# relection coefficient?

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

1. ### timGuest

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
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 AndersonGuest

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. ### timGuest

Thanks,my sanity is restored ( for now)...

4. ### John WoodgateGuest

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 WescottGuest

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 AndersonGuest

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 AndersonGuest

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 BloggsGuest

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. ### Paul BurridgeGuest

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. ### Paul BurridgeGuest

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 AndersonGuest

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. ### Paul BurridgeGuest

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 McCammonGuest

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. ### Paul BurridgeGuest

No, he need only be concerned with voltages. The phase shift and
magnitude differences between applied and reflected voltages.