34 votes
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Stationary vs non-stationary signals?

There is no stationary signal. Stationary and non-stationary are characterisations of the process that generated the signal. A signal is an observation. A recording of something that has happened. A ...
A_A's user avatar
  • 10.7k
20 votes

Stationary vs non-stationary signals?

@A_A's good answer misses one point: stationarity or nonstationarity are generally only applied to stochastic signals, not deterministic signals. In general, when statistical tests are applied for ...
Peter K.'s user avatar
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11 votes
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Why is $A\cos(2\pi f_ct)$ a non-stationary process?

A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where ...
Dilip Sarwate's user avatar
8 votes

If the mean of a random process is constant, does it imply the process is first order stationary?

In the usual sense of the term, first-order stationarity means that the first-order distribution of all the random variables is the same: each $X_t$ has the same CDF, and so the same pdf (or pmf) too ...
Dilip Sarwate's user avatar
6 votes
Accepted

Why is $\sin(t)$ a stationary process?

$\sin(t)$ is no random process, because there's nothing random about it. You could add a random amplitude to get a random process: $$x(t)=A\sin(t)\tag{1}$$ This is a random process because $A$ is a ...
Matt L.'s user avatar
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5 votes
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Cross-correlation or cross-covariance of non-zero mean signals

What are reasons to choose for cross-correlation or cross-covariance when comparing signals with non-zero mean? Well, part of the issue is that cross-correlation as defined in your equation: $$(f \...
Peter K.'s user avatar
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4 votes

Why is $A\cos(2\pi f_ct)$ a non-stationary process?

In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, ...
Marcus Müller's user avatar
4 votes
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LTI filtering for wide-sense stationary process

It's basically just about using the definitions and doing the math: $$\begin{align}R_{WW}[n]&=E\{W^*[k]W[k+n]\}\\ &=E\left\{\sum_mU^*[k-m]h^*[m]\sum_lU[k+n-l]h[l]\right\}\\ &=\sum_m\...
Matt L.'s user avatar
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4 votes
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How to find the output mean and autocorrelation of a non-linear system

The OP's updated working is incorrect. Following up what Hilmar suggested gives \begin{align} Y(t) &= a\left(X(t)\right)^2\\ &= a\left(S(t) + N(t)\right)^2\\ &= a\left(S(t)\right)^2 + 2aS(...
Dilip Sarwate's user avatar
3 votes
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Why doesn't law of large numbers apply to this stationary time-series?

I believe that you are thinking that each value of $X_t$ is determined by a different realisation of $Y$, which in this example is not true. Suppose that $Y$ is the value that comes out from a dice ...
Tendero's user avatar
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3 votes
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Autocorrelation of Addition of Two Independent Signals

You're correct as the Cross Correlation function vanishes. This has the implicit assumption the process has zero mean (Actually, at least one of them). Namely, in order to have $ {R}_{XY} \left( \tau \...
Royi's user avatar
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3 votes
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Doubt about wide sense stationary random process

The mean of $X(n)$ is always $\mu_X =0$, because the noise has zero mean. Thus we should check whether the autocorrelation corresponds to a WSS process. $$\begin{align} R_X(n_1,n_2) &=\mathbb{E}[...
Tendero's user avatar
  • 5,010
3 votes

Stationary signal: time-domain vs frequency domain

Your definitions are not correct. For a Strict Sense Stationary process (signal) the joint distribution of your process' value for all instants of time must be independent of time, in other words if ...
Mohammad M's user avatar
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3 votes
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Z-transform of difference equations and stability of a process

The paper deals in the lag operator $L$, rather than $z^{-1}$ as DSP people tend to use. As a result, all the DSP results you know are inverted. So "inside the unit circle" becomes "outside the unit ...
Peter K.'s user avatar
  • 25.7k
3 votes

Is there any computational method to prove whether a series is stationary or not?

1 Proof You're misusing the word proof. Remember that a proof, even one led with stochastic methods, always leads to an absolute "If A, then B, no doubt" statement. Since your $x$ is a realization ...
Marcus Müller's user avatar
3 votes
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Is $A\cos(\omega t+\theta)$ a Gaussian random process?

I'm pretty sure that even algebraically, $Z$ will not be Gaussian. You're multiplying a Gaussian by a random variable that has a distribution that is effectively the histogram of a (co)sine function. ...
Peter K.'s user avatar
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3 votes
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How to find a variance of sample sequence

$\sigma_r^2=\sigma_y^2\sigma_v^2$ iff 1) $\mathbb{E}\{y\} = \mathbb{E}\{v\}=0$. AND 2) $y$ and $v$ are independent, OR uncorrelated and their squares ($y^2$ and $v^2$) are also uncorrelated. Proof:...
Sofiane's user avatar
  • 178
3 votes

Why is a random process strictly stationary when its joint Probability density function is time invariant?

Why is a random process strictly stationary when its joint probability density function is time invariant? This query, taken from the title of the question cannot be answered because there is no Why. ...
Dilip Sarwate's user avatar
2 votes

What is the difference between wide sense and strict sense stationary processes?

A process is stationary if: its mean is a constant value: $\mu_x(t)=\mu x$ its MSV(mean square value) is a constant value. its variance is a constant value. $\sigma^2_x(t)=\sigma^2_x$ its ...
switch03's user avatar
2 votes

What is the difference between Kalman filter algorithm and stationary Kalman filter algorithm?

I’m not sure what you mean by the stationary Kalman filter, but it seems to be what I would call the steady-state Kalman filter. If that is the same thing, then you just solve for the Kalman gain at $...
Peter K.'s user avatar
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2 votes
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Converting a non-stationary random process into a WSS process by adding a random phase

It is not true that you can convert any non-stationary random process into a WSS random process by just adding a random phase. What is true is that you can make a (wide-sense) cyclostationary process ...
Matt L.'s user avatar
  • 89.5k
2 votes

Decimator effect on wide sense stationary input

The decimator by integer $M$ can be shown to be the following block: $$ x[n] \longrightarrow \boxed{ \downarrow M } \longrightarrow y[n] = x[Mn] $$ Assuming that the input $x[n]$ is WSS it has the ...
Fat32's user avatar
  • 28.1k
2 votes
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Is a pulse of white noise still properly described as stationary?

The pulsed process is not stationary. You can use the term "quasi-stationary" to describe a process that looks stationary in short time scales (e.g. during silence, or in the middle of a pulse).
MBaz's user avatar
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2 votes
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Understanding kurtogram parameters

[Given figures are from A new improved Kurtogram and its application to planetary gearbox degradation feature analysis, Xianglong Ni et al.] This $k$ in the graph (and the $K$ in the Table) ...
Laurent Duval's user avatar
2 votes

How to find the output mean and autocorrelation of a non-linear system

Put your second equations into your first equation, express $Y(t)$ as a function of $S(t)$ and $N(t)$ Apply the definition for mean and autocorrelation. Simplify and solve
Hilmar's user avatar
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1 vote

Identifying whether or not a cyclostationary signal is noisy or not using the cyclic autocorrelation

You're on the right track. The cyclic autocorrelation can be estimated by inverse Fourier transformation of the cyclic periodogram. If $X_T(t, f)$ is the Fourier transform of your data block (indexed ...
Chad Spooner's user avatar
1 vote

Decimator effect on wide sense stationary input

just a WSS signal $x[n]$ passed through an $M$-fold decimator (a multirate signal processing block where $M=2$, used in fractional rate conversion), no filter follows the decimator. So $y[n] = x[Mn]$ ...
Marcus Müller's user avatar

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