1a. Ground must be lifted at at least one end. You can always isolate both ends, but lifting the transmitter (unless via transformer) is likely the hardest, so I'll assume receiver instead.
To be clear, this device will be both a receiver and a transmitter. So it will be easiest to lift the ground at the amplifier itself rather than the GPSDO source or the test equipment destination(s).
Also, to be clear, by "lift" I mean anything other than hard earth grounding between the cable and circuit ground. As noted, the "lift" may be conditional e.g. over a modest voltage range, or with some resistance, and in any case resuming a low impedance at AC.
The box itself should be grounded, for reasons as given, and also to set a ground on the transmitting sides. I don't mean "lift" in the sense of physically cutting a ground pin, as it often means in unfortunately-too-common parlance!
Unless you mean something bidirectional in which case you may need a more involved solution, but I understand a "distribution amplifier" to have one input and multiple outputs. In which case, that one input can have all the signal conditioning needed, and the rest is hard grounded, easy peasy.
I agree that a coax feedthrough cap would be ideal here. There is a picture in Ott's book of an XLR connector with a ring of radially connected ceramic caps that I find fascinating. I'm not sure I understand how your alternative solution provides similar function though. The shield would have to be bypassed to the chassis, but how would that connection be made without considerable inductance from component leads?
Yes, like that.
You do it without component leads, obviously!
Or enough in parallel that it's enough not to care.
Ceramic chips are cheap and affordable, and seam connections can be made between panels and PCBs with EMI gaskets.
It also doesn't necessarily have to be coronal, it just needs to surround the signal trace well enough to provide shielding. For example, consider a connector at the end of a rectangular PCB. The center pin transitions to a trace which goes along the board (preferably on an inner layer, surrounded by planes). The whole end of the board can be slotted across (except for the trace), with bypass caps stacked across the ground slot, top and bottom (preferably) to provide the AC bypass with DC isolation. Then the ground return (for EMI purposes) path mates to earth-ground somewhere just past the bypass caps, while the signal proceeds over, now being surrounded by the local ground potential.
Descriptions are probably ineffective, and anyway Ott provides diagrams; they should match this description, no? Or do you have any particular concerns with them, or just need help explaining like, equivalent circuits of geometries or whatever..?
I hadn't really thought about transformers like this in the past, but it is neat that (for the 1:1 turns ratio case) they really do just act like a unity gain differential amplifier. The voltage difference across the primary winding appears across the secondary. Connect one end of the secondary to ground, and you have a differential to single-ended conversion.
Yes. Well, it's not amplifying, there's no isolation, you need a terminator on the load side and you want to avoid weird impedances, nonlinearities or backfeeding up the line -- there are of course always two waves on the line, at all times, one in each direction -- more it's an extension of the line itself; which is what's particularly emphasized with
transmission line transformers, as having predictable characteristics and relatively easy design.
It's also not perfectly differential, that's what the shields are for of course; coax isn't balanced so we need to take care of that. If it were twisted pair, we'd use a single winding with no shield (or one shield for an unbalanced output), and indeed that's what Ethernet for example does. They also add a CMC (just a bunch of twisted pair done up on a choke core) for better immunity.
A key insight is that a twisted-pair transformer is an isolation transformer
either way you wire it.
Obviously, as a CMC, it doesn't isolate down to DC, but over a given bandwidth, it has at least some specified CM impedance, and so allows a degree of freedom with relatively little current flow. The upper bandwidth however can be essentially infinite, limited by the spacing of the turns of transmission line, core characteristics, and the transmission line itself (as a finite width TL has some cutoff frequency, above which non-TEM00 modes appear). Note that CM bandwidth is limited, while signal bandwidth is "infinite".
As a transformer proper, it isolates down to DC (CM), but signal bandwidth is limited. In particular, in a similar way that the CMC upper cutoff is limited by geometry and losses, the signal bandwidth here is limited by winding length. (We can model the transmission line as two ideal ports with some CM impedance between them; in a given instant, the ports have TL impedance, so each port acts in series between the source and load. The equivalent circuit is one loop: source, Rs, port 1, RL, load, port 2. After one propagation delay, the ports receive each others' signals, and interference occurs; this causes a notch at the 1/2 wave frequency and harmonics. It stands to reason that, if we had a lossy transmission line, i.e. that was terminated in the middle, we could get unlimited bandwidth, if we also don't mind 6dB insertion loss.) The difference in relevant parasitics is due to the even/odd mode the signal is applied in: one case the transmission line is simply inline between source and load, the other it's a 1/2 wave stub across them.
Even more generally, we can model the transformer as a 4-port where each pin is a port with respect to some reference plane, and by considering pairs and phases of those ports, we can understand more deeply the two cases above, and which parasitics dominate which modes.
I am definitely not up to the task of winding an electrostatically shielded transformer for the input and each of the outputs. Is the degradation in CMRR due solely to the interwinding capacitance in the transformer?
As above, it's a bit more complicated than that, but that is the LF asymptotic case, yes. At transitional frequencies, you also have to worry about the delayed common mode. That is: drive both primary pins high; instantly, the secondary pins both go up, then one delay later, the propagated waves bring them back down. For an ideal terminated TL, the step response is a rect pulse. So, the impulse response is a complementary pair of deltas, so the frequency response is a sine wave starting at zero (so, rising ~proportionally from there, hence ~capacitive CM response), peaking at the resonant frequency, then valleys and peaks at harmonics.
Tim