Based on hamming code (7 , 4) . Given the code 0 1 1 1 1 0 0, if we redid the parity bit generation and all three were wrong, then which data bit was transmitted incorrectly? (d1, d2, d3 or d4?)
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- A block of bits with n rows and k columns uses horizontal and vertical parity bits for error detection. Suppose that exactly 4 bits are inverted due to transmission errors. Drive a mathematical expression of n and k for the total number of different cases where these 4 error bits will be undetected. Show each important step of your derivation. (Hint: the error bits will be undetected if all the parity bits computed from the block remain unchanged.)a. Compute for the data efficiency of transmitting if the data format is a General Synchronous Data Frame. b. For an asynchronous data transmission (wherein 1 byte of data is accompanied by a specific number of auxiliary bits), how many start and stop bits are there if the data efficiency in transmitting is ? The number of start bits is equal to the number of stop bits.With the use of a Checksum Method, show the transmission of “m” and “n” from the sender to destination side. The following block is sent: 00001111 11110000 11001100 10101010. With the use of a two-dimensional parity check and assuming that no single/multiple-bit or burst error have occurred, what will be received in the receiver side? NOTE: Put a space after each frame.
- A)Convert the data word OxAF96 into Hamming Code using even parity B)Transmitter sent OxAF96. Assume that the data word received at the receiver was OxBF96. How does the receiver detect an error bit and correct the bit? Show the working steps.1. One way of detecting errors is to transmit data as a block of n rows of k bits per row and add parity bits to each row and each column (for this problem k is >=5). The bit in the lower-right corner is a parity bit that checks its row and its column. This scheme will detect burst errors of length 5? (A) TRUE (B) FALSE 2. According to Shannon, if the amount of Signal is increased on a transmission medium, everything else constant, the expected theoretical bitrate would (A) increase (B) decrease (C) remain the same (D) cannot determine 3. In nonpersistent CSMA, if a transmitter ready to transmit senses the channel is busy, the transmitter waits a random amount of time, and senses the line again. If it’s still busy it repeats the process, otherwise it transmits the data. (A) True (B) FalseGiven are the following codewords for an error control coding scheme data 000 011 100 codeword 00011000 00000111 10101000 . a.What is the minimum Hamming distance? b. What is the coding rate for this scheme? c. what raw data rate would be required if we need a throughput of 6 Mpbs? d. in a different situation, the number of data bits to transfer is 5, and the minimum hamming distance is dmin=5. find the maximum coding rate for this situation
- in hamming coding system(7,3) shown as ; 3 test bit and 7 total bit (test+data) and test bits are shown as (c1,c2,c3...) or (p1,p2,p3...) bit series coded in haming(15,4) is recieved as 0011001110111000(p1,p2,....,d11) on reciever side according to this bit series; what is the (d1,d2,...d11) ?Consider the example below that shows both the payload and the two-dimensional parity bits (shown in blue), and where exactly one of the payload or parity bits shown has been corrupted. In this example, odd parity is used.When bit stuffing is used, is it possible for the loss, insertion, or modification of a single bit to cause an error not detected by the checksum? If not, why not? If so, how? Does the checksum length play a role here?
- Suppose the data bits to be encoded in a (7,4) Hamming code are: 1100 Question 2a) What would the code word be (note, it will be a 7 bit sequence)? Question 2b) If upon receipt, the receiver redoes the parity bit generation to test for correctness of the data bits transfer, and finds that all three parity bits are showing up incorrectly, which data bit was incorrectly transmitted (your answer is one of d1, d2, d3, or d4)?In ___________ there is always a transition at the middle of the bit, but the bit values are determined at the beginning of the bit. If the next bit is 0, there is a transition; if the next bit is 1, there is none. a. Manchester b. differential Manchester c. both (a) and (b) d. neither (a) nor (b)In CRC, if the dataword is 5 bits and the codeword is 8 bits, how many 0s need to be added to the dataword to make the dividend? What is the size of the remainder? What is the size of the divisor? What is the maximum effect of a 2-ms burst of noise on data transmitted at the following rates?