Jim,
PLEASE read this not in the tone of a friendly discussion on theory. There is
NO repreat NO meaness in this response. I enjoy a good argument (logical
definition of argument). I can be swayed with logic and will admit it when I
am in error, and that is often enought the case.
> But Ronald, the R in R+/-jX in the coax cable's characteristic impedance
> is what is 50 ohms, not the loss resistance of the wire! The R which
> is 50 ohms in the characteristic impedance statement is the "effective"
> R which would be calculated form the "apparent" resistance resulting
The R in R+/-jX IS the loss resistance of the wire. A lossless t-line is
expressed solely in C and L and consists only of X. Z=SQT(R**2 + X**2). A
lossy line throws in the series R loss
> from the necessity of the line appearing to the RF signal as a series
> of inductors, each of which has a shunt capacitor across to the other
> conductor of the line. The series inductor limits the rate at which
> the shunt capacitor can charge; this sets up a particular voltage
> and current relation of the line which is entirely controlled by
> the line geometry; in this way the line has an "apparent" resistance,
> which is called the characteristic impedance ( in the old days,
> it was called the surge impedance reflecting the idea of the limitation
> to the line capacitance charging time).
Agree 100%.
>
> This R of the line impedance that the RF sees when the reactance terms
> are equal and opposite (that is the antenna system is "tuned") is NOT
> the resistance that causes RF signal loss. The R that causes loss is
> approximately the resistance measured with a DC ohm meter.
Along with dielectric losses.
> This is the R that is reduced with parallel shielded lines.
Yes but only 1/2 the signal.
> Forget that one is using a piece of coax
> as an easy way to get a shielded line; it is not wired up as, and
> does not behave as coax cable does, that is the entire point of
> using four pieces of coax in a parallel/parallel configuration to
> realize a very low loss transmission line.
Yes the entire point of using 4 lines for a PARALLEL TWIN-LEAD feed.
>
> Were the R that causes signal loss in a 50 ohm characteristic
> impedance transmission line really 50 ohms, a 1.5kW signal
> travelling down the line would try to dissipate the entire
> 1500 watt signal in the line; nothing would reach the antenna.
> Note: 50 ohm coax line fed from and delivering to matched 50
> ohm source and load impedances will have an rms voltage on the
> line of 274 volts and an rms current flow of about 5.5 amps, so
> 5.5 amps squared X 50 ohms = 1500 watts. Obviously this is
> not the way the transmission line operates!
Very true!! It also shows that the R in R+/-jX is not the characteristic
impedance either as you state in your first comment. The Z HAS to be VASTLY
composed of reactive elements in order not to disipate power.
>
>
> And, BTW, the signal does not make a round trip when traversing
> parallel shielded lines, it does travel through loss resistances
> which are equal, and in parallel, so the loss is cut in half
> using two lines, and to 1/4th using four lines.
In an unbalanced line all the signal is presented on the center conductor
referenced to the shield. On a balanced feed 1/2 the signal is presented on
one conductor and 1/2 on the other conductor 180deg out of phase. 0.5V**2*R +
0.5V**2*R = V**2*R. The R would be constant between the two cases.
In the two pair scenario the loss would be 1/2 since the above condition would
occur in parallel. As well the impedance would be 1/2.
--
73 de KK1L ex N1PBT...ron (rrossi@btv.ibm.com) <><
Ron Rossi H/P SRAM Engineering -- IBM Microelectronics
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