The following is an excerpt from the book Statistical Digital Signal Processing and Modelling (Monson Hayes)

...consider a random process consisting of a random phase sinusoid in white noise $$ x(n) = A\sin(n\omega_{0} + \phi) + v(n)$$ where $\phi $ is a random variable that is uniformly distributed over the interval $[-\pi, \pi]$, and $v(n)$ is white noise with a variance of $\sigma_{v}^{2}$. The power spectrum of $x(n)$ is

$$ P_{x}(e^{j\omega}) = \sigma_{v}^{2} + \frac{1}{2}\pi A^{2}[u_{0}(\omega - \omega_{0}) + u_{0}(\omega + \omega_{0})]$$

Therefore it follows from Eq. (8.23) that the expected value of the periodogram is $$E\{{P_{per}(e^{j\omega})}\} = \frac{1}{2\pi}P_{x}(e^{j\omega})*W_{B}(e^{j\omega})$$ $$= \sigma_{v}^{2} + \frac{1}{4}A^{2}[W_{B}(e^{j(\omega - \omega_{0})}) + W_{B}(e^{j(\omega + \omega_{0})})]$$

  1. I am unable to figure out how he obtained the expression for power of the sinusoidal term. Shouldn't it be just $\pi^{2}A^{2}$ multiplied by the bracket term?
  2. Considering that $W_{B}$ represents fourier transform of Bartlett window, how is the convolution expression being evaluated here?

The power spectrum $P(e^{j\omega})$ is the Fourier transform of the autocorrelation function of $x(n)$:


If you evaluate $(1)$ you should get


Taking the Fourier transform of $(2)$ leads to the given expression for $P(e^{j\omega})$.

The convolution is obtained by noting that

$$u_0(\omega-\omega_0)\star W_B(e^{j\omega})=W_B(e^{j(\omega-\omega_0)})\tag{3}$$


$$\frac{1}{2\pi}\sigma_v^2\star W_B(e^{j\omega})=\sigma_v^2\frac{1}{2\pi}\int_{-\pi}^{\pi}W_B(e^{j\omega})d\omega=\sigma_v^2\tag{4}$$


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