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#1
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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. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.498 / Virus Database: 297 - Release Date: 7/8/03 |
#2
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![]() Jim, And also only 50% modulation... 'Doc |
#3
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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%. ============= "...but I admitted I was wrong, Like a man! Something you and QRM have a problem with. You guys are wrong and you both know it and are both too small to admit it." ---- Twistedhed ---- ============= -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
#4
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![]() 'Doc wrote: Jim, And also only 50% modulation... 'Doc How do you arrive at that figure? |
#6
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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. |
#7
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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...) |
#8
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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. |
#9
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: Scott Unit 69 wrote:
: Although they teach AM wrong, like the question on the amateur exam. Is the Amateur Exam question published by the ARRL..? How about telling us what the question is as stated and what the correction is. It's pretty hard to publish large books and get everything right. Did they correct it later on...? : For AM look he : http://www.cbtricks.com/~kd6tas/amsig.htm Bill's web page bores me to tears... : 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. have to see if a copy can be dug up... cheers skipp http://sonic.ucdavis.edu |
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