Transmission line and measuring Zo

Discussion in 'Homework Help' started by TheRedDevil18, Aug 21, 2016.

  1. TheRedDevil18

    Thread Starter New Member

    May 23, 2015
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    [​IMG]

    [​IMG]

    3.2) I made an equation for the input voltage and input current
    Vi = Vs*(Zi)/(Zi+Rs+Rm)

    Ii = Vs/(Zi+Rs+Rm)

    I am a bit confused about what voltages V1 and V2 are measuring, and what do they mean by using the phase ?, I don't know the phase

    3.3) After some research I found this equation for a twisted pair cable

    [​IMG]

    Though i'm not sure what to use for Er. On the site it says
    εR=(c/v)^2

    but I don't know the velocity of the signal

    http://www.sigcon.com/Pubs/edn/TwistedImpedance.htm
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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  3. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    You measure V1 and V2, then calculate:
    i2=(V1-V2)/Rm

    then calculate the line input impedance:
    ZLin=V2/i2

    That's basically like Ohm's Law but with complex quantities.

    ZLin will in general be complex unless the load impedance is EXACTLY equal to the characteristic impedance which they usually call Zo. That can only happen with a complex load usually, but sometimes a resistive load of the right value sqrt(L/C) comes close (L is inductance per meter, C is capacitance per meter of the line).
    Because ZLin is complex, there will be a phase angle associated with it unless the load is perfectly the same as Zo, which it almost always is not.

    BTW, ZLin is calculated or measured with the load ZL connected if there is one. The formal calculation is:
    ZLin=Zo*(ZL+Zo*tanh(Y*x))/(Zo+ZL*tanh(Y*x))

    where Y is the propagation constant [LEFT OUT] given R,L,G,C are in units of per meter. LEFT OUT because it's a question there.

    Of course if ZL is a short then ZL=0, and if ZL is open then ZL=infinite.
     
    Last edited: Aug 22, 2016
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  4. TheRedDevil18

    Thread Starter New Member

    May 23, 2015
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    Yes, thank you

    And then to calculate Zo, I would use
    Zo = sqrt(Zoc*Zsc)

    And how would I know what value to choose for Rm when measuring the Zoc and Zsc ?
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hello again,

    Well in theory you can use anything other than zero or infinite Rm for a measurement, but of course you want to use something practical. For example, if your Zo is around 50 ohms you dont want to use 1Megohm and if your Zo is 1Megohm you dont want to use 50 ohms.

    Probably the best bet is to use something that gives you a good V1-V2 differential but does not reduce the input voltage to the line by too much. We want a good V1-V2 differential so that we can easily read the difference on a meter or scope, but we want to keep the line input voltage from dropping too much if possible.

    A typical real world cable 10 meters long might have the following ZLin at 1MHz with a 50 ohm resistive load using 16 digits of numerical precision for the calculations:
    Zlin1=50.0983353050726-j*0.015585194433412

    and from this we can see there is a very small imaginary part of about 0.016 so the load is pretty well matched to the line impedance, so this is almost like a regular 50 ohms resistor. Ask yourself how would you measure an unknown resistor value that is known to be 'around' 50 ohms. The answer is you would choose a series resistor Rm that gives you enough of a voltage difference reading so you can make a good estimate on the current with (V1-V2)/Rm yet at the same time not reduce the voltage to the unknown resistor too much (because operation of most equipment needs a voltage at least close to the normal operating voltage).
    So say 5 to 10 percent if possible. That would mean 2.5 ohms to 5 ohms, but if we could get away with 1 ohm that would be better, and if we could get away with 0.1 ohm that would be even better yet. In fact, if we could get away with using a current shunt like 50mOhms that would be even better, but this might be a stretch depending on the actual input current.
    If we apply 1v to the line we get about 1/50=0.02 amps, which may or may not be enough. If we can get away with 50v then we'd get 50/50=1 amps but that may stress the line too much. So the input voltage selection is important too and should be the normal operating voltage or close to it. Some transmission lines are made for high power but many are not. If the center conductor is AWG 24 gauge for example then we have to be careful not to apply too much current.

    The main point though is it is almost like measuring the resistance of an unknown resistor except there will be an imaginary part too which makes the input impedance a complex quantity.

    I should mention that if the generator resistance is small, then a resistance almost equal to the line impedance would be better because that is the normal way the line is used. So in that case something close to 50 ohms is better.
    If the generator had 5 ohms impedance though, then something close to 45 ohms would be better.
    If the generator is already 50 ohms then we would want to use a resistor as with as low a value as possible, or else parallel another resistor with the generator first and then use a resistor that brings the total resistance back to 50 ohms. This is theoretically the best way but probably isnt necessary.
    These of course assume the line will be used in the actual application with matching impedance source and load.


    Just for reference, the values used for the calculation above were (in units of per meter and at 1MHz):
    R=0.01
    G=1e9
    C=50pf
    L=125nH
    which gives Zo=50 at infinite frequency.
     
    Last edited: Aug 23, 2016
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