WEBVTT NOTE Transceiver Design, Filters, and DSP 1 00:00:00.000 --> 00:00:07.920 Making an SSB signal. Two stages, and the pool asks both. 2 00:00:07.920 --> 00:00:13.479 Question. What output is produced by a balanced modulator? 3 00:00:14.379 --> 00:00:18.929 Answer. Double-sideband modulated RF. 4 00:00:19.529 --> 00:00:47.326 A balanced modulator multiplies audio by a carrier and cancels the carrier, leaving two sidebands and no carrier, double sideband suppressed carrier. 5 00:00:47.326 --> 00:00:54.496 Question. What circuit is used to select one of the sidebands from a balanced modulator? 6 00:00:55.396 --> 00:00:58.396 Answer. Filter. 7 00:00:58.996 --> 00:01:16.196 A sharp filter removes one sideband. Balanced modulator plus filter gives single sideband, the carrier cancelled by the balance, the unwanted sideband removed by the filter. Recovering one. 8 00:01:16.196 --> 00:01:20.537 Question. How is a product detector used? 9 00:01:21.437 --> 00:01:28.591 Answer. Used in a single sideband receiver to extract the modulated signal. 10 00:01:29.191 --> 00:01:47.078 A product detector is a mixer supplied with a locally generated carrier at the frequency the transmitter suppressed. Multiply the incoming sideband by that reinserted carrier and the audio comes back out. 11 00:01:47.078 --> 00:02:04.965 This is why an SSB receiver's tuning must be accurate: the reinserted carrier's frequency is the reference, and an error of 100 Hz shifts every audio frequency by 100 Hz. It is why the RIT control exists. 12 00:02:04.965 --> 00:02:06.806 Frequency generation. 13 00:02:06.806 --> 00:02:14.226 Question. Which of the following is characteristic of a direct digital synthesizer (DDS)? 14 00:02:15.126 --> 00:02:21.806 Answer. Variable output frequency with the stability of a crystal oscillator. 15 00:02:22.406 --> 00:02:27.949 A DDS builds a waveform numerically from a single crystal reference. 16 00:02:27.949 --> 00:02:44.822 You get arbitrary frequency in tiny steps with crystal stability, a combination that used to be impossible, and the reason a modern radio can be both accurate and continuously tunable. Filter specifications. 17 00:02:44.822 --> 00:02:58.679 Four terms, each describing a different part of a filter's response, and the pool asks all four. Term, Insertion loss; What it describes, attenuation inside the passband. 18 00:02:58.679 --> 00:03:13.514 Term, Cutoff frequency; What it describes, where output power falls to half the passband level. Term, Ultimate rejection; What it describes, maximum attenuation outside the passband. 19 00:03:13.514 --> 00:03:20.606 Term, Bandwidth; What it describes, between the upper and lower half-power frequencies. 20 00:03:20.606 --> 00:03:27.547 Question. What term specifies a filter's attenuation inside its passband? 21 00:03:28.447 --> 00:03:31.487 Answer. Insertion loss. 22 00:03:32.087 --> 00:03:45.374 Even a good filter costs you something in the band you want. That cost is insertion loss, and it matters most in a receiver's front end, where it adds directly to the noise figure. 23 00:03:45.374 --> 00:03:54.614 Question. What is the frequency above which a low-pass filter's output power is less than half the input power? 24 00:03:55.514 --> 00:03:58.594 Answer. Cutoff frequency. 25 00:03:59.194 --> 00:04:08.754 Half power is −3 dB. The half-power point is the convention for every filter edge, which is why: 26 00:04:08.754 --> 00:04:16.050 Question. The bandwidth of a band-pass filter is measured between what two frequencies? 27 00:04:16.950 --> 00:04:21.830 Answer. Upper and lower half-power. 28 00:04:22.430 --> 00:04:32.230 Both edges at −3 dB. Any bandwidth figure quoted at a different point is measuring something else and should say so. 29 00:04:32.230 --> 00:04:41.230 Question. What term specifies a filter's maximum ability to reject signals outside its passband? 30 00:04:42.130 --> 00:04:45.530 Answer. Ultimate rejection. 31 00:04:46.130 --> 00:05:02.090 How far down the stopband gets, eventually. A filter with steep skirts and poor ultimate rejection lets very strong signals through from far away. Receiver sensitivity. 32 00:05:02.090 --> 00:05:21.413 Question. Which parameter affects receiver sensitivity? A. Input amplifier gain. B. Demodulator stage bandwidth. C. Input amplifier noise figure. D. All these choices are correct. 33 00:05:22.313 --> 00:05:25.833 Answer. All these choices are correct. 34 00:05:26.433 --> 00:05:41.098 All three enter the same calculation. Noise figure is how much the receiver adds; bandwidth is how much noise it admits; antenna gain is how much signal arrives. 35 00:05:41.098 --> 00:05:52.393 A low-noise preamplifier in front of a wide filter gains you less than narrowing the filter does. Digital signal processing. 36 00:05:52.393 --> 00:06:02.636 Question. Which of the following is an advantage of a digital signal processing (DSP) filter compared to an analog filter? 37 00:06:03.536 --> 00:06:09.488 Answer. A wide range of filter bandwidths and shapes can be created. 38 00:06:10.088 --> 00:06:14.927 An analogue filter's response is fixed by its components. 39 00:06:14.927 --> 00:06:32.248 A DSP filter's response is a set of numbers, so one piece of hardware provides every bandwidth from 50 Hz to 3 kHz, with shapes no analogue filter of reasonable cost could achieve. Software-defined radio. 40 00:06:32.248 --> 00:06:45.888 Question. What is the phase difference between the I and Q RF signals that software-defined radio (SDR) equipment uses for modulation and demodulation? 41 00:06:46.788 --> 00:06:49.828 Answer. 90 degrees. 42 00:06:50.428 --> 00:07:05.948 Two versions of the signal, ninety degrees apart, in-phase and quadrature. Together they capture both the amplitude and the phase of the signal, which is complete information about it. 43 00:07:05.948 --> 00:07:14.131 Question. What is an advantage of using I-Q modulation with software-defined radios (SDRs)? 44 00:07:15.031 --> 00:07:22.151 Answer. All types of modulation can be created with appropriate processing. 45 00:07:22.751 --> 00:07:40.431 Because I and Q carry everything, any modulation is a matter of processing rather than circuitry. One hardware front end handles SSB, FM, digital voice, and modes not yet invented. 46 00:07:40.431 --> 00:07:58.734 Question. Which of these functions is performed by software in a software-defined radio (SDR)? A. Filtering. B. Detection. C. Modulation. D. All these choices are correct. 47 00:07:59.634 --> 00:08:03.154 Answer. All these choices are correct. 48 00:08:03.754 --> 00:08:18.156 That is the definition. What was a chain of tuned circuits becomes a program, and the hardware reduces to converting between RF and numbers as accurately as it can. 49 00:08:18.156 --> 00:08:35.983 The SDR lab in the practicum track is where this stops being abstract. Check yourself. What two stages generate an SSB signal from audio, and what does each remove? A filter is specified as 2.4 kHz wide. 50 00:08:35.983 --> 00:08:44.940 Between which two points? Why can one SDR front end handle modes that did not exist when it was built? 51 00:08:50.940 --> 00:09:04.305 A balanced modulator cancels the carrier, leaving double sideband; a filter removes the unwanted sideband. The upper and lower half-power (−3 dB) frequencies. 52 00:09:04.305 --> 00:09:16.740 It captures I and Q, 90 degrees apart, which together contain the signal's full amplitude and phase information. Everything after that is software.