find the power of a random process?

Question

I know all the steps expect the last step i don't know how to evaluate the integral

can someone show me the step that lead to the answer to be A^2/2

Answer 1

We compute $$ \begin{align*} \lim_{T\to\infty}\frac1{2T}\int_{-T}^T\mathbb E[X^2(t)]\mathsf dt &=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T\mathbb E\left[\frac{A_0^2}2(1 + \cos(2(\omega_0t+\theta)\right]\mathsf dt\\ &=\frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T (1 + \mathbb E[\cos(2(\omega_0t + \theta))])\mathsf dt\\ &= \frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T \left(1 + \mathbb E\left[\frac12\left(e^{i2(\omega_0t+\theta)} + e^{-i2(\omega_0t+\theta)} \right) \right]\right)\mathsf dt\\ &= \frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T \left(1 + \frac12\left(e^{i2\omega_0t}\mathbb E\left[e^{i2\theta}\right] + e^{-i2\omega_0t}\mathbb E\left[e^{-i2\theta}\right] \right) \right)\mathsf dt\\ &= \frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T \left(1 +\frac12 e^{i2\omega_0t}\cdot\frac{2i}\pi + \frac12 e^{-i2\omega_0t}\cdot\frac{-2i}\pi\right)\mathsf dt\\ &= \frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T\left(1 + \frac i\pi e^{i2\omega_0t} - \frac i\pi e^{-i2\omega_0t} \right)\mathsf dt\\ &= \frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \left[t + \frac 1{2\pi\omega_0}e^{i2\omega_0t} +\frac1{2\pi\omega_0}e^{-i2\omega_0t} \right]_{-T}^T\\ &= \frac{A_0^2}2\lim_{T\to\infty}\frac1{2T}\left[t + \frac1{\pi\omega_0}\cos(2\omega_0 t) \right]_{-T}^T\\ &= \frac{A_0^2}2\lim_{T\to\infty}\frac1{2T}\left(T-(-T) +\frac1{\pi\omega_0}(\cos(2\omega_0t)-\cos(-2\omega_0t) \right) \\ &= \frac{A_0^2}2\lim_{T\to\infty}\frac{2T}{2T}\\ &= \frac{A_0^2}2. \end{align*} $$