Autocorrelation of cos(200πt) at τ = π/4

Power and energy spectral density

• Energy spectral density (ESD): for an energy signal, $\Psi_x(f) = |X(f)|^2$ (J/Hz); it shows how energy is distributed over frequency, and $E = \int\Psi_x(f)df$.
• Power spectral density (PSD): for a power signal, $S_x(f) = \lim_{T\to\infty}\frac{|X_T(f)|^2}{T}$ (W/Hz); it shows how average power is distributed over frequency, and $P = \int S_x(f)df$. Each is the Fourier transform of the corresponding autocorrelation function.

Autocorrelation of $x(t) = \cos(200\pi t)$ at $\tau = \pi/4$

$x(t)$ is periodic ($f_0 = 100$ Hz, $T_0 = 0.01$ s), so it is a power signal and we use the time-average autocorrelation:

$$\begin{aligned} R_x(\tau) &= \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}\cos(200\pi t)\cos(200\pi(t-\tau))\,dt \\ &= \lim_{T\to\infty}\frac{1}{2T}\int_{-T/2}^{T/2}\left[\cos(200\pi\tau) + \cos(400\pi t - 200\pi\tau)\right]dt \\ &= \frac12\cos(200\pi\tau) \end{aligned}$$

(the second term averages to zero).

At $\tau = \pi/4$ s:

$$\begin{aligned} R_x\!\left(\frac{\pi}{4}\right) &= \frac12\cos\left(200\pi\cdot\frac{\pi}{4}\right) = \frac12\cos(50\pi^2) \\ &= \frac12\cos(493.48\ \text{rad}) = \frac12\times(-0.9689) \\ &\approx -0.484 \end{aligned}$$

($493.48$ rad reduces to $493.48 - 78\times2\pi = 3.392$ rad, and $\cos 3.392 = -0.9689$.)

Answer: $R_x(\tau) = \frac12\cos(200\pi\tau)$, so $R_x(\pi/4) = \frac12\cos(50\pi^2) \approx -0.484$ W (normalised). For comparison $R_x(0) = 0.5$ W = average power.