I've been looking at a useful way to represent Doppler shift from a satellite passing over a ground station. I've calculated the Doppler shift frequency values at 1-second interval for the duration of each pass (a pass maybe 600s or so). Doppler shift from satellites looks like this:
The closet mathematical function to this that I've come across is the logistic function. You can see they look quite similar (although the form of the curve is a mirror image):
It seems to me that I should be able to define a logistic curve that should match the Doppler shift values (or at least get a line of best fit that's very close). The form of the logistic equation I've been working with is this:
$$ f(x) = \frac L {1 + e^{-k(x-x_0)}} $$
I believe that the following values are appropriate for the constants:
$L$ - The maximum Doppler shift times $2$. I've found this by trial and error. The maximum Doppler shift is where the line in the first image crosses the $y$ axis. $x_0$ - The pass duration/$L$. Equally worked this out by trial and error.
Then I've tried to iterate and find a value for x whereby the error is minimal. The problem is after enough iterations the error becomes constant at around 50,000%. Is there a basic rule here that I'm overlooking? Thanks!
The problem is with your choice of $f(x) = \dfrac{L}{1+e^{-k(x-x_0)}}$.
If $k > 0$, then we have $\displaystyle \lim_{x \to \infty}\dfrac{L}{1+e^{-k(x-x_0)}} = L$ and $\displaystyle\lim_{x \to -\infty}\dfrac{L}{1+e^{-k(x-x_0)}} = 0$.
Similarly, if $k < 0$, then we have $\displaystyle \lim_{x \to \infty}\dfrac{L}{1+e^{-k(x-x_0)}} = 0$ and $\displaystyle\lim_{x \to -\infty}\dfrac{L}{1+e^{-k(x-x_0)}} = L$.
However, the graph appears to satisfy $\displaystyle \lim_{x \to \infty}f(x) = -L$ and $\displaystyle \lim_{x \to -\infty}f(x) = L$.
Therefore, you should use a different model, such as $f(x) = \dfrac{2L}{1+e^{k(x-x_0)}}-L$.
If $k > 0$, then $\displaystyle \lim_{x \to \infty}f(x) = -L$ and $\displaystyle \lim_{x \to -\infty}f(x) = L$, which matches the graph better.