Simulation of a Frequency ramp
I waited a bit to see if someone else takes the challenge, but since there are no answers yet, I'm providing mine now.


   For a given signal $m(t)=\cos(\phi(t))$ the instantaneous frequency is given by $$\nu(t)=\frac{\phi'(t)}{2\pi}\tag{1}\\$$One pitfall here is that if we write $m(t)=\cos[2\pi f(t)t]$, $f(t)$ is generally not the instantaneous frequency (unless $f(t)$ is constant). We want the following instantaneous frequency:$\nu(t)=\begin{cases}F_1,&amp;t&lt;t_1\\F_1+\frac{F_2-F_1}{t_2-t_1}(t-t_1),&amp;t_1&lt;t&lt;t_2\\F_2,&amp;t&gt;t_2\end{cases}\tag{2}\\$From $(1)$ and $(2)$, in the time interval $[t_1,t_2]$ the phase $\phi(t)$ must be chosen as $$\frac{\phi(t)}{2\pi}=F_1t+\frac{F_2-F_1}{t_2-t_1}\left(\frac{t^2}{2}-t_1t\right)+C\tag{3}$$Choosing $C=0$, and since $\phi(t)/(2\pi)=f(t)\cdot t$, we get from $(3)$ $$\begin{align}f(t)&amp;=F_1+\frac{F_2-F_1}{t_2-t_1}\left(\frac{t}{2}-t_1\right)\\&amp;=F_1-\frac{\Delta F}{\Delta t}t_1+\frac12\frac{\Delta F}{\Delta t}t,\quad t_1&lt;t&lt;t_2\tag{4}\end{align}\\$$ with $\Delta F=F_2-F_1$ and $\Delta t=t_2-t_1$.