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So I was trying to program vibrato, or freqency modulation, naively using the model: $$\cos((\omega + \alpha\cos(\omega' t))t)$$ Where $\alpha \lt \omega$ and $\omega' \ll \omega$. For practical purposes, assume $\omega=200\,\mathrm{Hz},\ \alpha=5\,\mathrm{Hz},\ \omega'=1\,\mathrm{Hz}$ so that $$195\,\mathrm{Hz}<\omega + \alpha\cos(\omega' t)<205\,\mathrm{Hz}.$$

However in practice this looks and sounds completely wrong. It starts out sounding like vibrato, but the modulation amplitude increases indefinitely:

enter image description here

After some Googling, I find that I'm supposed to use the model: $$\cos(\theta(t))$$ where in general, $$\omega(t)=\frac{d\theta(t)}{dt}$$ So I set it up as: $$\theta(t)=\int\omega + \alpha\cos(\omega' t)\,dt$$ and I get the desired result:

enter image description here

Question is, being that I'm a little rusty with math, how can the first approach be analyzed to show why it behaves the way it does?

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  • $\begingroup$ The parenthesis in your post don't match the parenthesis in your title. Could you please clarify? $\endgroup$
    – DanielV
    Jan 23, 2015 at 3:47

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Let $\theta(t) = \omega t + t \alpha \cos (\xi t)$, then $\theta'(t) = \omega + \alpha \cos (\xi t) - t \alpha \xi \sin (\xi t)$, so you can see that the instantaneous frequency is unbounded.

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  • $\begingroup$ Interesting, but what's the significance of $\theta'(t)$? $\endgroup$
    – Anthony
    Jan 23, 2015 at 4:07
  • $\begingroup$ Sorry, I see that's the derivative with respect to $t$. $\endgroup$
    – Anthony
    Jan 23, 2015 at 4:08

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