Electronics Demystified, 2nd edition Stan Gibilisco Explanations for Quiz Answers in Chapter 9 1. In a frequency-modulated (FM) signal, the instantaneous carrier "swings" above and below the unmodulated-carrier frequency. We call the maximum extent of this "carrier swing," usually expressed as a positive-or-negative (±) quantity, the carrier deviation (or simply the deviation as long as we know the context). The correct choice is C. 2. If we want to determine the frequency fMHz (in megahertz) of a radio-frequency (RF) signal in terms of its free-space electromagnetic (EM) wavelength wft (in feet), we use the formulafMHz = 984 / wft In this case, we're told that wft = 12.0, so fMHz = 984 / 12.0 = 82.0 MHz The correct choice is C. 3. At first glance, Fig. 9-17 looks like the arrangement for a push-pull RF amplifier. However, when we scrutinize the labeling for the pairs of terminals, we can see that this circuit is intended as a balanced modulator to obtain double-sideband (DSB) output from an RF carrier and an audio input signal. Unfortunately, there's a major wiring error in the circuit as shown in Fig. 9-17. The transistor collectors should both connect to the bottom end of the output transformer primary winding, not to opposite ends of that winding. The correct choice is D. 4. To make the circuit of Fig. 9-17 perform properly, we must connect the collector of the top transistor to the bottom end of the transformer primary winding. That way, the transistor outputs will appear in parallel with each other, and the currents from both collectors will be forced through the transformer primary winding in order to reach the positive power-supply pole. (If we connect both collectors to the top end of the transformer primary, we'll still get a DSB signal, but it will go straight to the power supply, bypassing the transformer and never appearing at the output terminals.) The correct choice is B. 5. At a frequency of 150 MHz, radio signals don't propagate well in the F2-layer or surface-wave modes. However, tropospheric bending commonly occurs for EM waves at 150 MHz. The correct choice is B. 6. To determine the free-space wavelength wm (in meters) of an RF signal in terms of its frequency fMHz (in megahertz), we use the formulawm = 300 / fMHz In this case, we're told that fMHz = 2.00, so wm = 300 / 2.00 = 150 m The correct choice is A. 7. Figure 9-18 illustrates a form of pulse modulation. The pulse widths, which represent their durations, are all the same. Therefore, we know that this illustration doesn't portray pulse-width modulation (PWM), also known as pulse-duration modulation (PDM). We can rule out choices C and D. This figure doesn't show pulse-code modulation (PCM) because we see no indication of digital restrictions on any of the pulse characteristics; we can rule out choice A. However, there's an obvious variation in the spacing, representing the time interval, between adjacent pairs of pulses. That variation gives away the fact that it's pulse-interval modulation (PIM). The correct choice is B. 8. An FM signal with a deviation of ±5.0 kHz has 4 times the "carrier swing" of an FM signal with a deviation of ±1250 Hz. The easiest way to get a signal with a deviation of ±5.0 kHz from a signal having a deviation of ±1250 Hz is to put the entire signal through a frequency quadrupler. We can build that system using a nonlinear amplifier that's not biased beyond cutoff or pinchoff (class-AB1, class-AB2, or class-B will work), thereby getting an output rich in harmonics. We can tune the output LC circuit to the fourth harmonic of the input signal frequency. That action gives us an output signal whose frequency equals 4 times the input-signal frequency, and whose deviation equals 4 times the input-signal deviation. The correct choice is D. 9. In the amplitude-versus-frequency graph of Fig. 9-19, we see only one sideband, and it lies above the carrier frequency. The carrier is suppressed, so Fig. 9-19 constitutes a so-called spectral display of an upper-sideband (USB) signal. The correct choice is A. 10. In Fig. 9-19, the parameter marked X shows the difference (that is, the amount of frequency span) between the highest signal frequency and the lowest signal frequency. By definition, that's the bandwidth. The correct choice is D.