Networking Basics Questions and Answers

Answer: D

The program is a network operating system.

A network operating system runs on computers and servers and provides the services that let them communicate and share resources over a network: file and print sharing, user accounts and security, and network protocols. Novell NetWare and Windows Server are examples.

System software is the broad category, application software serves end users' tasks, and a particular Windows desktop operating system is not the general term asked for.

Answer: A

The line speed is 1.608 Mbps.

Each frame carries one 10-bit sample from each of the 20 channels plus one framing bit, and there are 8000 frames a second:

\[ \begin{aligned} \text{bits per frame} &= 20 \times 10 + 1 = 201 \\ \text{line speed} &= 201 \times 8000 \\ &= 1\,608\,000 \text{ bps} = 1.608 \text{ Mbps} \end{aligned} \]

Leaving out the framing bit would give 200 x 8000 = 1.6 Mbps, which is not offered either. 1.2 and 1.204 Mbps come from miscounting the bits per frame, and 3.406 Mbps is about twice the true rate.

Answer: C

The device is called the head end.

In a dual cable broadband system, computers transmit on cable 1 towards the head end at the root of the cable tree. The head end transfers the signal to cable 2, which carries it back down the tree to all the computers.

'Tail end', 'after one hop' and 'round trip' are not names of this device; round trip describes a delay, and a hop is a link traversed in a network.

Answer: A

The figure is 4 kHz.

A telephone voice channel is allocated 4 kHz of bandwidth: the usable voice band is roughly 300 to 3400 Hz, about 3.1 kHz, with guard bands making up the 4 kHz slot used in frequency-division multiplexing. This 4 kHz is also why PCM samples voice at 8000 times a second, twice the highest frequency.

3 kHz approximates the voice band alone, and 1 or 2 kHz is too narrow for intelligible speech.

Answer: A

The transmitted pattern is 101101011.

10110101 contains five 1s. With even parity, the parity bit makes the total number of 1s even, so it must be 1, appended at the end: 10110101 followed by 1.

101101010 would have an odd count of 1s, which is odd parity. Options C and D change the first bit of the data, which the parity scheme never does.

Answer: B

PCM samples the voice signal every 125 microseconds.

Voice channels are limited to about 4 kHz, and the Nyquist criterion requires sampling at twice the highest frequency, 8000 samples per second:

\[ \frac{1}{8000} \text{ s} = 125\ \mu\text{s} \]

The other intervals correspond to sampling rates of about 13 300, 5700 and 4400 per second, none of which is the telephone standard.

Answer: B

The key gives (B) 3.

The usual answer is three transmission modes: simplex (data flows one way only), half duplex (both ways, but one direction at a time) and full duplex (both ways at once). Every standard classification of a data communication circuit contains these three.

Some data communications texts add a fourth, full/full duplex, in which a station sends to one station while receiving from another. That is why 4 can be argued, but the key and the common classification give 3. Two and one would leave out one of the three basic modes.

Answer: B

This is a point-to-point topology.

A point-to-point configuration links exactly two stations over a dedicated circuit, and is used where high-speed digital information has to be transferred between only two stations.

A multipoint topology connects three or more stations over one circuit. Simplex and duplex are transmission modes (the direction of transmission), not topologies, so they do not describe a network linking only two stations.

Answer: B

This is return to zero (RZ).

In a return-to-zero code the binary pulse is active for only part of the bit time, less than 100%, and the signal returns to zero for the rest of the bit. In non-return-to-zero (NRZ) codes the pulse stays at its level for the full bit time.

Unipolar RZ and unipolar NRZ are particular forms that also specify polarity; the defining term for an active time shorter than the bit time is RZ itself. NRZ is the opposite case.

Answer: C

Analog transmission uses 75 ohm cable.

75-ohm coaxial cable (broadband coax) is used for analog signals such as cable television and broadband networks.

50-ohm coax (baseband) was used for digital transmission, as in the original Ethernet. 25 and 100 ohm are not the standard impedances of coaxial cable for these uses; 100 ohm is typical of twisted pair.