About phase unwrapping I am not too familiar with phase-related problems so I need some suggestions or corrections in a problem that might be an old but easy one. Here is my problem: given a complex signal:
$$y(t) = e^{j\phi(t)}$$
where $\phi(t)$ is the instantaneous phase. Now we have the observations $y(t)$. Can we somehow reconstruct $\phi(t)$?

I understand the phase will be wrapped into $-\pi$ to $\pi$ when we calculate the phase from $y(t)$. For example, let $\phi(t) = 130t+150t^2 + 0.2, t\in[0,1]$ with a sampling frequency $f = 512$. When I try to reconstruct the phase from $y(t)$ using unwrap(angle(y(t))) in Matlab, i.e., the unwrapped phase, the result is shown in the following figure:



The second example is that $\phi(t) = 450t-100t^3 + 0.2, t\in[0,1]$. The result is



The reconstructed phases are different from the original ones. Does this mean that the phase cannot be uniquely reconstructed? Could anyone show me some explanations or suggestions? Thank you!

Edit: the code is as follows.

clear all 
clc
N = 512;
t = (0:N-1)/N;
% phi = (130*t + 150*t.^2) + 0.2;
phi = (450*t - 100*t.^3) + 0.2;
s = exp(2*pi*1i*phi);
phase_rcst = unwrap(angle(s))/(2*pi);
figure
subplot(2,1,1)
plot(phi)
title('Phase of the original signal')
subplot(2,1,2)
plot(phase_rcst)
title('Unwrapped phase')
The problem comes from sampling the phase. For the given phase function, the changes in the phase from one value $t_i$ to the next one $t_{i+1}$ can become too large, so phase unwrapping will not work properly. If you simply removed the factor $2\pi$ in the exponent, the problem would be solved already (note that the factor shouldn't be there anyway according to definition of $y(t)$ in your first formula):

s = exp(1i*phi);

If you want to keep the factor $2\pi$, then use more sampling points.