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Dirac delta function has a continuous argument, but Kronecker delta function has a discrete argument. Your example is a discrete signal so Kronecker delta is used.

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That seems fine. If you integrate your PSD over all frequencies you get a $1$ at $-f_0$ and $+f_0$ and zero everywhere else. $1+1 = 2$ so the total integral will come out to be $A^2/2$ which matches your time domain number. Yes, the PSD is also the magnitude squared of the Fourier Transform, i.e. $$PSD(f) = X(f) \cdot X^*(f)$$ where $X(f)$ is the Fourier ...

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The difference here is that statistical autocorrelation assumes a stationary power signal (so basically an infinite periodic signal) and does a normalisation to $[-1,1]$ and that autocorrelation just takes the finite signal as it is and does the classical convolution of the signal with itself, leading to just one point of perfect overlap and decreasing ...

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Yes. Mathematically, the limiting value of the autocorrelation function $R_X(\tau)$ of a wide-sense-stationary process is $$\lim_{\tau\to\infty}R_X(\tau) = \lim_{\tau\to-\infty}R_X(\tau) = \mu_X^2$$ where $\mu_X$ is the mean of the process. So it is not too surprising that res1 and res2 have identical shapes but res2 is shifted upwards in comparison to res1 ...

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Actually, we have two kinds of autocorrelation functions. One is defined for stochastic signals and the other for deterministic ones. If $x(t)$ is a stochastic signal, then its autocorrelation function will be $$R_x(t+\tau,t)=\Bbb E\{x(t+\tau)x^*(t)\}$$where $\Bbb E(\cdot)$ denotes the mathematical expectation. If x(t) is a deterministic signal, then its ...

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