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Amplitude modulation - why is ωc >> ωm necessary?

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karakov
#1
Feb11-14, 02:02 PM
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i just want to ask why is ωc >> ωm is necessary for a good amplitude modulation
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berkeman
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Feb11-14, 02:25 PM
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Quote Quote by karakov View Post
i just want to ask why is ωc >> ωm is necessary for a good amplitude modulation
Welcome to the PF.

Where have you seen that condition referred to (textbook/web page/lecture notes/etc.)? Is it discussed there?
karakov
#3
Feb11-14, 02:36 PM
P: 6
we admit that condition . but why??

berkeman
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Feb11-14, 02:45 PM
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Amplitude modulation - why is ωc >> ωm necessary?

Quote Quote by karakov View Post
we admit that condition . but why??
What do you mean? You use that condition in some calculations? Can you show us the calculations?
karakov
#5
Feb11-14, 03:06 PM
P: 6
i have in a lesson about amplitude modulation :
technically to have a good modulation .two conditions should be considered : M< 1 and ωc >> ωm (at least ten times )
berkeman
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Feb11-14, 03:09 PM
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Quote Quote by karakov View Post
i have in a lesson about amplitude modulation :
technically to have a good modulation .two conditions should be considered : M< 1 and ωc >> ωm (at least ten times )
Correct. Can you see why M < 1 is necessary to avoid distortion of the modulating waveform that rides on the carrier waveform? There are good pictures that you can find with a Google Images search on Amplitude Modulation.

The requirement that ωc >> ωm is a little more subtle. Think about what the modulated carrier looks like when ωm gets close to ωc. What happens if they are equal?
karakov
#7
Feb11-14, 03:24 PM
P: 6
it is going to lead to distortion of the modulating waveform ?!
berkeman
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Feb11-14, 03:29 PM
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Quote Quote by karakov View Post
it is going to lead to distortion of the modulating waveform ?!
Yes, when you try to demodulate it with a simple rectifier, the recovered signal will be distorted compared to the original modulating signal. With a little Google searching, you may be able to find a graphical waveform calculator to show the distortion as the two frequencies get too close...
karakov
#9
Feb11-14, 03:48 PM
P: 6
I really can't find wavesform calculator .so tell me what happens ?
berkeman
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Feb11-14, 05:05 PM
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Quote Quote by karakov View Post
I really can't find wavesform calculator .so tell me what happens ?
You didn't look for very long...
karakov
#11
Feb11-14, 05:17 PM
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ok whatever just tell me what happens ?? please.
berkeman
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Feb11-14, 05:23 PM
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Quote Quote by karakov View Post
ok whatever just tell me what happens ?? please.
No, this is your schoolwork, and you should be able to figure this out. Especially if it is part of an assignment, we can give you hints, but we are not allowed to do your work for you.
sophiecentaur
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Feb12-14, 04:03 AM
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Quote Quote by karakov View Post
i just want to ask why is ωc >> ωm is necessary for a good amplitude modulation
It depends on what you mean by "good".

You can AM 'anything onto anything', in fact; the formula still applies. This question is both theoretical and practical, I think. If the modulating frequency is higher than that of the carrier, where would the sidebands turn up? What sort of sense would a simple demodulator make of that? (You might be familiar with the subject of Sampling and aliasing, which is along the same lines)
A practical consideration, (and practical ones are sometimes difficult to find in text books) is that it is difficult to ensure that the usable pass band of a channel will be better than, say 10% of the carrier frequency. If the frequency response of the channel is asymmetrical then what will happen to the sidebands? (Amplitude response is the easy one to consider but Phase response will also be relevant)
These comments should be enough to help you come up with a reasonable answer to your question. I suggest you should do some thinking about the problem before putting values into a calculator. A calculator doesn't tell you anything about why things happen but some sketches of the spectrum could give you some serious insight here..


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