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Assuming that we have a continuous signal $v(t)$

$$v(t) = \int_0^t g(u)\, \mathrm du\tag1$$

where $g(t)$ is white noise.Then take the derivative of it.

$$\frac{\mathrm d}{\,\mathrm dt} v(t) = g(t)\tag2$$

Then sample $v(t)$ so that we get discrete-time model, and calculate the finite-difference instead of derivative

$$ v\left( (n+1) T_s\right) - v\left(nT_s\right) = \tilde g \left(nT_s\right) \cdot T_s \tag3$$

where $T_s$ is sampling interval.

I'm told that $\tilde g(nT_s)$ is the low-passed white noise, so how to prove it?

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    $\begingroup$ you seem to be mixing lower and upper case – so please confirm: Is your $V(t) = v(t)$? $\endgroup$ – Marcus Müller Apr 19 at 17:54
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    $\begingroup$ by the way, your eq. $(2)$ is slightly wrong! $\frac{\mathrm d\,V}{\,\mathrm dt} (t) = G(t) - G(0)$. $\endgroup$ – Marcus Müller Apr 19 at 17:59
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    $\begingroup$ in your last sentence, you mean $\tilde G$, not $G(nT_s)$, right? $\endgroup$ – Marcus Müller Apr 19 at 18:08
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    $\begingroup$ By the way, this feels a lot like homework; so, what have you tried so far? $\endgroup$ – Marcus Müller Apr 19 at 18:10
  • $\begingroup$ i'm not voting to close, but i'll be watching this question. i hope to see it revised to make some sense. $\endgroup$ – robert bristow-johnson Apr 19 at 23:53
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A continuous-time white noise process $\{X(t)\colon -\infty < t < \infty\}$ is a hypothetical construct that we can treat (in the simplified versions that we use on dsp.SE) as a zero-mean wide-sense stationary process with autocorrelation function $K\delta(\tau)$ where $\delta(\cdot)$ is the Dirac delta. More strongly, all the random variables $X(t)$ are mutually independent, not just uncorrelated. Note that the power spectral density is $S_X(f) = K, \infty < f < \infty.$ Now, the stochastic integral $$Y(t) = \int_0^t X(u) \,\mathrm du$$ gives us a random variable whose mean we can find as \begin{align} E[Y(t)] &= E\left[\int_0^t X(u) \,\mathrm du\right]\\ &= \int_0^t E[X(u)] \,\mathrm du\\ &= 0 & \scriptstyle{E[X(u)] = 0~\text{as per definition of white noise}} \end{align} and whose variance is \begin{align} \operatorname{var}(Y(t)) &= E[(Y(t))^2]-\left(E[Y(t)]\right)^2\\ &= E[(Y(t))^2]\\ &= E\left[\int_0^t X(u) \,\mathrm du \int_0^t X(v) \,\mathrm dv\right]\\ &= \int_0^t \int_0^t E[X(u)X(v)] \,\mathrm du \,\mathrm dv\\ &= \int_0^t \left[\int_0^t K\delta(u-v) \,\mathrm du \right] \,\mathrm dv\\ &= \int_0^t K \,\mathrm dv\\ &= Kt \tag{1} \end{align} which shows us immediately that if we collect the $Y(t)$'s to form a random process $\{Y(t)\colon 0 \leq t < \infty\}$, then this process is not wide-sense stationary; its variance increases with time instead of being constant as is needed for wide-sense stationarity.

Turning to sampling from continuous-time processes to form a discrete-time process, note that it is not possible to sample a white noise process $\{X(t)\colon -\infty < t < \infty\}$ to extract a single random variable, say $X(5)$, from it in part because white noise is an abstraction and in part because we (as engineers) must model reality and the reality is that an actual honest-to-goodness physical sampler doesn't sample instantaneously at $t$ but grabs a small chunk (say from $t-\varepsilon/2$ to $t+\varepsilon/2$) of whatever it is sampling, and it might be set up to report the average value $$ \frac{1}{\varepsilon} \int_{t-\varepsilon/2}^{t+\varepsilon/2} x(u) \,\mathrm du$$ of the waveform as the sample value. For deterministic signals, this is of little importance as long as $\varepsilon$ is small in comparison to how fast $x$ can change (for small $\varepsilon$, $\int_{t-\varepsilon/2}^{t+\varepsilon/2} x(u) \,\mathrm du \approx x(t)\cdot \varepsilon$ for continuous signals $x$) but for white noise, $$ \frac{1}{\varepsilon} \int_{t-\varepsilon/2}^{t+\varepsilon/2} X(u) \,\mathrm du$$ is a zero-mean random variable with variance $\displaystyle\frac K\varepsilon \gg K$.

But the OP is not interested in sampling the white noise process but rather in looking at the random variables $$Z[n] = Y((n+1)T_s) - Y(nT_s) = \int_{nT_s}^{(n+1)T_s} X(u) \,\mathrm du$$ which can be easily shown to be independent zero-mean random variables of variance $KT_s$. That is, $\{Z[n]\colon 0 \leq n < \infty\}$ is a discrete-time white noise process: a collection of independent identically distributed random variables with mean $0$ and variance $KT_s$.

Now consider the random process $\{\hat{X}(t)\colon -\infty < t < \infty\}$ that is obtained by passing our original white noise process $\{X(t)\colon -\infty < t < \infty\}$ through an ideal lowpass filter (unit gain in the passband) of bandwidth $W = \frac{1}{2T_s}$ so that the power spectral density $S_{\hat{X}}(f)$ is given by $$S_{\hat{X}}(f) = S_X(f)\cdot \operatorname{rect}\left(\frac{f}{2W}\right) =\begin{cases}K, & -W < f < W,\\ 0, &\text{otherwise},\end{cases}$$ and autocorrelation function $$R_{\hat{X}}(\tau) = K(2W)\operatorname{sinc}(2W\tau) = KT_s\operatorname{sinc}\left(\frac{\tau}{T_s}\right).$$ This process $\{\hat{X}(t)\colon -\infty < t < \infty\}$ is called a low-pass or band-limited white noise process (meaning that the power spectral density is constant within the frequency band of interest), and is not an abstraction as white noise (of infinite bandwidth and infinite power) is. So, we can talk of sampling this process instantaneously if we like --effectively assuming that the real-life sampler is returning exactly $\hat{X}(t)$ as the sample value, and the samples $\hat{X}(nT_s)$ all are zero-mean independent random variables with variance $KT_s$. Note that this is precisely the description of the $\{Z[n]\}$ process, that is, we can consider $\{Z[n]\}$ to be samples from a white noise process that has been low-pass filtered to bandwidth $\frac{1}{2T_s}$ and then sampled at $T_s$ second intervals (i.e. the Nyquist rate).

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  • $\begingroup$ Thanks for your attention. $\endgroup$ – user8873052 Apr 24 at 9:14
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Working with your definitions:

$$ v \left( \left( n + 1 \right) {T}_{s} \right) - v \left( n {T}_{s} \right) = \int_{0}^{ \left( n + 1 \right) {T}_{s} } g(u) du - \int_{0}^{ n {T}_{s} } g(u) du = \int_{ n {T}_{s} }^{ \left( n + 1 \right) {T}_{s} } g(u) du $$

So basically we have integration (Which is a Low Pass Filter) of White Noise over a Time Interval of $ {T}_{s} $.

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  • $\begingroup$ Well...It's actually a small part of a signal model which should be solved by EFC. It seems that I paid much attention to the proof of EFC and frazzled my brain...I consider about something like s(nTs) = s(t) *sigma δ(t-nTs)and its Fourier transform..Thank you very much. $\endgroup$ – user8873052 Apr 20 at 0:52

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