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[AMPS] The Worldwide, No Holds Barred, SWR Quiz.

To: <amps@contesting.com>
Subject: [AMPS] The Worldwide, No Holds Barred, SWR Quiz.
From: jono@enteract.com (Jon Ogden)
Date: Mon, 4 Oct 99 16:07:37 -0500
Gilmer, Mike wrote:

>> Source impedance is 50 Ohms.  Load impedance is 50 Ohms.  The
>>transmission line is a length where even you agree that the transformed
>>impedance at the end of it is 50 Ohms.  So where is the impedance
>>mismatch?????????
>> 
>
>At the load end.  93 ohms to 50 ohms.  Seems clear to me!!!!!!!!!

IF THE SOURCE WAS 93 OHMS, YES!  But you are dealing with a 50 OHM source!

----------                                                  ------------
|        |         ----------------------------             |          |
| 50 Ohm |---------|   93 Ohm 1/2 wave coax    |-------------| 50 Ohm   |
| Source |         ----------------------------             | load     |
----------                                                  ------------

You have to think of the entire system.  The 93 Ohm line acts as an 
impedance transformer.  The impedance at the input of that line is 50 
Ohms.  Think of the 50 Ohm load and the 1/2 wavelength 93 Ohm line as 
that black box.  If the impedance is 50 Ohms at the input to the 93 Ohm 
cable, we have no mismatch!  It matters not what happens between the 93 
Ohm line and the 50 Ohm cable.


>2. Isn't the fact that Zload not equal Zcable what creates the standing
>waves on the cable?

That isn't what the question really is.  There likely are standing waves 
on the 93 Ohm cable.  What the question was, what the SWR is at the 1/2 
wave point in a 50 Ohm system.

It's called an impedance transformer.  

Do you use stacked Yagis in your station, Mike?  Do you know anyone that 
does?  The stacking harness for a 2 stack array uses 75 Ohm cable.  Yet 
the antenna feedpoints are 50 Ohms.  Gee, where is the SWR then?   You 
transmit into your array and you see 1:1 SWR. Same principle, Mike.

>
>Are Rich and Jon saying that the mismatch of the 50 ohm SWR bridge to
>the 93 ohm cable (at the near end) create standing waves that cancel the
>original ones (created at the far end) and yield a 1:1 condition?

That's probably what happens as you have a dual mismatch situation.  The 
reflected waves are likely out of phase and cancel.  I've not done the 
voltage and current analysis on this, but that's likely what happens.

The beauty of using the Smith Chart is that you don't have to do all that 
math.  You can solve the problem graphically instead.

73,

Jon
KE9NA



--------------------------------------------------------------------------
The Second Amendment is NOT about duck hunting!


Jon Ogden

jono@enteract.com
www.qsl.net/ke9na

"A life lived in fear is a life half lived."


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