In an asenkron transmission a character size is 6 bits. When 1700 characters are transmitted, if total overhead amount is 5,1 Kbit, which answer is correct? а. 1 CDeri Dote
Q: The rate osferring a frame of 58500 bits in a network is 160 bps. Calculate the taken time in…
A: The calculation is given in step 2.
Q: In a simplex communication by using STP cable, which type of change can reduce the received signal…
A: The correct answer of the questions is option ("E") "ALL"
Q: what ways is NRZ Coding unsuitable for transmitting s
A: Lets see the solution.
Q: A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the…
A: The Answer is
Q: what is the throughput if the system produces the 1000 frames/sec then A slooted ALOHA network…
A: The slotted ALOHA calculation is
Q: what is a modem connection speed in kbps, if the circuit symbol rate is 13 thousand of symbols per…
A: Given, symbol rate 13000 symbols/second and since each symbol carries 3 bits, that is 3 bit /…
Q: A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n…
A: Introduction :Given , Implementation of "go back n "link bandwidth = 20 Kbps Tp = 400 ns n= 10 Frame…
Q: How long does the asynchronous transmission of a 150 KByte (1 KByte = 1024 bytes) file take via a…
A: Solution:-- 1)The given question has required for the solution which is to be provided as the…
Q: 000. the signal to noise ratio is usually 2165. calculate maximum capacity of the noisy channel and…
A: Capacity can be calculated as- Capacity C=B log2(1+SNR) = C= 4000 log2(1+2165)= 4000 log2 2165= 4000…
Q: Q3/ Given that the bit sequence [1 0 1 1 0 0 1 0] is to be transmitted. Draw the resulting waveform…
A: Draw the sequence of [1 0 1 1 0 0 1 0] 1) Unipolar RZ 2) Polar RZ 3) AMI
Q: what the total time required to send 5000 bytes across a 3000 kilometers link at 1 Mbps?use a signal…
A: The transmission time (Tt) = L/B = Length of the data / Bandwidth The propagation time (Tp) = d /…
Q: If the bandwidth of the network is 10Mbps, which is used to transmit the data between two nodes…
A: Given Bandwidth B=10Mbps=10*106bps=107bps Distance L=200m Propagation speed P=2*108m/sec
Q: Assume that signals propagate at 300000 km/s in wireless pure ALOHA network which are a maximum of…
A: In random access or contention methods, methods, no station station is superior to a different…
Q: A 3000km long trunk operates at 1.536 Mbps is used to transmit 64 byte frames. If the propagation…
A: The Answer is
Q: what is the throughput if the system produces the 1000 frames/sec then A pure ALOHA network…
A: throughput of ALOHA network is calculated in step 2.
Q: what is the throughput if the system produces the 250 frames/sec then A slooted ALOHA network…
A: The Slotted ALOHA calculation is
Q: A data communication has the following delays computed as follows: Transmission delay is 1 second.…
A: Total delay in sending one data packet or End to End time = Transmission delay + Propagation delay…
Q: The rate of transferring a frame of 47000 bits in a network is 970 bps. Calculate the taken time in…
A: Transmission rate = frame rate * no of bits in a slot. Max frame rate is calculated by single…
Q: 17. A 1 KM 1long CSMA/CD with a propagation 10Mbps delay 5usec has bandwidth. Repeaters are not…
A: Here is the solution with an explanation:-
Q: you are using a connection with transmission rate of 64000 bps and a 3200 Km/s propagation speed,…
A: DATA GIVEN:- Data to carry = 1 bit Propagation speed =3200 km/s Distance = 1600 km TO FIND:- Total…
Q: A computer on 6 Mbps network is regulated by token bucket. The token bucket is filled at the rate of…
A: Introduction :Given , Implementation of token bucket technique on a network.Bandwidth of the network…
Q: A signal can travel either forward (from port 1 to port 2) or backward (from port 2 to port 1) and…
A: Solution: Given, A signal can travel either forward (from port 1 to port 2) or backward (from…
Q: 16. Which one is not transmission impairment? a) Distortion b) Attenuation c) Signal d) Noise e)…
A: According to answering policy we can answer only 3 parts, for remaining parts please resubmit again…
Q: Multiplying a polar NRZ signal with a carrier frequency , generates O a. ASK O b. FSK Oc. PSK O d.…
A: Dear Student, When we multiply a polar NRZ with a carrier frequency PSK is generated, ASK and FSK…
Q: n a full duplex communication by using UTP cable, Pr power is measured as 15 mw and Adb as 10,79 dB.…
A: The correct answer to this question is " ( C.) minumum pc:8.33mw NEXTdb: 10 db ACRdb:17.79dB "
Q: what is the throughput if the system produces the 500 frames/sec then A pure ALOHA network transmits…
A: The Answer is
Q: H.W: Binary data are transmitted at rate of 10 bit/sec, over a microwave link having a BW of…
A: Introduction: In this question we have the transmitted binary data rate , Receiver i/p and BW. We…
Q: UTP cables are bulkier and more expensive when compared to STP cables Select one: O True O False
A: Actually, STP stands for Spanning Tree Protocol.
Q: A.9. We need to send 256 kbps over a noiseless channel with a bandwidth of 16 kHz. How many signal…
A: Rate given =256 kbps Bandwidth =16khz signal levels have to find We can use formula to find this…
Q: how much bandwidth would required to transmit a DS-1 signal (1.544 Mb/s) using a four-level code…
A: GIVEN: how much bandwidth would required to transmit a DS-1 signal (1.544 Mb/s) using a four-level…
Q: what is the throughput if the system produces the 250 frames/sec then A pure ALOHA network transmits…
A: The answer is in step 2:
Q: If the probability pk of a message is 1/8 then the information in bits is ____________
A: 1. 3 bits 2. True 3. ARQ
Q: The data [0 1 0 0 0 1 0 1 1 1 0 0 1 1 1] is generated in an information signal with a speed of 64…
A: DSSS- Direct Sequence Spread Spectrum It is a type of spread range transmission which uses spreading…
Q: Computer A, B C, D and E is arranged in a ring topology. Calculate the propagation delay of sending…
A: All workings are below Explanation: Detailed explanation:
Q: Calculate the utization of a 100 m long 100 Mbps coaxial CSMACO system that transmits 10 Kb packets.…
A: Given that, Distance= 100 m Bandwidth= 100 Mbps Length of packet= 10 KB Propagation speed= 2*108…
Q: In QAM, two carriers with a phase difference of 90 degrees is used Select one: O True O False
A: QAM uses both amplitude and phase components to produce a kind of modulation that may give high…
Q: . How many check bits? Suppose you want to send 3500 characters of data. a) How many check bits…
A: we will solve this question in step No. 2
Q: Determine Vo in the network shown. 4 k2 2 k2 6 mA 2 k2 12 k2 3 kQ O + A IO
A: To add resistance in series do normal arithmetic addition To add resistance in parallel do…
Q: A common measure of transmission for digital data is the baud rate, defined as the number of bits…
A:
Q: We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate…
A: given data 1 MHz bandwidth SNR channel 63 find bit rate and signal level? Solution C = B log…
Q: If we need to send data 3 bits at a time using MFSK (Multilevel Frequency Shift Keying) modulation…
A: Answer: Our instruction is answer the first three part from the first part and . I have given…
Q: Binary data to be sent over the network is 11011001 (8 bits) . Use the Hamming code method, to…
A: In this question, we have to calculate code word (12 bits) that will be determine by the Hamming…
Q: 200 km 300 km ten-car caravan toll booth 1 toll booth 2 toll booth 3 (aka 10-bit packet) (aka link)…
A: How long until caravan is lined up before 3rd toll booth?
Q: Given the main frequency of the wideband channel to be 250MHZ with a bandwidth of 200MHZ, what will…
A: Given Data : Main frequency = 250 MHz Bandwidth = 200 MHz Number of subchannels = 4
Q: In which modulatio technique is bit rate equal to baud rate? a. ASK b. BPSK c. all of the choices…
A: option D FSK, modulation technique is bit rate equal to baud rate as it generally uses two different…
Q: A system is using Manchester to transfer 10-Mbps data. What are the average signal rate in kbaud?
A: here I written Step by step solution. I hope you like it.
Q: 4) A coaxial cable has a bandwidth of _________ of megahertz. i. 100 ii. 150 iii. 1000 iv. 10000
A: Question 4) A coaxial cable has a bandwidth of _________ of megahertz. i. 100 ii. 150 iii. 1000 iv.…
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- The data [0 1 0 0 0 1 0 1 1 1 0 0 1 1 1] is generated in an information signal with a speed of 64 kbit / s. This information mark [1 0 0 1 0 1] is transmitted using GPSK modulation with the spectrum code division multiple access (DSSS-CDMA) method, which is spread directly with the sequence, using the spreading sequence . 1- Show the bit sequence generated after spreading. 2- Show that the disseminated information can be retrieved at the receiver.In a selective-reject automatic repeat request (ARQ) error control protocol, the bit error rate (BER) is assumed to be BER= 10 ^-8 and the frames are assumed to be 1000 bits long. The probability that it will take exactly 11 attempts to transmit a frame successfully is: (a) 0.99 (b) 99x10 ^-22 (c) 10x10^ -6 (d) 1000x10 ^-5 The probability of receiving a frame in error is approximately: (a) 10^ -3 (b) 10^ -5 (c) 10^ -6 (d) 10^ -2Please 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,…
- 1)If the probability pk of a message is 1/8 then the information in bits is ____________ 1 bit. 3 bits. 2 bits. None of them. 2) Bipolar RZ line codes ensures both zero DC, and synchronization at every bit. FALSE TRUE 3) In _______, a receiver can use an error-correcting code that automatically corrects certain errors. FEC Huffman coding ARQ Line coding 4) Coaxial cable is more susceptible to external noise than fiber optics cable. FALSE TRUE 5) Which of the following requirement must be met for fast communication? High channel capacity. High sampling rate. High transmitter power. Large signal bandwidth.Plagiarism and wrong answers will downvotes Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions) for sliding window protocol using selective repeat on a heavily loaded 50 kbps satellite channel with data frames consisting of 40 header and 3960 data bits. ACK frames never occur. NAK frames are 40-bits. The error rate for data frames is 1 percent and the error rate for NAK frames is negligible. The sequence numbers are 8-bits.How long does the asynchronous transmission of a 150 KByte (1 KByte = 1024 bytes) file take via a serial null modem cable with a transmission speed of U baud (U bit/sec)? Here, no parity, one start bit, 2 stop bits and eight data bits are to be used. Explain your answer or give the calculation method and determine the transmission time for U = 19,200 baud.
- An end station sends 135,200 bits into a Frame Relay network in 200 milliseconds. The CIR is 400,000 bits per second and the maximum transmission rate is 1.544 Mbps. A. Has the user exceeded the maximum transmission rate? (show your work, then answer yes or no)B. If every frame is exactly 50 octets long, how many of these frames will have the DE bit set to 1?Dear Writer this is my calculation solution.could you please double check it and let me know it if it soccrect or should i proceed with your solution: Please provide justification if correction is need and correct answer.Thank you in advance. 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…This is my solution, Please check the solution and let me know If my solution is correct:Please adjust your decisiona nd provide corrected version if it is needed. 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…
- Any errors in the following scenario would be greatly appreciated. The sender sends "001011010" to the recipient. For each piece of information, it utilizes 8-bit data with odd parity and 1-bit (LSB) parity. If devices utilize even parity, why don't they use odd parity as well? The communication protocols used in Tesla vehicles may be found on the company's website. Provide a schematic if you have one.Suppose an 802.11b station is configured to always reserve the channel with the RTS/CTS sequence. Suppose this station suddenly wants to transmit 1,000 bytes of data, and all other stations are idle at this time. As a function of SIFS and DIFS, and ignoring propagation delay and assuming no bit errors, calculate the time required to transmit the frame and receive the acknowledgment.Dear Writer could oyu please kindly explin why i get this soltuion and it is different that your solution? The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.025ms 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 * 0.025ms + 8ms = 0.05ms + 8ms = 8.05ms The transmission time of the frame from A to D is 8.05 ms. b) To calculate the efficiency of the CSMA/CD protocol in this LAN, we can use the formula: Efficiency = (1 / (1 + 5tprop / tframe)) tprop is the propagation delay tframe is the time it takes to transmit a frame. To calculate tprop, we need to determine the time it takes for a signal to propagate through the 5 Km distance between Station A and Station D. The speed of propagation is 200,000 Km/sec The propagation delay is: tprop = distance /…