计算机网路基础第二章习题

2.“Fast” Ethernet operates 10 x faster （100Mbps）than regular ethernet. Explain why the following changes were made.

(a) Encoding changed to 4B/5B

(b)CAT-5 cable has more twists

(a). it is hard to tell the bits apart, as 15 zeros look much like 16 zeros unless you have a

very accurate clock.So, use 4B/5B method. The five bit patterns are chosen so that there will never be a run of more than three consecutive 0s

(b).More twists result in less crosstalk and a better-quality signal over longer distances,

making the cables more suitable for high-speed computer communication, especially 100-Mbps and 1-Gbps Ethernet LANs.

4.If a binary signal is sent over a 3-KHz channel whose signal-to-noise ratio is 20dB, what is the maximun achievable data rate?

because 1db = 10 logS/N , so 20db means S/N = 100.

Shannon limit = 3 * log2(1+S/N) ≈ 19.975 kbps

Nyquist limit = 2 * 3 = 6 kbps.

So the maximun achievable data rate is 6kbps.

6.What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?

advantages:It can handle much higher bandwidth than copper.It has lower attenuation.It also has advantages of not being affected by bower surges, electromagnetic interference, or

power failures.It is thin and lightweight. Fibers do not leak light and are difficult to tap.

disadvantages:Fiber is a less familiar technology requiring skills not all engineers have, and fibers can be damaged easily by being bent too much. Since optical transmission is

inherently undirectional, two-way communication requires either two fibers or two

frequency bands on one fiber. Fiber interfaces cost more than electrical interfaces.

9.Is the Nyquise therom true for high-quality single-mode opticle fiber or only for copper wire?

The Nyquist theorem is a property of mathematics and has nothing to do with technology.

So, the Nyquist theorem is true for all media.

15.What is the minimum bandwidth needed to achieve a data rate of B bits/sec if the signal is transmitted using NRZ and Manchester encoding?Explain your answer.

NRZ : the signal completes a cycle at most every 2 bits. So, bandwidth is at least B/2 Hz.

Mancherster: The signal completes a cycle in every bit, So bandwidth is at least B Hz .

20,Is an oil pipeline a simplex system, a half-duplex system, a full-duplex system, or none of the above? What about a river or a walkie-talkie-style communication?

It is half-duplex ..The river is like simplex system. a walkie-talkie-style communication is like half duplex system.

22.A modem constellation diagram similar to Fig.2-23 has data points at the following coordinates:(1,1) (1,-1) (-1,1) (-1,-1). how many bps can a modem with these parameters achieve at 1200 symbols/second?

One baud has 4 value , so the bit rate = baud rate* 2.so data rate is 1200*2 = 2400 bps. 30.What is the difference, if any between the demodulator part of a modem and the coder part of a codec?(After all, both convert analog signals to digital ones.)

A demodulator accepts a modulated sine wave only but a coder accepts an arbitrary analog

signal.

36.What is the available user bandwidth in an OC-12c connection?

12*87=1044 columns

The first column is taken up by path overhead. So 1044 -1 =1043 columns.

1043 * 9 * 8 * 8000 = 600.768Mbps.

44.Suppose that A,B and C are simultaneously transmitting 0 bits, using a CDMA systrm with the chip sequences of Fig.2-28(a). What is the resulting chip sequence?

A: +1 +1 +1 -1 -1 +1 -1 -1

B: +1 +1 -1 +1 -1 -1 -1 +1

C: +1 -1 +1 -1 -1 -1 +1 +1

A+B+C = +3 +1 +1 -1 -3 -1 -1 +1