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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 ...

11

Having Ethernet and your antenna co-located is not an ideal situation. But then most amateur antenna situations involve compromises. The general idea of the following recommendations is to take as many precautions as practical to minimize the interference possibilities. I recommend that your Ethernet cable to your WAP be a CAT6 shielded cable (STP). While ...

10

An antenna matching network (aka "tuner") does not affect the conditions of the load (antenna) or the transmission line between the load and the matching network. The matching network transforms the impedance "looking into" the transmission line to a more desirable value, typically 50$\Omega$ for ham applications. If the network manages to achieve such a "...

8

For many modulations, the modulation is very slow compared to the propagation delay of the feedline. For example, SSB is typically limited to no more than 4 kHz. That corresponds to a wavelength of of 75 km. As long as the feedline is significantly shorter than this, then the delay due to the feedline is negligible. It may be easier to understand ...

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 ...

7

The SWR is related to the reflection coefficient $\Gamma$: $$\Gamma = {Z_L - Z_0 \over Z_L + Z_0 }$$ $$\text{VSWR} = {1+|\Gamma| \over 1 - |\Gamma|}$$ where: $Z_0$ is the feedline impedance, usually 50 ohms, and $Z_L$ is the load (nominally, the antenna) impedance. The reflection coefficient is a complex number and thus takes into account the phase ...

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

This is a bit like saying, "it's not the fall that kills you, but the sudden stop at the end." Technically, it's not the standing waves per se that cause the damage. However, standing waves imply a mismatched load possibly outside the transmitter's specifications, which implies a potential for excessive heating or voltage in the transmitters, which can lead ...

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

It can be done! Remember that a transmission line consists of some self-inductance per unit length, and some capacitance per unit length, and the ratio of these determines the line's characteristic impedance: $$Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$$ Intuitively then, a lumped element implementation might look like some series inductance and some ...

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

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

There is no such thing as a minimum power level at which a conductor will start radiating. If it radiates at all, then the radiation will be proportional to the power applied. Therefore, the maximum usable power in this circumstance is the power at which such radiation will interfere with (or damage) the receiver, and in order to know it you must test ...

6

If you fed a dipole not with a sine wave but with an impulse, you'd see a wave travel down the feedline, to the feedpoint, and then "ring" several times in the dipole at the dipole's resonant frequency. The ringing doesn't continue forever: on each oscillation this wave loses some energy to radiation. If the feedline is terminated on the non-...

5

Ideally, it doesn't matter where on the constant VSWR circle you are. The transmitter is an ideal voltage source with 0Ω impedance, essentially a short. There's no way to couple power into a short, so whatever power is reflected from the antenna gets reflected back when it reaches the transmitter, so all you need to worry about is the impedance presented to ...

5

At 8W, the power from the radio is 9.03dB. With the cable losses of the receiving station being 1.5dB (1.41W), and the antenna receiving 1.733e-8W, it appears the received signal will never make it to the receiving station's radio. This seems to be the crux of your misunderstanding. You can't convert watts into decibels, or decibels into watts. A ...

5

Given the matched loss of the feedline and the SWR at the transmitter, we can calculate the SWR at the antenna in three simple steps. First convert the SWR at the transmitter to the corresponding magnitude of the reflection coefficient (Gamma), or MRC for short within the context of this answer. The MRC is the magnitude of the complex ratio of the reflected ...

5

VSWR is related to the magnitude of the reflection coefficient $\Gamma$: $$\text{VSWR} = {1+|\Gamma| \over 1-|\Gamma|}$$ The reflection coefficient can be calculated from the load impedance $Z_L$ and the characteristic impedance $Z_0$: $$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$ The reflection coefficient, like the load impedance, a complex number with ...

5

Depends on how much current you mean by no current. But in general, unless the dipole is actually perfectly symmetrical, in a perfectly symmetric environment, above a symmetric ground, with the feed line at exactly 90 degrees all the way into the far field, and/or with the radio equipment and their ground connections also exactly 90 degrees perpendicular ...

5

When making a coax choke balun, does the size of the ferrite matter, and if so, why? Simplest answer: If the choke impedance is low enough to allow some common mode power through, then there is a possibility of overheating. The bigger cores either dissipate heat better or provide higher impedance. Another way to say the same thing: As long as the choke ...

5

So yeah, this image is confusing. There's actually a section on the talk page questioning its correctness. I've considered a few times just deleting it, because it's so damn confusing. The arrows are the AC current which flows through the antenna and the source. I'm not sure we can say that. Notice the arrows are at any instant of equal magnitude at all ...

4

I'm afraid that it is not correct that there is no energy transfer back to the source after an initial period. Wherever you have a mismatch on a transmission line, there is a reflection (at least partial) back to the source. A percentage of power is "actually" reflected back. It can be separated from the transmitted power using a suitable device like a ...

4

Perhaps some of my confusion comes from an audio background, where one has "balanced" and "unbalanced" cabling in a way that is easier for me to understand: an unbalanced cable is simply a signal and ground wire, while a balanced cable actually has a ground wire and two conductors whose voltages are opposite. That's not entirely accurate. What makes a ...

4

Assuming the systems have no internal loss. Reflected power ratio for system 1 is 0.1, for system 2 is 0.01. $= \Gamma^2 = 10^{-RL/10}$ Power transmitted ratio for system 1 is 0.9, for system 2 is 0.99 $= 1-\Gamma^2$ The first reflection back from each system will be the largest and subsequent reflections will be heavily attenuated, so we approximate by ...

4

A "Guanella balun" is really just a common-mode choke. simulate this circuit – Schematic created using CircuitLab For differential-mode currents, the current on one half of the transmission line is equal but opposite the current on the other half. Likewise, the magnetic fields due to these currents is equal but opposite, so they cancel, and so there'...

4

Firstly I would note that your initial statement isn't true. Indeed, the resistance of a wire increases with length, but resistance isn't loss in the same sense of transmission lines where loss is measured in power or a ratio of powers. Consider: simulate this circuit – Schematic created using CircuitLab  I_\text{source} = I_{R1} = I_{R2} = \sqrt{1 ...

4

The first equation describes a wave propagating in a conductor. The second equation describes a wave propagating in a dielectric. Electromagnetic waves don't propagate well in good conductors. If the conductor is good, that means a small electric field in the conductor can result in a large current. With the large current comes large ohmic losses, and thus ...

4

It is helpful to understand the basic functioning of a coaxial cable. But first there are two important phenomenons that must be understood in order to proceed. Skin Effect When direct current (time invariant current) passes through a conductor it tends to uniformly use the entire cross sectional area of the conductor. When alternating current (time ...

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