On 4 mar, 03:28, "Dan Finn" wrote:
"Wimpie" wrote in message

...

Hello Dan,

Can it be that Bob's situation is better with respect to losses?

75 Ohms load in a 50 Ohms system gives reflection coefficient of
(75-50)/(75+50) = 0.2 (voltage or current reflection).

We agree at this point. My SWR of 1.5 was calculated by (1 + rho)/(1-rho) =
1.2/.8 = 1.5. Given that the SWR = 1.5 Vmax/Vmin = 1.5. The mismatch
produces a reflected to forward voltage "gain" on the transmission line of
1.5 to 1. In dB, this gain is expressed as 10log(Vout/Vin). Since this is a
ratio, the impedance cancels, and we know that power is proportional to
voltage squared, thus Ploss would be 2 X 10log(Vout/Vin) = 20 log(1.5) =
3.6dB. If this seems like a "lot" when compared to your 0.18dB, intuitively
it seems right to me since the impedance mismatch is actually fairly large,
i.e the 75 ohm coax has 50% higher characteristic impedance than the 50 ohm
load, even though it ultimately does not make a big difference when compared
to the gain of the back end of the receiver that converts microvolt signals
to millivolt levels.

So the power loss in this transition = 0.2^2 = 0.04 (4%, 0.18 dB).

If I made a mistake, this would explain why. I need to find a reference that
has the derivation showing that rho**2 = power lost. Since rho is unitless
and power is in units of watts, it seems unlikley on the face of it unless
something is being cancelled out that I am unaware of. If you have such a
reference handy, please post it. Otherwise I will try to find it myself. It
has been about 40 years since I took my last RF course so maybe you can
understand if I am a bit rusty ;-)

The absolute worst-case situation would be when you have a quarter
wave line between a 50 Ohms source and a 50 Ohms load. The 75 Ohms
quarter wave line with 50 Ohms load will show a real impedance of
75^2/50 = 112.5 Ohm. This results in a reflection coefficient of
0.38, hence a power loss of 0.15 (0.7 dB). Of course when the line is
0.5 lambda long, there will be no additional loss.

I think that, other than resistive losses, the length of a transmission line
has absolutely nothing to do with the SWR (and power loss due to mismatch)
OTHER THAN the impedance discontinuity of the match at the antenna
feedpoint. In the above example I assumed a perfect match and was only
considering the mismatch of the load to the characteristic impedance of the
coax.



Best regards,

Wim

73 de AI4QJ

Hello Dan,

To help you a bit: Assume a source with 1Veff EMK and 50 Ohms output
impedance.

Calculate the power dissipated in a load of 50 Ohms. When you do this,
you will find 5mW dissipated into the load. 0.5Veff will be across the
load.

Now, do the same for a 75 Ohms load, you will find 4.8 mW (that is
0.18 dB below 5mW). So 0.2mW (4%) is reflected back to the load. Check
you SWR instrument, probably it gives both SWR and reflect power on
the scale.

Note that this calculus is for linear systems, hence not valid for
most power amplifiers because these want to see a 50 Ohms load, but do
not show a 50 Ohm output impedance in general.

Regarding length of transmission lines. When using a 75 Ohms line in a
50 Ohms system, you have two transitions where reflections occurs; at
the load and at the source. These reflection will interfere and
depending of the amount of wavelength between the load and source,
constructive or destructive interference occurs. Search for quarter
wave transformer. When you put two, quarter wave transformers (with
same cable impedance) behind each other, the input impedance will be
equal to the load impedance (assuming loss free cable).

When the VSWR in the cable becomes excessive (high voltage and high
current regions), the effective cable loss will increase (with respect
to the published values).

Best regards,

Wim
PA3DJS
www.tetech.nl (Dutch)