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Old July 10th 03, 04:52 AM
JJ
 
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'Doc wrote:

Jim,
And also only 50% modulation...
'Doc


How do you arrive at that figure?

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Old July 10th 03, 06:11 AM
Jim Hampton
 
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Default Modulation percentage

A 1 KW carrier will be modulated 100% with 500 watts of high level
modulation. You end up with 1,000 watts of carrier, 250 watts upper
sideband, and 250 watts of lower sideband. Most inefficient.

73 from Rochester, NY
Jim


"Marco Ferra" wrote in message
om...
Salut

If an audio signal (cosine) with 4 KHz of frequency and 10 V of
amplitude modulates a carrier of 100 MHz, with a percentual modulation
of 85%, does the carrier have 11.76 V of amplitude?

Supposing that the transmitter power is of 1 kW how can I know the
power of the carrier and sidebands? How can I know this values in
Watt, dBW and dBm?

If I just hadn't fluke the exame...

Thank everybody in advance, Marco.



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Old July 10th 03, 07:58 AM
'Doc
 
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Jim,
And also only 50% modulation...
'Doc
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Old July 10th 03, 11:24 AM
JJ
 
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'Doc wrote:

JJ,
500 / 1000 = 50%
'Doc


That is not correct. To 100% modulate an rf carrier the audio
power must equal 50% of the unmodulated carrier power. It takes 50
watts of audio to 100% modulate 100 watts of carrier. This is
assuming high level modulation where the audio is applied to the
output stage. At 100% modulation power is increased 1.5 times the
unmodulated carrier power. So for 100 watts with 100% modulation,
the output would be 150 watts, with 100 watts in the carrier and
25 watts in each sideband. Get a book and look it up.


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Old July 10th 03, 12:06 PM
Frank Gilliland
 
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In , 'Doc wrote:



Jim,
And also only 50% modulation...
'Doc


Wrong. Jim was correct: It takes 500 watts to modulate a 1000 watt carrier to
100%.



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Old July 10th 03, 04:46 PM
JJ
 
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'Doc wrote:

JJ,
Got a book, even read it. You might do the same...
'Doc


I did, some time ago and here is what I found. Just for
you....from "The Radio Amateur's Handbook".

This is concerning the amount of audio power to modulate a carrier.

"The modulator must be capable of supplying to the modulated r.f.
stage sine-wave audio power equal to 50 per cent of the d.c. plate
input. For example, if the d.c plate power input to the r.f. stage
is 100 watts, the sine-wave audio power output of the modulator
must be 50 watts."
"With a sine-wave modulating signal, the average power in a 100
per cent modulated wave is one and one-half times the value of
unmodulated carrier power; that is, the power output if the
transmitter increases 50 per cent with 100 per cent modulation."

From "Elements of Radio", by William Marcus
"It can be calculated mathematically that the average power of the
modulated carrier is one and one half times that of the
unmodulated carrier. Since this 50 per cent increase in power must
come from the modulator, the audio frequency power must be ONE
HALF the unmodulated carrier."

In other words Doc, for 100% plate modulation of a 1000 watt
carrier you need 500 watts of audio. Your calculation that 1000
watts carrier divided by 500 watts audio equals 50 per cent
modulation proves you know nothing about the subject.

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Old July 10th 03, 05:52 PM
Skipp
 
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Que Tal Marco...

: If an audio signal (cosine) with 4 KHz of frequency and 10 V of
: amplitude modulates a carrier of 100 MHz, with a percentual modulation
: of 85%, does the carrier have 11.76 V of amplitude?

If you use a carrier of 100MHz, we would assume you to be in the common US
FM Broadcast band. Many people refer to the audio signal as a sine wave.
(Unless we want to talk cofunctions... probably not.)

: Supposing that the transmitter power is of 1 kW how can I know the
: power of the carrier and sidebands? How can I know this values in
: Watt, dBW and dBm?

At 100% modulation or at 85%..?

You can know the values by assuming a reference or knowing what the
reference is. dB might assume a starting point, reference power level
supplied/measured by you.

dBm often refers to a power level referenced to a milliwatt value across a
termination resistance. Other examples to research might be dBc, dBrnc and
similar labels.

: If I just hadn't fluke the exame...
: Thank everybody in advance, Marco.

Assuming you are working with an AM signal, one of... if not the most
accurate way to measure the complex output signal would be with a scope
properly connected.

The ARRL Handbooks provide a measure of reference information... some of
the vintage handbooks are better at describing AM which was most popular
in the 50's through the 70's.

cheers
skipp

(el gusto es mio...)
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Old July 10th 03, 06:17 PM
Scott Unit 69
 
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The ARRL Handbooks provide a measure of reference information... some of
the vintage handbooks are better at describing AM which was most popular
in the 50's through the 70's.



Although they teach AM wrong, like the question on the amateur exam.

For AM look he
http://www.cbtricks.com/~kd6tas/amsig.htm

This is the correct information which can be verified by reading
chapter 4 in the RF Technician's Handbook, 2nd edition, by
Cotter W. Sayre, 621.38.
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Old July 10th 03, 10:29 PM
'Doc
 
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JJ,
500 / 1000 = 50%
'Doc
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