. What is the result of scrambling the sequence 11100000000000 using the following scrambling techniques? Assume that the last non-zero signal level has been positive. a. B8ZS p. HDB3 (The number of nonzero pulses is odd after the last substitution
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- Which encoding technique solves the loss of synchronization because of long 0’s and 1’s? A. Manchester B. Psuedoternary C. Polar-NRZ D. Polar-RZ2. (a) What is a single error and burst error? Explain with an example. (b) Find the Hamming code of the following bitstream using even parity. 11011111010111010101110111000010Assume that the channel introduces an error pattern “110100000110111011110111010100000001110100011010111011001000011”(Each bit transmitted 3 times). What is received? (:-Use Majority voting algorithm). Can the error be detected? Convert detected code into ASCII message
- ________ provides synchronization without increasing the number of bits.For the FSM below, assume to start from state S0 and to use the indicated encoding. The following sequence of inputs are going to be selected one after the other at times t0, t1, t2, t3, and t4. Use the FSM to indicate the next state bits and output Y after each time point. t0 -> AB=00 t1 -> AB=10 t2 -> AB=11 t3 -> AB=11 t4 -> AB=01 Indicate your answer using 3 bits for each time point. For example, if AB=10 at time t0, your answer would be 101, where the first two bits are for the next state (10 -> S2) and the third bit is for the output value YQUESTION 14Assume (+) is a bit interval with positive voltage, (-) is a bit interval with a negative voltage, and '0' is a bit interval with zero voltage, write the bits transmitted if the following signal was encoded using:a. Bipolar-AMIb. Pseudoternary 0+0-+0-+00-
- Choose the right answer: 1- A line encoding method wherein the positive half interval pulse is followed by a negative half interval pulse for transmission of ‘1’ is a.Bipolar NRZ b.NRZ-M c.bi-phase-level d.RZ-AMI 2- In block coding, the number of valid codewords is equal to a.2^n + 2^k b.2^n c.2^k d.2^n - 2^k 3- Which statement is correct for a simple parity-check code? a.can detect an odd-number of errors. b.can detect two errors. c.can detect no errors d.can detect an even-number of errors. 4- In modulo-2 arithmetic, adding 1 and 1 will have an outcome of a.0 b.2 c.1 d.11 5- The remainder in cyclic redundancy checking represents the a.The CRC b.The error c.The divisor d.The syndromeIf ASCII “A” and “a” are combined into a 16-bit pattern and then appended with a FCS (Frame Check Sequence) generated by a CRC polynomial of X5+X4+X+1. Determine the FCS and the output if NRZI encoding is applied. You can assume the idle state is ‘1’ and data is processed from LSB to MSB.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.
- About error control:a. Write down the bit series of the polynomial x10 + x7 + x4 + 1 b. If there is a message M = 110010 and the remaining bit of division in CRC Technique is R =001, what is the shape of the transmitted bit series if SCS-4 . is usedbits ?Beginning with one start bit followed by 21 data bits, 2 parity bits and 1 stop bit: a) What are the total number of bits in the frame? b) What would the overhead and throughput be (answer in percent)?4. The Hamming (7,4) encoded sequence 1001000 was received. If the number of transmission errors is less than two, what was the transmitted sequence?