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 is assumed that each sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps

1 Answer

Answer :

(C) 128 kbps

Related questions

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 : 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 : What is the minimum bandwidth required for broadband connection: a) 128 kbps b) 256 kbps c) 512 kbps d) None of These

Last Answer : b) 256 kbps

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 : 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 : 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 : 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 : What is size of the IPv6 Address? A. 32 bits B. 64 bits C. 128 bits D. 256 bits

Last Answer : C. 128 bits 

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 : 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 : Each ________in a SONET framecancarry a digitized voice channel. A) bit B) byte C) frame D) none of the above

Last Answer : byte

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 : A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. What is the bit rate? (A) 100 Kbps (B) 200 Kbps (C) 400 kbps (D) 1000 Kbps

Last Answer : (C) 400 kbps

Description : Suppose that the one-way propagation delay for a 100 Mbps Ethernet having 48-bit jamming signal is 1.04 micro-seconds. The minimum frame size in bits is: a. 112 b. 160 c. 208 d. 256

Last Answer : d. 256

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 : A dynamic RAM has refresh cycle of 32 times per msec. Each refresh operation requires 100 nsec and a memory cycle requires 250 nsec. What percentage of memory’s total operating time is required for refreshes? (A) 0.64 (B) 0.96 (C) 2.00 (D) 0.32

Last Answer : Answer: D Explanation: in 1ms : refresh = 32 times Memory cycle = 1ms/250ns = 106ns/250ns = 4000 times Therefore, % of refresh time = (32 x 100ns)/(4000 x 250ns) = 3200ns/1000000 x 100% = 0.32%

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 : In ISDN, what is the basic access B channel data rate? A. 32 kbps B. 64 kbps C. 144 kbps D. 16 kbps

Last Answer : B. 64 kbps

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 : Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide. (A) 30,000 (B) 32,700 (C) 33,000 (D) None of the above

Last Answer : (B) 32,700

Description : Define the term 'sample rate'. A. Snapshots of sound are taken as the wave cannot be represented as a series of continuous changes. B. The number of samples taken each second C. The number of bits used per second of audio D. The number of bits available for each clip

Last Answer : B. The number of samples taken each second

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 : A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32

Last Answer : (3) 15 and 14

Description : For a sample rate of 40 kHz, determine the maximum analog input frequency A. 30 kHz B. 40 kHz C. 20 kHz D. 10 kHz

Last Answer : C. 20 kHz

Description : The baud rate of a signal is 600 baud/second. If each signal unit carries 6 bits, then the bit rate of a signal is ................. (A) 3600 (B) 100 (C) 6/600 (D) None of the above

Last Answer : (A) 3600 

Description : If the bandwidth of a signal is 5 KHz and thelowest frequency is 52 KHz, what is the highest frequency? A) 5 KHz B) 10 KHz C) 47 KHz D) 57 KHz

Last Answer : 57 KHz

Description : For the 8-bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory? (A) 10011001 (B) 00011001 (C) 00111000 (D) 11000110

Last Answer : (B) 00011001

Description : A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ? (A) 2 Mbps (B) 60 Mbps (C) 120 Mbps (D) 10 Mbps

Last Answer : (A) 2 Mbps Explanation: In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), ... second (Mbps) or gigabits per second (Gbps). Here, Throughput = 15000 x 8000/60 = 2 Mbps 

Description : What is the number of levels required in a PCM system with S/N ratio of 40 dB? A. 64 B. 128 C. 256 D. 512

Last Answer : B. 128

Description : DES works by using a. permutation and substitution on 64 bit blocks of plain text b. only permutations on blocks of 128 bits c. exclusive ORing key bits with 64 bit blocks d. 4 rounds of substitution on 64 bit blocks with 56 bit keys

Last Answer : a. permutation and substitution on 64 bit blocks of plain text 

Description : An IPv6 address is _________ bits long A. 32 B. 64 C. 128 D. none of the above

Last Answer : C. 128

Description : a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become: (1) 128 times (2) 256 times (3) 512 times (4) 64 times

Last Answer : 512times

Description : 2. In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become: (1) 128 times (2) 256 times (3) 512 times (4) 64 times

Last Answer : 256 times

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 : How much space will be required to store the bit map of a 1.3 GB disk with 512 bytes block size ? (A) 332.8 KB (B) 83.6 KB (C) 266.2 KB (D) 256.6 KB

Last Answer : (A) 332.8 KB

Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps

Last Answer : (B) 2 Mbps

Description : What is the size of an IP address? A. 64 bit B. 128 bit C. 16 bit D. 32 bit

Last Answer : D. 32 bit

Description : What is the probability that a randomly selected bit string of length 10 is a palindrome? (A) 1/64 (B) 1/32 (C) 1/8 (D) ¼

Last Answer : (B) 1/32

Description : The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits

Last Answer : (C) 16 bits Explanation: VirtualAddress = 32 bits PageSize = 4KB = 12 bits therefore : VPNTag = 20 bits, OffsetTag = 12 bits TLBEntryLength = VPNTag = 20 bits TotalTLBEntries = 64, 4-way implies ... therefore : TLBIndex = 4 bits TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits

Description : The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64

Last Answer : (D) 64

Description : The SP is of ___ wide register. And this may be defined anywhere in the ______. a) 8 byte, on-chip 128 byte RAM. b) 8 bit, on chip 256 byte RAM. c) 16 bit,

Last Answer : a) 8 byte, on-chip 128 byte RAM.

Description : If the frequency spectrum of a signal has a bandwidth of500 Hz with thehighest frequency at 600 Hz, what shouldbe the samplingrate, according to the Nyquist theorem? A) 200 samples/s B) 500 samples/s C) 1000 samples/s D) 1200 samples/s

Last Answer : 1200 samples/s

Description : Consider a disk queue with I/O requests on the following cylinders in their arriving order: 6,10,12,54,97,73,128,15,44,110,34,45 The disk head is assumed to be at cylinder 23 and moving in the direction ... . The disk head movement using SCAN-scheduling algorithm is: (1) 172 (2) 173 (3) 227 (4) 228

Last Answer : (2) 173

Description : The period of a signal is 10 ms. What is its frequency in Hertz ? (A) 10 (B) 100 (C) 1000 (D) 10000

Last Answer : (B) 100

Description : What do the following numbers have in sequence - 512-256-128-64?

Last Answer : It is a geometric progression with common ratio 0.5

Description : Find the work done in an inductor of 4H when a current 8A is passed through it? a) 256 b) 128 c) 64 d) 512

Last Answer : b) 128

Description : How many hostsare attached to eachof the local area networks at your site? A. 128 B. 254 C. 256 D. 64 E. None of the above

Last Answer : 254

Description : 256 128 ? 192 96 240 80 a) 64 b) 156 c) 96 d) 128 e) 176

Last Answer : The series is ×0.5, ÷1, ×1.5, ÷2, ×2.5.... Answer: d)