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When a bandlimited random process whose PSD \begin{equation} S(\omega) = \begin{cases} \frac{N_0}{2} & -10B<\omega<10B\\[2ex] 0 & \text{otherwise.} \end{cases} \end{equation} is sampled at a rate of $B$ samples per second, what would be the PSD of the output sequence. As per my understanding, we cannot use Fourier transform as the signal energy is infinite. And since the sampling frequency is less than Nyquist rate aliasing occurs. In the post, Matt explains how to relate PSDs of CT random process $x(t)$ and DT random process $y_k$ by the equation \begin{equation} S_y\left(e^{j\omega T}\right) = \frac{1}{T}\sum_{k=-\infty}^{\infty}S_x\left(\omega-\frac{2\pi k}{T}\right) . \end{equation} My doubt in the above equation is about the $1/T$ factor. When the signal is sampled and the power is integrated at the output from frequencies $0$ to $B/2$ it is $10BN_0$ and not $10BN_0/T$ [from electrical circuit analysis].

Can anyone please explain what am I missing?

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  • $\begingroup$ i wish you would stick with convention and call bandwidth "$B$" and call sample rate "$f_\text{s}$". $\endgroup$ – robert bristow-johnson Dec 28 '16 at 4:34
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The power of a discrete-time WSS process $x[n]$ is given by the value of its auto-correlation function at lag $k=0$, which can be evaluated by integrating the power spectrum:

$$P_x=R_x[0]=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\omega'})d\omega'\tag{1}$$

where $\omega'=\omega T$ is the normalized frequency in radians, and $T$ is the sampling period. Using the unnormalized frequency $\omega$ in $(1)$ gives

$$P_x=\frac{T}{2\pi}\int_{-\pi/T}^{\pi/T}S_x(e^{j\omega T})d\omega\tag{2}$$

Note the factor $T$ in $(2)$, which will cancel with the factor $1/T$ in the expression for the power spectrum in your question.

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First of all, with a sampling rate of $B$ samples per second the sampling period is $T=1/B$. So the PSD of the discrete process is

$$S_y\left(e^{j\omega}\right) = {B}\sum_{k=-\infty}^{\infty}S_x\left({\omega}{B}-{2\pi k}{B}\right)$$ Regarding your second question, it seems you are missing the frequency normalization part. Notice that generally a scaled version of $\omega$ (with a factor of $1/T$) is on the RHS. So for example a signal bandlimited to $\omega_0$ becomes bandlimited to $\omega_0T$:

$$S_y\left(e^{j\omega}\right) = \frac{1}{T}\sum_{k=-\infty}^{\infty}S_x\left(\frac{\omega}{T}-\frac{2\pi k}{T}\right)$$ Therefore, while the magnitude is divided by $T$, the integration bound should be appropriately scaled (from $0$ to $BT/2$) and power remains the same.

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  • $\begingroup$ Sampling rate is $B$ $\endgroup$ – Manikanta S Dec 27 '16 at 13:41
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The power in a continuous-time band-limited white noise process $\{X(t)\colon -\infty < t < \infty\}$ with power spectral density $$S_X(f) = \begin{cases} \frac{N_0}{2} & -10B<f<10B\\[2ex] 0 & \text{otherwise.} \end{cases} $$ is given by $$P = \int_{-\infty}^{\infty} S(f)\, \mathrm df= \int_{-10B}^{10B}\frac{N_0}{2} \, \mathrm df = 10N_0B$$ while the autocorrelation function $R_X(t)$ is given by $$R_X(t) = \int_{-\infty}^{\infty} S(f)\exp(j2\pi f t)\, \mathrm df= 10N_0B \operatorname{sinc}(20Bt)$$ where the sinc function is defined as $$\mathrm{sinc}(\tau) = \begin{cases}\frac{\sin(\pi \tau)}{\pi \tau}, & \tau \neq 0,\\ 1, & \tau = 0,\end{cases} $$ Note that the autocorrelation function has value $0$ at (nonzero-)integer multiples of $(20B)^{-1}$ seconds, and so, samples that are $B^{-1}$ seconds apart are uncorrelated random variables. Specifically, $$\{X[n]\colon n \in \mathbb Z\} = \{X(B^{-1}n)\colon n \in \mathbb Z\} = \{X(nT)\colon n \in \mathbb Z\} $$ where $T = B^{-1}$ is a discrete-time random process consisting of uncorrelated identically distributed random variables with mean $0$ and variance $10N_0B$. You can work out the power spectral density from this. (You will likely need to do some translation from Fourier transforms with respect to $f$ to Fourier transforms with respect to $\omega$.

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