1. TV channels are 6 MHz wide. How many b/s can be sent if four-level digital signals are used? Assume a noiseless channel.
Using Nyquist theory, we can sample 12 million times/sec. Four level signals provide 2 bits per sample. This gives a total data rate of 24 Mbps.
2. It is desired to send a sequence of computer screen images over an optical fiber. The screen is 480x460 pixels, each pixel being 24 bits. There are 60 screen images per second. How much data rate is needed?
480x460x24x60 bps, which is 318 Mbps.
3. Is Nyquist theory true for optical fiber, or only for copper wire?
Yes for both. It has nothing to do with technology or transmission media.
4. 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?
lf = c, for l = 1cm and c = 3x108 m/s, we get 30 GHz. For l = 5 m, we get 60 MHz. Thus the band covered is 60 MHz to 30 GHz.
5. Why has the PCM sampling time been set at 125 µsec?
This corresponds to a sampling rate of 8000 samples/sec. According to the Nyquist theory, this is the sampling frequency needed to capture all the information in 4 kHz channel.
6. Compare the maximum data rate of a noiseless 4-kHz channel using: a) Analog encoding with 2 bits per sample. b) The T1 PCM system.
In both cases 8000 samples/sec are possible. With dibit encoding, two bits are sent per sample. With T1, 7 bits are sent per per sample. The respective data rate are 16 kbps and 56 kbps.
7. What is the difference, if any, between the demodulator part of a modem and the coder par of a codec? (After all, both convert analog signals to digital ones).
A coder accepts an arbitrary analog signal and generates a digital signal form it. A demodulator accepts a modulated sine wave only and generate a digital signal.
8. Compute the bit rate for a 1000-baud 16-QAM signal.
A 16 QAM signal means that there are four bits per signal element since 2**4 = 16. Thus, 1000x4= 4000bps.
9. Compute the baud rate for a 72,000 bps 64-QAM signal.
A 64 QAM signal means that there are six bits per signal element. Thus, 72,000/6=12,000 baud.
10. Explain the difference between data rate and baud rate?
Data rate is the capacity of a channel in bits per second. Baud rate represents the number of times the line condition (i.e., frequency, amplitude or phase) changes each second. data rate = n bits x baud rate, where n is equal to the number of bits one signal represents.
11. 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 7 x 24 = 168 of the 193 bits in a frame. The overhead is 25/193= 13%
12. How many lines can a TSI switch handle if the RAM access time is 50 nsec?
If one switch has n lines, it takes 100 ns (R+W) to switch each frame. Suppose we have a frame time of 125 µsec, we get 0.1n = 125 or n = 1250 lines.
13. How many bits of RAM buffer does a TSI switch need if the input line samples are 10 bits and there are 80 input lines?
The buffer must contain 80 x 10 = 800 bits.
14. Does time division switching necessarily introduce a minimum delay at each switching stage? If so, what is it?
Yes. The entire frame must be stored before transmitted, since the last slot in may be the last slot out. The the delay is one frame time. For T1, it is 125 µsec.
15. How many crosspoints are needed if we use a crossbar switch to connect 1000 phones in a small town?
Assume a full duplex lines, then we have (1000x1000 - 1000)/2 = 499,500 crosspoints. We are assuming there are no self-connections.
16. Why does fiber-optic cable have greater capacity than copper-based media?
Fiber is immune to EMI. Fiber supports higher frequencies with out signal reflections or noise. Fiber frequencies can be multiplexed to increase bandwidth considerably.
17. Compare and contrast MMF with SMF.
See class notes.
18. Compare and contrast GEO and LEO satellite communications systems.
GEO satellites orbit at altitude of 22,000 miles. Speed of satellite is same as of earth. Total global coverage with 8 satellites. LEO satellites orbit at altitude of 300 to 1200 miles. LEO satellite speed is much greater than GEO. Total global coverage takes 48 satellites.
