On Sat, 14 Apr 2007 22:31:37 GMT, Owen Duffy wrote:

Walter Maxwell wrote in
:

Summarizing reflection coefficient values at stub point with stub in
place:
Line coefficients: voltage 0.5 at +120°, current -60° (y = 1 + j1.1547)
Stub coefficients: voltage 0.5 at -120°, current +60° (y = 1 - j1.1547)
Resultant coefficients: voltage 0.5 at 180°, current 0.5 at 0° WRONG
Resultant coefficients: voltage 1.0 at 180°, current 1.0 at 0° CORRECT

Walt,

Though admittance or impedance at a point on the mismatched line are
calculated from the underlying Zo and the reflection coefficient
corrected for line loss, they are easier to work in than the raw
reflection coefficient.

Depends on the instrumentation available for obtaining the raw data.

It is easier to explain why the stub is located at a position where Yn'=1
+jB than where Gamma=0.5120 (assuming lossless line). It is relatively
obvious that where Yn'=1+jB, a shunt reactance of -jB from a s/c or o/c
stub will leave Yn=1 which is the matched condition.

Re your worked solution (above), I agree that the normalised admittance
looking into 30deg of line with load 16.667+j0 is about 1-j1.1547 (not
the different sign).

Yes Owen, you're right. I added the y values at the last moment, and didn't catch the errors. Both the line
and stub signs are reversed. Sorry 'bout that.

I make the normalised admittance looking into the stub about 0+j1.15 (and
the reflection coefficient about 0.5-98, how do you get 1+j1.15?

Normalized y looking into the stub directly is y = 0 + 1.1547, but looking at the stub while on the line at
the 30° point is y = 1 + 1.1547. To view the stub separately on the line the line is terminated in 50 ohms,
because the real component of the line impedance at the match point is 50 ohms.

The addition of the two normalised admittances 1-j1.15 + 0+j1.15 gives 1
+j0 which is the matched condition.

Of course.

The design is correct, the stub results in a match at the stub connection
point (irrespective of what is connected on the source side of the
point), but I can't understand your maths above (allowing for the sign
error that I think you have made).

The resultant coefficients are obtained by simply adding the voltage coefficients and the current
coefficients, as in the adding of the line and stub admittances.

Is the reflection coefficient explanation a clearer explanation than
using admittances?

Not at all, Owen, but as I said in the original post, the instrument I acquired early in my time with the RCA
antenna lab was the PRD-219 reflectometer. At that time I considered it the best instrument for measuring
transmission line circuitry at VHF, and because it delivered readings in reflection coefficient I became
somewhat more efficient in my thinking process using that mode.

Walt





















Owen

BTW, my line loss calculator solutions (http://www.vk1od.net/tl/tllc.php)
for Belden 8262 RG58 (you said RG53, but you probably meant RG58) a

(Note some symbols arent supported by plain ascii and appear as '?'.)

Load to Stub connection:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.030 metres
Zload 16.67+j0.00 ?
Yload 0.059999+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 29.97 ?, 0.083 ?
Line Loss (matched) 0.059 dB
Line Loss 0.149 dB
Efficiency 96.63%
Zin 22.12+j24.55 ?
Yin 0.020258-j0.022480 ?
Gamma, rho, theta, VSWR (source end) -2.44e-1+j4.29e-1, 0.493,
119.6?, 2.950
Gamma, rho, theta, VSWR (load end) -5.00e-1+j4.03e-3, 0.500, 179.5?,
3.000
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884

Stub:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.685 metres
Zload 100000000.00+j0.00 ?
Yload 0.000000+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 49.02 ?, 0.136 ?
Line Loss (matched) 0.096 dB
Line Loss 40.574 dB
Efficiency 0.01%
Zin 0.50-j43.44 ?
Yin 0.000265+j0.023019 ?
Gamma, rho, theta, VSWR (source end) -1.37e-1-j9.69e-1, 0.978, -
98.0?, 90.720
Gamma, rho, theta, VSWR (load end) 1.00e+0+j1.07e-8, 1.000, 0.0?, inf
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884

Walter Maxwell wrote in
:

....
Re your worked solution (above), I agree that the normalised
admittance looking into 30deg of line with load 16.667+j0 is about
1-j1.1547 (not the different sign).

Yes Owen, you're right. I added the y values at the last moment, and
didn't catch the errors. Both the line and stub signs are reversed.
Sorry 'bout that.

Ok.

I make the normalised admittance looking into the stub about 0+j1.15
(and the reflection coefficient about 0.5-98, how do you get 1+j1.15?

Normalized y looking into the stub directly is y = 0 + 1.1547, but
looking at the stub while on the line at the 30° point is y = 1 +
1.1547. To view the stub separately on the line the line is terminated
in 50 ohms, because the real component of the line impedance at the
match point is 50 ohms.

You have a junction where three current paths appear in parallel, we can
add the admittances of each of those paths.

We are agreed that admittance of the load+30deg line is 1-j1.15, and that
of the stub is 0+j1.15, so the only place the additional 1+j0 can come
from is the source+line branch.

If that is the case, then your explanation of the stub (which I assume to
be a steady state explanation because you are talking about frequency
domain admittances), depends on the source admittance (or impedance). If
the equivalent source impedance at the junction figures in the calcs, you
are saying that the VSWR on the line from source to junction depends on
the source impedance... I thought we got over that error.

My view is that the stub in shunt with the 30deg line+load results in an
equivalent impedance of approximately 50+j0 at the junction, irrespective
of what is on the source side of the junction.

Owen