# Very simple AM question

Discussion in 'Wireless & RF Design' started by count_volta, Nov 10, 2009.

1. ### count_volta Thread Starter Active Member

Feb 4, 2009
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Hey, we have began to touch a little on radio and amplitude modulation in my Electronics lab in college. Its very interesting to me, but I have some questions.

So we have this AM signal below called signal(t)

Here is the equation of how this happens.

signal(t) = A[1+B*message(t)]*cos(ωt) where the carrier wave is a cosine.

The carrier wave is in the middle. I understand how the carrier wave is mathematically forced to stretch and compress itself according to the amplitude of the message wave.

But what I don't understand is where is the frequency information of the message signal? Since the frequency of the carrier wave is constant, and only its amplitude changes, how does it retain the frequency of the message?

For example speech is composed of an infinite number of frequencies (fourier series). How does signal(t) retain these frequencies?

When demodulated, the message signal has to be unaltered (including its frequency content) so how does this happen?

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
The 'message' is also sin or cos wave.

Thus you have the product of two sinusoids.

A bit of trig will show you that this introduces new frequencies that did not exist in the original.

3. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
This information is fully contained in the sidebands. As it turns out, you can't change the amplitude of a sine wave without changing the SHAPE of a sine wave....the results being SUMS and DIFFERENCES between the carrier and the modulation information.

(Actually, voice doesn't contain an INFINITE number of frequencies....just a LOT!)

If you have a 1 MHZ carrier and A.M. modulate it with a single sine wave of 1000 Hz, you will generate TWO signals in addition to the carrier, one at 999 KHZ and one at 1.001 MHZ. Interestingly the power at precisely 1MHZ remains unchanged. (Unless there's a problem with your transmitter!)

Eric

4. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Nope. The carrier frequency remains unchanged. You ADD two new "carriers" with the modulation process. The total transmitted power is the sum of the power in the carrier and the two new sidebands.

5. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Okay this really helps. Thanks.

So the transmitted signal would be something like:

signal(t) = A[1+B*message(α*t)]*cos(β*t) =

A*cos(βt) + AB*message(αt)*cos(βt)

Then you use trig identities and etc.

But if this is true, then how does a demodulator recover just message(αt) and leave out the other stuff which was mathematically introduced? Through filtering? And what exactly does the diode in a demodulator do? Why do you need only the top half of the signal?

You see in our lab we have to simulate a demodulator in Pspice. And then build one on a breadboard. When my lab partner and I tried simulating the demodulator we ran into small problems because I think we were never properly explained how the whole system works, i.e. AM modulator and demodulator.

Is there a good explanation somewhere online? And I mean an explanation like the AAC ebook where people who are totally new to this topic can understand it.

Last edited: Nov 10, 2009
6. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295

There are actually TWO very different ways of demodulatiing A.M., one called simple detection, and one called synchronous demodulation. Synchronous demodulation is almost a mirror image of the modulation process, whereas simple detection is VERY different, but almost universal, because of its simplicity.

In a simple A.M. detector (rectifier), you are only looking at the TOTAL transmitted power, no effort is made to discriminate between upper or lower sideband, or carrier. What you do is simply RECTIFY the R.F. signal, which gives you a voltage that follows the modulation "envelope". This envelope that has both upper and lower sideband information, follows the original modulating signal. (This is how a crystal radio works).

The advantage of this method is that you don't need to worry about the phasing reference...the carrier does that for you...it follows along for the ride.

eric

7. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Sighs, I still don't quite get it. Will take me a bit to figure it out. But you made some things clear already, so its progress.

8. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Here's a good little site dedicated to the topic:

9. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Cool, it explains many other things too. Thanks.