WEBVTT NOTE Software-Defined Radio Internals 1 00:00:00.000 --> 00:00:01.880 Direct sampling. 2 00:00:01.880 --> 00:00:08.680 Question. What is meant by "direct sampling" in software defined radios? 3 00:00:09.580 --> 00:00:19.620 Answer. Incoming RF is digitized by an analog-to-digital converter without being mixed with a local oscillator signal. 4 00:00:20.220 --> 00:00:30.469 No mixer, no intermediate frequency, no local oscillator. The antenna feeds an ADC and everything after it is arithmetic. 5 00:00:30.469 --> 00:00:46.900 It is the architecture of most modern HF SDRs, and it removes every image-response and mixer-spur problem the superheterodyne has, at the cost of demanding an extremely good converter. Sampling. 6 00:00:46.900 --> 00:00:54.180 Question. How frequently must an analog signal be sampled to be accurately reproduced? 7 00:00:55.080 --> 00:01:01.743 Answer. At least twice the rate of the highest frequency component of the signal. 8 00:01:02.343 --> 00:01:19.735 The Nyquist criterion. Sample at less than twice the highest frequency and the information is not merely degraded, it is aliased, appearing as a different, plausible-looking, entirely wrong frequency. 9 00:01:19.735 --> 00:01:33.095 Question. What aspect of receiver analog-to-digital conversion determines the maximum receive bandwidth of a direct-sampling software defined radio (SDR)? 10 00:01:33.995 --> 00:01:36.595 Answer. Sample rate. 11 00:01:37.195 --> 00:01:59.213 Directly from Nyquist: bandwidth is at most half the sample rate. A 122.88 MHz converter covers up to about 61 MHz, which is why HF direct-sampling receivers are practical now and microwave ones are not. Bit depth. 12 00:01:59.213 --> 00:02:21.642 What is the minimum number of bits required to sample a signal with a range of 1000 volts? Answer: 10 bits, for 1-volt resolution, 2 to the power of 10 = 1024 ≥ 1000. Bit depth sets dynamic range, roughly 6 dB per bit. 13 00:02:21.642 --> 00:02:35.635 A 16-bit converter gives about 96 dB; a 24-bit one about 144. That is why bit depth is the specification that matters on a crowded band: 14 00:02:35.635 --> 00:02:47.555 Question. What sets the minimum detectable signal level for a direct-sampling software defined receiver in the absence of atmospheric or thermal noise? 15 00:02:48.455 --> 00:02:54.375 Answer. Reference voltage level and sample width in bits. 16 00:02:54.975 --> 00:03:10.495 Together they set the size of one least-significant bit, the smallest change the converter can represent, and nothing smaller can be received at all. Decimation and aliasing. 17 00:03:10.495 --> 00:03:14.895 Question. What is the function of decimation? 18 00:03:15.795 --> 00:03:21.915 Answer. Reducing the effective sample rate by removing samples. 19 00:03:22.515 --> 00:03:35.854 Having sampled the whole HF spectrum, you probably want 3 kHz of it. Decimation throws away samples to reduce the rate to something the rest of the processing can handle. 20 00:03:35.854 --> 00:03:42.010 Question. Why is an anti-aliasing filter required in a decimator? 21 00:03:42.910 --> 00:03:52.190 Answer. It removes high-frequency signal components that would otherwise be reproduced as lower frequency components. 22 00:03:52.790 --> 00:04:09.445 The filter must come before the decimation. Reducing the sample rate lowers the Nyquist limit, so signals that were fine at the original rate are now above it, and they will fold down into your passband as aliases. 23 00:04:09.445 --> 00:04:22.131 Filter first, then decimate. Getting that order wrong produces signals that are not there, which is the most confusing class of bug in DSP. Transforms and filters. 24 00:04:22.131 --> 00:04:28.145 Question. What function is performed by a Fast Fourier Transform? 25 00:04:29.045 --> 00:04:34.845 Answer. Converting signals from the time domain to the frequency domain. 26 00:04:35.445 --> 00:04:50.565 Samples in, spectrum out. The FFT is what draws the waterfall, and it is why an SDR displays a whole band at once where a superheterodyne shows one frequency. 27 00:04:50.565 --> 00:04:58.325 Question. What type of digital signal processing filter is used to generate an SSB signal? 28 00:04:59.225 --> 00:05:03.025 Answer. A Hilbert-transform filter. 29 00:05:03.625 --> 00:05:10.410 Question. Which method generates an SSB signal using digital signal processing? 30 00:05:11.310 --> 00:05:16.897 Answer. Signals are combined in quadrature phase relationship. 31 00:05:17.497 --> 00:05:34.153 The phasing method, done in software. A Hilbert transform shifts every frequency component by exactly 90 degrees; combining the original and the shifted version in quadrature cancels one sideband. 32 00:05:34.153 --> 00:05:39.337 No physical filter, and the sideband choice is a sign change. 33 00:05:39.337 --> 00:05:50.057 Question. What kind of digital signal processing audio filter is used to remove unwanted noise from a received SSB signal? 34 00:05:50.957 --> 00:05:54.195 Answer. An adaptive filter. 35 00:05:54.795 --> 00:06:07.131 Adaptive, it adjusts its own response continuously based on what it is receiving, which is what lets it distinguish a steady tone or hiss from speech. 36 00:06:07.131 --> 00:06:16.835 It is also why pushing it too hard distorts the wanted signal, from the General controls lesson. FIR filters and taps. 37 00:06:16.835 --> 00:06:23.315 Question. What is the function of taps in a digital signal processing filter? 38 00:06:24.215 --> 00:06:30.095 Answer. Provide incremental signal delays for filter algorithms. 39 00:06:30.695 --> 00:06:44.215 A FIR filter is a chain of delays, each output multiplied by a coefficient and the results summed. Each delay-and-coefficient stage is a tap. 40 00:06:44.215 --> 00:06:53.388 Question. Which of the following would allow a digital signal processing filter to create a sharper filter response? 41 00:06:54.288 --> 00:06:57.568 Answer. More taps. 42 00:06:58.168 --> 00:07:13.392 More taps, more coefficients, more precise control of the response, at the cost of processing and latency. It is the DSP equivalent of adding sections to an analogue filter. 43 00:07:13.392 --> 00:07:22.312 Question. Which of the following is generally true of Finite Impulse Response (FIR) filters? 44 00:07:23.212 --> 00:07:31.155 Answer. FIR filters can delay all frequency components of the signal by the same amount. 45 00:07:31.755 --> 00:07:40.934 Linear phase. Every frequency is delayed equally, so the waveform's shape is preserved rather than smeared. 46 00:07:40.934 --> 00:07:55.346 Analogue filters generally cannot do this, and it is why a DSP filter can be very sharp without the ringing a comparably sharp analogue filter produces. Check yourself. 47 00:07:55.346 --> 00:08:11.731 You want to receive up to 30 MHz by direct sampling. Minimum sample rate? You decimate a signal by 8 without filtering first. What happens? Why does a FIR filter's linear phase matter for CW? 48 00:08:17.731 --> 00:08:31.300 At least 60 MHz, twice the highest frequency component. Everything above the new Nyquist limit aliases into your passband, appearing as signals that are not there. Filter before decimating. 49 00:08:31.300 --> 00:08:42.931 It delays every frequency component equally, so keying transitions keep their shape rather than ringing, which is what a sharp non-linear-phase filter does to CW.