Description : What is the data rate of the ISDN Basic access B channel? A. 192 kbps B. 32 kbps C. 64 kbps D. 144 kbps
Last Answer : C. 64 kbps
Description : Basic speed rate of digital system. A. 144 kbps B. 1,544 kbps C. 64 kbps D. 2,048 kbps
Description : Broad Band ISDN handles data rate of about (A) 64 kbps (B) 100 mbps (C) 5.4 mbps (D) 2.048 mbps
Last Answer : (A) 64 kbps
Description : Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it ... sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps
Last Answer : (C) 128 kbps
Description : What equation defines the composition of an ISDN basic access line? A. B+2D B. B+D C. 2B+2D D. 2B+D
Last Answer : D. 2B+D
Description : What is the transmission rate of a GSM cellular system? A. 64 kbps B. 240 kbps C. 128 kbps D. 270 kbps
Last Answer : D. 270 kbps
Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______. a. 64 Kbps b. 32 Mbps c. 32 Kbps d. 64 MbpS
Last Answer : c. 32 Kbps
Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ............... (A) 64 Kbps (B) 32 Mbps (C) 32 Kbps (D) 64 Mbps
Last Answer : (C) 32 Kbps
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 : 12 voice channels are sampled at 8000 sampling rate and encoded into 8 bit PCM word. Determine the rate of the data stream. a. 354 kbps b. 750 kbps c. 768 kbps d. 640 kbps
Last Answer : c. 768 kbps
Description : Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milli seconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is (1) 20 (2) 10 (3) 30 (4) 40
Last Answer : (2) 10
Description : Which protocols adopted the standards of HDLC (High Level Link Control) protocol? a. X.25 protocol as LAPB (Link Access Procedure Balanced, V.42 protocol as LAPM (Link Access Procedure for ... Services Digital Network) protocol as LAPD (Link Access Procedure on D channel) d. All the above
Last Answer : d. All the above
Description : What does the noise weighting curve shows? A. Noise signals measured with a 144 handset B. Power levels of noise found in carrier systems C. The interfering effect of other frequencies in a voice ... a reference frequency of one kilohertz D. Interfering effects of signals compared with a 3-kHz tone
Last Answer : C. The interfering effect of other frequencies in a voice channel compared with a reference frequency of one kilohertz
Description : What is the maximum theoretical data rate of the above problem if one transponder is used for binary transmission? A. 36 Mbps B. 72 Mbps C.18 Mbps D. 144 Mbps
Last Answer : B. 72 Mbps
Description : A noiseless 3 KHz Channel transmits bits with binary level signals. What is the maximum data rate? A. 3 Kbps B. 6 Kbps C. 12 Kbps D. 24 Kbps
Last Answer : 6 Kbps
Description : The speech coding rate in D-AMPS is ____. A. 8 kbps B. 13 kbps C. 15 kbps D. 20 kbps
Last Answer : A. 8 kbps
Description : In GSM cellular system, the speech coding rate is ____. A. 13 kbps B. 8 kbps C. 15 kbps D. 20 kbps
Last Answer : A. 13 kbps
Description : What should be minimum requirement of random access memory (RAM) for internet access: a) 8 MB b) 16 MB c) 32 MB d) 64 MB
Last Answer : c) 32 MB
Description : A 32-bit address bus allows access to a memory of capacity(a) 64 Mb (b) 16 Mb (c) 1Gb (d) 4 Gb 2.Which processor structure is pipelined? a) all x80 processors b) all x85 processors c) all x86 processors
Last Answer : c) all x86 processors
Description : The conversion of digital signal into analog for purposes of transmitting into the telephone line is done through ______. A. ISDN B. Radio C. RS232C D. Modem
Last Answer : D. Modem
Description : These are used to connect non-ISDN equipment ot ISDN line. A. Digipeaters B. Terminal adapters C. Local repeaters D. Terminal repeaters
Last Answer : B. Terminal adapters
Description : What is the channel symbol rate in Bluetooth for each user? a) 270.833 Kbps b) 1 Gbps c) 100 Mbps d) 1 Mbps
Last Answer : d) 1 Mbps
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 : Main basic components of a data communication are composed of the following. A. Computer, modern and router B. Computer, bridge and gateway C. Transmitter, channel and receiver D. Transmitter, computer and modem
Last Answer : C. Transmitter, channel and receiver
Description : For a space wave transmission, the radio horizon distance of a receiving antenna with a height of 64 meters is approximately A. 8 km B. 32 km C. 64 km D. 256 km
Last Answer : B. 32 km
Description : What bandwidth is needed to support a capacity of 128 kbps when the signal power to noise power ratio in decibels is 100? A. 19224 Hz B. 3853 Hz C. 19244 Hz D. 3583 Hz
Last Answer : B. 3853 Hz
Description : What is the molecular mass of a gas that effuses through a small hole at twice the rate as oxygen gas at the same temperature. (The molecular mass of O2 is 32.) w) 8.0 atomic mass units x) 16 atomic mass units y) 48 atomic mass units z) 64 atomic mass units
Last Answer : ANSWER: W -- 8.0 ATOMIC MASS UNITS
Description : A train, 1800 meters long running at the rate of 204 km/hr will cross a platform in: A) 16 B) 32 C) 45 D) 64
Last Answer : ANSWER: B Explanation: Speed of the train = 204 km/hr. We have to find the time in seconds, so convert the speed in the unit of m/sec. Therefore 204 km/hr = 204 x 5/18 m/s = 1020/18 m/s Length ... m/sec Time = distance/speed = 1800/(1020/18)sec = (1800*18/1020)sec = 32 sec (approx.)
Description : The ASCII encoding of binary data is called a. base 64 encoding b. base 32 encoding c. base 16 encoding d. base 8 encoding
Last Answer : a. base 64 encoding
Description : You are measuring noise in a voice channel with a Lenkurt 601A, F1A weighting network and a flat meter. Your meter reads -47dBm. What is this reading in dBa? A. 77 dBa B. 35 dBa C. 38 dBa D. 32 dBa
Last Answer : C. 38 dBa
Description : You are measuring noise in a voice channel at 7 dB test point level. The meter reads -56 dBm (FIA weighted). What is the reading in dBrnc? A. 20 B. 32 C. 35 D. 25
Last Answer : C. 35
Description : The signal in a channel is measured to be 23 dB while noise in the same channel is measured to be 9 dB. The signal to noise ratio therefore is __________. A. 9/23 B. 23/9 C. 32 dB D. 14 dB
Last Answer : D. 14 dB
Description : The ISDN channel D designates _____ which contains control information
Last Answer : Data
Description : The ISDN channel B designates _____
Last Answer : Bearer
Description : The midrange frequency range of sound is from A. 256 to 2048 Hz B. 2048 to 4096 Hz C. 512 to 2048 Hz D. 16 to 64 Hz
Last Answer : A. 256 to 2048 Hz
Description : At what power level does a 1 KHz tone cause zero interference (144 weighted)? A. 90 dB B. 90 dBm C. -90 dBm D. -90 dBm
Last Answer : D. -90 dBm
Description : Determine from the following radio frequency that falls under the very high frequency band of the radio spectrum. A. 345.00 MHz B. 144.50 MHz C. 235.50 MHz D. 450.00 MHz
Last Answer : C. 235.50 MHz
Description : ECE Board Exam March 1996 At what power level does a 1 kHz tone cause zero interference (144 weighted) ? A. -90 dB B. -90 dBm C. 90 dBm D. 90 dB
Last Answer : B. -90 dBm
Description : A pure ALOHA Network transmits 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is ______. a. 0.384 b. 0.184 c. 0.286 d. 0.58
Last Answer : b. 0.184
Description : A pure ALOHA Network transmit 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is .............. (A) 0.384 (B) 0.184 (C) 0.286 (D) 0.586
Last Answer : (B) 0.184
Description : Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B ... What is the optimal window size that A should use? a. 20 b. 40 c. 160 d. 320
Last Answer : b. 40
Description : Nominal voice channel A. 20 to 20 KHz B. 16 to 16 KHz C. 3 to 3 KHz D. 4 KHz
Last Answer : D. 4 KHz
Description : You are measuring a voice channel at a -4 dB test point level, the meter reads -73 dBm (pure test tone) convert the reading in dBmCO. A. 16 B. 18 C. 22 D. 12
Last Answer : C. 22
Description : A quadrature signaling have ________ possible states. A. 16 B. 8 C. 32 D. 4
Last Answer : D. 4
Description : Maximum data transfer rate of the optical fiber is……….? a. 50 kbps b. 1000 kbps c. 1000 Mbps d. None of the these
Last Answer : c. 1000 Mbps
Description : How much packet data rate per user is supported by W-CDMA if the user is stationary? a) 2.048 Kbps b) 100 Mbps c) 2.048 Mbps d) 1 Gbps
Last Answer : c) 2.048 Mbps
Description : Triple DES uses a. 168 bit keys on 64-bit blocks of plain text b. Working on 64-bit blocks of plain text and 56 bit keys by applying DES algorithm for three rounds. c. Works with 144 bit ... algorithm once. d. Uses 128 bit blocks of plain text and 112 bit keys and apply DES algorithm thrice.
Last Answer : b. Working on 64-bit blocks of plain text and 56 bit keys by applying DES algorithm for three rounds.
Description : EBCDIC can code up to how many different characters? A) 256 B) 16 C) 32 D) 64
Last Answer : Answer : A
Description : What is the base of the Octal Numeral System? (1) 8 (2) 16 (3) 32 (4) 64
Last Answer : (1) 8 Explanation: The octal numeral system is the base-8 number system and uses the digits 0 to 7. As it uses only eight digits (0 through 7) there are no numbers or letters used above 8.
Description : EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64
Last Answer : 256