On Thu, 28 Aug 2003 07:20:28 GMT, Richard Clark
wrote:

The scenario begins:

"A 50-Ohm line is terminated with a load of 200+j0 ohms.
The normal attenuation of the line is 2.00 decibels.
What is the loss of the line?"

Having stated no more, the implication is that the source is matched
to the line (source Z = 50+j0 Ohms). This is a half step towards the
full blown implementation such that those who are comfortable to this
point (and is in fact common experience) will observe their answer and
this answer a

"A = 1.27 + 2.00 = 3.27dB"

"This is the dissipation or heat loss...."

we then proceed:

"...the generator impedance is 100+0j ohms, and the line is 5.35
wavelengths long."

Beware, this stumper has so challenged the elite that I have found it
dismissed through obvious embarrassment of either lacking the means to
compute it, or the ability to simply set it up and measure it. It
takes two resistors and a hank of transmission line, or what has been
described by one correspondent as:
There is no institutionalized ignorance, just a
lot of skepticism regarding the reliability of the
analysis methods and the measurement methods.

Hi All,

One emailer provided the following solution:
Total loss = 2.0 + 1.26639 + 1.57316 = 4.83955 dB
which varies from the citation by about 0.06dB.

The math offered:
1.Effective length of 50-ohm coax is 5.35 l – 5.0 l = 0.35 l = 126°.
2.Using W2DU program HP2, Page 15-35, Reflections, 1st ed.,
input impedance on 50-ohm line = Zin =31.10025 – j26.14159 ohms.
3.Using W2DU SWR program (Appendix 4, Page Appendix 23.7),
mismatch between the 100 -ohm generator and impedance
ZL = 31.10025 –j26.14159 = 3.456687:1
4.The problem now simplifies to determining the insertion loss
due to the mismatch between the 100-ohm source resistance
R of the generator and load ZL = 31.10025 – j26.14159 ohms.
5.Input voltage reflection coefficient rin = 0.55125
6.Input power reflection coefficient r2in = 0.551252 = 0.30388
7.Input power transmission coefficient t2 = (1 – 0.30388) = 0.69612
8.Power transmitted is then 69.612%, insertion loss expressed in dB, 1.57316 dB.
9. Using ITT terminology, P/Pm = 1.43654, which is also 1.57316 dB.

As I said, all achievable through references that many here have
available to them, but fail to read, or practice at the bench through
the angst of low self confidence or being too imbued with the kulture
of institutionalized ignorance. There are other sources that approach
this problem through more elegant means, but they are equivalent to
reaching for the stars when NONE here can pick the fruit from their
own orchard.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"As I said, all achievable through references that many have available
to them---."

Richard Clark`s question was:
"A 50-Ohm line is terminated with a load of 200+ j0 Ohms. The normal
attenuation of the line is 2.00 decibels. What is the loss of the line?"

The ARRL Antenna Book is a reference designed for radio amateurs and
available to many..

(Eq 13) on page 24-9 gives: SWR = R/Zo, or 4 in Richard Clark`s
question.

(Fig 13) on page 24-10 gives a reflected power of 36 W from a 4:1
mismatch when 100 W is applied.

(Fig 14) on page 24-11 gives an additional 1.2 dB loss when an SWR of 4
applies to a cable with a matched loss of 2 dB. Total is 3.2 dB loss.

The 19th edition of the "ARRL Antenna Book" eliminates most of the
arithmetic needed to calculate loss added by SWR. Just use the
convenient figures on pages 24-10 and 24-11.

Next to Terman`s "Radio Engineering", I think the ARRL`s "Antenna Book"
is the best reference you can have on your shelf.

Best regards, Richard Harrison, KB5WZI