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'Doc wrote: Jim, And also only 50% modulation... 'Doc How do you arrive at that figure? |
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. --- 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 |
Jim, And also only 50% modulation... 'Doc |
'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. |
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! =----- |
'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. |
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...) |
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. |
JJ, 500 / 1000 = 50% 'Doc |
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'Doc wrote: JJ, Got a book, even read it. You might do the same... 'Doc What book is that? Go get The RF Technician's Handbook, by Cotter W. Sayre, 621.38. Read that, and you might be an expert. : |
JJ, I knew enough to make you quote it! (LOL) Sometimes it's fun to argue the 'wrong' side. Also saves a lot of typing (sorry 'bout that), and people remember it longer... 'Doc |
In , 'Doc wrote:
JJ, I knew enough to make you quote it! (LOL) Sometimes it's fun to argue the 'wrong' side. Also saves a lot of typing (sorry 'bout that), and people remember it longer... 'Doc Lame. Hey Doc, maybe you should take a closer look at my sig.... ============= "...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! =----- |
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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 |
Scott didn't you say hams bored you and you would never be a Ham?
Guess Ill have to dig up that post. "Scott Unit 69" wrote in message ... : 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...? This question was noted in one of the Elements I took on Sunday. I don't remember if it was in 3 or 4. I took them back to back. : For AM look he : http://www.cbtricks.com/~kd6tas/amsig.htm Bill's web page bores me to tears... No one said technical stuff was the funny pages... : 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... It is a very good book. Anyone interested in radio should read it. |
"sheik yerbooti" wrote:
wrote in message ... That sig file would get you flamed for its length in some towns, Twist. ;) Your too much of a pussy and kiss ass to say anything to him . I've told him how much I like webtv, I had one, so I can say. You're too much of a retard to be sociable to anyone, leg humper. -- GO# 40 ------------------------------------------------------------- http://NewsReader.Com/ 50 GB/Month |
wrote in message ... "sheik yerbooti" wrote: wrote in message ... That sig file would get you flamed for its length in some towns, Twist. ;) Your too much of a pussy and kiss ass to say anything to him . I've told him how much I like webtv, I had one, so I can say. You're too much of a retard to be sociable to anyone, leg humper. You never let m e down as usual, you have succesfully chased me around again, as usual. |
wrote in message ... "sheik yerbooti" wrote: wrote in message ... That sig file would get you flamed for its length in some towns, Twist. ;) Your too much of a pussy and kiss ass to say anything to him . I've told him how much I like webtv, I had one, so I can say. You're too much of a retard to be sociable to anyone, leg humper. -- GO# 40 Never a bucket of water around when you need one, LMAO!!!! Landshark -- Try these to learn about newsgroup trolls. http://www.io.com/~zikzak/troll_thesis.html http://members.aol.com/intwg/trolls.htm |
I sent an off-group message explaining this. The reasoning is quite simple.
The audio signal must supply enough voltage to just cut off the DC power supply at maximum negative peaks. This would imply that at positive peaks it will double (for a brief instant) the power supply voltage. Double the voltage will cause (by E squared / R power formula) the *peak* power to be 4 times the unmodulated carrier. Now, consider the audio voltage. It's *peak* voltage equals the DC voltage to the final. Hmmmm ... the RMS voltage of that audio signal is only .707 (square root of two divided by two) of its' peak voltage. Feeding into the same impedance, it will develop only half of the power (E squared over R or .707 times .707); therefore, you only need 1/2 of the unmodulated carrier power in audio power to modulate at 100%. This leads to the numbers in my original response. 73 from Rochester, NY Jim "Jim Hampton" wrote in message ... 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 --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.500 / Virus Database: 298 - Release Date: 7/10/03 |
I had replied concerning this post off-group, but it appears that I probably
should have posted in the group. In that transmitter being high-level am modulated, we have a power supply. In order to modulate to 100%, the audio signal will have to exactly cancel the DC supply at 100% negative peak. It will also add to the DC supply at 100% positive peak causing the final to see twice the DC supply voltage. Now, with a given impedance, what will the power of a sine wave be? Consider that it's peak voltage will equal the DC supply voltage. That means its' RMS voltage will be .707 (square root of two over two) its' peak voltage. The power that this RMS voltage will develop at this point will be (E squared over R) .707 times .707, or .5 .... one half of the DC carrier power. Now, at the positive peak, we see twice the supply voltage for an instant. E squared / R (we don't know the resistance value, but we can see where the power is going). This implies a *peak* power of (2*2) 4 times carrier power. Hope this helps :) 73 from Rochester, NY Jim "Jim Hampton" wrote in message ... 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 --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.500 / Virus Database: 298 - Release Date: 7/10/03 |
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