# Autoconvolution vs Autocorrelation

I have a serious doubt about Autocorrelation and AutoConvolution.
My understanding is,

## Autocorrelation

Autocorrelation is nothing but correlation of the signal and it's delayed version.
Example : Autocorrelation of 2 functions f(t) & f(t+1).

## Autoconvolution

Autoconvolution is combination of the two signals where the output signal will be a modified version of a signal in term of another signal with every point/unit.

## My Question

Are two signals are the same if their auto-correlation functions are the same?
Are two signals are the same if their auto-convolution functions are the same?

Are two signals are the same if their auto-convolution functions are the same?
Almost. Look at the autoconvolution in the frequency domain where the autoconvolution of $$x$$ (with itself) gives us $$(X(f))^2$$ in the frequency domain while the autoconvolution of $$-x$$ (with itself) gives us $$(-X(f))^2 = (X(f))^2$$. So, given an autoconvolution function, there are two (very related) signals $$x$$ and $$-x$$ that have the same autoconvolution function.
Are two signals are the same if their auto-correlation functions are the same?
Not quite. Now we are given $$|X(f)|^2$$ in the frequency domain and there are many different factorizations possible. For example, if $$y(t)$$ is a signal such that values taken in by its Fourier transform always lie on the unit circle in the complex plane (for every $$f$$, $$|Y(f)|=1$$) then $$|X(f)Y(f)|^2 = |X(f)|^2$$ and so $$x\star y$$ has the same auto-correlation function as $$x$$. Note that $$x(t)$$ and $$x(t-\tau)$$ (which is a delayed version of $$x(t)$$) have the same autocorrelation function ($$Y(f)$$ happens to be $$\exp(-j2\pi f \tau)$$ here). Another factorization replaces $$X(f)$$ by $$X^*(f)$$ which tells us that $$x(t)$$ and $$x(-t)$$ (which is just $$x(t)$$ running backwards in time) have the same autocorrelation function.