Network Design

A. Symbol rate and noise level

B. Symbol rate and number of symbols

C. Symbol rate and Baud rate

D. Symbol rate and bandwidth

E. Symbol rate and signalling rate

A. Noise power and signal power

B. Signal to noise ratio and signalling rate

C. Bandwidth and signal power

D. Bandwidth and signal to noise ratio

E. Bandwith and number of symbols

A. Hertz (Hz)

B. Bits per second (bps)

C. Baud (Bd)

D. Baud per symbol (bps)

E. Decibels (dB)

A. The power at which the signal is transmitted

B. Attenuation over the distance of transmission

C. Thermal noise

D. Amplification of the received signal

E. Distortion caused by frequency response characteristics of the medium

A. Attenuation is a reduction in signal power over the distance travelled

B. Attenuation is a form of noise and reduces data rates

C. Attenuation the power level at which a signal is transmitted

D. Attenuation the power level at which a signal is received

E. Attenuation is measured in Hertz (Hz)

A. Thermal noise is caused by environmental factors such as switching or mains hum

B. Distortion is cause by other signals in close proximity

C. Interference is caused by vibrations at an atomic level and is always present

D. Crosstalk (inter-symbol interference) is caused by delayed versions of the signal itself

E. Crosstalk (inter-channel interference) is caused by the frequency response characteristics of the medium

A. Bandwidth

B. Signal to noise ratio

C. Data rate

D. Maximum channel capacity

E. Symbol rate

A. The high frequencies used by the signal

B. Thermal noise

C. Resistance and other impedance's

D. The insulating sheath around the cable

E. Interference from other signals in the cable

A. Thermal noise

B. Interference

C. Crosstalk

D. Attenuation

E. Distortion

A. Electrons vibrating and absolute zero

B. Thermal noise and the number of symbols

C. Signalling rates and distortion at high frequencies

D. The Baud criterion and information content

E. Finite bandwidth and the presence of noise

A. Coaxial cable: 2×108 m s-1

B. UTP: 2×108 m s-1

C. Multimode optical fibre: 2× 10 to the power of 8 m s-1

D. Monomode optical fibre: 3× 10 to the power of 8 m s-1

E. Radio: 3× 108 m s-1

A. Reflection

B. Refraction

C. Propagation

D. Diffraction

E. Scattering

A. -10 dB

B. 0 dB

C. +2.78 dB

D. +3 dB

E. +10 dB

A. The wavelength increases

B. The wavelength decreases

C. The wavelength stays the same

D. The wavelength changes phase

E. The wavelength's energy becomes unstable

A. Monomode fibre is made of a single strand of glass, while multimode fibre is made of many strands

B. Monomode fibre can only carry a single signal, while multimode fibre can carry many signals

C. Monomode fibre is made of glass with only one refractive index, while multimode fibre is made of glass with multiple refractive indices

D. Monomode fibre has a single coloured coating, while multimode fibre has a striped coating

E. Monomode fibre has a much narrower core than multimode fibre, so power losses are lower

A. The twisting of the pairs reduces the impact of external noise through cancellation

B. A protective jacket surrounds the twisted pairs of wire

C. Each pair is twisted by a different amount to reduce crosstalk

D. It has a braid or foil shield to reduce interference

E. Different grades are available with different frequency and attenuation characteristics

A. The plastic insulating sheath prevents interference reaching the copper conductors inside

B. The braiding prevents interference reaching the internal conductor

C. The dielectric between the braiding and the conductor is the main means of reducing interference

D. The twisting of the cable uses cancellation to reduce interference

E. The copper meterial of the internal conductor is resistant to interference

A. Straight-through cable must be used

B. Crossover cables must be used

C. At least one device must have an internal crossover in the interface

D. An odd number of crossovers is required, which may be present in cable and/or the device interfaces

E. An even number of crossovers is required, which may be present in cable and/or the device interfaces

A. Fibre has higher bandwidth

B. Fibre has lower attenuation

C. Fibre infrastructure is easier to install

D. Fibre is more resistant to interference

E. Fibre does not require grounding on inter-building runs

A. Infrastructure for wireless infrastructure is generally cheaper, since cables are not required

B. Attenuation increases with the square of the distance, rather than distance

C. Radio is less prone to interference and noise

D. Since radio bandwidth is infinite, data throughput is generally higher with wireless media

E. There is no limit to transmit power on unlicensed radio bands

A. The signals are travelling at a faster speed from transmitter to receiver

B. The data is being transmitted at a faster rate (in bits per second)

C. A carrier wave is being modulated, rather than a signal being sent directly on the medium

D. A `local loop' phone line is being used to transmit the data

E. One end of the transmission path is in somebody's home

A. x and y

B. y and z

C. x and z

D. x and t

E. π and y

A. By the direction (positive or negative) of the transition

B. By the presence or absence of an additional transition

C. By the transition being in the same or the opposite direction as the previous one

D. By the amplitude of the transition

E. By varying the exact time of the regular transition

A. 1

B. 2

C. 4

D. 8

E. 16

A. The number of symbols

B. The number of bits per symbol

C. The number of symbols per bit

D. The number of bits

E. The number of modulation systems used

A. 3 symbols, differing in amplitude

B. 3 symbols, differing in attenuation

C. 8 symbols, differing in amplitude

D. 8 symbols, differing in attenustion

E. 8 symbols, differing in phase

A. Compresses 4 bytes into 5 bits to increase data transfer speeds

B. Transmits 4 bits, then pauses for 5 bits to allow other stations to transmit

C. Encodes 4 bits as 5 symbols to increase data throughput

D. Encodes 4 bits as 5 bits to include a parity bit for error detection

E. Encodes 4 bits as 5 bits in order to increase the number of transitions for better synchronisationE. Encodes 4 bits as 5 bits in order to increase the number of transitions for better synchronisation

A. They generally use two signal levels, one positive and one negative

B. They can use any number of signal levels, all of which must be positive

C. They use three signal levels, including zero, which cannot be repeated more than once

D. They only use a single signal level, which is not zero

E. They can use any number of signal levels, but when a zero is repeated, a line code is required

A. 1

B. 2

C. 3

D. 4

E. 8

A. 100 Mbps Ethernet over copper

B. IEEE 802.11 over a radio physical layer

C. Dialup modem access to the Internet

D. Radio and television broadcast

E. ADSL home Internet access

A. The number of bits expected to be in error in a typical PDU

B. The number of bits expected to be in error per second

C. The proportion of bits expected to be in error

D. The probability of an PDU containing an error

E. The base-2 logarithm of the packet length

A. Error detection codes and error correction codes

B. Idle-RQ and continuous RQ

C. Selective repeat and go-back-N

D. Feedback error correct and forward error correction

E. Checksums and CRCs

A. 18%

B. 37%

C. 80%

D. 100% media utilisation

E. It depends on the normalised delay-bandwidth product

A. Data transmission rate

B. Typical PDU length

C. Physical length of the transmission link

D. Propagation speed of medium

E. Bandwidth of the physical channel

A. The speed at which errors can be detected or corrected

B. The number of PDUs which can be checked each second

C. The number of bits used for the code in each PDU sent

D. The proportion of errors which can be detected or corrected using the code chosen

E. The proportion of the number of bits of original data to the number of bits of data plus code sent

A. Idle RQ

B. Cyclic Redundancy Check (CRC)

C. 2D parity

D. Forward error correction

E. Checksums

A. They are based on polynomial division

B. They can only detect errors with an odd number of incorrect bits

C. They are used in LAN protocols

D. They are used in WAN protocols

E. They can be implemented in hardware

A. They can detect single bit errors

B. They can correct single bit errors

C. They can detect 2-bit errors

D. They are robust in the face of multiple errors

E. They cannot be implemented in hardware

A. The transmission time

B. The propagation time

C. The normalised delay-bandwidth product

D. The sliding window size

E. The length of an average PDU

A. Standard and idle-RQ

B. Go-back-N and selective repeat

C. Sliding window and synchronised window

D. Error detection only and error correcting mode

E. Synchronisation and acknowledgement

A. Reservation Aloha

B. CSMA

C. Token Passing

D. CSMA/CA

E. CSMA/CD

A. CSMA always has a higher media utilisation

B. CSMA checks that no other stations are transmitting before transmitting data

C. CSMA has a collision detection function

D. CSMA cannot operate in a radio environment

E. CSMA requires that the value of a is known

A. The sum tTX + tPROP

B. The difference tTX - tPROP

C. The product tPROP × tTX

D. The quotient tPROP / tTX

E. The quotient tTX / tPROP

A. 1-persistent CSMA

B. p-persistent CSMA

C. Nonpersistent CSMA

D. CSMA/CD

E. CSMA/CA

A. CSMA, because collisions will prevent correct communication

B. CSMA/CA, because collisions are unavoidable with radio

C. CSMA/CA, because of random backoff errors

D. CSMA/CD, because of hidden nodes and the duplex transmission requirement

E. CSMA/CD, because JAM signals cannot be transmitted on a radio link

A. Unslotted versions generally perform better than slotted versions

B. Persistent CSMA performs best at high loads

C. Nonpersistent CSMA outperforms slotted Aloha at high loads

D. CMSA performs best when a is large

E. CSMA always performs better than token passing

A. The reserved phase of Reservation Aloha

B. TDMA

C. FDMA

D. CDMA

E. CSMA

A. It waits for a random backoff period after a collision is detected

B. It waits for a random backoff period after the medium becomes free

C. It can anticipate when other stations are about to transmit, and holds back data until it has seen these transmissions take place

D. It uses separate frequencies for transmitting and acknowledgments

E. It retransmits immediately with probability p

A. Collisions may not be noticed at higher transmission rates

B. The interframe gap becomes too short for the processor speed to keep up with

C. Hidden nodes are more likely at higher data transmission rates

D. Random numbers required for the backoff time calculation cannot be produced quickly enough

E. The increase in a means that utilisation of the medium will fall unless propagation times are also reduced

A. Token passing has a higher media utilisation than Aloha

B. Token Ring is a well-known implementation of token passing

C. Token passing is a deterministic media access protocol

D. Stations may only transmit when they are in possession of an electronic `token', which is passed in a special PDU

E. Token passing is less complex than the CSMA or Aloha protocols