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Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station $k$ is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where $\gamma_k\psi_k$ never gets close to $\omega_k$ (so you're only modulating the frequency and not completely disrupting it). If the interesting signal $\psi_k$ is a pure note at frequency $\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing $\omega$. (The number of sidebands is controlled by how large $\gamma_k$ is.)

The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!)

To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage $f$ that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency $\omega_0$ is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at $\omega_0$ and of width $\gamma$. Choose $\gamma$ to match the spectral width of the typical radio station, and you've got a fantastic filter!

EDIT: After doing some looking up, I find that the $\psi_k$ here is not exactly the audio signal the station is trying to encode, but rather something like its average over the interval $[0,t]$, so it is equivalent to it up to simple mathematical operations performed at the decoder.

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Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station $k$ is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where $\gamma_k\psi_k$ never gets close to $\omega_k$. \omega_k$(so you're only modulating the frequency and not completely disrupting it). If the interesting signal$\psi_k$is a pure note at frequency$\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing$\omega$. (The number of sidebands is controlled by how large$\gamma_k$is.) The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!) To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage$f$that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency$\omega_0$is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at$\omega_0$and of width$\gamma$. Choosing Choose$\gamma$to match the spectral width of the typical radio station, and you've got a fantastic filter.! 1 Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station$k$is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where$\gamma_k\psi_k$never gets close to$\omega_k$. If the interesting signal$\psi_k$is a pure note at frequency$\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing$\omega$. (The number of sidebands is controlled by how large$\gamma_k$is.) The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!) To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage$f$that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency$\omega_0$is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at$\omega_0$and of width$\gamma$. Choosing$\gamma\$ to match the spectral width of the typical radio station, you've got a fantastic filter.