# Why the angle of a real signal in frequency is continuous?

This original signal is shown below (time domain): It is a real signal. According to the signal processing theory, the form part of the angle is conjugate to that of the latter part. That is, $\sphericalangle X(e^{j\omega}) =\sphericalangle X^*(e^{-j\omega})$ when $x[n] \in \mathbb{R}$.

However, I implement X = angle(fft(x)); and plot(X) in MATLAB. The figure is shown below: The phase is continuous! Could anyone explain the reason? Thanks in advance!

## 2 Answers

Your real signal is a sine wave. The amplitude is smoothly shifting in a manner consistent with delay.

If you picked a signal that was real white noise, the phase would have abrupt transitions

X = angle(fft(x))

is numerically very questionable. If your signal is periodic in exactly the length of vector x you're observing, then the (theoretical) DFT would be zero in all but two bins. You can't really take the angle of $0$.

If your signal has a period that doesn't exactly fit in x, then you're observing the phase of two superimposed sinc pulses, as that's what a rectangular windowing leads to.

A sinc is for all points (but the maximum) defined as $\frac{\sin ax}{bx}$ (with $a$ and $b$ depending on who you ask and how you scale), and if you look at that, it definitely does have at least a strongly linear component.