I've recently been thinking about filters where the feedback loop does not have a delay and have found as long as the filter is linear, deriving a transfer function is fairly trivial. Today I thought I'd try and calculate the output of a sinusoidal oscillator where the output is passed through a gain and bias before being sent to its own frequency control, again with no delay.
While it was fairly easy to write this implicitly
$$y\left(t\right)=\sin\left(2\pi\int_{0}^{t}\left(g\cdot y\left(\tau\right)+b\right)d\tau\right)$$
It's been much harder, at least were my maths is at, to isolate the output $y\left(t\right)$.
After stumbling around for a while I've managed to get to this point
$$y\left(t\right)=2\cdot\frac{\sqrt{1-\left(\frac{g}{b}\right)^{2}}\tan\left(\pi\cdot b\cdot\sqrt{1-\left(\frac{g}{b}\right)^{2}}\cdot t\right)-\frac{g}{b}}{\left(\sqrt{1-\left(\frac{g}{b}\right)^{2}}\tan\left(\pi\cdot b\cdot\sqrt{1-\left(\frac{g}{b}\right)^{2}}\cdot t\right)-\frac{g}{b}\right)^{2}+1}$$
I think this is almost right but I've noticed that firstly, it doesn't start from zero. I think this is because I ended up converting my original equation to a differential equation at some point and lost the integral limits but I'm not sure how to fix this.
The second issue is that it definitely doesn't work when the gain exceeds the bias, $g\gt{b}$, i.e. Through Zero Frequency Modulation.
Is there a more general equation that intersects the origin or starts with a set phase at $t=0$? Is there one that can do TZFM and if so without going via complex numbers? If so how exactly would I go about deriving it?
I've come to the conclusion that this is the equation I was looking for,
$$y\left(t\right)=\frac{b\cos\left(2\pi\sqrt{b^{2}-g^{2}}t-atan2\left(\sqrt{b^{2}-g^{2}}, g\right)\right)-g}{b-g\cos\left(2\pi\sqrt{b^{2}-g^{2}}t-atan2\left(\sqrt{b^{2}-g^{2}},g\right)\right)}$$
$atan2$ has to be implemented for complex inputs but otherwise, this seems to do exactly what I was looking for.
While I'm not sure I could explain exactly how I got here I believe my initial mistake was early on.
I started out by taking the arcsin of both sides before applying implicit differentiation.
$$\arcsin\left(y\left(t\right)\right)=2\pi\int_{0}^{t}\left(g\cdot y\left(\tau\right)+b\right)d\tau$$
$$\frac{1}{\sqrt{1-y\left(t\right)^{2}}}dy\left(t\right)=2\pi\left(g\cdot y\left(t\right)+b\right)dt$$
Rearranging I was then able to take the integrals of both sides reintroducing my initial bounds.
$$\frac{1}{2\pi\left(g\cdot y\left(t\right)+b\right)\sqrt{1-y\left(t\right)^{2}}}dy\left(t\right)=dt$$
$$\frac{1}{2\pi}\int_{y\left(0\right)}^{y\left(t\right)}\frac{1}{\left(g\cdot y'+b\right)\sqrt{1-y'^{2}}}dy'=\int_{0}^{t}d\tau$$
Taking some liberties, with a $\frac{0}{0}$ that I possibly shouldn't have, I was able to get the answer to these integrals and rearrange to get $y\left(t\right)$ in terms of t.
$$\frac{1}{\pi\sqrt{b^{2}-g^{2}}}\left(\arctan\left(\frac{\frac{b\left(1-\sqrt{1-y\left(t\right)^{2}}\right)}{y\left(t\right)}+g}{\sqrt{b^{2}-g^{2}}}\right)-\arctan\left(\frac{g}{\sqrt{b^{2}-g^{2}}}\right)\right)=t$$
$$y\left(t\right)=2\frac{\left(\frac{\sqrt{b^{2}-g^{2}}}{b}\tan\left(\pi\sqrt{b^{2}-g^{2}}t+\arctan\left(\frac{g}{\sqrt{b^{2}-g^{2}}}\right)\right)-\frac{g}{b}\right)}{\left(\frac{\sqrt{b^{2}-g^{2}}}{b}\tan\left(\pi\sqrt{b^{2}-g^{2}}t+\arctan\left(\frac{g}{\sqrt{b^{2}-g^{2}}}\right)\right)-\frac{g}{b}\right)^{2}+1}$$
From here with a lot of trial and error, that I can't quite remember, I was able to get my final answer.
What would be really nice is to have managed to get the answer as an infinite sum of sin waves to demonstrate the harmonic effect of the self-modulation but alas the reciprocal nature of my solution has left me stumped. If anyone can shine a light on this I'd be most grateful.
As for the TZFM issue, I had simply overlooked the fact that with self-modulation the oscillator would inevitably stall as it crosses the zero point. So that aspect of my original equation was correct.