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Author Topic: SOLVING POWER DECIBELS  (Read 5413 times)
K0YNE
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« on: April 29, 2009, 09:21:01 PM »

ABOUT 7 YEARS AGO I POSTED ON EHAM AN ARTICLE THAT I THOUGHT WAS INTERESTING IN
DETERMINING CHANGES IN RF WATTS OF POWER. MOST OF THE TIME WE DEAL WITH A GAIN OF
3 DB WHICH TRANSLATES IN A DOUBLING OF THE POWER. CONVERSELY, A LOSS OF 3 DB MEANS A
HALVING OF THE POWER. WHAT DO YOU DO TO SOLVE THE EQUATION WHEN A GAIN OF 1 OR 2
DB'S ARE NEEDED? 

EXAMPLE:
 
A TRANSMITTER HAS 100 WATTS OUT. THE FEED LINE HAS A 3 DB LOSS. THE ANTENNA
HAS A 4 DB GAIN.

WHAT IS THE EFFECTIVE RADIATED POWER OF THE ANTENNA?

A. 200 WATTS
B. 186 WATTS
C.  90 WATTS
D. 126 WATTS

OF COURSE THE ANSWER IS 126 WATTS.

THE KEY TO THE EQUATION IS THE NUMBER 1.26.....
WHEN 1.26 IS MULTIPLIED BY THE POWER OF 100 WATTS.
IN THIS EXAMPLE OF126 WATTS. THIS EQUALS THE 1 DB GAIN IN THE
FEEDLINE AND ANTENNA CONFIGURATION.  THE LOSS IN THE
FEEDLINE WAS OVERCOME SLIGHTLY BY THE ANTENNA'S GAIN OF
1 DECIBEL.

IN THE EXAMPLE, IF A 1 DB LOSS WAS INCURRED THEN THE 100 WATTS
WOULD BE DIVIDED BY 1.26.  AND THE EFFECTIVE RADIATED POWER
WOULD HAVE BEEN 79 WATTS.
TO FIGURE A LOSS OF 2 DB'S, DIVIDED 100 BY 1.26 EQUALS
79 AND AGAIN DIVIDE 79 BY 1.26 EQUALS 62
 THE FRACTIONS WERE
NOT COMPUTED, BUT THE GENERAL IDEA IS TO REMEMBER
1.26 IN DETERMINING HOW MUCH DIFFERENCE IN POWER
GAIN OR LOSS,YOU HAVE WITH VARIOUS ANTENNAS AND
FEEDLINES.
LASTLY COVERING GAINS, 100 WATTS TIMES 1.26 EQUALS 126 WATTS,
AND 126 WATTS TIMES 1.26 EQUALS 158.76 AND GOING ANOTHER
DB , 158.76 TIMES 1.26 EQUALS 200 WATTS.
 YOU CAN CHECK THIS OUT TO VERIFY ITS ACCURACY.
SINCE THE REAL NUMBER OF 1.26 MAY HAVE NUMBERS IN THE THOUSANDS
PLACE, INSTEAD OF JUST 1.26 A SLIGHT DIFFERENCE MAY RESULT.
JUST A SHORT CUT WHEN YOU FIGURE THE WATTS AND ANTENNA GAINS
AND LOSSES IN THE EXTRA CLASS TESTS. PAUL K0YNE







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KCWA
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« Reply #1 on: May 01, 2009, 10:38:01 AM »

Great article Paul....... Have printed it out for future reference....... thanks, Woody..KCWA
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