# Coaxial Cable loss

1. Jan 6, 2014

### nelectrode

Hey Guys,

How do I work out Transmission loses and input impedance for coaxial cable?

I’ve been given this website to use http://fermi.la.asu.edu/w9cf/tran/

I do have the characteristic impedance of 75Ω and frequency and I am asked to work the attenuation in dB/100ft and “input impedance for a 150Ω load resistance”
So I am not sure How to complete the input data.
Cable type: RG59
Frequency in Mhz: 1
Cable length in feet: 100
Load reactance: How I find that? Do I have to use Z=R+JX?

Thanks

2. Jan 6, 2014

### davenn

hi there

all cable manufacturers have datasheets for their cable
this always includes a table showing
losses usually per 100ft or 100 metres at a given frequency, sometimes it may state a shorter length
10 metres or 30 ft

for standard RG59 the loss at 1MHz is going to be less than 1dB for 100 ft ( 33 metres)

resistance or impedance ??

you cant put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

Your load impedance must equal your cable impedance for best power transfer, else you will end up with a large reflected power component

Cheers
Dave

3. Jan 6, 2014

### tfr000

I think that's the idea here. The impedance changes depending how many wavelengths you go down the transmission line.
Old-school, done on a "Smith Chart", but I guess that Java app is supposed to do the same thing.

4. Jan 6, 2014

### Averagesupernova

This depends on the length of the transmission line in relation to the wavelength. An insignificant portion of a wavelength in transmission line length will not cause any problem at all. While we are on the subject of transmission line length what about the matching of the transmitter to the transmission line? Suppose the transmitter output impedance is 150 ohms and the load is 150 ohms but connected with 75 ohm line? If the line length is a small portion of a wavelength then no problem. If it is a half wavelength, then the line will appear as if it is not there at all minus some loss. I've only scratched the surface of this subject.

5. Jan 7, 2014

### sophiecentaur

It's not usual to find that both transmitter and antenna are well matched to the transmission line. The amplifying device is often operated just for 'best' power and the antenna would be matched to make sure most of that power goes up the spout. (In the same way that a 'good' battery has a very low output resistance but you would never dream of putting a 0.05Ohm (max power) load on it.)
Another reason for matching the antenna is that a mismatch produces standing waves which can the maximum voltage on the amplifier output. This can harm the device.

To deal with your particular problem. The operating frequency is 1MHz, so cable loss will be negligible for 100ft of feeder. If you look it up on the cable spec (here, for instance) you can see just how low it is at 1MHz!!

If you use the Applet they give you, you just need to put in the load impedance (which is 150Ω), cable impedance and length to find the impedance that the transmitter will 'see'. I think it will give you all the answers you want. (It does the difficult bits for you.)

6. Jan 7, 2014

### davenn

the problem is we don't cut coax lengths to a specific portion of a wavelength of the freq used in a particular system. We cut the coax to what is needed to get from the TX connector to the antenna connector ( ignoring any cavity filters, circulators that may be also inline). This means the coax is always just some random length depending on the height of the antenna up the tower.
Because of this we match the TX (RX) output (input) to the coax characteristic impedance and then that of the coax to the antenna.. If we don't, then we have the problems that Sophie and I have already stated concerning standing waves ( SWR) and the resulting high levels of reflected power back to the transmitter.
This can cause 2 common problems

1) as Sophie stated it can damage the transmitter output stage
in a lot of transmitter equip, protection is provided by measuring the reflected power level and it is used to lower the TX output power level till the problem is resolved

2) That reflected signal mixes in the TX output stages to produce intermodulation products that then get sent to the antenna and transmitted and can cause all sorts of hell with other comms services in the area

Matching TX, Xmission line and antenna is always the best method

Dave

7. Jan 7, 2014

### Baluncore

In theory, a load mismatch is not line length dependent if the source impedance is matched to the line. That is because there will be no energy re-reflected again when returned to the source, since the source is correctly matched to the line.

Real sources such as transmitters will reflect the energy, because the output stage will be adjusted to a different impedance that reflects all the reverse energy. The transmatch will be adjusted to minimise absorption of energy by the transmitter by maximising forward power.

8. Jan 8, 2014

### sophiecentaur

I wondered what you were getting at there but I see what you mean; it's a strictly theoretical comment. A well terminated source will kill reflections but at what cost to the requirements for the transmitter design? It's hard enough to build a transmitter that will provide the right power at the right cost and efficiency, without also requiring it to look like 50Ω .

9. Jan 8, 2014

### sophiecentaur

@davenn
Tx line reflections will also affect the frequency response (short lines) and introduce echoes / ringing (long lines) which is particularly relevant to analogue TV and tall masts.

10. Jan 8, 2014

### Baluncore

The OP was a transmission line question. The RG59 cable with Zo = 75Ω suggests a receive system with a probable maximum power of one mW. There is no need for a transmitter. Academic exercises like this are often based on theory, not practice.

That may be true for an inexperienced amateur, but it is not true for a competent professional.

[Generalisation] Amateurs place a tuner between the transmitter and the line, then call it an antenna tuner, but it is really a transmatch. Professionals tune the line to the antenna so as to avoid reflected energy on the line. [/Generalisation]

11. Jan 8, 2014

### sophiecentaur

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%. That would hardly be a "professional" design.

12. Jan 8, 2014

### nelectrode

Hey,

Thanks a lot, but as Baluncore stated this is purely a theoretical question,

so I basically have to use the Java app to work out my values.

I do know the following only:

Type: RG-59
Characteristic impedance: 75 Ω
frequency: 1Mhz

have to find:

Matched attenuation(dB/100ft): ?
Matched attenuation (dB/100m):?
input impedance for 150Ω load resistance: ?

so on the app at http://fermi.la.asu.edu/w9cf/tran/

I have:

Cable type: RG-59
Frequency: 1Mhz
cable length in feet: 100
Load resistance: Do I have to put 150Ω here?
Load reactance: I thought it is going to be 75Ω but that's seems to be wrong.

Thanks

13. Jan 8, 2014

### sophiecentaur

Load reactance will be zero if you're told 150 Ohms (i.e. no reactance in series)
That applet seems to do all you need.

14. Jan 8, 2014

### nelectrode

That's what I thought but the values don't really match the one here:

http://www.embedded.com/print/4402915

So what is input Impedance for a 150Ω load resistance?

Because the app gives a Z and Z0?

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15. Jan 8, 2014

### sophiecentaur

Z0 should be 75 Ohms but the loss per foot in the cable (this is a real cable they're discussing) will introduce some effective reactance, I guess. It seems to be about 2%, which is quite a lot but I think that's what it's about. RG59 isn't high quality cable. Put 75 Ohms into the 'user defined' option cable parameters and it gives you the ideal lossless cable with Z0 of 75 Ohms
The Zin is what you will see, looking into the transmitter end of the cable when you put your 150 Ohms at the other end. The line acts as a transformer to change the load impedance, as seen at the transmitter.
Applets are sooo useful - as long as yo have an idea what they're doing for you.
You guys have it so easy these days. All we had was a hand calculator or, before that, a Smith chart and china graph pencil,

Last edited: Jan 8, 2014
16. Jan 8, 2014

### Baluncore

I disagree. Under what restrictive conditions is your statement true. Please explain.
Maximum power transfer occurs when the transmitter is matched to the line.
Are you actually suggesting matching by using a resistive network?

17. Jan 8, 2014

### sophiecentaur

This can be confusing but, if your source impedance is the complex conjugate of the load impedance then you do get maximum power. Butttttt - you dissipate half of the power in the source resistance. (The melting anode or collector of your output stage) That is the 'downside' consequence of the maximum power theorem. Doing it is not a good idea when you have a 500kW transmitter. (Class C with 90% efficiency is favourite when the modulation system is suitable)

You don't need to use a resistive network for the matching (that would be plain loopy). But any form of matching for maximum power will have the above effect. It's not done in any system that I can think of except for distribution amplifiers, where the power is low and the marked gain needs to be relied on.
That goes for audio power amps too, they are usually designed as a voltage source.

18. Jan 8, 2014

### Baluncore

nelectrode.
Are you neglecting velocity factor?

Attenuation is simply specified by the physical length of the cable.
Phase shift is specified by electrical length = physical length / velocity factor.

Attenuation.
Your source will provide energy that is attenuated along the cable to the load.
The mismatch will reflect energy.
The reflected energy will be attenuated on it's way back to the source.

Input Impedance.
The line length must be corrected for velocity factor, maybe about 66% for a 75 ohm foam line.
The electrical line length, (phase shift), can then be calculated from the transit time.
The reflected wave and forward wave phase combine to form the apparent Zin.

19. Jan 8, 2014

### sophiecentaur

The applet works all that out for you. (Effective length etc.)
If you accept that the Z0 is the value of termination to eliminate reflections then the effect of loss in the feeder is to introduce the requirement for a bit of reactance. Mr Smith and his chart doesn't allow for that to be included easily, although I'm sure it can be done by someone really familiar with using the chart.

20. Jan 8, 2014

### Baluncore

Yes, I think you are confused.

The reactive component does not waste real power. It is a “reactive” recycling of circulating energy.
RF energy efficiency of close to 100% is possible with a matched and non-reactive transmission line.

Playing reflective ping-pong with energy on a lossy line can be very expensive. By avoiding reflected energy on the line, the transmitter will see a non-reactive load and so does not have to dedicate supply current to the untwisting of output stage phase.

Resistive “Pi” or “T” pads are often used when coupling broadband microwave signal modules to reduce the reactive effect of other module's mismatch, so they stabilise the module and it's specifications. I agree resistive pads should never be used to match power transmitters.

I'm glad I don't need to worry about a 500kW transmitter. I only have a total of 75 kW here and I do my best not to use it. Unfortunately, if I run the anodes too cool, the “getter” on their surface does not function so well and the cathode emission begins to fall.

21. Jan 8, 2014

### Averagesupernova

I think Baluncore and sophie are barking up different trees.
-
Sophie says that an output stage with a very low Zout relative to the load will dissipate little power in said output stage compared to the load. I would say this is correct. Notice that sophie did NOT say that a reactive component was involved. Sophie also said that this is most efficient. This is where it gets questionable. Maximum power transfer does not imply most efficient. It implies that given an output that has a Zout of R, then you can load said output with a load of R and no matter how you adjust the load you will not get more power to dissipate in the load. Raising the load resistance will cause less power to be dissipated in the load but will NOT cause more power to be dissipated in the source. Lowering the load resistance will cause less power to be dissipated in the load but will cause MORE power to be dissipated in the source.
-
As I have stated in a different thread, introducing a reactive component in the load is hard on the output stage since when the most voltage is dropped across the output transistor (zero crossing) there is more current passing through the device compared to if there were a resistive load. Voltage and current are not in phase, so this naturally occurs.

22. Jan 8, 2014

### sophiecentaur

If you had a 50 Ohm load connected directly to a transmitter with a 50 Ohm internal resistance, you would agree that was a perfect match (I hope). It is possible to match any load to any source impedance (at least in principle), so you could achieve the equivalent of the above - if you wanted to. But, under those conditions, you would have equal power dissipated in the transmitter as the power dissipated in the load. That is 50% efficiency and not very kind to the output device.
If you have any doubt about the Maximum Power theorem then you can look it up. I don't need to find it for you.
Very few power distribution systems use a matched arrangement (name one, in fact). Car batteries would all melt, given a 0,05 Ohm load and all the Generating Company's equipment would fly off its beds with such a high current demand (the equivalent of several times the present National Load). I could ask you what the source resistance of your mains supply is? Quick calculation: The volts probably dip by, perhaps 5V when you take 50A - so that means 0.1 Ohms. Would you connect a 0.1 Ohm load and expect the equipment to survive (without those fuses in the way?)
Whatever it is that you know about the practicalities of transmitters, you need to realise that, when you 'tune for max power', the max power you are tuning for is set by the practical limits of the amplifying device you are using and not how well it's matched.
You have stated that you 'back off' your transmitter to stop it knackering itself. That is the case for all transmitter designs. I suggest that you don't in fact, have any way of measuring the output impedance of your transmitter so you aren't in a position to assert whether or not you are providing it with a conjugate match.

There's nothing special in the principles involved, whether it's a battery, generator of transmitter.

23. Jan 8, 2014

### sophiecentaur

I agree (and have stated) that a perfect match is NOT efficient. It involves losing half the power from your source.
I took the case of a Class C transmitter, which can be 90% efficient and which cannot be matched - for the above reason. The actual efficiency of output stages is more to do with the physics of the devices themselves and the need to 'control' the current flowing through them whilst avoiding too much voltage drop. But that's another issue.
I agree that reactive components are rather a red herring.
My post ,above, states the case about not matching source to load in power systems.

24. Jan 8, 2014

### Averagesupernova

Yeah sophie I was thinking about car batteries, 1000 watt subwoofer amplifiers, the source impedance of a typical electrical service, etc. also. None of them worry about loading to satisfy maximum power transfer. Nor should they. In car audio, preoutputs have a Zout of a couple hundred ohms maybe up to 1000 I believe but the Zin at the amplifier is typically 10K. Interesting point there also, they don't worry about transmission line impedance either. My comment in a previous post about insignificant portions of a wavelength would apply here.

25. Jan 8, 2014

### Baluncore

nelectrode, first let me apologise for the sound of battle in the background.

Resistance and output impedance share the same dimensions, but they are certainly not the same thing.

The difference in the barking up trees appears to come from the fact that a 50 ohm internal resistance in a low frequency power supply, is quite different to the output impedance of an amplifier driving a 50 ohm transmission line. Zout is actually the slope of the RF amplifier's load line, the ratio of the voltage to current in the output stage.

A power distribution grid needs a very low resistance so it does not vary in voltage with variable load conditions. If you could match your electric light to the grid, you would get most of the power available at the time. That would not be good.