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?

• AM and FM are different enough that I think you should pick one or the other and edit the post to ask about just that one. – DanielSank Jan 1 '17 at 16:54
• @DanielSank thank you for your reply. I am not looking for the detailed explanation of how AM and FM work, just what principles they use to transfer the 2 different types of information, for example for digital it would be: 1. Frequency of wave pulses 2. Amplitude of waves Perhaps iit would be helpful if I stated this in my question? I thought my question already suggested that it was the principles I was after... – 21joanna12 Jan 1 '17 at 17:01
• I'm telling you that, knowing how AM and FM both work, I'm disinclined to try to write an answer to this question because covering both cases in generally would make the answer pretty long. AM and FM transfer loudness and frequency in very different ways. I might try, but it would be easier to do one at a time. – DanielSank Jan 1 '17 at 17:15
• @Daniel are the two really that different in their Fourier fundamentals? Both require a bandwidth of some 10 kHz or so to transmit the signal, and both have stations separated by more than that, so the essentials of the tuning circuit are very similar. – Emilio Pisanty Jan 1 '17 at 19:03
• @EmilioPisanty Saying that FM takes 10 kHz of bandwidth oversimplifies things. The bandwidth of FM is determined by the dynamic range of the sound to be encoded. Loud sounds take up more bandwidth, and can spill into the next radio channel and upset the FCC. In the past some classical music stations intentionally reduced the volume so that a higher dynamic range could be encoded. The result was a lower signal to noise ratio, and there was hiss. (I don't hear that too much today. I don't know if that's because of dynamic range compression or noise reduction technology.) – garyp Jan 1 '17 at 19:49

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.

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 $$Q = 10 \text{MHz} / 10 \text{kHz} = 1,000 \, .$$
• If you need more details, i.e. on how a rectifier and filter gets rid of the carrier oscillations and turns $s(t)$ into $m(t)$, let me know. – DanielSank Jan 1 '17 at 18:00