2.A noiseless 8-kHz channel is sampled every 1 msec. What is the maximum data rate?Solution:According to: Maximum data rate=: 2B log2 V bits/secAssume: log2 V=4the maximum data rate =2Hlog2 V b/s=2*8k*4=64kbpsAssume: log2 V=16the maximum data rate =2Hlog2 V b/s=2*8k*4=256kbpsSo, the maximum data rate of the noiseless channel is decided by how many bits there is every sample.3. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20 dB,what is the maximum achievable data rate?Solution: According to Shannons theorem:the maximum data rate= B log2 (1 + S/N) According to : 10lgS/N = 20 dB So:S/N =100 log2 1016.658So: the maximum data rate = B log2 (1 + S/N)3k*6.658=19.98Kbps4. What signal-to-noise ratio is needed to put a T1 carrier on a 100-kHz line?Solution: According to Shannons theorem:the maximum data rate= B log2 (1 + S/N)T1 carriers bandwidth= 1544000 = 100000log2(1+S/N)So: log2(1+S/N)=15.4415S/N= 215110lg (2151)=45dB5. (P124)What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?Answer: Fiber optics can handle much higher bandwidths than copper. This alone would require its use in high-end networks. Due to the low attenuation, repeaters are needed only about every 50 km on long lines, versus about every 5 km for copper, resulting in a big cost saving. Fiber also has the advantage of not being affected by power surges, electromagnetic interference, or power failures. Nor is it affected by corrosive chemicals in the air, important for harsh factory environments. It is thin and lightweight. For new routes, fiber wins hands down due to its much lower installation cost. Finally, fibers do not leak light and are difficult to tap. These properties give fiber good security against potential wiretappers.On the downside, 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 unidirectional, two-way communication requires either two fibers or two frequency bands on one fiber. Finally, fiber interfaces cost more than electrical interfaces.Table (come from 袁子超)PropertyFiberWires(copper)DistanceLong(tens of km)Short(100s of m)BandwidthVery HighModerateCostExpensiveCheapConvenienceHardEasySecurityHard to tapEasy to tapInterferenceNoYes17. (P146)What is the minimum bandwidth needed to achieve a data rate of B bits/sec if the signals transmitted using NRZ, MLT-3, and Manchester encoding? Explain.Solution:With NRZ, the signal may cycle between the positive and negative levels up to every 2 bits (in the case of alternating 1s and 0s). This means that we need a bandwidth of at least B/2 Hz when the bit rate is B bits/sec.In MLT-3, the signal(+,0,-) completes a cycle at most every 3 bits, thus requiring at least B/3 Hz to achieve B bits/sec data rate.Manchester encoding mix the clocks signal with the data signal by XORing them together so that no extra line is needed. So it requires twice as much bandwidth as NRZ because of the clock.(2B).21. A modem constellation diagram similar to Fig. 2-23 has data points at (0, 1) and (0, 2).Does the modem use phase modulation or amplitude modulation?Answer: Amplitude modulation, because amplitude is variable.22. What is the maximum bit rate achievable in a V.32standard modem if the baud rate is 2400 and no error correction is used?Solution: according to C=B*log2Vno error correction is used, So: C=2400* log232 =12000bps23. In a constellation diagram, all the points lie on a circle centered on the origin. What kind of modulation is being used?Answer:If all the points lie on a circle centered on the origin, so they have the same amplitude, so no use of amplitude modulation. In the constellation never use frequency modulation, so, here the encoding is a pure phase modulation.25. Ten signals, each requiring 4000 Hz, are multiplexed onto a single channel using FDM. What is the minimum bandwidth required for the multiplexed channel? Assume that the guard bands are 400 Hz wide.Solution:There needed a 400HZ between two sub-channel ,so: 4000 * 10 + 400 * 9 = 43600 Hz26.Why has the PCM sampling time been set at 125 sec?Solution: (mainly from 周粵嫻)A sampling time of 125 sec(1000000s/8000) corresponds to 8000 samples per second. According to the Nyquist theorem, this is the sampling frequency needed to capture all the information in a 4 kHz channel, such as a telephone channel.27.What is the percent overhead on a T1 carrier? That is, what percent of the 1.544 Mbps are not delivered to the end user? How does it relate to the percent overhead in OC-1or OC-768 lines?Solution：T1:193-24*7=25 25/19312.95%OC-1: (51.84-49.536)/51.844.44%On OC-768 line，the percent overhead is same as OC-1 line.28. Compare the maximum data rate of a noiseless 4-kHz channel using:(a) Analog encoding (e.g., QPSK) with 2 bits per sample.(b) The T1 PCM system.Solution:(a) log2 V=2the maximum data rate=2Hlog2 V=2*4k*2=16kbps(b) log2 V=7 (1bit for control)the maximum data rate=2Hlog2 V=2*4k*7=56kbps30. (P165)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.)Answer:The coder of a codec to accept arbitrary analog signal and produce digital signal. While the demodulator of a modem only receiving modulated sine wave ( or cosine ), to produce a digital signal.(Pay attention: codec=coder + decoder; modem=modulator + demodulator)33. In Fig. 2-40, the user data rate for OC-3 is stated to be 148.608 Mbps. Show how this number can be derived from the SONET OC-3 parameters. What will be the gross,SPE, and user data rates of an OC-3072 line?Solution:(1)The first three columns of each frame are reserved for system management information, so the OC-1s SPE is 87*9*8*800=50.112Mbps. The first column of the SPE is the path overhead (i.e., the header for the end-to-end path sublayer protocol),so the user data is (87-1)*9*8*800=49.536MbpsSo the OC-3s user data is 49.536*3=148.608Mbps(2)The OC-1s gross is 90*9*8*800=51.84MbpsSo the OC-3072s gross=51.87*3072=159252.48Mbpsthe OC-3072s SPE=50.112*3072=153944.064Mbpsthe OC-3072s user data=49.536*3072=152174.592Mbps34.What is the available user bandwidth in an OC-12c connection?Solution: (from 袁子超 and reference key)As an aside, when a carrier, such as OC-3, is not multiplexed, but carries the data from only a single source, the letter c (for concatenated) is appended to the designation, so OC-3 indicates a 155.52-Mbps carrier consisting of three separate OC-1 carriers, but OC-3c indicates a data stream from a single source at 155.52Mbps. so there is only one path overhead.The OC-12c frames are 12*90 = 1080 columns of 9 rows. Of these,12*3 = 36 columns are taken up by line and section overhead. This leaves an SPE of 1044 columns. One SPE column is taken up by path overhead, leaving 1043 columns for user data. So, the available user bandwidth is:1043*9*8*8000=600.768Mbps41. Suppose that A, B, and C are simultaneously transmitting 0 bits, using a CDMA system with the chip sequences of Fig. 2-28(a). What is the resulting chip sequence?Solution：A+B+C=(-3 -1 -1 +1 +3 +1 +1 -1)Because they transmitting 0 bits, so the negative of its chip sequence =(+3 +1 +1 -1 -3 -1 -1 +1)Or: (from 周粵嫻)(+1 +1 +1 -1 -1 +1 -1 -1)( +1 +1 -1 +1 -1 -1 -1 +1)(+1 -1 +1 -1 -1 -1 +1 +1)(+3 +1 +1 -1 -3 -1 -1