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“Max. data rate = 2B log 2 V bits/sec”, we can sample = 2 (6MHz) log 2 (4) = 24 million times/sec. four-level signals = 24 Mbps.

2. 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?

S/N = 10^2 = 100. So max data rate = (3000Hz)log2(1+100)bits/sec = 3000 * 6.643 = 19.92kbps

3. When the data-rate in T1 is 1.544 Mbps, what signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?

B = 50,000 Hz.    50,000 log 2 (1 + S/N) = 1.544 × 10^6 => log 2 (1 + S/N) = 30.88 => S/N = (2^30.88) -1 => In dB, S/N = 10 log10 (S/N) = 10 log10 ((2^30.88) -1) = 92.95 dB.

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

Fiber has many advantages over copper. It can handle much higher bandwidth than copper. It is not affected by power surges, electromagnetic interference, power failures, or corrosive chemicals in the air. It does not leak light and is quite difficult to tap. Finally, it is thin and lightweight, resulting in much lower installation costs. There are some downsides of using fiber over copper. It can be damaged easily by being bent too much. And fiber interfaces cost more than electrical interfaces.

5. Radio antennas often work best when the diameter of the antenna is equal to the wavelength of the radio wave. Reasonable antennas range from 1 cm to 5 meters in diameter. What frequency range does this cover?

c = 3x 10^8 m/s.   For lamba =1 cm, we get 30 GHz, since, f = c/ lamba(3 x10^8 m/s) /0.01m = 30 x 10^9 /s = 30GHz.      For lamba =5 m, we get 60 GHz, since, f = c/ lamba(3 x10^8 m/s) /5m = 60 x 10^6 /s = 60GHz.  so range is from 60 MHz to 30 GHz.

6. . A simple telephone system consists of two end offices and a single toll office to which each end office is connected by a 1-MHz full-duplex trunk. The average telephone is used to make four calls per 8-hour workday….

Since the average telephone is used to make four calls per 8-hour workday, each telephone makes (4/8) or, 0.5 calls/hour at 6 minutes each. telephone occupies a circuit for 3 minutes/hour,we can have 20 telephone/hours or, twenty telephones can share a circuit, although having the load be close to 100%. Since 10% of the calls are long distance, it takes 200 telephones to occupy a long-distance circuit full time. interoffice trunk has 1,000,000/4000 = 250 circuits multiplexed onto it. With 200 telephones per circuit, an end office can support 200 × 250 = 50,000 telephones.

9. A modem constellation diagram (see Fig. below) has data points at the following coordinates: (1, 1), (1, −1), (−1, 1), and (−1, −1). How many bps can a modem with these parameters achieve at 1200 symbols/second?

there will be 4 different distinct symbols, therefore, each symbol can indicate 2 bits. In this way, 1200symbols/second will be equivalent to (1200 × 2) or, 2,400 bits/second or, 2,400bps.

10. 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.

ten 4000Hz signals. we need nine guard bands to avoid any interference. the minimum bandwidth required is (4000 × 10 + 400 × 9) or, 43,600 Hz.

11. Why has the PCM sampling time been set at 125 μsec?

A sampling time of 125 μsec (micro-sec) corresponds to 8000 samples per second, because, in 1 second (or, in 10^6 μsec) we sample (10^ 6 /125) or, 8000.   Max. data rate = 2B log2 V bits/sec = 2 × 4,000 × log2 (2) bits/sec = 2 × 4000 × 1 = 8,000.

12. 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?

the end users get 8 × 24 = 192 of the 193 bits in a frame.  overhead is 1 / 193 = 0.5%. so at least 0.5% of the 1.544 Mbps are not delivered to the end user.

13. Three packet-switching networks each contain n nodes. The first network has a star topology (see the Figures) with a central switch, the second is a (bbidirectional) ring, and third is Fully interconnected…..?

13. In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in an adjacent cell. If 840 frequencies are available, how many can be used in a given cell?

Each cell has six neighbors. If the central cell uses frequency group A, its six neighbors can use B, C, B, C, B, and C, respectively. In other words, only three unique cells are needed. Consequently, each cell can have (840/3) or, 280 frequencies.

14. Sometimes when a mobile user crosses the boundary from one cell to another, the current call is abruptly terminated, even though all transmitters and receivers are functioning perfectly. Why?

Frequencies cannot be reused in adjacent cells, so when a user moves from one cell to another, a new frequency must be allocated for the call. If a user moves into a cell, all of whose frequencies are currently in use, the user’s call must be terminated.

16. Consider a different way of looking at the orthogonality property of CDMA chip sequences. Each bit in a pair of sequences can match or not match. Express the orthogonality property in terms of matches and mismatches. 

When two elements match, their product is +1. When they do not match, their product is -1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match

17. A cable company decides to provide Internet access over cable in a neighborhood consisting of 5000 houses. The company uses a coaxial cable and spectrum allocation allowing 100 Mbps downstream bandwidth per…Describe what the cable company needs to do to provide this guarantee.

A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a fiber node.