# Finding the autocorrelation of a sine wave.

The autocorrelation of sin(t) is defined as $$\displaystyle \int_{-\infty}^{\infty} \sin(t+\tau)\sin(t)d\tau$$

I've tried using the Wiener-Khinchin theorem which says that

$$Corr(g,g)\Longleftrightarrow|G(f)|^2$$

I've tried reverting the FT squared of the sine wave, and don't see any solution.

I can't find a derivation online for this. So I came here wondering if anyone of you could offer a hint; or perhaps go through a derivation.

One of my problems is calculating the inverse fourier transform of:

$$\frac{1}{4}(\delta(f+\frac{1}{2\pi})-\delta(f-\frac{1}{2\pi}))^2$$

In words: What is the inverse transform of the modulus squared of the FT of the sine function?

Thanks!

• Your integrand is periodic, hence the integral can only exist if it is 0. This might help a bit Nov 8, 2015 at 1:19
• Supposedly it is a cosine function. I've done it numerically and it is in fact a sinusoidal wave.
– DLV
Nov 8, 2015 at 1:43
• A cosine function is a sinusoidal function, just with a different phase ;) Nov 8, 2015 at 1:58
• Yeah, I definitely have no problem with that.... Sinusoidal often means wavy. Not sine wave. Either way it doesn't matter, I wouldn't like to discuss something like that.... Whats the autocorrelation of the sine function?
– DLV
Nov 8, 2015 at 2:05

The autocorrelation is a function of $\tau$, not $t$ The function is $Aut(\sin(t),\tau)=\displaystyle \int_{-\infty}^{\infty} \sin(t+\tau)\sin(t)dt$

where $t$ is a dummy variable.

Expand the sum in the argument of the sine function: $$\displaystyle \int_{-\infty}^{\infty} \sin(t+\tau)\sin(t)dt=\int_{-\infty}^\infty (\sin t \cos \tau+ \sin\tau \cos t)\sin(t)dt\\=\int_{-\infty}^{\infty} (\sin^2t\cos \tau + \sin t \cos t \sin \tau) \;dt$$

Over an interval of $2\pi$ in $t$, the second term averages out to $0$ and the first term becomes $\frac 12\cos \tau$, which is what you seek

• This doesn't seem right. The factor of $\frac{1}{2}$ comes from dividing through half the interval, which the autocorrelation formula doesn't do. I'm seeing other derivations of the Wiener-Khinchin theorem that do do this but the formula seems to be different. books.google.com.mx/…
– DLV
Nov 8, 2015 at 3:35
• It comes because the average value of $\sin^2$ over a period is $\frac 12$ Nov 8, 2015 at 3:56
• Hmm. I don't see why this integral is averaging. It's not dividing over an interval.
– DLV
Nov 8, 2015 at 3:58
• $\int_{-\pi}^{pi} \sin^2(x) dx = \pi$ for instance. Not $\frac{1}{2}$. The average will be $\frac{1}{2}$ though. But that integral isn't taking the average.
– DLV
Nov 8, 2015 at 4:27
• No, $\int_{-\pi}^{\pi} \sin^2(x) dx = \pi$ but the length of the interval is $2 \pi$, so the average value is $\frac 12$. That is the same factor $\frac 12$ You can see it from $\sin^2 x=\frac 12(1+cos 2x)$ The cosine term averages to zero. Nov 8, 2015 at 4:29