as 48 ns, which answer is correct? a. Data: 110000111110 Modulation Rate: 0,5 Kbaud Encoding: Manchester O b. Data: 0 10001000011Modulation Rate: 0,04 Kbaud Encoding: Differantial Manchester O c. Data: 10 1000100001 Data Rate: 0,25 Gbps Encoding: Differantial Manchester O d. Data: 110000111110 Modulation Rate: 0,04 GBaud Encoding: Manchester O e. Data: 10 1000100001 Modulation Rate: 0,02 Gbps Encoding: Manchester
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- The data word 10101110001101 is given for transmission. Draw the following encoding formats. NRZ-L NRZ-I Bi-phase Manchester RZ Differential ManchesterFor the above given signal each bit is separated by a vertical line. If duration of a bit is given as 48 microsec, which answer is correct? a. Data: 1101100 10100 Modulation Rate: 0,02 Mbps Encoding: Manchester b. Data: 100 100011000 Modulation Rate: 0,04 MBaud Encoding: Manchester c. Data: 100 10001100 0 Modulation Rate: 0,5 Gbaud Encoding: Manchester d. Data: 110110 010100 Data Rate: 0,25 Mbps Encoding: Differantial Manchester e. Data: 101100101001 Modulation Rate: 0,04 Gbaud Encoding: Differantial ManchesterSignal 16 14 10,06 8,01 5,73 5,52 4,15 1 3 6. Sample Number The given signal is digitized by using PCM using 4 bits. Which one is the correct data? a. 010010100111100101011011 O b. 010101100100100010101011 ০ ८. 010010001001010010100111 O d. 011001000111010110111010 O e. 101101101010011110101011 DO D t 2o
- An 8-bit byte with binary value 10101111 is to be encoded using an even-parity Hamming code. What is the binary value after encoding? ANS: The encoded value is 101001001111 Why is this answer longer from binary value? Please solve this question with details. Please explain why you use that method.Manchester encoding guarantees frequent clock synchronization by changing signal values. What is the maximum number of bits which may be encoded without a signal change? (Note: The way this question has been asked. you are not to include the bit which includes a signal change in your count)The modulus for the field GF(25) is x5+x2+1 and its elements are all five-bit strings. Calculate each of the following and enter your answer in exactly five bits (Example, 01101). 1. 11111 + 00001 2. 00011 x 00011 3. 01100 x 01100
- IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent 0.2 assuming a version of this format.Consider the example below that shows both the payload and the two- dimensional parity bits (shown in green), and where exactly one of the payload or parity bits shown has been corrupted. In this example, odd parity is used. - 11000100 001111000 01010010010101101 00110111 100011001 11110110011011111 10111011 110011100 00010011 100110000 Please indicate the row and the column location of the bit that has flipped from its original value.What are the CRC bits (R) associated with the payload containing first four bit 1101 and last four bit is the last digit of your reg number (i.e. 5, payload = 1101 0101). Suppose that the 4-bit generator (G) is 0101, and r = 3
- Find the Bitwise AND of the following bit strings: 01 1011 010111 0011 1101 Question 11 options: Find the Bitwise OR of the following bit strings: (same strings as previous question) 01 1011 010111 0011 1101 Question 12 options: Find the Bitwise XOR of the following bit strings: (same strings as previous question) 01 1011 010111 0011 1101 Question 13 options:In a communication, the transmitted data is 1 00001001000 and the received data is 10001110100 0. For error detection, 2D Parity algorithm is used with (4 row and 3 column). Which answer is correct? O a. Error is only detected by column, but cannot be corrected O b. Error cannot be detected O c. Error is detected by row and column, but cannot be corrected O d. Error is only detected by row, but cannot be corrected O e. Error can be detected and corrected An analog signal having frequency as 5,30 KHz is digitized with 5 bits and stored in a file. If the file size is 1,43 Mbit, which answer is correct? O a. Date Rate = 95,4 Kbps Duration of the Signal = 22 minute O b. Date Rate = 53 Kbps Duration of the Signal = 27 minute O c. Date Rate = 42,4 Kbps Duration of the Signal = 34 minute O d. Date Rate = 74,3 Kbps Duration of the Signal = 19 minute O e. Date Rate = 56,6 Kbps Duration of the Signal = 27 minuteGiven the codeword=01110101111101010, find if it has an error bit or none. If it has an error bit, indicate the bit position. Data message has 12 bits.