How many characters per sec (7 bits + 1 parity) can be transmitted over a 2400 bps line if the transfer is synchronous (1 start bit and 1 stop bit)?
300
Description : How many characters per sec (7 bits + 1 parity ) can be transmitted over a 2400 bps line if the transfer is synchronous (1 start and 1 stop bit)?
Last Answer : 1. 300 2. 240 3. 250 4. 275 Ans- 2400/(7+1+2) 240
Description : How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit ? a. 600 b. 800 c. 876 d. 1200
Last Answer : b. 800
Description : ECE Board Exam March 1996 What is the transmission rate of a system for transmitting the output of a microcomputer to a line printer operating at a speed of 30 lines/minute? Assume that the line printer has 8 bits ... and prints out 80 characters/line. A. 800 bps B. 400 bps C. 320 bps D. 640 bps
Last Answer : C. 320 bps
Description : An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are ............... and .............. (A) 4000 bauds \ sec & ... (B) 2000 bauds \ sec & 1000 bps (C) 1000 bauds \ sec & 500 bps (D) 1000 bauds \ sec & 4000 bps
Last Answer : (D) 1000 bauds \ sec & 4000 bps Explanation: Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate. Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps
Description : Usually, it takes10-bits to represent one character. How many characters can betransmittedat a speed of 1200 BPS? A. 10 B. 12 C. 120 D. 1200 E. None of the above
Last Answer : 120
Description : Iridium satellites can provide data communications of up ____ bps data rate. A. 1200 B. 4800 C. 2400 D. 3840
Last Answer : C. 2400
Description : What arethe most commonly used transmission speeds in BPS usedin datacommunication? A. 300 B. 1200 C. 2400 D. 9600 E. None of the above
Last Answer : 9600
Description : In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps ? (A) 10 bps (B) 100 bps (C) 1000 bps (D) 10000 bps
Last Answer : (C) 1000 bps
Description : How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame? a. 10,000 bytes b. 12,000 bytes c. 15,000 bytes d. 27,000 bytes
Last Answer : b. 12,000 bytes
Description : An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element ? (A) 0.336 bits/baud (B) 3 bits/baud (C) 120,00,000 bits/baud (D) None of the above
Last Answer : (B) 3 bits/baud
Description : Assume that we need to download text documents at the rate of 100 pages per minute. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. Then the required bit rate of the channel is ... ...... (A) 1.636 Kbps (B) 1.636 Mbps (C) 2.272 Mbps (D) 3.272 Kbps
Last Answer : Answer: Marks given to all
Description : Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometre) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable? a. 8000 b. 10000 c. 16000 d. 20000
Last Answer : d. 20000
Description : A device is sending out data at the rate of 2000 bps. How long does it take to send a file of 1,00,000 characters ? a. 50 b. 200 c. 400 d. 800
Last Answer : c. 400
Description : A device is sending out data at the rate of 2000 bps. How long does it take to send a file of 1,00,000 characters ? (A) 50 (B) 200 (C) 400 (D) 800
Last Answer : (C) 400 Explanation: 1,00,000 characters = 1,00,000 x 8 bits = 8,00,000 bits 8,00,000 bits/2000 bps = 400 seconds
Description : Which method is used to detect double errors and pinpoint erroneous bits. a. Even parity method b. Odd parity method c. Check sum method. d. All of these
Last Answer : c. Check sum method.
Description : Each check bit is grouped with the information bits as specified by a__ a. Parity check code b. Parity check matrix c. Parity check bit d. All of these
Last Answer : b. Parity check matrix
Description : __is divided into a number of fields and is represented as a sequence of bits: a. instruction b. — instruction set c. instruction code d. parity code
Last Answer : a. instruction
Description : Assuming 8 bits for data, 1 bit for parity, I start bit and 2 stop bits, the number of characters that 1200 BPS communication line can transmit is
Last Answer : Assuming 8 bits for data, 1 bit for parity, I start bit and 2 stop bits, the number of characters that 1200 BPS communication line can transmit is 12CPS.
Description : Given code word 1110001010 is to be transmitted with even parity check bit. The encoded word to be transmitted for this code is (A) 11100010101 (B) 11100010100 (C) 1110001010 (D) 111000101
Last Answer : (A) 11100010101
Description : If delaysare recorded as 10 bit numbers in a 50 router network, and delay vectorsare exchanged twicea second,how muchbandwidth per fill duplex line is occupied by the distributedrouting algorithm? A. 500 bps B. 1500 bps C. 5 bps D. 1000 bps
Last Answer : 1000 bps
Description : What is the average calorie intake of rural India for measuring poverty line? (a) 2400 calories per person per day. (b) 2800 calories per person per day. © 2100 calories per person per day. (d) None of these.
Last Answer : (a) 2400 calories per person per day.
Description : When we move from the outermost track to the innermost track in a magnetic disk, then density (bits per linear inch) (1) decreases (2) increases (3) remains the same (4) either remains constant or decreases
Last Answer : increases
Description : What is the commonly used unit for measuring the sped of data transmission? 1) Bits per second 2) Nano seconds 3) Characters per second 4) Mega Hertz
Last Answer : 1) Bits per second
Description : What is the commonly used unit for measuring the speed of data transmission ? (1) Mega Hertz (2) Characters per second (3) Bits per second (4) Nano seconds
Last Answer : Bits per second
Description : How do you calculate the number of bits of a body of text in ASCII? A. Number of characters * 7 B. Number of characters (including spaces) *7 C. bits in Huffman * 7 D. bits in Huffman / 7
Last Answer : B. Number of characters (including spaces) *7
Description : Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is __________. a. 400 bps b. 800 bps c. 1600 bps d. 3200 bps
Last Answer : d. 3200 bps
Description : Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is ............... (A) 400 bps (B) 800 bps (C) 1600 bps (D) 3200 bps
Last Answer : (D) 3200 bps Answer: D Explanation: The multiplexer is shown in the Figure. Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each ... channel, the frame rate must be 100 frames per second. The bit rate is 100 32 = 3200 bps.
Description : If it costs Rs.2400 to fence a square field at the rate of Rs.6 per m, find the length of the side and thearea of the field?
Last Answer : {\displaystyle \Theta } (N2) algorithm, it would take on the order of 1012 steps to multiply two one-million-word numbers. Numerous algorithms have been developed to efficiently perform arithmetic operations on numbers stored with arbitrary ... The worst case is Θ
Description : The population of India is about 1.2 billion. Take the average consumption of energy per person per year in India as 30 Mega Joules. If this consumption is met by carbon based fuels and the rate of ... tons (B) 540 million metric tons (C) 5400 million metric tons (D) 2400 million metric tons
Last Answer : (B) 540 million metric tons
Description : A Bunker has a leak which would empty the completely filled bunker in 20 hours. If the bunker is full of water and a tap is opened which admits 4 litres of water per minutes in the bunker, how many litres does the bunker holds? A) 2400 B) 4500 C) 4800 D) 7200
Last Answer : C Leak emptied the bunker per hour = 100/20 = 5% of water per hour; Quantity of water emptied per hour = 4*60 = 240; Thus, 5% = 240 liter; Hence, capacity of water, 100% = 240*100/5= 4800 liter.
Description : Some amount out of Rs.8500 was lent at 5% per annum and the remaining was lent at 3% per annum. If the total simple interest from both the fractions in4 years was Rs.1500, the sum lent at 5% per annum was a) Rs. 2400 b) Rs. 2200 c) Rs. 2000 d) Rs. 6000
Last Answer : D Total simple interest received , I = Rs.1500 Principal , p = 8500 period, n = 4 years Rate of interest, r = ? Simple Interest, I=PNR/100 1500 = ( 8500 4 r )/ 100 = ... total amount is Rs.8500. Therefore, the amount lent at 5% per annum (part1 amount) = 8500x12/17 = 6000
Description : Avalanche photodiode receivers can detect bits of transmitted data by receiving a. 100 photons b. 200 photons c. 2000 photons d. 300 photons
Last Answer : 200 photons
Description : What is the usualnumber of bits transmitted simultaneously inparalleldata transmission used by microcomputers? A. 16 B. 9 C. 8 D. 4 E. None of the above
Last Answer : 9
Description : The _____ layer uses data compression to reduce the number of bits to be transmitted. a. presentation b. network c. data link d. application
Last Answer : a. presentation
Description : Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is _____. a. 2 sec b. 3 sec c. 4 sec d. 1 sec
Last Answer : a. 2 sec
Description : A terminal multiplexer has six 1200 bps terminals and ‘n’ 300 bps terminals connected to it. If the outgoing line is 9600 bps, what is the value of n ? (A) 4 (B) 8 (C) 16 (D) 28
Last Answer : (B) 8
Description : If it takes a transmitted signal 1m sec to go up to the target and come back after reflection, how far is the transmitter from the target? A) 150 Km B) 150 m C) 15 Km D) 1500 m
Last Answer : If it takes a transmitted signal 1m sec to go up to the target and come back after reflection, how far is the transmitter from the target? A) 150 Km B) 150 m C) 15 Km D) 1500 m
Description : In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1- bit delay in this network is equivalent to: a. 500 metres of cable. b. 200 metres of cable. c. 20 metres of cable. d. 50 metres of cable
Last Answer : c. 20 metres of cable.
Description : How many bits are required to represent one nibble ?
Last Answer : 4-bits are required to represent one nibble.
Description : Four bits are used for packet sequence numbering in a sliding window protocol used in computer network. What is maximum window size? A) 4 B) 8 C) 15 D) 16
Last Answer : Four bits are used for packet sequence numbering in a sliding window protocol used in computer network. What is maximum window size? 15
Description : How many bits Internet address is assigned to each host on a TCP/IP Internet which is used in all communications with host? A) 16 bits B) 32 bits C) 48 bits D) 64 bits
Last Answer : How many bits Internet address is assigned to each host on a TCP/IP Internet which is used in all communications with host? 32 bits
Description : Will a DELL Dimension 2400 be able to hold Windows Vista or Windows 7?
Last Answer : A P4 can handle it, Win7 is less resource intensive than Vista and supposedly on par with XP. However, I’d suggest at least 1GB of RAM. If you don’t have it already, your computer probably uses DDR RAM, which is stupid expensive since it hasn’t been in mass production for years.
Description : Which error detection method consists of just one redundant bit per data unit? A) Simple parity check B) Two-dimensionalparity check C) CRC D) Checksum
Last Answer : Simple parity check
Description : A clothing store that sets their advertising budget by following the major competitor and adding an additional 15 per cent is using the method 1. percentage of sales 2. arbitrary allocation 3. objective and task 4. competitive parity 5. None of these
Last Answer : competitive parity
Description : Method of managing advertising budget at a certain percentage of sales price per unit or forecasted sales of products is classified as A. percentage of sales method B. affordable method C. competitive parity method D. objective and task method
Last Answer : A. percentage of sales method
Description : A DMA controller transfers 32-bit words to memory using cycle Stealing. The words are assembled from a device that transmits characters at a rate of 4800 characters per second. The CPU is fetching and executing instructions at an average ... of the DMA transfer? (A) 0.06% (B) 0.12% (C) 1.2% (D) 2.5%
Last Answer : Answer: B
Description : What is framing? A. Is concerned of synchronous system B. Refers to parallel transmission C. Is concerned with the boundaries between characters D. Is concerned with individual bits
Last Answer : C. Is concerned with the boundaries between characters
Description : How many bits are used to represent Unicode, ASCII, UTF-16, and UTF-8 characters?
Last Answer : Ans. Unicode requires 16 bits and ASCII require 7 bits. Although the ASCII character set uses only 7 bits, it is usually represented as 8 bits. UTF-8 represents characters using 8, 16, and 18 bit patterns. UTF-16 uses 16-bit and larger bit patterns.
Description : What term is used for the representation of two numeric characters stored in eight bits?
Last Answer : Packing or packed data.
Description : The basic idea behind Huffman coding is to A. compress data by using fewer bits to encode fewer frequently occuring characters B. compress data by using fewer bits to encode more ... frequently occuring characters D. expand data by using fewer bits to encode more frequently occuring characters
Last Answer : B. compress data by using fewer bits to encode more frequently occuring characters