| Teach Yourself Electricity and Electronics, 5th edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 27 |
| 1. In order to calculate the efficiency Eff of an antenna when we know the
radiation resistance RR and the loss resistance RL, we
must divide the radiation resistance by the total resistance (the sum of the radiation
resistance and the loss resistance). Mathematically, the formula is Eff = RR / (RR + RL) If we want to calculate the efficiency as a percentage, we should use the formula Eff% = 100 RR / (RR + RL) In this example, we've been told that RR = 100 ohms and RL = 50 ohms. Therefore Eff% = 100 x 100 / (100 + 50) rounded off to the nearest percentage point. The correct choice is (d). |
| 2. In this situation, we have RR = 12 ohms and RL
= 36 ohms. Therefore Eff% = 100 x 12 / (12 + 36) rounded off to the nearest percentage point. The correct choice is (b). |
| 3. A receiving loopstick antenna exhibits its greatest sensitivity off the sides of the coil, that is, in a plane perpendicular to the coil axis. The correct choice is (b). |
| 4. We can calculate the physical span of a half wavelength in free space using the
formula Lft = 492 / fo where Lft represents the straight-line distance in feet, and fo represents the frequency in megahertz. If we set fo = 25.0 MHz, then we have Lft = 492 / 25.0 The correct choice is (a). |
| 5. If we construct an antenna with a straight wire radiator, the length Lft
(in feet) at a design frequency fo (in megahertz) for a center-fed,
half-wavelength dipole equals approximately Lft = 467 / fo If we want the antenna to work at fo = 25.0 MHz, then Lft = 467 / 25.0 The correct choice is (b). |
| 6. In free space, the radiation resistance of a center-fed, straight wire antenna decreases as we make it shorter than a full wavelength, as we can see in the graph of Fig. 27-1B. All of our choices here are less than a full wavelength, so the correct answer is the smallest value, which appears at (a). |
| 7. When we examine the graph of Fig. 27-1A, we'll see that the radiation resistance of a base-fed vertical antenna, starting out at very short heights, increases as the height approaches a half wavelength. Of the choices given here, (d) represents the antenna height at which the radiation resistance is the greatest. |
| 8. If we want to construct a quarter-wavelength vertical antenna over ground and feed
it at the bottom, we can calculate its height Lm, in meters, using the
formula Lm = 75v / fo where v equals the velocity factor and fo equals the design frequency in megahertz. We've been told that v = 93%, or 0.93. We also know that the design frequency is 18 MHz. Therefore, we have Lm = 75 x 0.93 / 18 which we can round off to 3.9 m. The correct choice is (c). |
| 9. If we want to get the highest possible efficiency from a base-fed vertical antenna over lossy ground, the single most important thing we can do is maximize its radiation resistance. We can do that by cutting the antenna to a height of a half wavelength (or close to it). The correct choice is (c). You might suspect that choice (d) has some merit. In a sense it does, because we'll have to employ an antenna tuning network of some sort to make a half-wavelength vertical antenna work well with a common transmission line such as coaxial cable. However, unless we maximize the radiation resistance by tailoring the physical height of the antenna, all other measures (including the use of a transmatch) won't do us much good. |
| 10. A circular waveguide must measure at least 0.6 free-space wavelength in diameter
to perform reasonably well. Recall the formula for determining a quarter wavelength in
free space, given the frequency in megahertz: Lm = 75 / fo where Lm represents the straight-line distance in meters, and fo represents the frequency in megahertz. If we want to find the span of a full wavelength, we can multiply 75 in the above equation by 4, getting Lm = 300 / fo If we want to input the frequency in gigahertz instead of in megahertz, then we should divide 300 in the preceding formula by 1000, getting Lm = 0.30 / fo Now we can input the frequency, which we're told equals 7.5 GHz. That gives us a free-space wavelength, in meters, of Lm = 0.30 / 7.5 which is 4.0 cm. Six-tenths of that span is 4.0 x 0.6 = 2.4 cm, represented by choice (b). That's the minimum diameter that a circular waveguide must have if we want it to work reasonably well at 7.5 GHz. |
| 11. A J pole constitutes a modified Zepp, oriented vertically and constructed so that the feed line runs straight down from the bottom end of the radiating element. The correct answer is (b). |
| 12. In a parasitic array, we normally make the reflecting element a little longer than the driven element, so that the reflector resonates at a slightly lower frequency. The correct answer is (c). |
| 13. If we mount a short vertical antenna at the earth's surface where the soil conducts poorly, and if we want the system to have reasonable efficiency nevertheless, we must put a lot of effort into minimizing the ground loss resistance. Only one practical solution exists among the choices available here: Install a large number of ground radials. The correct answer is (d). |
| 14. A quarter-wavelength ground-plane antenna with four quarter-wavelength horizontal radials has a feed-point impedance of about 37 ohms, purely resistive. The correct choice is (a). |
| 15. Let's recall the formula for determining antenna efficiency Eff%
in percent: Eff% = 100 RR / (RR + RL) where RR represents the radiation resistance and RL represents the loss resistance. We've been told that RL equals three-quarters (75%) of the total resistance RR + RL. Therefore, RR represents only one-quarter (25%) of RR + RL. We have a fraction in which the numerator is 25% of the denominator, so the value of the whole fraction must equal 25%. We've just figured out that Eff% = 25%. The correct choice is (a). |
| 16. In a parasitic array, we always connect the driven element directly to the transmission line. Any existing directors, as well as the reflector, constitute parasitic elements, which we never connect directly to a transmission line. The correct choice is (a). |
| 17. The loop, as described here, is a small loop. This type of antenna receives best from directions in the plane perpendicular to the loop axis. That's the plane in which the loop actually lies. Choice (a) is the right answer. |
| 18. Ground-plane antennas are designed for use with coaxial-cable transmission lines. The correct choice is (c). Some amateur radio operators (myself included) have used parallel-wire transmission lines with ground-plane antennas. Such systems work fairly well, but parallel-wire line is meant for use with balanced antennas such as dipoles or loops, not with unbalanced antennas. The ground-plane constitutes an unbalanced antenna. In my experimental system, I observed significant feed-line radiation (an undesirable condition) because of the balanced-to-unbalanced design mismatch. Poor design notwithstanding, I made a few ham-radio contacts with that system! |
| 19. A hobbyist can construct a folded dipole (a), a ground plane (b), or a closed loop (c) and get it to transmit signals efficiently at 7 MHz. However, a dish antenna at 7 MHz would be far too bulky and massive for any hobbyist to build! The correct choice is (d). |
| 20. An isotropic antenna exhibits an effective power loss with respect to a half-wave dipole. All of the other antenna types mentioned here provide some gain with respect to a dipole. The correct choice is (d). |