[ARC5] dB Power Apples and dB Voltage Oranges, was Re: Selectivity Ratings...

Bruce Long coolbrucelong at yahoo.com
Tue Oct 29 18:15:38 EDT 2013 

There seems to always be a lot of confusion when using Decibels with respect to 10 log(a/b) or 20 log(a/b)
This stuff confused me as well until I realized Decibels are only properly used to present a power ratios.

So in my way of thinking the only proper decibel formula is 10 Log (P1/P2)  where P1 and P2 are power levels in Watts

Transmitter harmonic rejection is then 10 Log (P1/P2) with the harmonic power P1 and the total transmitter power as P2

It is also possible to specify P2 as a given, specific power level 
examples  dBm is the logarithmic power ratio with respect to 1 mW
                dBw is the logarithmic power ratio with respect to 1 Watt

dBc is the logarithmic power level with respect to the carrier power

Of course power ( in Watts) is equal to the RMS voltage squared, divided by the load resistance

This means 10 Log (P1/P2) can also be written as  10 Log (V1^2/R1  /   V2^2/R2)  where V1^2/R1 is the power dissipated in resistor R1 by the voltage V1

If R1 = R2  then the resistance symbols in the equations above cancel  and the voltage squared function can be brought through the Log function as a factor of 2 leading to  20 Log(V1/V2) but this is still a power ratio even though it looks like a voltage ratio.

Two things to keep in mind here.  20Log(V1/V2) is only proper if both voltages are measured across the same value of load resistance a rule often violated to much confusion.  Second the formula still represents the ratio of two powers even through the formula uses voltage instead of watts.


Modern professional usage it seems to me conforms to the "power only" rule stated above.  I have only ever found one exception to this usage and it was obviously a mistake.

However early usage-- up to sometime in the 60's it seems to me often violated the "Power ratio" rule.  Often it was assumed the reader would know what was meant as 20 log (v1/V2) was often used even in cases where R1 was not equal to R2 as in receiver gain.  I think part of this is there was no easy way to measure RF power directly until HP introduced their first milliwatt meter in the early 60's so the use of voltage gain was quite comman.

So my advice is always to think of dBs as power ratios and tread carefully when dealing with technical literature or common usage pre-1960's

Hope this is helpful   bruce


                






On Tuesday, October 29, 2013 4:56 PM, Christopher Bowne <aj1g at sbcglobal.net> wrote:
 
When you are measuring the selectivity of a circuit, you typically are looking at a voltage measurement.  dB change relative to a reference voltage follows the equation:

dBx = 20 X log(vx/vref)

so a doubling of a measured voltage relative to a reference voltage to a results in a 6 dB increase.

Power changes follow the equation dBx = 10 X log (px/pref)

so a doubling of power will result in a 3 dB increase.

So if you are running 100 watts output to you antenna, and then put your 1KW output linear amplifier on service, you power output will increase by 10 dB.

The S-meter reading  at a receiving station, however, assuming that it has a 6 dB per S unit calibration that is measuring a voltage, will show an increase of 20 dB, or over 2 S units.  Same result for a field strength meter.

I think I just convinced myself that, as far as the effects observed on a receiving station's S meter are concerned, adding 10 dB of power gain with an amplifieris twice as effective as adding 10 dB of antenna pattern gain, since  unless I am missing something, antenna gain measurements are made comparing differences in received voltages of a field strength meter.

It all goes back to the power equation Power in Watts = Voltage Squared/Resistance.








________________________________
From: D C _Mac_ Macdonald <k2gkk at hotmail.com>
To: Leslie Smith <vk2bcu at operamail.com> 
Cc: ARC-5 List <arc5 at mailman.qth.net> 
Sent: Tuesday, October 29, 2013 11:56 AM
Subject: Re: [ARC5] Selectivity ratings of ARC-5 receivers.


You have to remember that if you double the voltage applied to a fixed resistance/impedance you will also get double the current giving FOUR times the power (6 dB)!


73 - Mac, K2GKK/5 in OKC


Sent from my iPad

> On Oct 29, 2013, at 0:08, "Leslie Smith" <vk2bcu at operamail.com> wrote:
> 
>  Hello Ken,
>  I'll be a goose and make a public fool of myself.
>  Dad always said, "Remember son, there's no bigger fool than a public
>  fool."
>  The units of measurement 2x and 1000x are voltage (measured with a
>  volt-meter or CRO or similar meter)
> 
>  I understand 1 "S" unit equals 6 dB - and in my view this is too
>  small.  
> 
>  "My thinking (your thinking) is that in order to achieve a 3bD
>  increase in signal level at the receiver the transmitter [output
>  power] must be doubled"
>  That's correct, assuming your unit of measurement is power.
>  But at the receiver your unit of measurement is voltage.
>  dB = 20log(E2/E1)  so 20log(1000/1) is 20 * 3 = 60
>  For a 50% increase in voltage level 20 * log (2/1) = 20 * 0.3010 =
>  6dB.
> 
>  Sorry to be long winded
> 
> 
>  73 de Les Smith
>  vk2bcu at operamail.com
> 
> 
>> On Tue, Oct 29, 2013, at 14:31, Kenneth G. Gordon wrote:
>> Reading the ARC-5 maintenance manual, I see that the selectivity for,
>> say, 
>> the R-26 (3 - 6 MHz) receiver is listed at 7.3 Kc at 2X, and 26 Kc at
>> 1000X.
>> 
>> Now 1000X is 60 db, but isn't 2X 3db and not 6db?
>> 
>> Color me confused at this point.
>> 
>> My thinking is that in order to achieve a 3db increase in signal level at
>> the 
>> receiver, the transmitter must be doubled in output power, given
>> identical 
>> conditions.
>> 
>> Or am I all wet here?
>> 
>> Ken W7EKB
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