Re: [TenTec] Tennessee Dreamin'

["James Duffer" <dufferjames@hotmail.com>]

You measure field strength with a field strength meter.....no test leads a 
calibrated antenna, and I can't say what the impedance of "free space" would 
be where you have positioned the field strength measuring antenna/probe, but 
the key is whatever the impedance is (it will be somewhere around 377 ohms 
affected by objects, ground etc.) it should remain relatively constant 
between the two measurements.  First measurement at one power level, second 
measurement at double power level from the power source.

The first part of the paragraph, the logic was based on an old law called 
Ohm's law.  This basic law can be arranged in such a manner to show that:  
Power (watts) equal the voltage (volts) squared divided by the resistance in 
Ohms.  I just used the values as a matter of convenience.  Sorry that this 
confused you.

The impedance of free space is a  concept and that is one of the jobs of a 
horn type antennat to match the impedance of the waveguide to free space . 
Theoretical and kind of like an isotropic radiator in that it is an ideal 
situation.  Like the impedance of a half wave dipole in free space is 73 
Ohms.  In the real world there are things that affect this value.


> From: "Mike Hyder  -N4NT-" <mike_n4nt@charter.net>
> Reply-To: tentec@contesting.com
> To: <tentec@contesting.com>
> Subject: Re: [TenTec] Tennessee Dreamin'
> Date: Mon, 2 Aug 2004 21:53:53 -0400
>
> Maybe if you rewrite your response it'll make more sense to me.  As it is,
> my mind doesn't follow your logic.  By the way, where did you hook your
> meter to measure free space's impedance as 377 ohms?
>
> If we're guessing, though, I'd say that doubling ones power would seem to
> more than double the received signal voltage.  Maybe you're an engineer and
> can explain why it wouldn't or maybe someone else can answer my question.
>
> Mike N4NT
>
> ----- Original Message -----
> From: "James Duffer" <dufferjames@hotmail.com>
> To: <tentec@contesting.com>
> Sent: Monday, August 02, 2004 8:14 PM
> Subject: Re: [TenTec] Tennessee Dreamin'
>
>
> You didn't mention any change in impedance so assuming an impedance of 10
> Ohms, a voltage of 10 volts would then result of power of 10 watts.  Now
> doubling the power to 20 watts, and assuming no change in resistance the
> resulting voltage would be 14.14 volts so the answer to your question, if 
> it
> was a question, is no, the voltage did not quadruple, it only increased by
> the square route of the increase in power ratio.
>
> powe ratios result in a 3 dB change for doubleing, voltage ratios (assuming
> equal impedance) would be 6 dB for a doubling.  Now when power is doubled
> measured at the 50 ohm output, what would be the change field in strength
> (volts/meter) measured at the impedance of free space (377 Ohms) My guess
> would be 3 dB. What is your guess?
>
>
> >From: "Mike Hyder  -N4NT-" <mike_n4nt@charter.net>
> >Reply-To: tentec@contesting.com
> >To: <tentec@contesting.com>
> >Subject: Re: [TenTec] Tennessee Dreamin'
> >Date: Mon, 2 Aug 2004 19:47:08 -0400
> >
> >Since we measure signal strength in volts and power in watts and since
> >watts
> >vary with the square of the voltage, are your 6dB per S-unit dB of 
> voltage
> >difference or of power difference?  In other words, if I double my power 
> am
> >I quadrupling my volts?  Inquiring minds want to know.
> >
> >Mike N4NT
> >
> >____________________________
> >
> >It's even better than that. One S unit is 6db so 25 watts is one S unit
> >less
> >than 100 watts.  So an S9 signal at 100 watts is just a frog's hair under
> >S8
> >at 20 watts.
> >
> >Carl Moreschi N4PY
> >Franklinton, NC
> >
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>
>
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