# Why is $\sin(t)$ a stationary process?

I am trying to understand the meaning of the term Stationary Process. For example, I was told that $\sin(t)$ is a stationary process.

Could someone try to explain, in simple words, why is $\sin(t)$ (for example) is a stationary process?

$\sin(t)$ is no random process, because there's nothing random about it. You could add a random amplitude to get a random process:

$$x(t)=A\sin(t)\tag{1}$$

This is a random process because $A$ is a random variable. However, $x(t)$ is not stationary, but it is cyclostationary, i.e., its statistical properties vary periodically. You can make the process $x(t)$ stationary by adding a random phase:

$$\tilde{x}(t)=A\sin(t+\phi)\tag{2}$$

The phase $\phi\in [0,2\pi]$ is a uniformly distributed random variable that is independent of $A$. It can be shown that the statistical properties of $\tilde{x}(t)$ given by $(2)$ are independent of $t$, and hence, the process is stationary.

• Thank you for your answer. What does it mean that the statistical properties of $x(t)$ are independent of $t$? Could you explain in simple words what makes $Asin(t + \phi)$ stationary? for example, how come the enteries $C_{ij}$ of the correlation matrix depend only on $|i-j|$? Thanks! Jul 7, 2016 at 7:23
• I think I eventually got it. I multiplied the processes and used trigonometric identities of $sin(\alpha + \beta),cos(\alpha - \beta)$. Thanks!!! Jul 7, 2016 at 8:15
• @user135172: You could e.g., compute the mean and auto-correlation of $x(t)$, and you'll find out that they don't depend on $t$. This makes the process at least wide-sense stationary. In fact it is strictly stationary. Jul 7, 2016 at 8:23
• Perhaps you need to say that by random phase $\phi$ you mean a random variable that is uniformly distributed on $[0,2\pi)$. (Strictly speaking, all that is needed is that $$E[\cos \phi]=E[\sin\phi]=E[\cos 2\phi]=E[\sin 2\phi]=0$$ which holds for $U[0,2\pi]$ as well as for discrete random variables taking on each of the values $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ with probability $\frac 14$ but $U[0.2\pi]$ is good enough in this case. Jul 7, 2016 at 14:43
• It is wide-sense stationary, since the autocovariance can be expressed via a time difference. I believe the variance also converges to a constant for a single realization as time goes to infinity. The benefit of wide-sense stationary signals is that you can compute the stationary/steady-state response of deterministic and stochastic signals by superposition in LTI systems... Jul 8, 2016 at 9:02

While @Matt L. gave the perfect statistical signal processing answer, I'll dare a more mundane one. When analysing a signal with an oscilloscope, one can be interested in that the signal's amplitude spectrum does not vary over moving windows. So a sine is sort of stationary in frequency. Additionally, the signal is itself stationary in envelope (modulus $1$ for the analytic version of the signal).

Though not "correct", such mentions of stationarity are in use.