OT impedance matching

I thought you would only get standing waves if the length of the cable matched a multiple of the wavelength.

If the round trip length is such that the reflected wave returning to

smaller wave that falls on the load and less power into the load. Nothing you list above addresses this possibility. Just saying that energy is conserved does not explain what happens.

In this situation I believe the source will see the 75 ohm load as if the cable were not there. Consider what happens if the load were open. The incident wave travels down the cable drawing current from the source as if it were a 50 ohm load. On reaching the open end the wave is reflected with the same polarity and amplitude. This reflected wave reaches the source and the current drawn from the source stops.

With a 75 Ohm load the reflected wave will be smaller and the final current drawn will be what the load by itself would draw.

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Rick

You're welcome. It has a couple of annoying glitches, but I still found it so useful that I paid for the license.

I made a large mistake on the T values I gave you upthread. I used

10dB/meter for the coax loss and should have used .3dB/meter.

Revised numbers: Ant to T: about 0.36 wavelengths (electrical) Open stub: about .06 wavelengths (electrical) T to radio: whatever

My apologies for the mistake.

Nope. You get standing waves when either the source or the load is not matched to the cable. It doesn't need to be any specific multiple or sub-multiple of the electrical wavelength of the coax cable. Any length will suffice.

Note that the "V" in VSWR means voltage, which is the usual measure of max and mins in the standing wave pattern. VSWR = Vmax / Vmin When VSWR = 1 the coax is properly terminated.

I'm not totally sure of the following, but I'll take a chance that it's correct.

The problem is that you're using the voltage as an indication of the power at any point on the coax cable. That doesn't work because the impedance seen by the voltmeter is not constant along a badly terminated coaxial cable. The key is the current through the coax cable is the same at any point along its length. So, if looking at the impedance of a 1/4 wave electrical length of coax looks like a rather high impedance, and the current was constant, then the voltage at the same point will be rather high. Similarly, if the coax cable length were 1/2 wave electrical length, which looks like a rather low impedance, the corresponding voltage at that point will be low. Yet, if you calculate the power at these points using: Power = Voltage^2 / Impedance you'll find 100% of your input RF power, which will be the same at any point along the coax. Conservation of energy and all that.

The trap that everyone (including me) falls into is using a voltmeter as an indicator of power. For example, the common Bird 43 wattmeter is really a directional coupler feeding a voltmeter. It's only accurate with 50 ohms in and out. Put a complex source or load on the Bird wattmeter and the directional coupler looses directivity and the voltage reading is horribly inaccurate. Move the Bird 43 to various points along a transmission line that has a high VSWR, and you'll see the max and min voltages as in standing waves.

Well, I tried. Is the above any better?

If all else fails, a call to authority is always useful.

"Understanding SWR by Example. Take the mystery and mystique out of standing wave ratio."

At cable length multiples of 1/2 wave electrical, that's exactly what will be seen at the other end of a 50 ohm cable. Every 1/2 wave length of coax makes the load look like it was sitting right at the source. However, cable lengths in between will require a Smith chart to provide the complex impedance seen by the source.

Mot quite. The first problem is the length of the cable. A very short unterminated 50 ohm cable will not look like 50 ohms. Instead, it will more closely resemble an open circuit or a no load condition to your signal source. An infinitely long 50 ohm cable, will look like a 50 ohm load because the mistermination is so far away from the source, that the reflection will never be returned to the source.

The 2nd problem is that you're modeling the coax cable as if it were a water pipe full of standing water. You don't fill a coax cable full of RF and then take your time measuring the voltages and powers. RF, like water, flows through the cable. If the current ever stopped, so would the flow, and most of your model would not work. A loss less unterminated 1/2 wave (and multiples thereof) electrical cable will certainly reflect all the power back to the source. However, the current doesn't just stop when this happens. It continues to flow out of the xmitter, through the loss less coax, bounces off the open circuit, returns through the loss less coax, and is dissipated in the xmitter output stages, probably destroying them in the process. If the current stopped at the xmitter under these conditions, there would be no RF at the open circuit load for it to reflect.

If I change the original configuration to a perfect voltage source, 50 ohm equivalent series source impedance, 50 ohm coax, and a 75 ohm load, which is more realistic, 4% of the delivered power is reflected and then dissipated in the 50 ohm equivalent source impedance. That's because the source is matched to the transmission line so there's no

2nd reflection from the source. Not exactly the same, but close enough to function adequately without requiring a 50 to 75 ohm matching network.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

On a sunny day (Sat, 13 Sep 2014 19:48:54 -0500) it happened John S wrote in :

OK, got it, wil take some time to read all that stuff, interesting guy, have to read his other stuff too ..

On a sunny day (Sat, 13 Sep 2014 21:26:39 -0500) it happened John S wrote in :

No problem, had not build it yet, will try to work it out from the chart myself.

>

No, that isn't true. The current is *not* the same everywhere along its length.

Both voltage *and* current show standing wave patterns. Voltage maxima coincide with current minima, and the reverse.

Jeroen Belleman

I use a six hole ferrite bead and wind a 5:6 turns ratio transformer to get a 52*1.44=74.88 ohm match. No where near the loss of a minimum loss pad.

--
Anyone wanting to run for any political office in the US should have to 
have a DD214, and a honorable discharge.

Sure, but what's your flatness and bandwidth? A resitive minimum loss pad is flat down to DC with essentially unlimited bandwidth. For a mere 5.72dB loss, I don't have to worry about flatness and bandwidth. I'll take the loss.

CATV Minimum Loss Pad for 75 Ohm Measurements

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

It was flat across the 54 to 300 MHz range that I used it in the lab. (36 channel, with mid and super band.) There was no need for a DC specification.

--
Anyone wanting to run for any political office in the US should have to 
have a DD214, and a honorable discharge.