Convert reflection coefficient to Z
 

OK, I used to be able to do this years ago, but I can't seem to find the
right references now.

If an antenna reflection coefficient is measured at, for example. 0.333
at -100 degrees, how is Z calculated? I think I can do this with a smith
chart, but the result does not match my attempted calculations.

TIA
Wayne

Convert reflection coefficient to Z
 

On Thu, 05 Apr 2007 01:50:52 GMT, "Wayne" wrote:

OK, I used to be able to do this years ago, but I can't seem to find the
right references now.

If an antenna reflection coefficient is measured at, for example. 0.333
at -100 degrees, how is Z calculated? I think I can do this with a smith
chart, but the result does not match my attempted calculations.

TIA
Wayne


Hello Wayne,

The first reference I can give you is in "Theory and Problems of Transmission Lines," by Robert Chipman, in
Schaum's Outline Series, Page 128, Eqs. 7.9a and 7.9b.

If you don't have access to Schaum, I'll reconstitute the equations for you in Word, and send them via email.

In addition, I ran across these equations a day or so ago on this NG--I'll try to find them and direct you to
them on this NG.

Walt,W2DU

Convert reflection coefficient to Z
 

"Wayne" wrote in
news:0CYQh.11100$P84.5052@trnddc07:

OK, I used to be able to do this years ago, but I can't seem to find
the right references now.

If an antenna reflection coefficient is measured at, for example.
0.333 at -100 degrees, how is Z calculated? I think I can do this
with a smith chart, but the result does not match my attempted
calculations.

TIA
Wayne



The expression for Gamma that springs to mind is (Zl-Zo)/(Zl+Zo).

Rearranging the terms gives Zl=-Zo(Gamma+1)/(Gamma-1) doesn't it?

Owen

Convert reflection coefficient to Z
 

On Thu, 05 Apr 2007 01:50:52 GMT, "Wayne" wrote:

OK, I used to be able to do this years ago, but I can't seem to find the
right references now.

If an antenna reflection coefficient is measured at, for example. 0.333
at -100 degrees, how is Z calculated? I think I can do this with a smith
chart, but the result does not match my attempted calculations.

TIA
Wayne


Wayne, I just put the pertinent equations into an email addressed to , which I sent.
Is this your correct address? If not, please give me your correct address in an email to me at .

Walt, W2DU

Convert reflection coefficient to Z
 

Owen Duffy wrote in news:Xns99097DA18D708nonenowhere@
61.9.191.5:

Rearranging the terms gives Zl=-Zo(Gamma+1)/(Gamma-1) doesn't it?

That probably looks better written as Zl=Zo(1+Gamma)/(1-Gamma)

Owen

Convert reflection coefficient to Z
 

On Thu, 05 Apr 2007 02:21:06 GMT, Owen Duffy wrote:

"Wayne" wrote in
news:0CYQh.11100$P84.5052@trnddc07:

OK, I used to be able to do this years ago, but I can't seem to find
the right references now.

If an antenna reflection coefficient is measured at, for example.
0.333 at -100 degrees, how is Z calculated? I think I can do this
with a smith chart, but the result does not match my attempted
calculations.

TIA
Wayne



The expression for Gamma that springs to mind is (Zl-Zo)/(Zl+Zo).

Rearranging the terms gives Zl=-Zo(Gamma+1)/(Gamma-1) doesn't it?

Owen

Owen, l just now sent Wayne the equations from Chipman, using the eq editor in Word.

In case others would like the equations I'll try to format them here without the Word editor.

R/Zo = (1 - rho squared)/(1 + rho squared - 2 rho cos phi)

X/Zo = (2 rho sin phi)/(1 + rho squared - 2 rho cos phi)

Walt, W2DU

Convert reflection coefficient to Z
 

Walter Maxwell wrote in
:

On Thu, 05 Apr 2007 02:21:06 GMT, Owen Duffy wrote:

"Wayne" wrote in
news:0CYQh.11100$P84.5052@trnddc07:

OK, I used to be able to do this years ago, but I can't seem to find
the right references now.

If an antenna reflection coefficient is measured at, for example.
0.333 at -100 degrees, how is Z calculated? I think I can do this
with a smith chart, but the result does not match my attempted
calculations.

TIA
Wayne



The expression for Gamma that springs to mind is (Zl-Zo)/(Zl+Zo).

Rearranging the terms gives Zl=-Zo(Gamma+1)/(Gamma-1) doesn't it?

Owen

Owen, l just now sent Wayne the equations from Chipman, using the eq
editor in Word.

In case others would like the equations I'll try to format them here
without the Word editor.

R/Zo = (1 - rho squared)/(1 + rho squared - 2 rho cos phi)

X/Zo = (2 rho sin phi)/(1 + rho squared - 2 rho cos phi)

Walt, W2DU

Walt, I take that to use rho to mean to magnitude of the reflection
coefficient Gamma.

My formula appears correct for Gamma, and the answer to Wayne's example
is 36-j27.

There is an uglier formula with tanh terms in it that gives the impedance
at a distance along a line with a given propagation constant... but it is
much more complicated to calculate than my expression when the distance
is zero.

Owen

Convert reflection coefficient to Z
 

Walter Maxwell wrote in
:

....
R/Zo = (1 - rho squared)/(1 + rho squared - 2 rho cos phi)

X/Zo = (2 rho sin phi)/(1 + rho squared - 2 rho cos phi)
....

These appear to depend on Zo=Ro to be correct, perhaps they would be more
correctly expressed using Ro instead of Zo.

Owen

Convert reflection coefficient to Z
 

Owen and Walt--
Thanks for the info. Now things are working out right.
--Wayne

"Owen Duffy" wrote in message
...
Walter Maxwell wrote in
:

...
R/Zo = (1 - rho squared)/(1 + rho squared - 2 rho cos phi)

X/Zo = (2 rho sin phi)/(1 + rho squared - 2 rho cos phi)
...

These appear to depend on Zo=Ro to be correct, perhaps they would be more
correctly expressed using Ro instead of Zo.

Owen

Convert reflection coefficient to Z
 

Yes, that strange email address is correct. I received your word document.
Thanks very much.
Wayne
"Walter Maxwell" wrote in message
...
On Thu, 05 Apr 2007 01:50:52 GMT, "Wayne"
wrote:

OK, I used to be able to do this years ago, but I can't seem to find the
right references now.

If an antenna reflection coefficient is measured at, for example. 0.333
at -100 degrees, how is Z calculated? I think I can do this with a smith
chart, but the result does not match my attempted calculations.

TIA
Wayne


Wayne, I just put the pertinent equations into an email addressed to
, which I sent.
Is this your correct address? If not, please give me your correct address
in an email to me at .

Walt, W2DU