I'm trying to learn how AM radio communication works. If I understand correctly, an AM radio signal transmits on a fixed frequency, using the amplitude of the wave to transmit the frequency of the sound. However, this only carries the frequency of the source sound, not the volume. So if you're transmitting voice, how is a whisper quiet and a scream loud? Why wouldn't they be the same volume?
Electronic – How does AM radio transmit volume
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It's a bit unclear what exactly you are asking. I suppose you have seen a antenna like a dipole and it looks at first glance like just two wires that aren't connected to anything, therefore the question is how can current flow and power be drawn? If so, it would help if you clarified that. Some antennas, like a loop or folded dipole, are exactly the opposite in that they appear to be dead shorts at first glance.
In any case (if my interpretation of your question is correct), what you are missing is that antenna is no longer a open or short at its intended frequency. Antennas are not lumped systems, which means that different parts will be at different phases of the signal at the same time. In fact, antennas exploit this to help produce the large voltages and current it takes to radiate significant power.
Often resonance is involved. Fill a bathtub partway with water. Now put your hand in the middle and move it back and forth only a short distance in the end to end direction. Once you find the right frequency, you will see that you can get a lot of water to slosh back and forth despite only a relatively small motion of your hand. Note that at the peaks, the water at one end of the tub is high and the other low, and nothing is flowing. In between the water is roughly level but is flowing strongly in the middle. Also note that it takes very little force from your hand to cause this and keep it going, but you have to be moving your hand at just the right frequency. A little faster or slower and it doesn't work anymore.
That was resonance, and is exactly what is happening in a dipole. In the case of a dipole, that water level becomes voltage and the flow rate of the water becomes current. The feedpoint in the middle of the dipole is where a little current of just the right frequency is fed in, which causes resonant sloshing of charges back and forth in the antenna. The voltages created at the ends of the antenna can therefore be a lot higher than anything you put in.
This sloshing of current back and forth makes high voltages at the ends. Together with the high current in the middle, the antenna disturbs the local E and B fields in such a way that power is lost from the antenna into those fields. That power eventually organizes itself into a self-propagating wave we call radio.
Since real power is lost, the antenna must appear to have a resistive component from the driving circuit point of view. A ideal antenna used at exactly the right frequency will appear to be purely resistive, meaning all the power dumped into it gets transmitted. This happens very close to the best resonant sloshing frequency for most antennas. That also explains why antennas often only work well for a narrow frequency range.
So getting back to your question, the problem is you are analyzing the antenna at DC, which is completely irrelevant. You have to analyze antennas at the frequencies they are intended to radiate at. At those frequencies, a lot of other stuff happens so that they don't look like opens or shorts as they do at DC.
SSB is not how you describe but first regular AM as broadcast by the radio stations. This you appear to correctly describe except for the demodulation which you omit.
In AM the amplitude is modulated but not by so much that the carrier amplitude becomes zero at any point. This is regular AM and for instance a \$1V_{peak}\$ carrier might reach (due to the effects of modulation) peaks of 1.8V and troughs of 0.2V: -
A simple diode detector will demodulate this so this method is suitable for public broadcast radio stations because the radio receiver is simplified by using the diode detector.
Suppressed carrier AM is like regular broadcast AM but where the modulating signal pushes the carrier beyond the limits of broadcast AM. The result of this is that the carrier phase becomes reversed every half cycle of the modulating signal. A simple diode detector no longer produces any decent recovery of the audio signal. Here is a picture of what happens when AM is over-modulated (heading towards a suppressed carrier scenario) and you try and use a simple diode detector: -
Why is it called suppressed carrier - the carrier inverts every half cycle of the modualting wave and if you used a spectrum analyser there is no trace of the carrier despite the resultant modulated carrier being centred at the carrier frequency: -
Recovery of the modulating signal (i.e. the original audio signal) is much trickier and that is why this technique is not used in broadcast transmitters in order to keep receiver cost and/or complexity low.
SSB usually is a further modification to suppressed carrier where one of the sidebands is either totally or partially suppressed in order to create a smaller bandwidth in what is being transmitted - more power can be focussed into the single remaining sideband and with a more complex receiver it can be adequately received. To understand what SSB looks like you have to look at the frequency spectrum of DSBSC (Double sideband suppressed carrier) and imagine it is lopped off on one side: -
Best Answer
The audio signal to be transmitted varies the amplitude of the transmitted signal.
The frequency of the amplitude variations correspond to the modulating audio frequency.
The size of the amplitude variations transmit the volume - a loud sound will produce a large amplitude variation at the transmitter. The receiver will, in turn, produce a loud sound on receiving a large amplitude variation.