# Calculate autocorrelation of a sinus

I am a beginer in signal processing. I am currently studying a signal $$s(t) = \cos(h(t))$$, and I have to calculate its autocorellation. Since $$h(t)$$ is periodic of period $$T_s$$, I calculated the value of

$$R_s(\tau) = \frac{1}{T_s} \int_0^{T_s} \cos(h(t)) \cos(h(t-\tau)) dt$$

Honestly, this calculation is not easy (since $$h(t)$$ is also a cosinus), and in order to simplify my final document, I would like to use complex representation by writing $$s(t) = Re\left\{ e^{ih(t)} \right\}$$. However, This cannot work for the autocorrelation since $$Re\left\{ e^{ih(t)} e^{-ih(t-\tau)} \right\} = s(t)s(t-\tau) + Re\left\{ \sin(h(t)) \sin(h(t-\tau)) \right\}$$

So, do you know a way to calculate $$R_s$$ by calculating $$\frac{1}{T_s} \int_0^{T_s} e^{ih(t)} e^{-ih(t-\tau)} dt$$ ?

I was thinking about a relation like $$R_s(\tau) = \frac{1}{2} Re \left\{\frac{1}{T_s} \int_0^{T_s} e^{ih(t)} e^{-ih(t-\tau)} dt \right\}$$, but I cannot prove it :(

• Is this a cold remedy? Apr 23, 2020 at 16:42

$$\cos(h(t))=\frac{e^{jh(t)} + e^{-jh(t)}}{2}$$
Note that this yields an autocorrealtion of $$R_s(\tau)=\int_{0}^{T_s} \frac{e^{jh(t)} + e^{-jh(t)}}{2}\frac{e^{jh(t-\tau)} + e^{-jh(t-\tau)}}{2} dt$$