Aliasing below $f_s/2$
My example isn't in fact below $f_s/2$; I blundered with $f_s=N$. Adjusting this eliminates aliasing, for linear, exponential, and hyperbolic chirps - and, in fact, for any signal with $\phi '(t) &lt; f_s/2$.
This can be verified two ways:

For $\omega_\text{max} = \text{max}(\phi'(t))$ (assume $\phi(t)$ includes $2\pi$), the worst-case scenario is a straight line with slope $\omega_\text{max}$ (a curve can't go any steeper else it's no longer &quot;max&quot;, and flatter is &quot;more favorable&quot;). But we know if $f_\text{max} &lt; f_s/2$, a pure tone won't alias, thus the worst case is accounted for.

Aliasing occurs if the finite difference $\phi[n] - \phi[n-1] = \phi(n\Delta t) - \phi((n-1)\Delta t) \geq \pi$ for any $n$, where $\Delta t=$ sampling period, or increment of the t vector.  The greatest such difference is $\omega_\text{max} \cdot \Delta t$ - setting the inequality: $\omega_\text{max} \Delta t &lt; \pi \Rightarrow \omega_\text{max} &lt; \pi / \Delta t \Rightarrow f_\text{max} &lt; 1/(2\Delta t)= f_s/2$.


Thus if I ensured $f_s/2$ was met, phi could never exceed $\pi$.

Example in question, corrected: if we seek (tmin, tmax) = (0, log(511)), our $N$ must be large enough so that $\Delta t$ is small enough for $\text{max}(\phi'(t)) &lt; \pi$. The max is 2*pi*exp(log(511)) = 2*pi*511 = wmax, so max finite difference is wmax*dt &lt; pi -&gt; dt &lt; pi/wmax, and dt = t[1] - t[0] = (tmax - tmin)/N, thus (tmax - tmin)/N &lt; pi/wmax -&gt; N &gt; wmax/pi*(tmax - tmin) = 2*pi*511 / pi * log(511) = 2*511*log(511) = 6373.6.
N = 6374
t = np.linspace(0, np.log(511), N, endpoint=False)
phi = 2*np.pi * (np.exp(t) - 1)
x   = np.cos(phi)