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G8B

SIGNALS AND EMISSIONS

- Frequency changing; bandwidths of various modes; deviation; intermodulation

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G8B011 of 13

Which mixer input is varied or tuned to convert signals of different frequencies to an intermediate frequency (IF)?

Why In a superheterodyne receiver the mixer produces the sum and difference of the two frequencies fed into it, and the IF filter is built for one fixed frequency. Since the IF stage cannot move, the local oscillator is tuned so that the difference between it and the desired incoming signal always equals that fixed IF. Tuning the radio across the band is really just tuning the local oscillator: LO = signal frequency plus or minus the IF.
Watch out The RF input is whatever the antenna delivers, so you do not vary it to select a station; the beat frequency oscillator works later in the chain to make CW and SSB audible, and the image frequency is an unwanted response, not an input you tune.
Fixed IF, moving LO: turning the dial turns the local oscillator, not the IF.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B022 of 13

What is the term for interference from a signal at twice the IF frequency from the desired signal?

Why In a superheterodyne receiver the mixer combines the incoming signal with a local oscillator, and any signal that differs from the LO by the IF will pass through the IF filter. Besides the desired frequency there is a second one, offset from the desired signal by exactly twice the IF, that also produces the same IF output, so it gets heard right along with the wanted station. That unwanted response is called the image, and front-end selectivity (or a higher first IF) is what rejects it.
Watch out Quadrature response sounds technical but refers to signals 90 degrees apart in phase, as used in some detectors, not to an off-frequency signal sneaking through the mixer; 'mixer interference' and 'intermediate interference' are not standard terms.
Image = 2 x IF away from the desired signal. Picture a mirror image on the far side of the LO.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B033 of 13

What is another term for the mixing of two RF signals?

Why When two RF signals are applied to a nonlinear device such as a mixer, the output contains new frequencies: the sum and the difference of the two inputs (plus the originals and harmonics). That process is called heterodyning, and it is the basis of superheterodyne receivers, where an incoming signal is mixed with a local oscillator to produce an intermediate frequency.
Watch out Synthesizing is tempting because frequency synthesizers do generate signals, often using mixers and dividers, but the term refers to creating a desired output frequency from a reference, not to the act of combining two signals.
Hetero = different, dyne = power: two different frequencies combine to make sum and difference.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B044 of 13

What is the stage in a VHF FM transmitter that generates a harmonic of a lower frequency signal to reach the desired operating frequency?

Why A frequency multiplier is a stage deliberately driven into nonlinearity so its output is rich in harmonics, and a tuned circuit at the output selects the desired harmonic (2x, 3x, etc.) of the input. Classic VHF FM transmitters generate the signal at a low frequency where a crystal oscillator and reactance modulator work well, then multiply up to the operating frequency. A useful side effect is that the frequency deviation is multiplied by the same factor, so a small deviation at the oscillator becomes the full 5 kHz deviation on the air.
Watch out A mixer is the tempting pick, but mixing combines two different input frequencies to produce sum and difference products, not a harmonic of a single input. A reactance modulator is what applies the FM in the first place.
Multiplier = multiply the frequency (and the deviation) by a whole number; mixer needs two inputs, multiplier needs one.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B055 of 13

Which intermodulation products are closest to the original signal frequencies?

Why Mixing two signals f1 and f2 produces products at combinations like 2f1-f2 and 2f2-f1 (third order, an odd order). If f1 and f2 are close together, those odd-order products land just above and below the original pair, inside the same band or passband. Even-order products like f1+f2 or f2-f1 land near twice the operating frequency or down near the difference, far away from the originals and easy to filter out. That is why third-order intermodulation is the spec that matters for receiver and amplifier linearity.
Watch out Second harmonics are a form of even-order product, appearing at twice the input frequency rather than near it; intercept point is a measurement of how strong the distortion products are, not a frequency location.
Odd order = odd man close by. 2f1-f2 sits right next door; even-order products run far away.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B066 of 13

What is the total bandwidth of an FM phone transmission having 5 kHz deviation and 3 kHz modulating frequency?

Why FM bandwidth is estimated with Carson's rule: bandwidth = 2 x (peak deviation + highest modulating frequency). Here that is 2 x (5 kHz + 3 kHz) = 16 kHz. The factor of 2 appears because the sidebands spread out on both sides of the carrier.
Watch out The choice that says 8 kHz is just the sum of deviation and modulating frequency, forgetting to double it for both sidebands.
Carson: add deviation + audio, then double. 5 + 3 = 8, times 2 = 16 kHz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B077 of 13

What is the frequency deviation for a 12.21 MHz reactance modulated oscillator in a 5 kHz deviation, 146.52 MHz FM phone transmitter?

Why In a multiplier-chain FM transmitter, the deviation is multiplied by exactly the same factor as the carrier frequency. Here 146.52 MHz / 12.21 MHz = 12, so the oscillator only needs 5000 Hz / 12 = 416.7 Hz of deviation to produce 5 kHz at the output.
Watch out The choice of 5 kHz assumes deviation passes through the multipliers unchanged, but multiplying frequency by 12 also multiplies the instantaneous frequency swing by 12; 60 kHz is the result of multiplying instead of dividing.
Find the multiplication factor (146.52/12.21 = 12), then divide the final deviation by it: 5 kHz / 12 = 417 Hz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B088 of 13

Why is it important to know the duty cycle of the mode you are using when transmitting?

Why Duty cycle is the fraction of time a transmitter is putting out full power during a transmission, and it determines the average power the final amplifier, power supply and antenna components must dissipate as heat. SSB voice has a low duty cycle (roughly 20-25 percent) because power only peaks on voice syllables, while FM, RTTY, and digital modes like FT8 key the transmitter at full output continuously, a 100 percent duty cycle. Running a continuous-carrier mode at the same power setting you use for SSB can overheat the finals and exceed the rig's average power rating, so many manuals tell you to reduce power for those modes.
Watch out Overmodulation is an audio drive level problem, not a thermal one, and tuning or listening for break-in calls has nothing to do with how long the carrier is present.
Duty cycle = heat. FM/RTTY/FT8 are 100 percent on; back the power down or cook the finals.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B099 of 13

Why is it good to match receiver bandwidth to the bandwidth of the operating mode?

Why Receiver noise power is proportional to the bandwidth you let through, so a filter wider than the signal admits extra noise and adjacent interference while adding no more desired signal. Matching the filter to the mode's bandwidth (about 2.4 kHz for SSB, 500 Hz or less for CW, a few hundred Hz for many digital modes) passes all of the signal and rejects the rest, giving the best signal-to-noise ratio. Too narrow a filter is also bad because it chops off part of the signal and distorts it.
Watch out FCC rules set limits on transmitted bandwidth and symbol rates, not on how wide your receiver filter must be, so the 'required by rules' choice is wrong; receiver filtering has nothing to do with antenna impedance or power consumption.
Noise comes in with bandwidth: wider filter = more noise, same signal. Fit the filter to the mode.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B1010 of 13

What is the relationship between transmitted symbol rate and bandwidth?

Why A digital signal's occupied bandwidth grows in proportion to how fast the symbols change. Each symbol transition is a change in the waveform, and faster changes create sidebands spaced further from the carrier, so doubling the symbol rate roughly doubles the necessary bandwidth. That is why slow modes like PSK31 (about 31.25 symbols per second) fit in under 100 Hz, while fast packet or high-speed data modes need many kilohertz.
Watch out The choice saying bandwidth is half the symbol rate inverts the Nyquist idea: the minimum bandwidth needed is on the order of the symbol rate, not half of it, and real signals with filtering and keying sidebands take somewhat more.
Faster keying, fatter signal. Symbol rate up, bandwidth up.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B1111 of 13

What combination of a mixer's Local Oscillator (LO) and RF input frequencies is found in the output?

Why A mixer is a nonlinear device that multiplies two signals together, and the trigonometric identity for multiplying two sine waves produces components at the sum and the difference of the two input frequencies. So feeding in an RF signal and a local oscillator gives outputs at LO + RF and LO - RF, and a filter then selects whichever one you want as the IF. This is the basis of superheterodyne receivers and transmitters: a 14.2 MHz signal mixed with a 5.0 MHz LO yields 19.2 MHz and 9.2 MHz.
Watch out The arithmetic product is tempting because the mixer literally multiplies the two waveforms in the time domain, but multiplying in time does not create a signal at the product of the frequencies; in the frequency domain it creates the sum and difference.
Mixers multiply in time but add and subtract in frequency: out comes LO+RF and LO-RF.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B1212 of 13

What process combines two signals in a non-linear circuit to produce unwanted spurious outputs?

Why Any non-linear device (an overdriven amplifier, a corroded joint, a transistor biased into curvature) acting on two or more signals generates sums and differences of their frequencies and harmonics. When that happens where it was not intended, the resulting mix products are spurious emissions or spurious responses, and the process is called intermodulation. Third-order products at 2F1-F2 and 2F2-F1 are the troublesome ones because they land close to the original signals and fall inside the passband.
Watch out Heterodyning is the same physics used deliberately, a mixer combining a signal with a local oscillator to shift frequency; the pool reserves that term for the wanted result, while intermodulation names the unwanted one. Detection is recovering modulation from a carrier, and rolloff describes how a filter's response falls off past its cutoff.
Same physics, different intent: heterodyning is mixing you wanted, intermodulation is mixing you did not.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
G8B1313 of 13

Which of the following is an odd-order intermodulation product of frequencies F1 and F2?

Why The order of an intermodulation product is the sum of the absolute values of the coefficients on the two frequencies. For 2F1 - F2 that is 2 + 1 = 3, an odd (third-order) product. Third-order products are the ones that matter most in practice because they land very close to the original signals and fall inside the receiver or transmitter passband.
Watch out The choice with 3F1 - F2 adds up to 3 + 1 = 4 and 5F1 - 3F2 adds up to 5 + 3 = 8, so both are even-order products even though the individual coefficients look odd. Even-order products usually fall far from the originals and are easier to filter out.
Add the coefficients, not look at them: 2+1=3 is odd. 2F1-F2 is the classic third-order troublemaker.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
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