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relection coefficient?

Discussion in 'Electronic Design' started by tim, Sep 4, 2004.

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  1. tim

    tim Guest

    Maybe I'm going nuts but I ran across this in Carr's book" Practical
    Antenna Handbook". The formula is givin for the reflection coefficient
    as Load impedence minus Characteristic impedence divided by load
    impedence plus char. impedence. R=Z_l -Z_o / Z_l+Z_o. He then gives
    an example of a 50 ohm transmission line terminated in a 30 ohm
    resistive impedence and says this ...R=50-30/50+30 without any sign
    change in R. Shouldn't it be R= 30 -50/30+50 as 30 is the load?
    I would just ignore this as errata but I've seen this same mistake
    in several books so it makes me wonder if I'm missing something
    simple? thanks
     
  2. Ray Anderson

    Ray Anderson Guest

    tim wrote:
    Shouldn't it be R= 30 -50/30+50 as 30 is the load?
    The reflection coeffient is negative if Zl < Zo and positive if
    Zl > Zo

    For example:

    R = (Zl - Zo) / (Zl + Zo)

    Case 1 (Zl < Zo):

    Zo=50
    Zl=30

    So R = (30-50)/(30+50) = -20/80 = -0.25


    Case 2 (Zl > Zo):

    Zo = 50
    Zl = 70

    R = (70-50)/(70+50) = 20/120 = +0.1667


    Ray
     
  3. tim

    tim Guest

    Thanks,my sanity is restored ( for now)...:)
     
  4. The reflection coefficient is always positive, and so the equation
    should be written R = |(Z1 - Zo)/(Z1 + Zo)|. |...| is the 'modulus'
    sign, indication that the answer is to be taken as positive, even if it
    isn't.

    Note how (IMHO, of course) to deploy spaces in ASCII math, and how you
    normally don't need to indicate subscripts (underscores make the
    expressions longer, with little increase in clarity). You do
    superscripts with ^, of course.
     
  5. Tim Wescott

    Tim Wescott Guest

    The author was probably taking the absolute value. Keep in mind that
    the reflection coefficient is complex in general, so if Zl is imaginary:

    Zl = j25 (an inductor), Zo = 50, so

    R = (j25 - 50)/(j25 + 50) = -0.8 + j0.6.

    There. Have I restored your insanity?
     
  6. Ray Anderson

    Ray Anderson Guest

    John Woodgate wrote:

    ..
    Consider a TDR, an instrument that measures the reflection coefficient
    of a DUT.

    If the DUT is, for example, a transmission line which is open circuited
    at the far end then the reflected signal will be reflected from the open
    circuit in phase with the incident signal and its amplitude will be a
    function of the reflection coeeficient (a positive value in this case).

    Now consider the same TDR connected to a the same transmission line that
    is short circuited at the far end. The signal will be reflected from the
    short circuit out of phase with the incident signal (i.e. it will
    exhibit a negative reflection coefficient).

    The above example cite the extreme boundary conditions (open and short),
    however you will see the same effects for intermediate cases where the
    load is either > Zo or < Zo.

    The sign of rho (the reflection coefficient) does make a difference and
    it isn't common practice to only use only the absolute value of rho.

    See http://cnx.rice.edu/content/m11379/latest/ for a graphic explanation
    on how the reflection coefficient (both postive and negative values) are
    used in a "Bounce Diagram" to illustrate how an incident signal's
    amplitude varies as it bounces back and forth on a transmission line
    with various source and load impedances. For Bounce Diagrams to work you
    need to consider the sign of rho.

    -Ray
     
  7. Ray Anderson

    Ray Anderson Guest

    Ray Anderson wrote:
    ..
    Another thing or two occurred to me after I posted my last response.

    Carr's book and the discussion contained therein are primarily assuming
    the topic is analog RF (as were the comments made by John). In that
    frame of reference, talking about the magnitude of rho makes sense as
    the RF signal is probably being measured under steady-state conditions
    long after any transient effects (where the sign of rho does make a
    difference) have died out or stabilized. In these cases it is usually
    the power of a reflection that is of interest.

    In the case of digital signaling, the sign of rho is significant and
    can't be ignored. Usually the voltage and polarity of the reflection are
    of interest in these cases.

    -Ray
     
  8. Fred Bloggs

    Fred Bloggs Guest

    Yeah- it's definitely an error- but for purposes of computing antenna
    mismatch loss, which is 10 x LOG10(1-rho^2) dB, it makes no difference,
    unless you have some other use you're putting the coefficient to.
     
  9. I'm not sure that's correct. IIRC, the reflection in such
    circumstances will depend on the length of the transmission line. You
    can't take an arbitrary length of the stuff and expect to get zero or
    total reflection at any wavelength. Aside from that - again IIRC and
    it's been a while - the reflection coefficient is a positive value
    between 0 (matched condition) and 1 (total mismatch), therefore, it
    never goes negative.
     
  10. Correction: the magnitude component never goes negative; there's a
    phase component that may, though. No phase shift with an open tx line;
    180' shift with a short and various shades of grey in between.
    Thought I'd just correct myself before some jumped-up panjandrum like
    Active8 sticks his 2p worth in.
     
  11. Ray Anderson

    Ray Anderson Guest

    I may have not been entirely clear in regard to my statements about
    reflections from shorts or opens.

    In the example I submitted, I proposed either a perfect open or short at
    the far end of some transmission line of some arbritrary length. (Yeah,
    I know you can't in reality have a perfect open or short, but this is
    hypothetical). The reflection coefficient caused by the end-of-line
    discontinuity (i.e. open or short) will be either + 1 or -1 (or if you
    will a magnitude of 1 with either a 0 or 180 degree phase angle). The
    actual value of rho observed at the driving end of the line can (and
    probably will be) different due to line loss, line Zo etc. The power of
    the TDR is in its ability to discern the physical location of
    discontinuities along a transmission line.

    So, all in all I think we may actually be in violent agreement once
    we've gotten past the assumptions and semantics.


    -Ray
     
  12. That's easily done as I well know!
    Depends on the frequency. Yeah, perfect open/shorts are extremely hard
    to achieve at anything beyond VHF. That's why Y-parameters fell out of
    favour.
    I'm not familiar with the operation of TDRs. However, line loss comes
    down to two obvious factors: the intrinsic loss in dB per metre of
    line and (once again) the frequency. Nobody's mentioned what the
    frequency of the test is. At relatively low frequencies with low-loss
    line, line loss can be disregarded.
    Probably.
     
  13. Roy McCammon

    Roy McCammon Guest

    There is a voltage reflection and a current reflection, that are
    180 degrees opposite. If its not specified, then its legit to
    consider only the absolute value of the reflection coef. Personally,
    I use both positive and negative (and complex) values.
     
  14. No, he need only be concerned with voltages. The phase shift and
    magnitude differences between applied and reflected voltages.
     
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