21

Dispelling the Myth To begin with, the typical HF SWR meter does not have the ability to separately sample the forward and reverse power, voltage, or current. Any description of the device or its circuitry that suggests this capability is flawed. We can show this empirically with two different experiments. Experiment 1 Connect a 100 ohm resistor directly ...


13

In the absence of common-mode currents, then the optimum feedline length is 0, because a longer feedline only increases your feedline losses. These losses are due to the resistance of the wire, dielectric losses, etc. and are specified in dB per unit length in the coax datasheet. At VHF and up, these losses can be significant even at car lengths, especially ...


13

There's nothing inherently wrong with “looping up” extra coaxial cable. In fact, a neatly wound coil of coax can function as an air-core choke balun (“ugly balun”) which is useful for some antenna systems when located at the feed point of the antenna. The only disadvantage is the extra loss in the coax from the extra length. Loss in a coaxial cable, when ...


10

"RG" stems from old, obsolete military specifications, standing for "radio guide", and the number is actually arbitrary. Part of the reason they are obsolete is that they weren't specific enough - RG-6 can reasonably be assumed to have an 18 gauge solid center conductor, but beyond that the dielectric, velocity factor, or completeness of the shielding ...


10

The difference is the "Velocity factor". A 36cm long physical coax wire of this type is electrical 54.4cm long. Different types of wire have different velocity factors. https://en.wikipedia.org/wiki/Velocity_factor


9

Your question seems as much about psychology as much as technical concerns. We mainly favor the technical questions, but I'll take a stab at the psychological aspects also. All of you, please feel free to disagree with my conclusions! Coaxial cable is fairly inexpensive for many, compared to our time, even for LMR400. (Your mileage may vary.) For many, ...


8

Executive Summary Assuming: air dielectric (insulator) 50Ω characteristic impedance Then for round coax, make the inside diameter of the outside conductor 2.302 times larger than the diameter of the inside conductor. If the shield is square, and the inner conductor is still round, make the inside length of one side of the shield 2.134 times larger than ...


8

Balanced lines (of which twisted pair is a special type) really have an upper frequency limit; you can't use them to transport 1 GHz (well, you can, but the smallest variation in direction or distance would have catastrophic effects, and the conductor distance would get pretty small). This can be seen in technical practice: 100 Mbit/s Ethernet (Fast ...


8

Our friends in the UK and other parts of the world are now wondering how you could even begin to solder a PL259 connector with a torch (aka flashlight)! But in their vernacular, you of course are referring to a burning torch. In general, when you heat a metallic object with the hopes of applying solder, the heat will cause oxidation to form on the metallic ...


8

I will need a UHF/VHF diplexer on either end to suitably merge/split the signals from each antenna Yes, this is correct. A tangent: If you wanted to save some money by using mass-market parts, you could use 75 Ω power dividers (coax splitters) instead of diplexers. This has 3 dB loss because the signals are not directed exclusively to the intended ...


8

For practical purposes, all frequencies. If you cut a transmission line into infinitesimal segments, each segment can be modeled as: simulate this circuit – Schematic created using CircuitLab ($G$ is conductance, the inverse of resistance) The characteristic impedance is: $$ Z_0 = \sqrt{ R + j\omega L \over G+j\omega C } $$ ($\omega$ is the angular ...


7

The most important thing about the cable is how much loss are you willing to accept in the feed line. Figure out how much cable you will need, and then determine what the loss on said cable will be. As I mentioned on my website, here's a few good rules of thumb: Keep the power loss to no more than 3 db in the cable. If you can, use the same impedance as ...


7

It looks like you're looking for an intuitive, practical understanding rather than precise definitions, so I'll see what I can do with that, with my own recent learning. The reason you care about impedance matching is that impedance mismatches cause the signal to be partially reflected — some of the energy is going the opposite direction than you want it to ...


7

I'll explain the operation of that balun very briefly: for the differential mode (which by definition has equal but opposite currents on each conductor), each conductor induces an equal but opposite magnetic flux in the core. These magnetic fluxes cancel, and so the differential mode sees no inductance: it's as if the core and the windings aren't there. ...


7

The electrically shortened antennas often found on HTs are not simple pieces of wire, but coils: By Shootthedevgru at English Wikipedia, CC BY-SA 3.0 The coil adds inductance over the length of the antenna, making it appear electrically longer than it is physically. Without the inductance, the antenna would need to be approximately a quarter-wavelength ...


7

Terminating unused ports will never make things worse, and indeed is necessary to provide "ideal" behavior. Is ideal behavior really necessary? It depends on the application. Is your setup working now? If so, terminators aren't necessary :-) A typical splitter has these properties: It has an input port and two output ports, which we'll call A and B. It has ...


7

No, adding ferrite beads to choke currents on the outside of the shield of a coaxial cable does not affect its impedance or velocity factor. Impedance and velocity factor are determined by the inside construction of the cable: the outside diameter of the inner conductor, d, the inside diameter of the outer conductor (shield), D, and the magnetic permeability,...


7

Think of it first without the feedline: The half-wavelength of transmission line provides a 180 degree phase shift, and the (short) connection between the shield ends is "ground". That means it's at 0V relative to the environment, and also the voltage at the open ends of the coax center conductor will be equal in magnitude but opposite in sign relative to ...


7

My name is Alex, I'm the head of technical support at RigExpert. This is actually an interesting question. Our engineers could not give an exact answer why the schedule in the second case behaves strangely. I can offer to perform another experiment - to measure the parameters of the RG-8/U coax cable using the analyzer and the AntScope and AntScope2 software....


6

What I have done once for a dipole was use a SO-239 panel receptacle with a solder cup to make my own "coax to two wire" adapter. For example, I used something like this part from Mouser. That way, I had a whole piece of coax running to the antenna that could be weatherproofed, and then an exposed section that just had regular wire soldered to the receptacle....


6

The source of the SWR limit on the transmitter end is the losses in the feedline. In general, the higher the matched line loss, the lower the maximum SWR that will be present at the transmitter end. Since the SWR limit is the result of losses in the feedline, the efficacy of the 'high SWR fix' must be considered. The mechanism has to do with the trips that ...


6

It appears to be typical moisture ingress. The joint between the coax and the connector is not inherently waterproof. A damaged jacket can also be the source of moisture ingress. Normally the connector to coax joint must be covered with a self sealing adhesive tape or heat shrink tubing with an interior thermal adhesive in order to protect it from the ...


6

When the SWR is 1:1, the matched line loss of ordinary ladder line is lower than the matched line loss of ordinary coax because at HF, most of the loss is $I^2R$ loss, and the current magnitude is inversely proportional to the characteristic impedance of the feedline. Of course, a high SWR results in higher standing wave currents and thus increases the $I^2R$...


6

For well-designed coax, the EM fields are confined to the space between the inside of the braid and the center conductor, i.e. the dielectric insulation region which affects the velocity factor. Therefore, those beads have negligible effect on the differential signals. They do have an effect on the common-mode signals on the outside of the braid which has ...


6

According to this online coax loss calculator, the loss of 50 feet of new Belden 9258 RG-8X with an antenna with an SWR of 1.5 is 4.29 dB at 450 MHz (UHF), so the loss for 100 feet of the same coax would be 8.6 dB. If you switched to fancier coax such as LMR-400, the loss for 100 feet of coax would be 2.8 dB. So if you bought another 50' of RG-8X your coax ...


6

You need a switch. Connecting two transceivers directly to one antenna is a great way to destroy one of them — as soon as you transmit with one, the receive circuit of the other one will see several million times more power than it's designed to handle. If you only wanted to use 2m on the ham rig then you could probably use a diplexer to send the 2m ...


5

I'd recommend one of two solutions: 1. Don't throw it out the window. Most lightning damage comes not from direct strikes but nearby strikes, which can still induce large-ish voltages on the feedline: not enough to make lightning, but enough to damage things. A couple feet of air between the feedline and the tranciever will protect against this. If you do ...


5

Baluns are designed to be transformers (like 1:1 4:1, 6:1, etc.) or choke baluns, and both. For an antenna, the purpose of a choke balun is to create a high-impedance to common mode currents that would flow on the outside of coaxial cable shielding. These common mode currents can cause all kinds of problems such as RF in the shack, matching problems, and ...


5

So, you say: I don't care if the signals are delayed. I'm just tapping the line out to a pan adapter. I am most concerned about influencing the existing function of the receiver And that's exactly what you're going to do with this. Your Opamp circuit has a high impedance, which translates into "looks like an open end". That means that your tapping coax ...


5

OK, let me try to answer this, but this answer may also be qualified as unqualified. If you have a 50 Ohm receiver, and connect a perfectly (Z=R) 75 ohm antenna system, then your VSWR would be 1.5, and the "load mismatch attenuation" will be about 0.177dB. (with antenna system I include feedline) I doubt that you would actually notice this. However, you ...


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