For 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: 110110010100 Modulation Rate: 0,02 Mbps Encoding: Manchester b. Data: 1001000 0 0 0 Modulation Rate: 0,04 MBaud Encoding: Manchester c. Data: 10010001100 0 Modulation Rate: 0,5 Gbaud Encoding: Manchester d. Data: 1101100 10100 Data Rate: 0,25 Mbps Encoding: Differantial Manchester e. Data: 1 011001O1001 Modulation Rate: 0.04 Gbaud Encoding:

For 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: 110110010100 Modulation Rate: 0,02 Mbps Encoding: Manchester b. Data: 1001000 0 0 0 Modulation Rate: 0,04 MBaud Encoding: Manchester c. Data: 10010001100 0 Modulation Rate: 0,5 Gbaud Encoding: Manchester d. Data: 1101100 10100 Data Rate: 0,25 Mbps Encoding: Differantial Manchester e. Data: 1 011001O1001 Modulation Rate: 0.04 Gbaud Encoding:

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter8: Data And Network Communication Technology

Section: Chapter Questions

Problem 31VE

See similar textbooks

Similar questions

The data word 10101110001101 is given for transmission. Draw the following encoding formats. NRZ-L NRZ-I Bi-phase Manchester RZ Differential Manchester
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)
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 10001000011 Modulation 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
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.
Signal 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
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.
a. Huffman coding assigns codes to characters such that the length of the code depends on the relative frequency or weight of the corresponding character. Thus, it is a variable-length code. If the estimated frequencies for letters match the actual frequency found in an encoded message, then the length of that message will typically be less than if a fixed-length code had been used. Due to such advantages, RadiantRay Networking Sdn Bhd uses Huffman compression technique to encode the message before transmitting over the network to its customer. Suppose the message contains the following strings "a fast runner never be afraid of dark“ a. How many bits may be required for encoding the message without using Huffman coding? b. Create an Huffman table indicating the number of character bits, frequencies and the code used in transmission of the message (space is also counted as a character) c. What is the total size of the massage to be sent?
a. Huffman coding assigns codes to characters such that the length of the code depends on the relative frequency or weight of the corresponding character. Thus, it is a variable-length code. If the estimated frequencies for letters match the actual frequency found in an encoded message, then the length of that message will typically be less than if a fixed-length code had been used. Due to such advantages, RadiantRay Networking Sdn Bhd uses Huffman compression technique to encode the message before transmitting over the network to its customer. Suppose the message contains the following strings "a fast runner never be afraid of dark“ d. Construct the Huffman encoding tree e. What is the size of the table for encoding the message?
Add the following two binary words and find the sum and carry bit: A = 1100 _ 0101 , B = 1010 _ 1010.
IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost 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 -1.625 * 10-2 assuming a version of this format. Calculate the sum of 2.625*102 and 4.150390625 * 10-1 by hand, assuming both numbers are stored in the 16-bit half precision described above. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps.
Using an optimum symbol to bit sequence mapping code, give the bit sequence of any one of the adjacent symbols of the given symbol 1001.
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