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Before I go into my question, I first want to review the basics of sampling a signal and at the same time I build the basics of my questions so that they make more sense. I know I have asked couple of questions here, which made this post long. But these questions are strongly related to each other and I could not separate them in different posts.

Suppose continuous time signal $x(t)$ has frequency spectrum of $X(f)$. We sample $x(t)$ by sampling frequency $f_s = 1/ \Delta t$ to get the sampled version $x_s(n)$. So we have $x_s(n) = x(n \Delta t)$. The spectrum of the sampled signal $x_s$ is the scaled (by $f_s)$ and replicated spectrum of $x(t)$ as shown in the figure below:spectrum of the sampled signal

More specifically the spectrum of the sampled signal can be written in two different ways:

\begin{equation} \tag{1} X_s(f) = \sum_{n}x(n \Delta t)\,\, e^{-j \, 2 \pi f \, n \Delta t} \end{equation}

\begin{equation} \tag{2} X_s(f) = \frac{1}{\Delta t} \,\, \sum_n X(f-n \, f_s) \end{equation}

or given the same equations wrt $\omega$:

\begin{equation} \tag{3} X_s(e^{j\omega \Delta t}) = \sum_{n}x(n \Delta t)\,\, e^{-j \, \omega\, n \Delta t} \end{equation}

\begin{equation} \tag{4} X_s(e^{j\omega \Delta t}) = \frac{2\pi}{\Delta t} \,\, \sum_n X(\omega-n \, \frac{2\pi}{\Delta t}) \end{equation}

Equation (3) is basically the DTFT of the sampled signal $x_s$.

Now if we look at the definition of the energy of the signal:

\begin{equation} \tag{5} \sum_n |x_s[n]|^2 = \frac{\Delta t}{2\pi}\int_{0}^{\frac{2\pi}{\Delta t}} \mid X_s(e^{j\omega \Delta t}) \mid ^2 d\omega \end{equation}

Equation (5) is the Parseval's theorem which states that the energy amounts found in the time domain must be equal to the energy amounts found in the frequency domain.

Let's assume Equation (5) holds when the sampling frequency is $f_s$. Now we want to check what happens if we change the sampling frequency, let's say make it $2\, f_s$. Since we are gathering more samples (twice) in time domain, the amount of energy in time domain becomes doubled. So the left-hand side of Equation (5) is doubled when the sampling frequency is doubled. Now let's see what happens to the right-hand side of the Equation (5). When the sampling frequency is $f_s$, as it can be seen in the above figure, the spectrum is made of triangles with the peak value of $f_s$.

To calculate the energy in the frequency domain we only look at one period (i.e. from 0 to $f_s$ or from 0 to $\omega_s$), we square the spectrum (which makes its peak ${f_s}^2$) and calculate the area underneath it. When you double the sampling frequency, the peak of the triangle become $2 f_s$, when you square it it become $4{f_s}^2$, hence the area under the squared triangle in one period is scaled by 4 (note that the integral range of equation (5) also gets doubled, but that has no effect on the area). The output of integral which is scaled by 4 is now multiplied by the coefficient $\frac{\Delta t}{2\pi}$. Since $f_s$ is doubled, $\Delta t$ is halved, and therefore the right-hand side of Equation (5) is also doubled when the sampling frequency is doubled. This is consistent with the time domain result.

minor question 1) The Equation (5) considers the energy of the signals. How can we formulate it for the power? Well, in the time domain it's easy I guess, you take the average. So after summing up you divide by the total number of samples. So in this way regardless of the sampling frequency we should get a constant power. But what about the frequency domain?

I hope you are still following me, cause I haven't got to my main question yet.

Now instead of sampling a signal and make it discrete, let's consider generating samples of a random process like white Gaussian noise (AWGN). Let's assume I am generating $N$ samples of AWGN with variance $\sigma ^2$ in MATLAB by using randn() funtion i.e.:

\begin{equation} \tag{6} n_1 = \sigma * \text{randn}(N,1) \\ n_2 = \sigma * \text{randn}(N,1) \end{equation}

Note that, here I am not sampling a noise signal to get these samples. I am simply generating some random numbers to consume them later.

Now I want to add these noise samples to my clean signal samples. Let's assume I have two cases of clean signals, namely $y$ which has a sampling frequency of $f_s$ and $z$ which has a sampling frequency of $2f_s$. So we have

\begin{equation} \tag{7} y_n = y + n_1 \end{equation}

\begin{equation} \tag{8} z_n = z + n_2 \end{equation}

So this means that in $y_n$ I am using noise samples with the rate $f_s$ and in $z_n$ and I am using noise samples with the rate $2f_s$.

Since here we are dealing with random process, instead of frequency spectrum we have the power spectral density (PSD). Let's name the PSD of $y_n$ and $z_n$, $S_{yy}(\omega)$ and $S_{zz}(\omega)$ respectively. Also let's consider the PSD of noise in Equation (7) which is read with lower rate is $S_{nn1}(\omega)$ and the PSD of noise in Equation (8) which is read with higher rate is $S_{nn2}(\omega)$

Question 2) Now my question is that how are $S_{nn1}(\omega)$ and $S_{nn2}(\omega)$ different? More precisely if you want to draw them on the frequency axis how do they look? Does $S_{nn1}(\omega)$ span from 0 to $f_s$ and $S_{nn2}(\omega)$ span from 0 to $2f_s$? And if you calculate the area underneath them (which shows the power or variance of the noise) which one is greater? Do they have the same power?

Let's consider the auto-correlation and PSD for random processes. We know that PSD is the DTFT of the auto-correlation, more specifically:

\begin{equation} \tag{9} R_{nn}(m) = \frac{\Delta t}{2 \pi}\int_{0}^{\frac{2\pi}{\Delta t}} S_{nn}(\omega) e^{j\omega m \Delta t} d \omega, \end{equation} where $R_{nn}$ is the auto-correlation and $S_{nn}(\omega)$ is the PSD.

From Equation (9) we have: \begin{equation} \tag{10} R_{nn}(0) = \frac{\Delta t}{2 \pi}\int_{0}^{\frac{2\pi}{\Delta t}} S_{nn}(\omega) d \omega, \end{equation} which means the area under PSD (which is the power of the signal) is given by the auto-correlation at zero.

Question 3) Which one gives the power of the signal, the output of the above integral in Eq (10), or the output of the above integral multiplied by $\frac{\Delta t}{2 \pi}$?

Question 4) This is the continue of question 2, or maybe similar to it. I basically want to know how Equation (10) changes for noises $n_1$ and $n_2$? Cause obviously in Eq. (10), $\Delta t$ is changing for the noise when the sampling rate is changed. But how $R_{nn1}$ and $R_{nn2}$ or $S_{nn1}$ and $S_{nn2}$ change?

I appreciate if someone can mathematically explain this to me.

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    $\begingroup$ Just a quick comment: signals do not have a frequency response; only systems do. Signals have frequency spectrums. You may want to edit the question for clarity. $\endgroup$ – MBaz Jun 1 '18 at 2:11
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    $\begingroup$ Welcome on dsp stackexchange. Please consider breaking your question into chunks and ask in separate pieces. It’s a bit long as of right now. $\endgroup$ – Gilles Jun 1 '18 at 7:06
  • $\begingroup$ This subject is treated in detail under the typical name(s) of noise shaping within oversampling AD converters of many DSP books. You may refer to your favorite one... It's a long question and your liberal use of notation also makes it hard to follow at places... $\endgroup$ – Fat32 Jul 19 '18 at 21:38

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