WEBVTT NOTE Digital Signals: Rate, Bandwidth, and Codes 1 00:00:00.000 --> 00:00:04.040 Symbol rate, baud, and bit rate. 2 00:00:04.040 --> 00:00:10.383 Question. What is the definition of symbol rate in a digital transmission? 3 00:00:11.283 --> 00:00:17.266 Answer. The rate at which the waveform changes to convey information. 4 00:00:17.866 --> 00:00:24.306 Question. What is the relationship between symbol rate and baud? 5 00:00:25.206 --> 00:00:28.366 Answer. They are the same. 6 00:00:28.966 --> 00:00:42.766 Baud is symbols per second, not bits per second. They are equal only when each symbol carries one bit. This is the distinction the next question depends on: 7 00:00:42.766 --> 00:00:49.006 Question. How can data rate be increased without increasing bandwidth? 8 00:00:49.906 --> 00:00:54.946 Answer. Using a more efficient digital code. 9 00:00:55.546 --> 00:01:14.561 Bandwidth is set by symbol rate. Put more bits in each symbol and the data rate rises while the bandwidth does not. QPSK carries 2 bits per symbol, 16-QAM carries 4, 64-QAM carries 6, same bandwidth, six times the data. 10 00:01:14.561 --> 00:01:31.666 The price is noise margin: more states packed into the same signal space means the states are closer together and harder to tell apart. That trade is the whole subject of digital modulation design. 11 00:01:31.666 --> 00:01:38.334 Question. What is Quadrature Amplitude Modulation or QAM? 12 00:01:39.234 --> 00:01:50.034 Answer. Transmission of data by modulating the amplitude of two carriers of the same frequency but 90 degrees out of phase. 13 00:01:50.634 --> 00:02:05.408 Two quadrature carriers, each amplitude-modulated, the I and Q of the SDR lesson, used to carry data. Every combination of I and Q amplitude is a distinct symbol. 14 00:02:05.408 --> 00:02:13.790 Question. What is described by the constellation diagram of a QAM or QPSK signal? 15 00:02:14.690 --> 00:02:20.530 Answer. The possible phase and amplitude states for each symbol. 16 00:02:21.130 --> 00:02:39.009 Plot I horizontally and Q vertically and each valid symbol is a point. A clean signal shows tight dots; noise spreads them into clouds; distortion warps the pattern into recognisable shapes. 17 00:02:39.009 --> 00:02:55.758 A constellation diagram is a diagnostic instrument, and a blurred one tells you the link is close to failing before the errors appear. Bandwidth figures. There is a diagram here. 18 00:02:55.758 --> 00:03:03.192 Horizontal bars on a logarithmic frequency axis from 100 hertz to 10 megahertz. 19 00:03:03.192 --> 00:03:23.799 PSK31 occupies about 31 hertz, FT8 about 50 hertz, CW about 150 hertz, single sideband voice about 3 kilohertz, FM voice on VHF repeaters between 10 and 15 kilohertz, and analogue fast-scan television about 6 megahertz. 20 00:03:23.799 --> 00:03:38.196 Signal, 13 WPM Morse; Bandwidth, approximately 52 Hz. Signal, FT8; Bandwidth, 50 Hz. Signal, 9600-baud ASCII FM with 4800 Hz shift; Bandwidth, 15.36 kHz. 21 00:03:38.196 --> 00:03:44.916 Question. What factors affect the bandwidth of a transmitted CW signal? 22 00:03:45.816 --> 00:03:52.136 Answer. Keying speed and shape factor (rise and fall time). 23 00:03:52.736 --> 00:04:08.343 Both. Faster keying needs more bandwidth, as always, but so does sharper keying. A hard-edged keying envelope is rich in harmonics, from the Fourier lesson, and those harmonics are bandwidth. 24 00:04:08.343 --> 00:04:25.667 Which is why a 13 WPM signal is about 52 Hz rather than the few hertz its raw information rate would suggest, and why the keying-shape question recurs in the next lesson as key clicks. Narrow-bandwidth technique. 25 00:04:25.667 --> 00:04:33.762 Question. Why should the phase of a PSK signal be changed at the zero crossing of the RF signal? 26 00:04:34.662 --> 00:04:38.302 Answer. To minimize bandwidth. 27 00:04:38.902 --> 00:04:53.674 A phase change at a zero crossing is a smooth transition; one at a peak is a discontinuity, and discontinuities are broadband. Same principle as CW keying shape. 28 00:04:53.674 --> 00:05:01.074 Question. What technique minimizes the bandwidth of a PSK31 signal? 29 00:05:01.974 --> 00:05:06.375 Answer. Use of sinusoidal data pulses. 30 00:05:06.975 --> 00:05:24.015 Rather than switching abruptly between states, PSK31 shapes each transition as a raised cosine. No sharp edges, no harmonics, and the mode fits in 31 Hz. Codes. 31 00:05:24.015 --> 00:05:31.856 Question. Which digital code allows only one bit to change between sequential code values? 32 00:05:32.756 --> 00:05:35.287 Answer. Gray code. 33 00:05:35.887 --> 00:05:41.283 In a Gray code, adjacent values differ in exactly one bit. 34 00:05:41.283 --> 00:05:57.283 That matters wherever a value is read while it is changing, a shaft encoder, or a flash ADC, because a single-bit change cannot produce a wildly wrong intermediate reading. 35 00:05:57.283 --> 00:06:08.167 Ordinary binary going from 0111 to 1000 changes four bits at once, and a reading caught mid-change could be anything. 36 00:06:08.167 --> 00:06:12.971 Question. How does ARQ accomplish error correction? 37 00:06:13.871 --> 00:06:19.631 Answer. If errors are detected, a retransmission is requested. 38 00:06:20.231 --> 00:06:29.984 Detect and ask again, the General distinction from forward error correction, which adds redundancy instead. Mesh networks. 39 00:06:29.984 --> 00:06:36.196 Question. What type of addresses do nodes have in a mesh network? 40 00:06:37.096 --> 00:06:41.185 Answer. Internet Protocol (IP). 41 00:06:41.785 --> 00:06:48.217 Question. What technique do individual nodes use to form a mesh network? 42 00:06:49.117 --> 00:06:53.859 Answer. Discovery and link establishment protocols. 43 00:06:54.459 --> 00:07:14.510 An amateur mesh is a real IP network. Nodes discover their neighbours and establish links automatically, and routing follows, which is what makes the network self-healing without anyone configuring it. Check yourself. 44 00:07:14.510 --> 00:07:32.620 A mode uses 16-QAM at 1200 baud. What is its bit rate, and how does its bandwidth compare with QPSK at 1200 baud? Your constellation diagram shows fuzzy blobs instead of dots. What does that mean? 45 00:07:32.620 --> 00:07:37.979 Why is 13 WPM CW about 52 Hz wide rather than a few hertz? 46 00:07:43.979 --> 00:08:03.454 16-QAM carries 4 bits per symbol, so 4800 bit/s. Bandwidth is set by symbol rate, so it is the same as QPSK at 1200 baud, which carries only 2400 bit/s. Noise, spreading each symbol's received position. 47 00:08:03.454 --> 00:08:17.627 The link is degrading and will start producing errors before the picture becomes unreadable. Bandwidth depends on keying speed and rise/fall shape. 48 00:08:17.627 --> 00:08:26.979 The sharp edges of the keying envelope contribute harmonics well beyond the raw information rate.