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Homework Help: Reflections in Coaxial Cable

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data

    An annular sheet, of thickness t=1mm and resistivity ρ=0.5Ωm, connects the inner and outer conductors of an air spaced coaxial transmission line at a point on the line. A low-frequency signal is fed into one end of the line and the other is terminated by its characteristic impedance. Calculate the relative amplitude of the wave reflected from the resistive sheet.


    3. The attempt at a solution

    The annular sheet results in a change in the impedance of the line at the point where the sheet is located. We know that, in general:

    Z= sqrt((R+iωL)/(G+iωC)), R is resistance per unit length of cylinder, G is conductance per unit length of dielectric between cylinders

    therefore, in this case: impedance at annulus=Za= sqrt((iωL)/((ρ/t)+iωC)), where it is assumed that the resistance in the cylinders is negligible.
    I feel that the ρ/t term should be more complicated, but it is at least dimensionally consistent, and there doesn't seem to be enough information to do much else.


    r=(Za-Z)/(Za+Z), where r is the reflection coefficient, Za is the impedance at the annulus and Z is the characteristic impedance of the line.

    From these two equations it should be possible to calculate the reflection coefficient. However, in this case, absolutely no information is given about the line itself, the only values provided are t and ρ. Even after fiddling around with the equations, I haven't been able to get things to cancel out, leaving an expression for r in terms of t and ρ.
    Am I going about this in completely the wrong way?

    Any help is greatly appreciated.
  2. jcsd
  3. Nov 25, 2011 #2

    rude man

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    Since the line is terminated in its characteristic impedance, you can model the situation as a transmission line terminated in the annular shunt in parallel with a resistor of resistance R = Z0.

    So just calculate the parallel resistance of R0 and the sheet resistance, calculate the reflection coefficient, & away you go.

    If your instructor is a real b*st*rd he migh also want you to include the annulus' inductance. I would not. Since frequency was not specified you can assume that annuluar inductance is zero.

    I'm not 100% up on this right now but I would expect the answer to be a function of line length (to the annular shunt), frequency, and Z0. Not being given any of these parameters is a bit strange, I must confess.
  4. Nov 26, 2011 #3

    Thanks for the suggestion Rude Man. I had tried the approach that you mentioned, but I ran into difficulties due to the absence of parameters.

    However, after revisiting this problem, I realised that the difficulty was indeed due to the incorrect expression for the resistance of the annulus. It works out as (ρln(b/a))/(2πt), where a and b are the radii of the inner and outer cylinders respectively, which is the same as (Z/Z0)*(ρ/t), where Z0 is the impedance of free space (μ0/ε0)^0.5. From this you can see how the Z will cancel throughout the equation for the load impedance.
    I still think that I might have had to make the assumption that the annulus is much closer to the terminating resistance than the source, but I'm not sure. Anyway, I got the stated answer.

  5. Nov 27, 2011 #4
    p/t=500, Zo=377 so r = 0.14

    im getting 0.14.... how did you get 0.27?

    btw jim you at cambridge??
  6. Nov 28, 2011 #5
    The terminating resistance and annular resistance are in parallel, they both contribute to the load resistance. Taking this into account you will find r=-0.27.
    Incidentally, I found that analysing this problem in terms of incident and reflected waves gave the same answer, and also showed that r is independent of the position of the annulus. This isn't true in general, but holds here because the terminating resistance is equal to Z and so absorbs all incident EM waves.

  7. Nov 28, 2011 #6

    rude man

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    Jim - the cable can't be effectively terminated in its characteristic impedance by the annulus, because the stretch of cable from the annulus to the terminating resistor already puts Z0 = R0 in parallel with the annulus at the point of the annulus, so everything beyond that point is nonexistent from the viewpoint of the excitation - so unless Z = infinity for the annulus, the cable is not terminated in Z0.

    Furthermore, if it were, there would be zero reflection anyway, of course.
  8. Nov 29, 2011 #7
    rude man:

    Sorry, my explanation wasn't particularly clear.
    What I mean is that there are essentially 3 resistances we can consider: the resistance of the annulus alone (annular resistance), the resistance of the impedance we've connected across the end of the cable alone (terminal resistance), and these two in parallel (load resistance).
    My solution method was to calculate the load resistance from the terminal resistance and annular impedance, then use this to work out the reflection coefficient, as described. However, this is not correct in general, it only works in this instance because the terminal resistance is equal to the characteristic impedance of the line and so no EM waves are reflected from the terminal resistance.
    This problem can also be solved by considering the various EM waves propagating up and down the line and treating the annular resistance and terminal resistance as two separate interfaces. The voltages and currents can then be equated at these points and this allows the reflection coefficient of the annulus to be calculated. This approach gives the same answer as above, but is quite a bit more work. However, this method is valid even if the terminal impedance is not equal to the characteristic impedance of the line.
    If you're not convinced, try it yourself and see.

  9. Nov 29, 2011 #8

    rude man

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    So what did we conclude is the effective resistance of the annulus?

    (Yes, if the end-point hadn't been terminated in the cable Z0, the we would have had to (1) calculate the drive-point impedance at the annulus' location without the annulus in situ, in other words pretending the cable was the section from the annulus to the end-point only - this could of course have been complex Z, depending on the end-point termination - then (2) we would have used that driving-point impedance as the termination for the cable section from the original drive-point to the annulus section, placing the annulus R in parallel wit the aforesaid drive point impedance from (1).

    Since the cable was end-point-terminated in Z0 we were spared this agony; we have as our load Z(annulus)||Z0.

    I too want to compute the reflected wave but want to know what everyone agrees is the annulus resistance first.
  10. Nov 30, 2011 #9
    Rude man:

    As I pointed out in my earlier post,the annulus resistance is given by:


    Where a and b are the radii of the inner and outer cylinders respectively.

  11. Nov 30, 2011 #10

    rude man

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    OK, so we have a problem, don't we, since neither a, nor b, nor b/a are given. The answer is allegedly independent of a and b directly.

    Who volunteers to compute b/a as (hopefully) a function of Z0?
  12. Dec 1, 2011 #11
    The characteristic impedance of a coaxial transmission line is given by:

    Z= (Z0ln(b/a))/(2πt), where Z0 is the characteristic impedance of free space

    and so Zannulus = (Z/Z0)*(ρ/t)

    as stated in my previous post.

  13. Dec 1, 2011 #12

    rude man

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    According to my ITT "Reference Data for Radio Engineers, the formula for the chas. impedance of a single-inner-conductor coax line is Z0 = 60ln(b/a) ohms. Notice Z0 is not the free space impedance.

    Jim has determined that Z(annulus) = (ρ/2πt)ln(b/a). Therefore, ln(b/a) = Z0/60 and so

    Z(annulus) = (ρ/2πt)(Z0/60) = [0.5/(6.28*1e-3)]*(Z0/60) = 1.33Z0.

    Since Z0 is in parallel with Z we have
    Z(termination) = Z = Z(annulus)(Z0)/[Z(annulus)+Z0] = 1.33(Z0^2)/2.33Z0 = 0.57Z0.

    Reflection coefficient Γ = (Z-Z0)/(Z+Z0) = -0.43/1.57 = -0.27
    which is supposedly the right answer, so looks like Jim computed the correct value for
    Z(annulus) all right! :smile:
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