Q4: Draw the constellation diagram of 2*4 QAM signal then show the output amplitudes and phases of the signal when the following data stream are modulated 11001011111100100011
Q: Q5/ The following data ( 101000110101110 ) is arrived in the receiver side, if you know that the…
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Q: Q5/ The following data ( 101000110101110 ) is arrived in the receiver side, if you know that the…
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Q: For the above given signal each bit is separated by a vertical line. If duration of a bit is given…
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Q: The following sequence of bits 0100110111 is to be transmitted using QPSK modulation. Take these…
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Q: The method used to transmit digital data from multiple sources over a single communicationchannel is…
A: Answer: (b) multiplexing
Q: 3. Draw the QAM analog signal that transmits the following bit stream: 0 1010010 1 0 0 1 1 0 1 1 1 1…
A: The Answer is
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A: Given Data : Binary data stream: 000011110011 To find : Transmitted sketch using : Unipolar…
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A: As per our guidelines, only 3 sub parts will be answered. So, please repost the remaining questions…
Q: Q1: Where we can combined the well- known signals as modulated signal which represents the video…
A: The ask is to name the procedure which combine the modulated signal and the driving signal.
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Q: source encoding is 101 and the r is 2. what is the channel encoding?
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A: Here we have given the answer for the first question as per the guideline.
Q: Draw the constellation diagram of 2*4 QAM signal then show the output amplitudes and phases
A: EXPLANATION Below is the answer for the given question. Hope you understand it well. If you have any…
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Q: In data communications, we use periodic analog signals and a. Periodic digital signals b. Non…
A: In data communications, we commonly use periodic analog signals and non-periodic digital signals.
Q: For the following binary channel, p(x 1) =0.5 yl 0.8 0.9 xl 0.8 0.9 y2 Evaluate: P(Y/X), P(Y),…
A: Solutions: i) PY/X PY/X=py1/x1py2/x1py1/x2py2/x2 =0.8000.8 ii) PYPY=PX.PY/XP(X)=px1.px2…
Q: In ________, the amplitude of the carrier signal is varied to create signal elements. Both frequency…
A: Amplitude Shift Keying The amplitude of the resultant output depends upon the input data whether it…
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Q: Q1) Answer the following: 1-given frame is 1101011111 and generator is 10011 check status of…
A: ANSWER:-
Q: In ________, the phase of the carrier is varied to represent two or more different signal elements.…
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A: Solution :
Q: 3-make deconvolution between two signals as shown deconv. x(t) h(t) dncete stepgl discte tep ugal as…
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Q: 280. In asynchronous transmission, we send one start bit that is a. 0 b. 1 c. 10 d. 11
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Q: Q4: Find and plot the MSK signal for the data bits (10110101), assume that the scheme also uses two…
A:
Q: characters
A: 1) The LRC code using odd parity is: 10000110 2) The message to be transmitted by sender is:…
Q: Q3: Draw the digital signals for the bit string 01101101011 using NRZ-L, NRZI. Bipolar-AMI,…
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Q: pose you have the following modulated signal: Ø(t) = 6 cos(1600nt) + 22 cos(1800nt) + 6cos(2000t) e:
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Q: 198. Modulation is the process of a. performing data encryption b. converting digital signals to…
A: Modulation is a process of converting the signal from one form to another
Q: . The _____ of a composite signal is the difference between the highest and the lowest frequencies…
A: Given: - The _____ of a composite signal is the difference between the highest and the lowest…
Q: Q3: Draw the digital signals for the bit string 01101101011 using NRZ-L, NRZI, Bipolar-AMI,…
A: Draw the digital signals for the bit string 01101101011 using NRZ-L, NRZI, Bipolar-AMI,…
Q: Draw the waveforms for the following data stream corresponding to NRZ-I, Differential Manchester.…
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- Q4: Draw the constellation diagram of 2*4 QAM signal then show the output amplitudes and phases of the signal when the following data stream are modulated 11001011111100100011 .Networking question: Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of thei. NRZ-Lii. NRZ-Iiii. Manchesteriv. differential Manchester schemeusing each of the given data streams, assuming that the last signal level has been positive. From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level.Consider a 1Mbps transmission channel. The clock at the receiver has a drift of 1 second in one year. How long asequence of bits (or frame) can be sent before the clock drift could cause a problem? Assume that the sender and receiver are synchronized at the beginning of each frame and that they cannot resynchronize during the frame. Also, assume that the receiver samples the received signal at the middle of each bit duration to detect if it is 0 or 1.
- q11- This question involves estimation of error handling overhead for a communication link which has a BER of 10−2 Forward error correct (FEC) Frame: Total 1144 bits 1024 bit of data 120 FEC bits Error Detection + automatic repeat request (ED + ARQ) Frame: Total 1074 bits 1024 bit of data 50 CRC bits a) Our task is to find the approximate frame error rate (FER) for FEC in percentage Answer: Answer b) Estimate the approximate overhead for ED + ARQ in percentage Answer: AnswerNetworking price 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.For the following stream of digital signal which is being transmitted by bit rate 5 Kbps: 101100000000110 a) Draw the waveform for all line codes you studied. b) Determine if the line code is self synchronized or not. c) Determine if the code can be used with ac coupling or not d) Determine the Bandwidth of the line code e) Which code is used in magnetic recording f) Which code is used in LAN
- Networking question: Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of the i. differential Manchester scheme 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.Consider a full-duplex 256 Kbps satellite link with a 240 millisecond end-to-end delay. The data frames carry a useful payload of size 12,000 bits. Assume that both ACK and data frames have 320 bits of header information, and that ACK frames carry no data. What is the effective data throughput When using Stop and Wait?Please check my answer and correct the part which is need to be foxed.And provide justification for the answer. a) The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.0025ms Transmission time (Tt) = frame size/bandwidth = 10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms Total time for transmission = 2 * Tp + Tt(since the frame has to travel from A to D and then D to A) = 2 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms The transmission time of the frame from A to D is 8.005 ms. b) The efficiency of the CSMA/CD protocol is given by the formula: Tt is the transmission time of a frame C is the number of collisions, Tp is the propagation time of a signal from one end of the segment to the other. Efficiency = Tt / (C * 2 * Tp + Tt + Tp) Assuming that there are no other stations transmitting or attempting to transmit on the segment, the transmission from A to D will succeed without collisions. Therefore,…
- I88. 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...Consider a full-duplex 512 kbps link with length 36,000Km connecting two hosts. The data frames carry a useful payload of size 6000 bits. Assume that both ACK and data frames have 400 bits of header information, and that ACK frames carry no data. The propagation speed along this link is 3×108 m/sec. a) What is the RTT on this link? b) What is the effective useful data throughput, in kbps, when using Stop-and-Wait?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.