# Tag Info

34

But how come unbalancedness of coaxial cables have no problem in impedance balancing issues? The beautiful thing about coax is that the shield shunts mostly all external electric field interference to ground and the inner wire is largely unaffected. For an external magnetic field interference, a subtle thing happens; the current that flows in the shield ...

22

EM energy can travel through a waveguide in different modes, i.e. different ways for the electrical and magnetic standing waves to fit inside of the confining cavity. Normally propagation in coax is in the TEM00 fundamental mode, which means that there are no standing waves perpendicular to the direction of propagation. However, if the wavelength used is ...

14

Andy talks about how coax works in general, but another point is that video generally doesn't have the same SNR requirements as audio to begin with. Data with 8 to 10 bits per color channel provides very good pictures, and this represents an SNR of only 50 to 60 dB. On the other hand, to be considered "CD quality", audio must have at least 16 bits of ...

13

In a waveguide consisting of a thin hollow tube, there's a cutoff frequency below which it will not work, it will not propagate energy. Above that frequency, various different patterns of E and H field can propagate, each pattern being called a mode. If you run a wire down the middle of the tube, you add an extra mode, the coaxial or TEM mode, which ...

11

The main technical difference is how they reject interference. Twisted pair relies on the interference affecting both wires equally, generating common mode noise that is easily rejected by the differential receiver. This works well for magnetic interference down to very low frequencies. Coax cable relies on the magnetic interference inducing opposing ...

8

A completely separate reason that coax is favored for TV is frequency response. The losses associated with twisted pair rise rapidly with frequency, to the point where DSL modems struggle to use even the lowest 10 MHz of bandwidth on analog phone subscriber loops. For the same reason, high-speed Ethernet (1G, 10G and up) over twisted pair is limited to very ...

7

A 50 Ω line looks capacitive when driven from 600 Ω, and so they act like a low pass RC filter. The line needs to be matched. There are three ways you can match the line Source matching - drive the line from a 50 Ω resistive impedance. The easiest way to do this is to shunt a 54.5 Ω resistor (exact value, 51 or 56 will be close) across the output of your ...

6

All TVs are designed with a standard input impedance of 75 ohms. No need to measure. The cable is also 75 ohms, the length doesn't matter. Except the signal will get weaker if it has to travel further. As @Hearth points out older TVs may also have a 300 Ohm balanced feeder antenna connection. Very rare these days though, especially on a cable-ready set.

6

How is transmission line impedance selected? explains why transmission line impedance matters. The CATV industry deals with low-level signals, so it cares ONLY about loss and not at all about power-handling. That's why they chose 75Ω transmission lines.

5

Sure, just attenuate it (50-60dB). You don't want to saturate the receiver. As for collisions, they happen on air too (at different power levels). If you have a MIMO NIC with multiple SMAs, you can even play with the phase a bit.

4

Makes sense to me. You have an ideal source, then a resistance is added in series. The filter is terminated with a parallel resistance. Of course these resistors are not part of the filter, rather they are source and load resistances. Answers to questions: No No A Thevenin equivalent would be like depicted - ideal source + series resistance.

4

So, my impulse will run towards the load. Then it will reflect off and run to the 50-ohm connector on PCB, then it will reflect off from the connection of PCB line and 50-ohm connector despite any of my impedance matching actions. Assuming the trace on the PCB is short compared to the rise-time (or fall-time) of the pulse, then very soon after the ...

4

a 50 ohm transmission line acts like a capacitor when it's fed from a 600 ohm source. To get better performance change R2 and R1 to 56 ohms and R1 to 50 ohms For your signal generator that would mean placing a 56 ohm resistor in parallel with the start of the coaxial cable, and changing the load at the other end to 50 ohms too.

3

A transmission line terminated in a mismatched load of a given impedance will repeat that impedance every λ/2 (electrical length) back from the termination (e.g. here). You can think of your odd multiples of λ/4 transformer as consisting of a single λ/4 impedance transformer (using the calculations you've already described) plus an arbitrary number of λ/2 ...

3

I have a function generator with 600Ohm output impedance. I connect this output to a digitiser channel input of input resistance 1MOhm. I connect this via a 30m BNC cable Note - this isn't a transmission line problem because the maximum frequency of 1 MHz has a wavelength that is 300 metres Your shield (outer) and cable inner are magnetically coupled to ...

3

Your cable has approximately 100pF of capacitance per meter. 30 meters means 3 nF of capacitance if considered as one lump. In practice the capacitance is not a lump but distributed so effectively it is less. But still that load capacitance driven with 600 ohms means you have a RC low pass filter which can't pass high frequencies well. Another reason could ...

2

The coil as such doesn't have a current rating cause it doesn't care for the current that flows through your wire. The coil just recognizes the dI/dt and produces a corresponding voltage based on the coil's sensitivity. A very high dI/dt can cause high voltage peaks, therefore you need to specify the voltage rating of the coax cable, and even more important, ...

2

Coaxial cable "works" because the two currents (center conductor and shield), which flow in opposite directions, are equal in magnitude and share a common axis. This balance is what causes the magnetic field that is produced by those currents to cancel out perfectly outside of the cable. By reciprocity, it is also what makes it immune to the influence of ...

2

A 50 ohm connector is designed to be connected with 50 ohm cable, and 50 ohm track on the board. It's a description of the geometry and material of the dielectric used in the cable, connector and track. This controls their impedance, or the ratio of voltage to current of a signal that flows through them. You get best results when they are all the same ...

2

You need to model your lumped line in much smaller quantities like "per metre". At the moment you are modeling it as "per 100 metres" and that will give you severe problems above a fairly low frequency. Consider this 25 uH and 10 nF have a resonant frequency of: - $$\dfrac{1}{2\pi\sqrt{LC}} = 318 \text{ kHz}$$ And clearly that is the problem with using a ...

2

that's a 300 ohm outlet for connecting ladder-line, it's best to replace it with a co-ax outlet. it appears to also be a pass-through splitter, so if you need ther other TV outlet(s) to continure to work you'll need a splitter here too.

2

Starting from the basic formula for resistance: R = $\rho L/A$ consider the concentric cylindrical shells, and cable length G We want to calculate the resistance from one side to the other of a thin cylindrical shell of length G and thickness dr (current passing radially through the shell). So L in this case is the infinitesimal radius change dr We will ...

2

If you can't put parallel termination of the signal at the receiver side, then you can use source-series termination to match the impedance at the source side. Also suppose that the FPGA output impedance is less than 50ohm. I have learned it from an IBIS model. Therefore in theory I can match an impedance with one series resistor on the source side. You ...

1

On the Farnell site, you want to display the 'Extended Attribute' for 'coaxial termination', so that you can compare the name with the picture. Then you can 'filter' the termination type that matches the picture you want. Farnell calls the type of termination you have displayed in the picture in your post a 'board edge' termination: it is designed for ...

1

Simplified method of using line transmission theory, "abaque de Bewley". The schematic is simple : Generator Ugen with internal Zin -> coaxial line of lenght=L, Zc -> Load ZL. Zc: characteristic impedance of the line (generally resistive, example 50 Ohm coaxial) Zin: operational impedance of the generator (for simplicity, resistance) ...

1

Seems to me that you are going to a lot of bother when this is likely to work for you: -

1

For maximum power transfer to the antenna you should match your output circuit to the impedance that your circuit "sees" at the output: cable + antenna. Coaxial cables have an impedance of 50/75/100 Ohm no matter how long the cable it is. Antennas have an impedance of 50/75 Ohm only if you install the antenna according to the manufacturer's ...

1

OK, so the normal formula for resistance, or the way you find it, is to integrate along the path that the current takes. If current flows in the +x direction, you integrate over x. Each section you integrate has a resistance of rho / A * dx. A is the cross-section area of the conductor. USUALLY in these problems, A is uniform. A constant. That can be pulled ...

Only top voted, non community-wiki answers of a minimum length are eligible