# Instananeous frequency of a chirp signal is halved?

 P: 10 How did you arrive at your formula for the instantaneous frequency? I suspect that it might be off. If I calculate the instantaneous frequency using a different approach, with the analytical signal 'y' using your code: dy = diff(y)/0.01; dy(501) = 0; instfrq = imag(dy./y); I get something that tracks 2*x instead. Here I used the imaginary part of the log-derivative of the analytic signal: $\hat{y} = A e^{i\varphi(t)}$ $\frac{d\hat{y}}{dt}=i \frac{d\phi(t)}{dt} A e^{i\varphi(t)}$ $\rightarrow \frac{d\phi(t)}{dt} = \frac{\frac{d\hat{y}}{dt}}{i \hat{y}}$