Coaxial Cable loss

  1. 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 resistance: 150
    Load reactance: How I find that? Do I have to use Z=R+JX?

    Thanks
     
  2. jcsd
  3. davenn

    davenn 3,546
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    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
     
  4. 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.
     
  5. 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.
     
  6. sophiecentaur

    sophiecentaur 13,532
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    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.)
     
  7. davenn

    davenn 3,546
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    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 :smile:

    Dave
     
  8. Baluncore

    Baluncore 2,576
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    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.
     
  9. sophiecentaur

    sophiecentaur 13,532
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    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 :wink:.
     
  10. sophiecentaur

    sophiecentaur 13,532
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    @davenn
    To add to your list:
    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.
     
  11. Baluncore

    Baluncore 2,576
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    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]
     
  12. sophiecentaur

    sophiecentaur 13,532
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    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.
     
  13. 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
     
  14. sophiecentaur

    sophiecentaur 13,532
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    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.
     
  15. 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?
     

    Attached Files:

  16. sophiecentaur

    sophiecentaur 13,532
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    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
  17. Baluncore

    Baluncore 2,576
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    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?
     
  18. sophiecentaur

    sophiecentaur 13,532
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    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.
     
  19. Baluncore

    Baluncore 2,576
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    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.
     
  20. sophiecentaur

    sophiecentaur 13,532
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    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.
     
  21. Baluncore

    Baluncore 2,576
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    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.
     
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