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(Svante) wrote in message

Now we're talking, these were the nuts and bolts I wanted to see. I
don't know if I think that this is easier than the RLC implementation,
but I guess that would depend on the computer implementation that is
available. One thing in the above explanation that seems akward is
that it seems like you have to have a TABLE of the s-parameters for
each and every frequency? In the RLC model it is easy to calculate the
impedances for ANY frequency. Is there any method to CALCULATE the
S-parameters, rather than having a large table?
Another clarification, the S-parameters above are they given as
magnitude/phase pairs?

Yes.

I assume that the two numbers that appear for
each S-parameter in some sense describe real and imaginary parts, or
amplitude/phase if you wish, is that correct?

Yes.

A circuit analysis program wouldn't use a two-port data file for simulating
a cable. That was just to show that transmission line theory works even at
small wavelenghts.

A circuit analysis program will have a set of rules that define the cable's
performance, at all frequencies.

The user would call up the type of cable, and specify a length for that cable.
The program would then look up the rules for that cable and calculate the
S-parameters (at each frequency) based on the cable's length, and the set of
loss rules the programer gave.

The way the program comes up with the S-parameters (for each frequency) is simple:

S21 and S12 are the loss (and angle) for that length of cable. This loss
is calculated based on the measured performance of a real cable, (when terminated
by a resistor of the characteristic impedance.)

S11 and S22 are set to -120 dB , 0.0 deg. S11 and S22 should be infinity if
the cable is terminated in it's characteristic impedance. My old computer doesn't
have an "infinity key", so I just use -120 dB at 0 degs.

My previous message was getting too long. I but wanted to show proof that a
two port, S-parameter file can simulate speaker cable accurately at 20 Hz.

A 40 ft length of line was choosen because that is slightly less
than 1 / 1,000,000 of a wavelength, at 20 Hz. People continue to believe
that transmission line theory can not accuratly produce a model of less than
1/100 W.L. I thought it would be good to show that transmission line theory can
model a cable that is less than one-millionth of a W.L.

Below is a two-port S-parameter data file, that will simulate 40 ft of
12 gage, speaker cable, at 20 Hz:

! 12 gage "speaker cable

# HZ DB S R 100

! Hz S11 deg, S21 deg, S12 deg, S22 deg

19 -120 0 -6.95e-3 -278.2e-6 -6.95e-3 -278.2e-6 -120 0
20 -120 0 -6.95e-3 -292.8e-6 -6.95e-3 -292.8e-6 -120 0
21 -120 0 -6.95e-3 -307.2e-6 -6.95e-3 -307.5e-6 -120 0

Two-port file

--------------
| |
------------|o o|----------
| | | |
Gen | | 4 Ohms
| | | |
------------|o o|----------
| |
--------------

The file was based on a cable resistance of 0.002 Ohms / ft, and a
characteristic impedance of 100 Ohms.

Running the analysis gave the following results:

20 Hz. Loss: 0.9804 (-0.17 dB) Z(in): 4.08 + j 0.0005

Now let us now check the results by using simple dc calculations.
We can use dc calculations to find the cable loss, at 20 Hz, because the
speaker cable's XL is so small it can be almost neglected.

R = 0.002 Ohms * 40 ft = 0.08 Ohms

-----0.08 Ohms-----
| |
Gen 4 Ohms
| |
------------------

Z(in) = 4.00 + 0.08 = 4.08 Ohms

Loss = 4/(4 + 0.08) = 0.9804 (-0.17 dB)

So, transmission line theory agrees with dc calculations.

Bob Stanton