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E8C

SIGNALS AND EMISSIONS

Digital signals: digital communication modes; information rate vs. bandwidth; error correction; constellation diagrams

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E8C011 of 15

What is Quadrature Amplitude Modulation or QAM?

Why QAM builds a signal from two carriers at the same frequency that are 90 degrees apart in phase, the in-phase (I) and quadrature (Q) components. Each carrier's amplitude is varied independently, and their sum produces a single signal whose amplitude and phase together encode several bits per symbol. Plotting the possible I and Q amplitude pairs gives the familiar constellation diagram, for example 16 points for 16-QAM carrying 4 bits per symbol.
Watch out The choice about shifting carrier and modulation phase describes the phasing method of generating SSB, which also uses 90 degree phase shifts but produces a single sideband analog signal, not a multi-level digital symbol scheme. The compression answers confuse modulation with data reduction; QAM adds no compression.
Quadrature = 90 degrees. Two carriers, I and Q, amplitudes varied, and the constellation dots show the result.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C022 of 15

What is the definition of symbol rate in a digital transmission?

Why Symbol rate, also called baud rate, counts how many times per second the transmitted waveform changes state (its amplitude, phase, or frequency) to carry information. Each of those states is a symbol, and depending on the modulation a single symbol can carry one bit (BPSK) or several bits (QPSK carries 2, 16-QAM carries 4). So bit rate equals symbol rate times the number of bits per symbol, and the two are equal only for simple binary keying.
Watch out Characters per second describes throughput or the character rate, which is further removed still, since overhead bits, framing, and error correction coding all sit between the symbols on the air and the characters delivered to the user.
Baud = changes per second on the air, not bits or characters. Bits = baud x bits per symbol.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C033 of 15

Why should the phase of a PSK signal be changed at the zero crossing of the RF signal?

Why Any abrupt change in a waveform creates high-frequency components, and the more abrupt the discontinuity the wider the resulting spectrum. If a PSK transmitter flips phase while the RF waveform is at a voltage peak, the signal jumps instantly from a large positive value to a large negative one, producing sidebands far from the carrier. Switching at the moment the waveform passes through zero makes the transition continuous in amplitude, so the occupied bandwidth stays narrow and the signal does not splatter onto neighboring frequencies.
Watch out Carrier suppression is a property of the balanced modulator or the chosen emission type, not of the timing of the phase reversals, and zero-crossing timing actually adds complexity to the modulator rather than simplifying it, so the all-of-these choice fails too.
Flip phase where the waveform is already at zero: no voltage step, no splatter, narrow bandwidth.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C044 of 15

What technique minimizes the bandwidth of a PSK31 signal?

Why PSK31 shifts phase 180 degrees to send data, and an abrupt phase reversal is a step function that splashes energy far outside the nominal bandwidth. To avoid that, PSK31 shapes each symbol with a smooth sinusoidal (raised-cosine style) amplitude envelope, taking the carrier down through zero as the phase flips. The result is a signal that fits in roughly 31 Hz with very little splatter, at a symbol rate of 31.25 baud.
Watch out Reed-Solomon is a real forward error correction code, but it is used in other systems, not as PSK31's bandwidth control; PSK31's character coding scheme is Varicode, and character coding affects throughput, not occupied bandwidth.
Sharp edges splatter, smooth curves don't: PSK31 rounds its pulses sinusoidally to stay about 31 Hz wide.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C055 of 15

What is the approximate bandwidth of a 13-WPM International Morse Code transmission?

Why For on-off keyed CW the bandwidth depends on the keying speed, not on the transmit frequency. The standard estimate is bandwidth = baud rate times a fading factor K, where K is about 5 for hand-sent Morse. A 13 WPM signal has a keying rate of roughly 10.4 baud (about 0.8 times the words per minute), so 10.4 x 5 gives approximately 52 Hz.
Watch out The choices equal to the WPM number or twice it come from just reading the speed off the question; they ignore the fading factor of 5 that the standard CW bandwidth estimate applies.
CW bandwidth is roughly 4 times the WPM: 13 WPM x 4 = 52 Hz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C066 of 15

What is the bandwidth of an FT8 signal?

Why FT8 is an 8-tone FSK mode with 6.25 Hz spacing between tones and a symbol rate of 6.25 baud. Eight tones spaced 6.25 Hz apart occupy 8 x 6.25 = 50 Hz, so each FT8 signal takes up only about 50 Hz of spectrum. That narrowness is why dozens of FT8 signals can share one SSB-wide slice of a band.
Watch out The 2.4 kHz figure is the width of the SSB receiver passband used to hear FT8, not the width of a single FT8 signal; many individual 50 Hz signals fit inside it.
8 tones x 6.25 Hz spacing = 50 Hz. FT8: eight tones, fifty hertz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C077 of 15

What is the bandwidth of a 4,800-Hz frequency shift, 9,600-baud ASCII FM transmission?

Why For FSK the pool uses the approximation bandwidth = baud rate + (1.2 x frequency shift). Here that is 9,600 + (1.2 x 4,800) = 9,600 + 5,760 = 15,360 Hz, or 15.36 kHz. The shift term accounts for the spread between the mark and space frequencies, while the baud term accounts for the sidebands created by how fast the signal keys.
Watch out The choice of 5.76 kHz is just the 1.2 x shift portion and the choice of 9.6 kHz is just the baud rate; both leave out half the formula, and 4.8 kHz is simply the shift itself.
Bandwidth = baud + 1.2 x shift. Both terms, always. 9600 + 5760 = 15.36 kHz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C088 of 15

How does ARQ accomplish error correction?

Why ARQ stands for Automatic Repeat reQuest. The receiver checks each received block against a checksum or CRC, and if the block fails the check, the receiver sends a NAK (or withholds the ACK) so the sender transmits that block again. This is error detection plus retransmission, not true forward error correction, so it needs a reliable return path and works best on links where the sender can hear the receiver, as in packet or PACTOR.
Watch out The choices about special binary or polynomial codes describe forward error correction (FEC), where redundant bits such as Hamming, Reed-Solomon or convolutional codes let the receiver rebuild the lost data with no reply from the sender at all.
ARQ = Automatic Repeat reQuest: "say that again." FEC fixes it locally; ARQ asks for a resend.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C099 of 15

Which digital code allows only one bit to change between sequential code values?

Why Gray code is an ordering of binary values arranged so that any two consecutive values differ in exactly one bit position. That property matters in shaft encoders, analog-to-digital converters and in mapping bits onto constellation points, because if the receiver or sensor picks the adjacent value by mistake, only one bit is wrong instead of several. In QAM and PSK constellation diagrams, Gray mapping means a symbol error between neighboring points causes just a single bit error, which greatly helps the error-correction decoder.
Watch out Binary Coded Decimal encodes each decimal digit as four bits, and EBCDIC and Extended ASCII are character sets for representing text; none of them controls how many bits change from one value to the next.
Gray = one bit changes at a time, so a neighbor slip costs only one bit error.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1010 of 15

How can data rate be increased without increasing bandwidth?

Why Bandwidth is set mainly by the symbol (baud) rate, not by how many bits each symbol carries. If you use a more efficient code, for example a higher-order modulation that packs more bits into each symbol, or a source code that wastes fewer bits, you move more data per second at the same symbol rate and therefore the same occupied bandwidth. Constellation diagrams show this directly: going from 4 points to 16 points doubles bits per symbol with no change in symbol rate.
Watch out Forward error correction is tempting, but FEC adds redundant check bits, so at a fixed symbol rate it lowers the net user data rate; it buys reliability, not throughput. Analog-to-digital resolution affects how finely an analog source is digitized, which actually creates more bits to send, not a faster channel.
Bandwidth follows baud, not bits. Pack more bits per symbol with a better code and the bandwidth stays put.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1111 of 15

What is the relationship between symbol rate and baud?

Why Baud is simply the unit for symbol rate: one baud equals one symbol (one signaling state change) per second. So symbol rate and baud rate are two names for the same quantity. Note this is not the same as bit rate, since a single symbol can carry several bits when multi-level modulation such as QPSK or QAM is used.
Watch out The choices offering a factor of two are confusing baud with bit rate; with QPSK, for example, the bit rate is twice the baud rate because each symbol carries 2 bits, but the baud rate itself never changes definition.
Baud = symbols per second, always. Bits per second is what varies with the code.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1212 of 15

What factors affect the bandwidth of a transmitted CW signal?

Why On-off keying is a form of amplitude modulation, so the transmitted CW spectrum is set by how fast and how abruptly the carrier is switched. Faster keying means more sidebands: bandwidth is roughly the baud rate times a shape factor K, with K near 3 for soft, well-rounded keying and near 5 for hard keying. A typical 25 WPM signal with proper rise and fall shaping occupies only about 50 to 60 Hz, while sharp, square keying edges spread energy far wider as key clicks.
Watch out IF bandwidth and Q describe the selectivity of a receiver or filter, not what the transmitter radiates, and modulation index applies to FM or PM, not to on-off keyed CW. Output power changes how loud the signal is, not how wide it is.
CW bandwidth = speed x shape. Click-free keying means soft rise/fall; BW is about 4 times the WPM-derived baud rate.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1313 of 15

What is described by the constellation diagram of a QAM or QPSK signal?

Why A constellation diagram is a polar/IQ plot where each point represents one symbol the modulator can transmit. The angle of a point from the origin is its carrier phase and the distance from the origin is its amplitude, so the whole set of points shows every phase/amplitude combination the scheme uses. QPSK has 4 points (4 phases, one amplitude, 2 bits per symbol); 16-QAM has 16 points spread over several amplitudes and phases, carrying 4 bits per symbol. In practice the plot is also used to judge signal quality, since noise and distortion smear the points into fuzzy clouds.
Watch out It does not count carriers or error-correction bits; a constellation applies to a single modulated carrier, and FEC overhead is a property of the protocol, not the modulation plot. Frequency response would be shown on a spectrum display, not an IQ plot.
Constellation = I/Q dots: angle is phase, distance is amplitude, one dot per symbol.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1414 of 15

What type of addresses do nodes have in a mesh network?

Why A mesh network is a data network in which each node routes traffic for its neighbors, forming a self-healing web of links. Amateur mesh systems such as AREDN and HSMM-Mesh run standard TCP/IP networking over Wi-Fi hardware retuned to ham bands, so every node is identified by an Internet Protocol address just like any computer on a LAN. Routing protocols then use those addresses to find the best path through the mesh.
Watch out Talk groups belong to DMR digital voice systems, where they select which conversation a repeater carries; they are not network node addresses. Email addresses and trust servers relate to other services (Winlink uses a trust/authentication concept), not to mesh node identification.
Mesh = a ham-band computer network, so nodes get IP addresses, just like any router on the Internet.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
E8C1515 of 15

What technique do individual nodes use to form a mesh network?

Why A mesh network is self-organizing: each node runs protocols that discover neighboring nodes within radio range, establish links with them, and share routing information so traffic can hop node to node. That automatic discovery and link establishment is what distinguishes a mesh from a fixed, manually configured network. Amateur mesh systems such as AREDN use exactly this approach, with routing protocols exchanging neighbor and route data continuously.
Watch out Store-and-forward digipeating describes older packet radio repeating, where a station simply retransmits a frame addressed through it; that alone gives no automatic route discovery. Forward error correction and Viterbi decoding are link-layer error control, not network formation, and code plugs belong to commercial trunked land mobile radios.
Mesh nodes must first find each other: think discover, link, route.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
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