7.1 Decibels

255

Solutions to Exercises in Chapter 6

Solution to Exercise 6.1 (p. 200)

In both cases, the answer depends less on geometry than on material properties. For coaxial cable, c = 1 1 √µd d . For twisted pair, c = √µ δ 2r cosh−1 + cosh − d 2r 1 d 2r .

Solution to Exercise 6.2 (p. 200)

You can find these frequencies from the spectrum allocation chart (Section 7.3). Light in the middle of the visible band has a wavelength of about 600 nm, which corresponds to a frequency of 5 × 1014Hz. Cable television transmits within the same frequency band as broadcast television (about 200 MHz or 2 × 108Hz). Thus, the visible electromagnetic frequencies are over six orders of magnitude higher!

Solution to Exercise 6.3 (p. 201)

As shown previously (6.11), voltages and currents in a wireline channel, which is modeled as a transmission line having resistance, capacitance and inductance, decay exponentially with distance. The inverse-square law governs free-space propagation because such propagation is lossless, with the inverse-square law a consequence of the conservation of power. The exponential decay of wireline channels occurs because they have losses and some filtering.

Solution to Exercise 6.4 (p. 202)

h1 d1 d2

R R R h2

Figure 6.43

Use the Pythagorean Theorem, (h + R)2 = R2 + d2, where h is the antenna height, d is the distance from the top of the earth to a tangency point with the earth’s surface, and R the earth’s radius. The line-of-sight distance between two earth-based antennae equals dLOS = 2h1R + h1 2 + 2h2R + h2 2 (6.66) As the earth’s radius is much larger than the antenna height, we have to a good approximation that dLOS = √2h1R + √2h2R. If one antenna is at ground elevation, say h2 = 0, the other antenna’s range is √2h1R.

Solution to Exercise 6.5 (p. 202)

As frequency decreases, wavelength increases and can approach the distance between the earth’s surface and the ionosphere. Assuming a distance between the two of 80 km, the relation λf = c gives a corresponding frequency of 3.75 kHz. Such low carrier frequencies would be limited to low bandwidth analog communication and to low datarate digital communications. The US Navy did use such a communication scheme to reach all of its submarines at once.

Solution to Exercise 6.6 (p. 203)

Transmission to the satellite, known as the uplink, encounters inverse-square law power losses. Reflecting off the ionosphere not only encounters the same loss, but twice. Reflection is the same as transmitting exactly what arrives, which means that the total loss is the product of the uplink and downlink losses. The geosynchronous orbit lies at an altitude of 35700 km. The ionosphere begins at an altitude of about 50 km. The amplitude loss in the satellite case is proportional to 2.8 × 10−8; for Marconi, it was proportional to 4.4 × 10−10. Marconi was very lucky.

Solution to Exercise 6.7 (p. 204)

If the interferer’s spectrum does not overlap that of our communications channel—the interferer is out-ofband—we need only use a bandpass filter that selects our transmission band and removes other portions of the spectrum.