How is information transmitted by radio waves (how radios work)? I am considering radios and have just learnt that radios work by having an RLC circuit with adjustable inductor inductance or capacitor capacitance to change the resonance frequency of the circuit, and thus allowing you to pick up different radio station wavelengths. I also learnt that the resonance of this circuit needs a high quality factor Q so that its resonance peak is thin and other stations are not picked up. 
So I understand the basics of how we select and transmit certain radio stations, but I don't understand how we can transmit all of the information that allows us to listen to music on a radio station. I think the only 2 bits of information needed are:


*

*frequency/pitch of sound

*loudness/amplitude if sound
I have read about FM and AM waves, but I don't really understand how they can work. Firstly, how can we transmit information by changing the wave frequency if we need a very narrow resonant peak? I get that frequencies that deviate slightly from the resonant frequency will still be picked up, but will they then not be picked up as a significantly lower amplitude wave due to the very narrow resonance peak of the RLC circuit? Is this not a problem?
In either case, changing the frequency or the amplitude of waves still only gives 1 piece of information, but we need two.
So I don't understand how radios can work via analog signal. Digital can carry the two pieces of information by the frequency of pulses (as opposed to the frequency of the wave being transmitted which remains the same) and the amplitude of the waves in each signal. But how can analog radio work?
 A: Consider the incoming electric field of the radio waves.
This field is a superposition of all broadcasts from stations near your receiver.
The job of the receiver is to pick out one of these transmissions and turn it into sound.
AM radio
Now consider an AM radio station transmitter.
Suppose the sound wave that station wants to transmit is represented by a function of time $m(t)$ where here $m$ is for "message".
Note that $m(t)$ includes all information about the sound, i.e. it includes frequency, amplitude... everything.
In an AM transmitter, we use a circuit to multiply $m(t)$ by a sinusoid, creating the transmitted signal
$$s(t) = m(t) \cos(\Omega t)$$
where here $s$ stands for "signal" and $\Omega$ is called the "carrier frequency".
Here we see the reason for the term Amplitude Modulation (AM): the message is a modulation of the amplitude of the carrier wave.
You can use trig identities or Fourier analysis to see that the spectral content of $s(t)$ is in the range $\Omega \pm \delta \omega$ where $\delta \omega$ is the highest frequency in $m(t)$.
The carrier frequency $\Omega$ might be in the tens of MHz range.
On the other hand, the actual message $m(t)$ would absolutely never have any frequencies above around 20 kHz because that's the upper range of human hearing.
In real life, $m(t)$ doesn't use up the full 20 kHz; useful speech and music don't need our full hearing range.
So now we see that the transmitted signal $s(t)$ is contained within some relatively narrow bandwidth, i.e. maybe a 10 kHz band centered at 10 MHz.
Therefore, a tuned circuit with a $Q$ of around 1,000 and centered at $\Omega$ picks up $s(t)$ but mostly nothing else.$^{[a]}$
Of course, we also have to enforce that the various stations' carrier frequencies are separated by more than their $\delta \omega$'s so that nobody's transmissions overlap with anyone else's.
So, the output of our tuned circuit is roughly just $s(t)$!
I say "roughly" because our tuned circuit isn't perfect, so we might pick up a bit of stuff  from other transmissions, but since it's farther away from the center of our tuned circuit the amplitude is suppressed.
Then, we just put the signal through a rectifier and a low pass filter so that the carrier oscillations are gone and we only get $m(t)$.
That's it!
Now we have the original sound message and we can put it into a speaker.
We don't have to think about amplitude and frequency separately: we have the entire original sound waveform.
$[a]$: $Q$ is the center frequency divided by the bandwidth, so
$$Q = 10 \text{MHz} / 10 \text{kHz} = 1,000 \, .$$
