Computer Network Questions and Answers

Answer: B

Correct answer: True

The statement is true. In a synchronous counter every flip-flop receives the same clock signal at the same instant, and combinational logic on the J and K inputs decides which stages actually change. The contrast is with an asynchronous or ripple counter, in which only the first flip-flop sees the clock and each subsequent stage is clocked by the output of the one before it. The difference is one of speed and of correctness. In a ripple counter the change propagates from stage to stage, so the delays accumulate and the total settling time is the number of stages multiplied by the propagation delay of one. During that interval the counter passes through transient false states, which can produce glitches if the outputs are decoded. A synchronous counter settles in one gate delay regardless of length, and never shows a false intermediate count. The price is more combinational logic, so ripple counters survive where speed does not matter and simplicity does.

Answer: C

Correct answer: 1.5 × 10⁻⁶ microns

The screen needs

\[ 480\times 640\times 24\times 60 \approx 4.42\times 10^{8}\ \text{bps} \]

and the spectral width for that bandwidth at \( \lambda = 1\ \mu\text{m} \) is

\[ \Delta\lambda = \frac{\lambda^{2}\Delta f}{c} = \frac{(10^{-6})^{2}\times 4.42\times 10^{8}}{3\times 10^{8}} \approx 1.5\times 10^{-12}\ \text{m} \]

which is \( 1.5\times 10^{-6} \) microns - a minute slice of the fibre's available spectrum.

Answer: B

Correct answer: 100 Base-FX

100Base-FX uses optical fibre, which reaches 2000 m. The copper varieties - TX and T4 - are limited to 100 m.

Answer: C

Correct answer: 30,000 GHz

Differentiating \( f = c/\lambda \) gives \( \Delta f = c\,\Delta\lambda/\lambda^{2} \):

\[ \Delta f = \frac{3\times 10^{8}\times 0.1\times 10^{-6}}{(1\times 10^{-6})^{2}} = 3\times 10^{13}\ \text{Hz} = 30{,}000\ \text{GHz} \]

That enormous figure is why a single fibre has so much more capacity than any copper medium.

Answer: A

Correct answer: n -1

With n sequence numbers the window is limited to n−1 frames. If it were n, a completely lost set of acknowledgements would leave the receiver unable to tell a retransmission from a new frame.

Answer: B

Correct answer: Reliability

Whether the network keeps working through failures - equipment faults, power cuts, natural disasters - is a question of reliability. Security concerns deliberate attack and performance concerns speed.

Answer: C

Correct answer: narrow band channel

A narrowband channel carries the least data per second, being defined by its small range of frequencies; voice-band channels are the next step up and wideband channels the fastest.

Answer: D

Correct answer: 45

A full-duplex line carries both directions, so one line serves each pair of routers:

\[ \binom{10}{2} = \frac{10\times 9}{2} = 45 \]

It would take 90 lines only if each were simplex and a separate one were needed for each direction.

Answer: C

Correct answer: both (a) and (b)

Manchester encoding sends a transition in the middle of every bit and never rests at zero volts, so it is a non-return-to-zero code, and it uses both positive and negative levels, so it is polar. Both descriptions apply.

Answer: B

Correct answer: Successive values differ in exactly one bit

In Gray code only one bit changes between consecutive values. That property matters in position encoders, where several bits changing at once could momentarily be read as a wildly wrong value.