Demonstrate by example (write down a few dozen arbitrary bit patterns; assume one start bit and a stop element of length one bit) that a receiver that suffers a framing error on asynchronous transmission will eventually become realigned.
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- A bit sequence is transmitted using Hamming code. What is the sequence of data bits transmitted for each of the following received streams: 1010 1100 1110, and 0101 0101 0101 0111 1111?Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of the i. NRZ-Iusing each of the given data streams, assuming that the last signal level has been positive. From thegraphs, guess the bandwidth for this scheme using the average number of changes in the signal level.A 15 bit data word along with parity bits received by the receiver is as follows: 101010 111111 011100 110100. Consider the two-dimensional odd parity error detection mechaniem to find out which bit is received in error? if, any. The bits shown in bold with underline are parity bits. Assume sender divides the data in a group of 5 bits each to calculate the parity bits
- By inspecting the LSB, you may quickly determine if a bit stream represents an odd or even decimal integer. Please explain why.Given 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 situationI88. Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of the i. Manchester using each of the given data streams, assuming that the last signal level has been positive. From thegraphs, guess the bandwidth for this scheme using the average number of changes in the signal level...
- A bit stream 10011101 is transmitted using CRC method. The generator polynomial is x^3 + 1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receivers end.Assume bit stream of 1110010 as data, polynomial X²+X as divisor. Using the Cyclic Redundancy Check (CRC) algorithm: a. Show in detail the calculations that need to be done by sender to generate sending bit stream. b. Show in detail the calculations that need to be done by receiver to check if the bit stream is received correctly.Suppose the sender and the receiver agree to use the bit pattern 01111110 to mark the beginning and the end of a frame. 1)The sender has the following bits to send. What does the sender actually send? 011110000111111011101111101 2) The receiver receives the following bits. What're the original data bits (note: this question has nothing to do with the previous question) 01111110111110111110001101111101000011111001111110
- Consider the following scenario: Frames of 12000 bits are sent over a 400kbps channel using a satellite whose propagation delay is 270ms. Acknowledgements are always piggybacked onto data frames. (That is, in the case of stop-and-wait, the next frame cannot be transmitted until the entire frame containing the acknowledgement is received.) Five-bit sequence numbers are used. What is the maximum achievable channel utilization for (a) Stop-and-wait. (b) Protocol 5 (i.e., Go Back N). (c) Protocol 6 (i.e., Selective Repeat)Consider the information sequence 1101 and suppose g(x) = x^3 + x + 1. The codeword of this information sequence is 1101001. Suppose a transmission error on the first bit (to the left). The error checking procedure at the receiver is able to detect the error. Select one: O True O False. Draw the graph of the NRZ-L, NRZ-I, Manchester scheme, differential Manchester scheme using eachof the following data streams, assuming that the last signa11evel has been positive. From the graphs,guess the bandwidth for this scheme using the average number of changes in the signal level.a. 00000000b. 11111111c. 01010101d. 00110011