How are sidebands generated in an AM signal? I can't understand how sidebands get generated, even after reading this wikipedia page: https://en.wikipedia.org/wiki/Sideband#Amplitude_modulation
This is how I picture Amplitude Modulation in its simplest form.
There is a LC circuit. A an inductor plus a capacitor work together to create an alternating current at a fixed frequency. This current is linked to an antenna and an electromagnetic wave is created at the same frequency. Now you can change the amplitude (AKA the power) of this frequency to encode data, and that's how AM radio works.
Now where is the sideband coming from? There is 1 RF coil, 1 Antenna which emits a single frequency. I imagine the sidebands are just garbage output by the coil when it goes off its designated frequency at times?
I have also seen this image:

From this post: Do Sidebands mean the frequency of an AM radio wave is not constant?
But I imagine these 3 waves are combined at transmission time to create a single frequency/wave. You don't actually emit 3 separate frequencies right?
 A: 
Now you can change the amplitude (AKA the power) of this frequency to encode data, and that's how AM radio works.

Correct. Let's say your carrier signal is $\cos\omega t$, and the message signal you want to encode is $\cos\omega_m t$ (where $\omega_m$ means the frequency of the message signal). Then the formula for the modulated signal (the carrier wave with its amplitude varied) is
$$x(t)=\cos\omega t\cos\omega_m t$$
From Euler's identity we know
$$\cos\omega t=\frac{e^{j\omega t}+e^{-j\omega t}}{2}$$
so
$$x(t)=\cos\omega t \cos\omega_m t=\left(\frac{e^{j\omega t}+e^{-j\omega t}}{2}\right)\left(\frac{e^{j\omega_m t}+e^{-j\omega_m t}}{2}\right)$$
Just doing algebraic manipulations this becomes
$$x(t) = \frac{e^{j\omega t}e^{j\omega_m t}+e^{j\omega t}e^{-j\omega_m t} + e^{-j\omega t}e^{j\omega_m t}+e^{-j\omega t}e^{-j\omega_m t}}{4}$$
Then using the rule for multiplying exponentials, $e^a e^b = e^{(a+b)}$ this becomes
$$x(t) = \frac{e^{j\omega t+j\omega_m t}+e^{j\omega t-j\omega_m t} + e^{-j\omega t+j\omega_m t}+e^{-j\omega t-j\omega_m t}}{4}$$
or
$$x(t) = \frac{e^{j(\omega+\omega_m) t}+e^{j(\omega-\omega_m) t} + e^{-j(\omega +\omega_m) t}+e^{-j(\omega-\omega_m) t}}{4}$$
or
$$x(t) = \frac{e^{j(\omega+\omega_m) t}+ e^{-j(\omega +\omega_m) t}}{4}+\frac{e^{j(\omega-\omega_m) t} +e^{-j(\omega-\omega_m) t}}{4}$$
Recombining the exponentials back into cosines:
$$x(t) = \frac{\cos(\omega+\omega_m) t}{2}+\frac{\cos(\omega-\omega_m) t}{2}$$
So what this shows is that just the act of varying the amplitude of the oscillating wave does inherently change the signal from a single frequency to a pair of side-band frequencies.
If you didn't fully modulate the signal (if the message signal were something like $1+a\cos\omega_m t$ with $a<1$) you would find some of the carrier frequency still in the modulated signal. If you use a message signal that contains multiple frequencies, like a voice signal or a real message, instead of a pure sinusoid, you'll find you get replicas of the spectra of the message signal on either side of the carrier, rather than discrete sideband frequencies.
A: To explain it in words: when a modulating signal is mixed with a carrier in an AM transmitter, sum-and-difference (carrier frequency plus signal frequency, and carrier frequency minus signal frequency) frequencies are generated. this defines the total bandwidth of the transmission i.e., how much of the radio spectrum is occupied by the transmitted signal when modulated at 100%.
This means that since the highest frequencies in the modulating signal generate the highest sum and the lowest difference frequencies in the transmitted signal, all the high-frequency content of the transmitted signal resides in the sidebands. This is why as you tune in to an AM broadcast station, you get more high-frequency content in your audio output if your receiver is tuned slightly off the transmitting frequency- because then the receiver is picking up more of the sideband and less of the carrier.
You can improve the high frequency response of an AM receiver by designing it with a broad bandpass so it scoops up more of the sidebands, but then the receiver will be unable to discriminate between two closely-spaced radio stations and you will hear the one you don't want in the background of the one you do want.
A: All three tones (lower sideband, higher sideband and carrier) are transmitted in so-called double sideband amplitude modulation (AM-DSB). It is possible to transmit and receive/demodulate only one of the sidebands, this is called single sideband (AM-SSB not using the other sideband nor the carrier that are not transmitted) but it is a lot more complicated to demodulate. The sidebands come from manipulating (modulating) the carrier tone by for example, an attenuator in line with the carrier tone generator whose attenuation is driven by the acoustic signal. If, say, the audio signal is not a pure tone then the sidebands are not tones either but just replicated and shifted to the side band location. (you should remove the "dc" and low frequencies from the audio before modulating it to t he tone, otherwise you confuse the demodulator.)
A: Not all the frequencies are generated in LC circuit. Mixing process can also make some. To further clarify, you can plot an exaggerated modulation. If you look at the dark blue plot, $\sin(x) \sin(5x)$, you can see that the troughs and crests are not evenly spaced and are not coincide with the ones in unmodulated signal (gray), $\sin(5x)$. As frequency is inverse of the distance between two identical location on the wave, suppose the crests for now, one can conclude that this wave contain other frequencies other than the carrier one ($\sin(5x)$). Although this point of view is not correct, it can give you some sense.

