Jim Kelley wrote:


Cecil Moore wrote:
Your example is the same as putting a load resistor on an open
transmission line, measuring the dissipated power, and then claiming the
same thing happens without the load resistor there.

No, it is more akin to presenting all the evidence.
Your approach is akin to rolling dice in the dark
where you are the only one allowed to report the
results. :-)

Are you willing to assert that the power being
dissipated in the circulator resistor didn't
make a round trip to the load and back even though
the actual delay is easy to measure?

Do the reflected waves that you see when looking
at yourself in the mirror contain any joules/sec?
--
73, Cecil http://www.w5dxp.com

Jeff wrote:
Change the source impedance to 100ohms and the picture changes to 5.5:1 at
the load and 4.75:1 at the source end.

Here's the equation for rho at the load.

rho = (Z0-Zload)/(Z0+Zload)

SWR = (1+rho)/(1-rho)

I don't see the source impedance in those equations.
--
73, Cecil http://www.w5dxp.com

Here's the equation for rho at the load.

rho = (Z0-Zload)/(Z0+Zload)

SWR = (1+rho)/(1-rho)

I don't see the source impedance in those equations.

Your analysis is fine if the source is matched to the coax, but you are
neglecting the mismatch at the source to coax interface.

If you used a TDR, for example, to look at the set-up you would see 2 points
of discontinuity, firstly at the 100 ohm source to 50 ohm cable interface,
and secondly at the cable to 200 ohm load. BOTH of these discontinuities add
to the overall mismatch as seen by the 100 ohm load.

Your application of the above equations neglects the first discontinuity.

73
Jeff

"
The solution is 4.9 dB.
snip
If anyone wants to challenge the 4.9 dB solution, they can impeach my
reference "Reference Data for Radio Engineers," (various editions). I
can supply other references that have been named in this group too,
but I would suggest with tackling one authority at a time.

73's
Richard Clark, KB7QHC

I have to admit to an error in my analysis of the problem, I made a mistake
with the attenuation of the line.

Re analysing with the correct values gives me the following:

Loss 4.78dB
S11 -5.25dB
Vswr as seen by the source 3.41:1

I think these values are close enough to Richard's answers to make little
difference. I have not got Reference Data for Radio Engineers to hand, but
it may be a graphical solution that could account for the slight
discrepancy. My figures were obtained using Ansoft Designer RF Cad package.

The error crept in because I did the analysis by using 5.35m of coax at 300m
to get the 5.35 wavelengths, and forgot to scale the attenuation value for
the coax.

73
Jeff

"Jeff" wrote in
.com:

If you used a TDR, for example, to look at the set-up you would see 2
points of discontinuity, firstly at the 100 ohm source to 50 ohm cable
interface, and secondly at the cable to 200 ohm load. BOTH of these
discontinuities add to the overall mismatch as seen by the 100 ohm
load.

Your TDR does not work in the steady state frequency domain space, and is
misleading you.

In the steady state, the (complex) ratio of forward voltage to reflected
voltage is determined solely by the load impedance and characteristic
impedance of the line.

In crude terms, during establishement of steady state, you can view that
a load end reflected wave which is then partially reflected at a
mismatched source end, will reach the load end and be reflected in the
same ratio as the earlier passes. The subsequent round trips as steady
state is approached do not change the (complex) ratio of forward voltage
to reflected voltage in the steady state.

I know you have support here for the assertion that source end mismatch
affects VSWR in the steady state, but you won't find it in reputable text
books.

Owen

Richard Clark wrote in
:

On Wed, 28 Feb 2007 08:11:30 GMT, Owen Duffy wrote:

Now, are you prepared to post your solution?

Hi Owen,


Your quick computation of 3.3 dB is suitably close to my reference's
first pass solution (3.27 dB), but it neglects the contribution of the
source's resistance.

The solution is 4.9 dB.

Reminding you that your question was "What is the loss in the line?",
check your own post.

Well, you posted an answer, not a solution. It wouldn't have been your
solution anyway, because it looks like it is copied straight out of a
book.

Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3
(which is the same as the problem you posed), they give the answer as
3.27dB.

I am happy that my answer rounded to 3.3dB is correct.

The source resistance has no influence over the line loss at all.

You posed this problem as difficult and one that no one has ever got
right. No wonder, you have a different answer to the the book!

Owen

Your TDR does not work in the steady state frequency domain space, and is
misleading you.

In the steady state, the (complex) ratio of forward voltage to reflected
voltage is determined solely by the load impedance and characteristic
impedance of the line.


What utter rot. The TDR indicates a discontinuity that is IN ADDITION to
the one that gives you your result. Ignoring that discontinuity will
certainly give you the wrong answer, regardless if which domain you are
working in!!!

Jeff

I am happy that my answer rounded to 3.3dB is correct.

The source resistance has no influence over the line loss at all.


Well it certainly does not agree with Ansoft Designer, the result it gives
is very close to Richards, and it shows a marked effect when you change the
source resistance.

With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB
and VSWR is 3.98:1
Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR
is 3.41:1

So a very well respected CAD package agrees with Richard at least!!

73
Jeff

Jeff wrote:
rho = (Z0-Zload)/(Z0+Zload)

SWR = (1+rho)/(1-rho)

Your application of the above equations neglects the first discontinuity.

The first discontinuity (inside the source) doesn't
have any effect on the SWR on the transmission line.
Tuners at the source present another discontinuity
and have no effect on transmission line SWR.
--
73, Cecil http://www.w5dxp.com

Your application of the above equations neglects the first discontinuity.

The first discontinuity (inside the source) doesn't
have any effect on the SWR on the transmission line.

The discontinuity is NOT "inside the source", it is at the source to coax
interface, and as such effects the VSWR that the source sees.

Tuners at the source present another discontinuity
and have no effect on transmission line SWR.

There are no tuners involved in the current example.

Jeff